Chapter 3: Gene Expression (Transcription), Synthesis of Proteins (Translation), and Regulation of Gene Expression

*The answer is D.* Only the coding region remains to be calculated 3 × 150 = 450.

A gene encodes a protein with 150 amino acids. There is one intron of 1,000 bps, a 5′ untranslated region of 100 bp, and a 3′-untranslated region of 200 bp. In the final processed mRNA, how many bases lie between the start AUG codon and the final termination codon? A. 1,750 B. 750 C. 650 D. 450 E. 150

*The answer is D.* A Southern blot is used to detect DNA by DNA-DNA hybridization. The bands can then be analyzed to determine if the child and the father share certain genes that the child could not have inherited from the mother as well.

A paternity dispute leads to the testing of a mother, a child, and the potential father. Which of the following tests was most likely performed on DNA from all three parties in order to ascertain paternity? A. Enzyme-linked immunosorbent assay B. Ligase chain reaction C. Northern blot D. Southern blot E. Western blot

*The answer is B.* The pathogenesis of cystic fibrosis is related to defective chloride transport in epithelial cells.

If a patient with cystic fibrosis were to be treated by gene therapy, which type of cells should be targeted as host cells? A. Germ cells B. Epithelial cells C. T cells D. Hemopoietic stem cells

*The answer is E.* Helicase binds to ssDNA at the origin of replication and moves into he replication fork. It then proceeds to separate and unqind the dsDNA.

A diagram illustrating the DNA replication process is shown below. Which of the points marked by letters represents the site of action of helicase in the following diagram?

*The answer is D.* It is the large ribosomal RNA that catalyzes peptide bond formation, using peptides and amino acids in the "A" and "P" sites on the ribosome. Destroying the secondary structure of this rRNA via the aberrant ribonuclease will limit the ability of the ribosome to create peptide bonds. The large, ribosomal RNA molecule is not essential for the initiation, termination, elongation (moving the ribosome along the mRNA after peptide bond formation has occurred), or tRNA activation and charging.

A eukaryotic cell line contains an aberrant, temperature- sensitive ribonuclease that specifically cleaves the large rRNA molecule into many pieces, destroying its secondary structure and its ability to bind to ribosomal proteins. This cell line, at the nonpermissive temperature, has greatly reduced the rates of protein synthesis. This rate-limiting step is which of the following? (A) Initiation (B) Termination (C) Elongation (D) Peptide bond formation (E) tRNA activation and charging

*The answer F.* In induction, a regulatory gene produces an active repressor, which is inactivated by binding to the inducer. The inducer prevents binding of the repressor to the operator rather than stimulating the binding of RNA polymerase. The structural genes are coordinately expressed. Transcription yields a single, polycistronic mRNA, which is translated to produce a number of different proteins. The regulatory gene is not a part of the polycistronic message.

Inducible bacterial operons exhibit which of the following properties?

*The answer is C.* The boy is displaying the symptoms of fragile X syndrome. Fragile X contains a triplet nucleotide repeat (CGG) on the X chromosome in the 5′ untranslated region of the FMR1 gene. The triplet repeat expansion leads to no expression of the FMR1 gene, which produces a protein required for brain development. Its function appears to be that of an mRNA shuttle, moving mRNA from the nucleus to appropriate sites in the cytoplasm for translation to occur. Depending on the level of expression of FMR1 (which is dependent on the number of repeats), the symptoms can vary from mild to severe. Less than 1% of cases of fragile X are due to missense or nonsense mutations; the vast majority are due to the expansion of the triplet repeat at the 5' end of the gene. Gene inactivation by methylation, or deletion, are not causes of fragile X syndrome. The syndrome was called fragile X because the X chromosome that carries the repeat expansion is subject to DNA strand break under certain conditions (such as lack of folic acid), which does not occur with normal X chromosomes. The area containing the repeat alters the staining pattern of the X chromosome, allowing this to be seen in a karyotype (as seen below). Fragile X is the most common inheritable cause of mental retardation. Males are more severely affected. In early childhood, developmental delay, speech and language problems, and autisticlike behavior are noticeable. After puberty, the classic physical signs develop (large testicles, long-thin face, mental retardation, large ears, and prominent jaw).

A 13-year-old exhibited developmental delay, learning disabilities, mood swings, and at times, autistic behavior when he was younger. His current physical exam shows a long face, large ears, and large, fleshy hands. His fingers exhibit hyperextensible joints. Examination of fibroblasts cultured from the boy showed abnormal DNA damage, but only in the absence of folic acid. This disorder has, at the genetic level, which one of the following? (A) A single missense mutation (B) A large deletion (C) An extended triplet nucleotide repeat (D) A nonsense mutation (E) Gene inactivation via methylation

*The answer is B.* DNA and RNA are composed of sequences of four bases arranged into codons composed of three sequential bases. Each codon calls for a particular amino acid except for one codon that signals the initiation of protein synthesis (AUG) and three that stop protein synthesis (UAA, UAG, and UGA). Changes in the DNA sequences are called mutations. Some changes in the genetic code can result in the formation of altered proteins, such as enzymes, channels, and structural proteins, which may lead to serious manifestations. There are several types of mutations: Base substitutions (point mutations), where one base is substituted with another base, are the most common type of mutation. These point mutations can be of three types: 1. Silent mutations (Choice D): These mutations result from codon base substitutions which code for the same amino acid. For example, a single base substitution in UCA to UCU will not result in any change in protein structure because of both codes for the same amino acid, serine. Silent mutations do not cause amino acid changes within proteins. 2. Missense mutations (Choice C): These mutations are characterized by base substitutions that result in the placement of an incorrect amino acid in a protein sequence. For instance, a change in the code from UUU to UCU changes the translated amino acid from phenylalanine (UUU) to serine (UCU) 3. Nonsense mutations: These mutations introduce a stop codon within gene sequences, resulting in the formation of shorter, truncated proteins. An example of a nonsense mutation is a mutational change in the codon UCA (serine) to UAA stop codon). Frameshift mutations result from deletion or insertion of bases that are not a multiple of three. As their name implies, frameshift mutations alter the reading frame of the genetic code, resulting in the formation of non-functional proteins. The patient in the vignette appears to have cystic fibrosis, a genetically inherited condition with hundreds of implicated mutations. The most common mutation is a codon deletion of the phenylalanine at position 508 in the CFTR protein. However, it appears that this patient's mutation is a frameshift mutation caused by a 2-base pair deletion in the CFTR protein.

A 3-year-old male is seen by his pediatrician with complaints or recurrent respiratory infections. The patient's mother has been exasperated with the boy's medical problems since coming to the country soon after he was born. She asks "why does my baby keep getting sick when all of his siblings are fine?" The pediatrician orders laboratory tests including genetic testing for a transmembrane chloride channel. The mRNA is isolated from cultured fibroblasts and cDNA is synthesized and amplified. The results are compared to the product obtained from a healthy sibling and are shown below. Which of the following is most likely responsible for the patient's condition? (A) Codon deletion (B) Frameshift mutation (C) Missense mutation (D) Silent mutation (E) Trinucleotide expansion

*The answer is C.* Fanconi anemia is the most common cause of inherited aplastic anemia and is an important cause of acute myeloid leukemia in children. Fanconi anemia is characterized by defective DNA repair that results in increased chromosomal breakage, rearrangements, and deletions. Patients with this condition are hypersensitive to alkylating agents and experience excessive chromosomal breakage and growth inhibition upon exposure. Radial ray abnormalities result in abnormal or absent radii or thumbs. Kidney malformations, hypogonadism, microcephaly, and high fetal hemoglobin concentration (relative to normal) are often seen as well.

A 10-year-old boy is brought to the family physician for bleeding gums. The child was born with small, wide thumbs and also has small eyes and pigmented patches on his abdomen. He has not met his cognitive milestones. Radiographic studies performed after birth demonstrated bifid thumbs and small kidneys. Laboratory results demonstrate pancytopenia and high fetal hemoglobin levels. The physician reminds the parents of his congenital condition and states that bone marrow transplantation is the suggested therapeutic modality for correcting the patient's hematologic problem. What congenital disorder is most consistent with this patient's presentation? A. Diamond-Biackfan anemia B. Edwards syndrome C. Fanconi anemia D. Hereditary spherocytosis E. Neurofibromatosis type 1

*The answer is E.* A base substitution at an intron-exon junction, which leads to the deletion of an entire exon is indicative of a splice site mutation.

A 10-year-old boy with severe progressive skin ulceration, decreased resistance to infection, and impaired cognitive ability has been diagnosed with a genetic deficiency of the enzyme prolidase. Mutation analysis has identified a single base substitution at the 3' end of intron 6 of the mutant allele as well as deletion of a 45-base exon (exon 7) in the prolidase cDNA. Which type of gene mutation was most likely inherited by this boy? A. Frameshift mutation B. In-frame mutation C. Missense mutation D. Nonsense mutation E. Splice site mutation

*The answer is A.* Dactinomycin binds to DNA and blocks the ability of RNA polymerase to transcribe genes, thereby blocking transcription. The drug does not bind specifically to ribosomes, transcription factors, or RNA polymerase II. It also does not interfere with DNA synthesis.

A 2-year-old has been diagnosed with a rhabdomyosarcoma and is placed on chemotherapy, including the drug dactinomycin. Dactinomycin exerts its effects by which of the following mechanisms? (A) Binding of the drug to DNA, thereby blocking RNA synthesis (B) Binding of the drug to ribosomes, thereby blocking translation (C) Binding of the drug to transcription factors, thereby blocking RNA synthesis (D) Binding of the drug to RNA polymerase II, thereby blocking RNA synthesis (E) Binding of the drug to DNA, thereby blocking DNA synthesis

*The answer is B.* F2 is most likely the father of this child. Restriction site polymorphisms are characteristic sites on individual allele that are recognized by restriction enzymes. These sites can be inherited along with the chromosomes on which they reside. Therefore, every fragment in the child's Southern blot should be found in either the father's or the mother's profile. The child could have received the 6-kb fragment from t he mother (M) and the 9-kb fragment from F2.

A 20-year-old mother is unsure of the paternity of her newborn son. To determine the father of her child, a genetic test based on DNA restriction fragment length polymorphism was performed. Blood was drawn from the four men suspected to be the father (Fl, F2, F3, F4) as well as from the mother (M) and the infant (C). DNA extracted from the samples was amplified using polymerase chain reaction and then treated with the restriction enzyme EcoRI. The resulting fragments were separated with gel electrophoresis and a Southern blot analysis was performed. According to the Southern blot shown in the image, which individual is most likely to be the father of this child? A. Fl B. F2 C. F3 D. F4

*The answer is A.* Characteristic symptoms of I-cell disease. Note release of lysosomal enzymes into serum, which would not be seen in the other deficiencies.

A 25-month-old Caucasian girl has coarse facial features and gingival hyperplasia and, at 2 months of age, began developing multiple, progressive symptoms of mental retardation, joint contractures, hepatomegaly, and cardiomegaly. Levels of lysosomal enzymes are elevated in her serum, and fibroblasts show phase-dense inclusions in the cytoplasm. Which of the following enzyme deficiencies is most consistent with these observations? A. Golgi-associated phosphotransferase B. Lysosomal α-1,4-glucosidase C. Endoplasmic reticulum-associated signal peptidase D. Cytoplasmic α-1,4 phosphorylase E. Lysosomal hexosaminidase A

*The answer is C.* Synthesis of RNA from DNA (transcription) occurs in the nucleus and is catalyzed by three types of RNA polymerases. Transcription leads to the formation of messenger RNA (mRNA), ribosomal RNA (mRNA), transport RNA (RNA) and small nuclear ribosomal proteins (snRNPs). Messenger RNA transfers the genetic code to the cytoplasm and serves as a template for protein synthesis. Synthesis of mRNA is catalyzed by RNA polymerase II and occurs in two stages. During the first stage, the RNA transcript (exact copy of the DNA template) is produced. In the second stage, the RNA transcript is converted into mRNA. Processing of the RNA transcript involves: 1. RNA capping: addition of methylated guanine nucleotide to the 5' end. 2. RNA polyadenylation: addition of several adenine nucleotides to the 3' end (poly-A tail) 3. RNA splicing removal of introns (non-coding regions). Splicing is performed by spliceosomes which consist of snRNPs plus proteins. Synthesis of snRNP also occurs in the nucleus, catalyzed by RNA polymerase II. *Educational Objective:* snRNPs (small nuclear ribonucleoproteins) are synthesized by RNA polymerase II in the nucleus. They help to remove introns from the RNA transcript and are thus necessary for synthesis of messenger RNA.

A 24-year-old female presents to her primary care physician with complaints of a rash that will not go away. The rash is present on the patient's face in a butterfly distribution sparing the nasolabial folds. She easily gets sunburned and thus avoids going to beach with her family. Additionally, she states that she is getting tired more and more easily. Laboratory studies show elevated antinuclear antibodies and anti-dsDNA levels. Further testing also reveals elevated anti-U1-snRNP liters. These reflect increased levels of antibodies to U1-snRNP molecules. snRNPs are involved in what critical cellular function? A. Charging RNA with amino acids B. Synthesizing Okazaki fragments C Removal of introns from RNA transcripts D Polyadenylation of RNA transcripts E Aiding in mRNA to exit the nucleus F. Allowing proper functioning of ONA ligate

*The answer is D.* This image depicts a berry aneurysm (red circle). Berry aneurysms are congenital and associated with several syndromes. In this patient, the combination of a berry aneurysm, and a history of easy bruising and bleeding suggests the vascular form of Ehlers-Danlos syndrome (EDS). EDS is a group of disorders result ing from defects in collagen synthesis and processing. Vascular EDS ( type IV) is associated with a defect in the formation of type III procollagen, a precursor of the collagen found in many tissues. Unlike the classical and hypermobility forms of EDS, vascular EDS manifests with thin skin with varicosities, particularly over the abdomen, abnormal facial structure, easy predisposition to bruising from minor trauma, and absence of large-joint hyperextensibility. Small-joint (more distal) hyperextensibility, however, may be present. In addition, patients are at high risk for spontaneous visceral rupture, such as rupture of the intestine or gravid uterus, as well as vascular abnormalities, such as the formation of arterial pseudoaneurysms, which are also at risk to rupture, but are unfortunately asymptomatic until they do. EDS IV is inherited in an autosomal dominant pattern, though up to 50% of cases arise from de novo mutations. In this patient, the combination of a positive family history, characteristic skin findings, and ruptured berry aneurysm make EDS IV the most likely diagnosis *Bottom Line:* Ehlers-Danlos syndrome (EDS) is a group of disorders resulting from defects in collagen synthesis and processing; vascular (type IV) EDS does not manifest with the typical joint hypermobility, but is clinically important due to the high likelihood of dangerous complications in patients with this disease. Its manifestations include characteristic skin findings, easy bruising, and predisposition to rupture of hollow organs or arterial aneurysms.

A 30-year-old woman is found unconscious in her home by her husband. On arrival in the emergency department, she is noted to have thin, translucent skin over her abdomen with prominent varicose veins and many bruises on her shins. He denies any major medical problems in her past, though does report that she always seemed to "bruise really easily." Her husband thinks the patient's mother may have had similar problems with bruising, and also that she had "some sort of surgery" when her intestine was found to "have a hole in it." The patient expires and an autopsy is performed showing an abnormality in the circle of Willis, similar to that shown in the image. Which of the following proteins was most likely defective in this patient? A. Fibrillin-1 B. Keratin 14 C. Type I procollagen D. Type III procollagen E. Type IV collagen

*The answer is A.* This patient has a urinary tract infection, and was likely prescribed a fluoroquinolone. Fluoroquinolones such as ciprofloxacin, levofloxacin, and ofloxacin inhibit DNA gyrase, also known as topoisomerase II, an enzyme that relaxes DNA supercoils. Tendonitis and spontaneous tendon rupture are rare but serious adverse react ions to fluoroquinolones.

A 34-year-old woman presents to her physician with urinary urgency and dysuria. Urinalysis shows gram-negative rods and a large number of WBCs. The patient is prescribed an antibiotic; however, she is cautioned to return if she experiences joint stiffness or pain, a known adverse effect of this medication. Which of the following enzymes is inhibited by the antibiotic she was most likely prescribed? A. DNA gyrase B. DNA polymerase C. Peptidyl t ransferase D. RNA primase E. Transpeptidase

*The answer is D.* To determine the genotype of the embryos, the genetic material must first be isolated and amplified. Polymerase chain reaction (PCR) is a laboratory procedure used to exponentially amplify DNA. Based on the primer sequences chosen, it results in the production of many copies of a desired fragment of DNA. PCR may be the initial step of a variety of diagnostic techniques, including Southern blot. Once the DNA is amplified, other laboratory techniques, such as gene sequencing, may then be implemented. The main advantage of PCR is speed. A standard PCR protocol typically takes only a few hours to complete, as opposed to any of the "blots," which take multiple days and may even require overnight steps.

A 37-year-old woman requests diagnostic testing before her frozen embryos are implanted. The physician agrees as the couple's first child had his early years complicated by failure to thrive and chronic respiratory symptoms. At the last hospital admission, the child 's symptoms were notable for persistent cough, dyspnea with thick sputum, loose and greasy stools, and a history of failure to pass meconium. He has since received the proper diagnosis and is being appropriately treated, but the parents hope to avoid their second child having the same problems, as both the patient and her husband are carriers of the disease. The laboratory isolates some of the DNA from each 8- to 10-cell embryo prior to implantation and proceeds to run tests to determine the genotype of the embryos. Which of the following laboratory procedures is the first step performed in determining the genotype of the embryos? A. Enzyme-linked immunosorbent assay B. Ligase chain reaction C. Northern blot D. Polymerase chain reaction E. Southern blot

*The answer is B.* The patient in this vignette suffers from xeroderma pigmentosum (XP), a rare autosomal recessive disease that results from loss-of-function (LOF) mutations in nucleotide excision repair (NER) genes, leading to a reduction or elimination of NER. Ultraviolet (UV) light can damage DNA, causing the dimer pyrimidine residues, particularly pathway is essential in removing these dimers, which can interfere with DNA processing NER pathway is deficient, as in XP resulting covalent bonds contribute to the adducts that impede transcription adducts in regions containing important tumor suppressor genes, such as p53, can then contribute to the development of multiple malignancies. For example, individuals afflicted by XP are at developing basal cell carcinoma (BCC), squamous cell carcinoma (SCC), and melanoma, and do so at a much younger age than individuals without the disease. The majority of individuals with XP perish before the age of 20, mostly from metastatic cases of SCC and melanoma. They generally have a history of severe sunburns, often starting with their first exposure to sunlight. The more sun-exposed areas of their skin are covered in many freckles, and they may have multiple types fo other lesions, such as actinic (solar) keratoses (of which this patient's biopsy findings are suggestive) and malignancies. Other findings associated with XP include corneal opacification, keratitis, iritis with synechia formation, and melanoma of the choroid. Approximately 20% of XP patients also suffer from some form of neurologic abnormality, the pathogenesis of which is poorly understand.

A 4-year-old girt is brought to the pediatrician by her mother to establish care. Her mother is mildly concerned about the numerous "freckles' her child has developed, as none of her other children have these. Otherwise, she has no concerns, and a review of systems is negative. On exam, the patients' corneas appear opacified with evidence of keratitis bilaterally. Her Skin is dry and scaly with prominent macular pigmentation of the arms, face, and neck. Multiple nevi are present, as well as several erythematous, scaly papules. Punch biopsy of the largest papule demonstrates a collection of atypical keratinocytes with hyperchromatic and pleomorphic nuclei extending from the basal layer toward the surface. There is also a hyperkeratotic stratum corneum with alternating areas of orthokeratosis and parakeratosis. Which of the following is responsible for the condition most likely underlying this patient's presentation? A. P16 loss of function B. Defective repair of pyrimidine dimers C. NFI loss of function D. BRAF gain of function E. Viral inactivation of RB

*The answer is E.* Western blotting is used to detect a target polypeptide or protein from within a mixed sample. Potential target proteins are separated by gel electrophoresis. The separated proteins are then transferred to a nitrocellulose membrane and probed with a primary antibody specific for the protein of interest. The membrane is then washed and treated with a (secondary) marked antibody that binds to the primary antibody and can be detected (eg, by colorimetry). For example, a serum sample from a patient with suspected HIV infection can be analyzed via Western blot to detect antibodies directed against specific viral proteins. Following separation of viral proteins by gel electrophoresis and protein transfer to a nitrocellulose membrane, the membrane is treated with the patient's serum. Patients who are HIV positive are likely to have antibodies that react with viral p24, gp41, and gp120f160. If 2 of these 3 bands are positive, the test is considered positive. Western blotting is similar to the enzyme-linked immunosorbent assay (ELISA) technique, however, in ELISA the patient's serum is tested directly, whereas in Western blotting the proteins are first separated by electrophoresis. (Choice B) Northern blots analyze mRNillL A sample containing a large number of mRNA molecules is separated vie gel electrophoresis. Separated bands are then transferred to a membrane and hybridized with a probe containing a nucleotide sequence complementary to the mRNA of interest. (Choice C) Southern blotting is used to analyze DNA sequences. DNA that is fragmented using restriction endonucleases is separated by gel electrophoresis and transferred to a nitrocellulose membrane. A radiolabeled DNA probe containing a sequence complementary to an area of interest is then used for hybridization. Restriction site mutations can be detected by Southern blotting because they alter DNA fragment lengths, thereby altering electrophoresis migration patterns. Microarray analysis is similar to Southern and Northern blotting but involves hybridization of a large number of probes at once (Choice A), The genomic DNA or cDNA being analyzed is labeled with a fluorescent tag and placed on a gene chip containing complementary sequences for a large number of genes. The degree of fluorescence corresponds to the mRNA expressed in the particular sample. (Choice D) Southwestern blotting is a technique that analyzes DNA binding proteins using principles of the Southern and Western blot techniques. DNA-binding proteins are recognized by their ability to bind specific oligonucleotide probes. *Educational Objective:* Western blotting is used to identify proteins, Northern blotting identifies specific RNA sequences, and Southern blotting identifies specific DNA sequences in an unknown sample.

A 43-year-old man is evaluated for progressive neuropsychiatric symptoms. A year ago, he began feeling depressed and having hallucinations. Five months later, he developed intermittent paresthesias and progressively worsening choreiform movements. myoclonus. and ataxia. These symptoms have not improved despite multiple hospitalizations, an extensive workup has been unrevealing. The patient is a slaughterhouse worker with extensive exposure to bovine offel. As part of the evaluation for prior disease, a tissue sample digested with protease is processed vie gel electrophoresis end transferred to filter paper. Antibodies to a specific prior protein ere added to the filter. Next, a marked protein that combines with the antibody-protein complex is used to determine whether the test is positive. Which of the following best describes this test? A. Microarray B. Northern blot C. Southern blot D. Southwestern blot E. Western blot

*The answer is D.* The condition described is xeroderma pigmentosum (XP), a genetic defect in the DNA excision repair pathway. In XP, defective excision repair results in the inability to repair thymidine dimers, which form in DNA exposed to ultraviolet light. XP is associated with dry skin, and individuals who have XP are much more likely to develop both melanoma and nonmelanoma skin cancers. Clinically, XP manifests as a sunburn-like reaction, and individuals with XP often die by age 30 years. XP is autosomal recessively inherited, which is demonstrated by the frequency of occurrence of XP in the vignette family.

A couple has four children. One of the children suffers from severe erythema and scaling on the sun-exposed areas of his body. The other three children and both parents are unaffected. The affected child later develops melanomas and squamous-cell carcinomas of the skin. The affected child most likely has a genetic defect affecting which of the following enzymes? A. Adenosine deaminase B. ATM kinase C. DNA mismatch repair pathway D. Endonuclease in the nucleotide excision repair pathway E. Hypoxanthine-guanine phosphoribosyltransferase F. Porphobilinogen deaminase (uroporphyrinogen-I-synthase)

*The answer is C.* Chromosome walking involves taking the "ends" of a probe (the DNA at the 5′ and 3′ ends of a probe) and using those as new probes in another library screen. The clones identified in this second screening of the library will hybridize to the "end" of the already identified clone, and should extend the DNA both in the 5′ and 3′ directions from that already represented by the first clone, as indicated in the figure below. (A) B and C (B) A and D (C) A and E (D) D and E (E) B and D

A genomic clone has been isolated and is going to be used for chromosome walking to obtain more clones of the gene. The regions of the clone to be used as probes in chromosome walking are which of the following?

*The answer is A.* Erythromycin is the antibiotic of choice for pertussis. It inhibits translocation.

A nasopharyngeal swab obtained from a 4-month-old infant with rhinitis and paroxysmal coughing tested positive upon culture for Bordetella pertussis. He was admitted to the hospital for therapy with an antibiotic that inhibits the translocation of the 70S ribosomes on the mRNA. This patient was most likely treated with A. Erythromycin B. Tetracycline C. Chloramphenicol D. Rifamycin E. Levofloxacin

*The answer is B.* The vignette describes gel electrophoresis. Gel electrophoresis uses an electric field to separate molecules based on their sizes. The negatively charged DNA migrates in the electric field toward the positive end. Smaller fragments move more rapidly through the gel. Bands of DNA can be visualized by staining the gel with fluorescent dyes such as ethidium bromide.

A scientist is performing restriction mapping of various mutant and wild-type genes involved in Drosophila fly wing shape. A specimen of interest is treated with restriction enzymes, pipet ted into a well of agarose gel, and subjected to an electric field, As the final step in the experiment, the gel is stained with ethidium bromide and visualized under ultraviolet light. What of the following is the best term for the described procedure? A. Enzyme-linked immunosorbent assay B. Gel electrophoresis C. Northern blot D. Polymerase chain reaction E. Sequencing F. Southern blot G. Western blot

*The answer is A.* This patient suffers from I-cell disease, which is caused by a failure of addition of mannose-6-phosphate by GlcNAc phosphotransferase on the Golgi apparatus. Without mannose-6-phosphate, lysosomal enzymes cannot be properly directed for inclusion into lysosomes and will instead be excreted by the cell. Thus lysosomal enzymes, including hexosaminidase, iduronate sulfatase, and arylsulfatase A, will be found in the extracellular, but not intracellular, space. I-cell disease is characterized by skeletal abnormalities, restricted joint movement, coarse facial features, and severe psychomotor impairment. Death usually occurs in the first decade of life.

A term child is delivered by spontaneous vaginal delivery without complications. Upon physical examination the child has bilateral hip dislocations, restricted movement in shoulder and elbow joints, and coarse facial features. Laboratory studies show that the activities of β-hexosaminidase, iduronate sulfatase, and arylsulfatase A are deficient in cultured fibroblasts, but are 20 times normal in the patient's serum. The primary abnormality in this disorder is associated with which of the following organelles? (A) Golgi apparatus (B) Lysosomes (C) Ribosomes (D) Rough endoplasmic reticulum (E) Smooth endoplasmic reticulum

*The answer is B.* The boy has MELAS (Mitochondrial myopathy, encephalopathy, lactic acidosis, and stroke), a neurodegenerative disorder due to a mutation in mitochondrial tRNA, leading to defective protein synthesis within the mitochondria. The component most often affected is complex 1 of the respiratory chain. The severity of the disease will be dependent on the extent of heteroplasmy (what percentage of the mitochondria codes for the altered tRNA). The more mutant mitochondria present, the more severe the symptoms.

A young boy exhibits myopathy, encephalopathy, lactic acidosis, and strokelike episodes. All of his siblings have some aspects of the same symptoms. The boy most likely has which type of mutation? (A) A defect in mRNA synthesis (B) A defect in mitochondrial tRNA production (C) A defect in mitochondrial rRNA production (D) A defect in cytoplasmic tRNA (E) A defect in cytoplasmic rRNA

*The answer is C.* eIF-2 designates a protein factor of the initiation phase in eukaryotic translation. The only event listed that would occur during this phase is placement of initiator tRNA in the P-site.

Accumulation of heme in reticulocytes can regulate globin synthesis by indirectly inactivating eIF-2. Which of the following steps is most directly affected by this mechanism? A. Attachment of spliceosomes to pre-mRNA B. Attachment of the ribosome to the endoplasmic reticulum C. Met-tRNAmet binding to the P-site D. Translocation of mRNA on the ribosome E. Attachment of RNA polymerase II to the promoter

*The answer is A.* Restriction fragment length polymorphisms (RFLPs) arising from variable numbers of tandem repeats (VNTRs) are the basis of the DNA fingerprinting technique. The process is: (1) isolation of DNA from parent/child or forensic specimens using blood, skin, or semen; (2) PCR amplification and radioactive labeling of DNA from variable regions in each sample; (3) separation of the variable DNA fragments by gel electrophoresis; and (4) comparison of the DNA fragment patterns among samples. Since numbers of arrays of repeats of two, three, or four base pairs (microsatellites) may vary from 5 to 100 at a particular chromosome locus, particular alleles may occur in less than 1% of the population. As a result, analysis of three loci, each with two alleles, can produce odds as high as (100)6 that the pattern matches a putative father or suspect as compared to a random person from the general population. Reverse transcription based upon RNA-directed DNA synthesis is not utilized in DNA fingerprinting, and the size distribution of undigested DNA reflects its integrity during isolation rather than individual identify. HLA typing uses antibodies to define the constellation of alleles from various loci in the HLA region on chromosome 6. The tendency for certain HLA alleles to occur together (associate) on the same chromosome as haplotypes greatly reduces their odds of identity when compared to DNA fingerprinting.

DNA fingerprinting is used for paternity testing and forensic identification of suspects. Which of the following is the most accurate description of DNA fingerprinting? a. DNA can be isolated from blood, skin, or sperm and analyzed for variable patterns of restriction fragments arising from tandemly repeated sequences (microsatellites) b. DNA is copied from blood, skin, or sperm RNA using reverse transcriptase and analyzed for the pattern of complementary DNAs c. DNA is isolated from blood, skin, or sperm and its fragment size distribution is analyzed by gel electrophoresis d. DNA is isolated from blood, skin, or sperm and hybridized with probes from the HLA locus to visualize HLA gene patterns e. DNA is isolated from blood, skin, or sperm, centrifuged to separate satellite DNA fractions, and analyzed by gel electrophoresis

*The answer is A.* Ultraviolet irradiation causes thymine dimers to form in DNA. Replication is inhibited in cells until the pyrimidine dimers are removed. Removal of the damaged areas occurs in two ways. The process can be simply reversed by a photoreactivating enzyme that cleaves the dimers and yields the original bases. Blue light is required for this. Alternatively, the dimer is removed. A UV-specific excinuclease nicks the dimer on its 5′ side. DNA polymerase I replicates the damaged sequence, while the damaged sequence swings out. Finally, the damaged piece is hydrolyzed by the 5′ to 3′ exonuclease activity of the DNA polymerase I. DNA ligase then joins the new piece to the original DNA at the cleavage site.

Following ultraviolet damage of DNA in skin a. A specific excinuclease detects damaged areas b. Purine dimers are formed c. Both strands are cleaved d. Endonuclease removes the strand e. DNA hydrolysis does not occur

*The answer is C.* Despite the great length of the chromosomes of eukaryotic DNA, the actual replication time is only minutes. This is because eukaryotic DNA is replicated bidirectionally from many points of origin. The hundreds of initiation sites for DNA replication on chromosomes share a consensus sequence called an autonomous replication sequence (ARS). Thus, while the process of DNA replication in mammals is similar to that in bacteria, with DNA polymerases of similar optimal temperatures and speed, the many replication forks allow for a rapid synthesis of chromosomal DNA. Proteins such as histones, which are bound to mammalian chromosomes, inhibit DNA replication or transcription. Dissociation of the protein-DNA complex (chromatin) and unwinding of DNA supercoils (followed by chromatin reassembly) is part of the replication process.

Given that the chromosomes of mammalian cells may be 20 times as large as those of Escherichia coli, how can replication of mammalian chromosomes be carried out in just a few minutes? a. Eukaryotic DNA polymerases are extraordinarily fast compared with prokaryotic polymerases b. The higher temperature of mammalian cells allows for an exponentially higher replication rate c. Hundreds of replication forks work simultaneously on each piece of chromosomal DNA d. A great many different RNA polymerases carry out replication simultaneously on chromosomal DNA e. The presence of histones speeds up the rate of chromosomal DNA replication

1. *The answer is B.* rRNA genes are transcribed by this enzyme in the nucleolus. 2. *The answer is A.* Less condensed chromatin, lighter areas in the nucleus. Darker areas are heterochromatin. 3. *The answer is A.* Polyadenylation of pre-mRNA occurs in the nucleoplasm. Generally associated with active gene expression in euchromatin.

Identify the nuclear location. 1. Transcription of genes by RNA polymerase I 2. Euchromatin 3. Polyadenylation of pre-mRNA by poly-A polymerase

*The answer is C.* When region C is removed from the DNA, there is a tremendous reduction in the expression of the reporter gene, indicating that this region of the DNA binds important positively-acting transcription factors for the expression of the reporter gene.

In order to better analyze the promoter region of a particular gene, this cloned region of the gene was placed in front of a reporter gene and the resultant vectors placed in eukaryotic cells to measure the expression of the reporter, using various deleted constructs. The results obtained were as follows: Referring to the figure above, which region binds stimulatory transcription factors?

*The answer is B.* A special DNA polymerase called telomerase is responsible for replication of the telomeric DNA. Telomerase contains an RNA molecule that guides the synthesis of complementary DNA. Telomerase is therefore an RNA dependent DNA polymerase in a category with reverse transcriptase. Telomerase does not require an RNA primer, initiating synthesis of the leading strands at 3′ ends within the telomeric DNA. Synthesis of the lagging strands uses primase, DNA polymerase III, and DNA polymerase I, as with the replication of other chromosomal regions.

Mammalian chromosomes have specialized structures with highly repetitive DNA at their ends (telomeres). Which aspect of telomeric DNA replication is different from that of other chromosomal regions? a. The DNA polymerase uses an RNA primer but does not degrade it b. The DNA polymerase contains an RNA molecule that serves as template for DNA synthesis c. The DNA polymerase must cross-link the 5′ and 3′ termini d. The DNA polymerase has a σ subunit that facilitates binding to repetitive DNA e. The DNA polymerase does not use an RNA template or primer

*The answer is B.* Decreased factor V secretion and a corresponding accumulation of cytoplasmic antigen suggest a defect in the translocation of the nascent protein to the endoplasmic reticulum. This implies a mutation in the N-terminal amino acid signal sequence required for targeting to the ER and encoded by the first exon of the gene.

Parahemophilia is an autosomal recessive bleeding disorder characterized by a reduced plasma concentration of the Factor V blood coagulation protein. Deficiency arises from a 12 base-pair deletion in the Factor V gene that impairs the secretion of Factor V by hepatocytes and results in an abnormal accumulation of immunoreactive Factor V antigen in the cytoplasm. In which region of the Factor V gene would this mutation most likely be located? A. 5′ Untranslated region B. First exon C. Middle intron D. Last exon E. 3′ Untranslated region

*The answer is C.* STR sequences are amplified using a PCR and analyzed by gel electrophoresis. Although RFLP analysis could potentially be used for this purpose, it is not the method of choice.

Paternal relationship between a man and infant can be best determined by the technique commonly referred to as DNA finger-printing. Which of the following sequences is most conveniently analyzed in a DNA fingerprint? A. Histocompatibility loci B. Centromeres C. Microsatellite tandem repeats (STRs) D. Restriction enzyme sites E. Single-copy sequences

*The answer is E.* In eukaryotic gene transcription, nuclear RNA polymerase II uses a DNA template to generate complementary mRNA, which is then processed and translated into protein. Eukaryotic genes have associated promoter and enhancer sequences. Promoter regions serve as binding sites for transcription factors and RNA polymerase II. There are two types of eukaryotic promoter regions: 1) The TATA, or Hogness, box is located approximately 25 nucleotides upstream from the gene being transcribed, and 2) The CAAT box is 70 to 80 bases upstream from the gene. In contrast to promoters, enhancers increase the rate of transcription initiation through protein binding and interactions with transcription factors bound to promoter sequences. Enhancers can be located upstream or downstream from the gene being transcribed, and may be near the gene or thousands of base pairs away. (Enhancers have been identified both within introns of the gene being transcribed as wail as on separate chromosomes). Repressor elements are similar to enhancer elements, but they decrease instead of enhance transcription rates. *Educational Objective:* Enhancers repressors may be located anywhere upstream, downstream or even within the transcribed gene. In contrast, promoter regions are typically located 25 or 70 bases upstream from their associated genes.

RNA polymerase II is the enzyme responsible for eukaryotic gene transcription. Gene-associated enhancer sequences can increase the rate at which transcription is initiated. Most enhancers are located where, with respect to the transcription start site? A. 25-30 base pairs upstream B At least 70 base pairs upstream C. Within the transcribed portion of the gene D. Downstream of the gene E. Variable locations

*The answer is C.* Pseudomonas and diphtheria toxins inhibit eEF-2, the translocation factor in eukaryotic translation.

Respiratory tract infections caused by Pseudomonas aeruginosa are associated with the secretion of exotoxin A by this organism. What effect will this toxin most likely have on eukaryotic cells? A. Stimulation of nitric oxide (NO) synthesis B. ADP-ribosylation of a Gs protein C. ADP-ribosylation of eEF-2 D. ADP-ribosylation of a Gi protein E. Stimulation of histamine release

*The answer is C.* In a general sense, the mechanism of protein synthesis in eukaryotic cells is similar to that found in prokaryotes; however, there are significant differences. Cycloheximide inhibits elongation of proteins in eukaryotes, while erythromycin causes the same effect in prokaryotes. Thus, one is an antibiotic beneficial to humans, while the other is a poison. Cytoplasmic ribosomes of eukaryotes are larger, sedimenting at 80S instead of 70S. While eukaryotic cells utilize a specific tRNA for initiation, it is not formylated as in bacteria. Finally, eukaryotic mRNA always specifies only one polypeptide, as opposed to prokaryotic mRNA, which may specify the synthesis of more than one gene product per mRNA.

Which of the following statements regarding eukaryotic cells is true? a. Formylated methionyl-tRNA is important for initiation of translation b. Single mRNAs specify more than one gene product c. Cycloheximide blocks elongation during translation d. Cytosolic ribosomes are smaller than those found in prokaryotes e. Erythromycin inhibits elongation during translation

*The answer is B.* About 30% of the DNA of humans and other mammals consists of repeated sequences. Repetitive DNA includes numerous families of genes like those for histones. Some families of repeated genes make identical mRNA molecules, suggesting that their multiple gene copies are needed to make adequate amounts of protein. Although many genes in bacteria produce a polycistronic mRNA that encodes several different peptides, all mRNAs in mammals encode a single peptide and are monocistronic. In addition, RNA is initially transcribed from proteinencoding genes as larger molecules called heterogenous nuclear RNA (hnRNA). These immature hnRNA molecules must be spliced to remove introns and chemically modified with 5′ caps and 3′-polyA sequences before reaching the cytoplasm as functional mRNA. The initial HnRNA transcript is colinear with its encoded protein within exons but not within introns. Mature mRNAs also have 5′ and 3′ untranslated regions that are not colinear with the encoded peptide.

Which one of the following statements correctly describes the synthesis of mammalian messenger RNA (mRNA)? a. Each mRNA often encodes several different proteins b. Several different genes may produce identical mRNA molecules c. There is colinearity of the RNA sequence transcribed from a gene and the amino acid sequence of its encoded protein d. The RNA sequence transcribed from a gene is virtually identical to the mRNA that exits from nucleus to cytoplasm e. Mammalian mRNA undergoes minimal modification during its maturation

*The answer is E.* The child has Prader-Willi syndrome, which is due to a deletion of a cluster of genes on chromosome 15, on the long arm. When this deletion is inherited from the father, Prader-Willi syndrome is observed. If the same deletion is inherited from the mother, an entirely different syndrome is observed, termed Angelman syndrome. The diagnosis can be confirmed by FISH analysis using a probe specific for the 15q11-13 region.

You have been following a newborn who first presented with hypotonia and trouble sucking. Special feeding techniques were required for the child to gain nourishment. As the child aged, there appeared to be developmental delay, and the child then gained a great interest in eating, and rapidly became obese. Developmental delay was still evident, as was hypotonia. A karyotype analysis of this patient would indicate which of the following? (A) A monosomy (B) A trisomy (C) A duplication (D) A chromosomal inversion (E) A deletion

*The answer is A.* As muscle works, and AMP levels rise, the muscle wants to preserve its ATP for muscle contraction, and not to use it for new protein synthesis. The increase in AMP levels leads to the activation of AMP-activated protein kinase, which will phosphorylate and inactivate eIF4E (eukaryotic initiation factor 4E), which is a necessary component in recognizing the 5′ cap structure of the mRNA to allow ribosome assembly on the mRNA (see the fi gure in the answer to the previous question). The activation of the AMP-activated protein kinase does not alter elongation or the termination of translation. It does not block overall transcription, either of mRNA or rRNA (although it may lead to an inhibition of ribosomal biogenesis as well as the transcription of certain specifi c genes).

Under conditions of active exercise, protein synthesis is reduced in the muscle. Under these conditions, which aspect of translation is inhibited? (A) Inability to initiate translation (B) Inability to elongate during translation (C) Inability to terminate translation (D) Inability to synthesize mRNA (E) Inability to produce rRNA

*The answer is A.* The codon for valine is GUG, so the anticodon is the base pair CAC.

A 2-year-old boy who recently emigrated from Somalia is brought to the physician because of a 1-day history of pain of his arms and legs. Physical examination shows pale mucous membranes and hepatosplenomegaly. Laboratory studies show a hemoglobin concentration of 8 g/dL. A peripheral blood smear shows sickle cells. Genetic analyses show a point mutation in the β-globin gene leading to a change of a GAG codon (glutamate) to a GUG codon (valine). Which of the following anticodons is most likely in the tRNA for valine? (A) CAC (B) CTU (C) CUC (D) GAC (E) GCC

*The answer is C.* Bacteria that develop resistance to antibiotics usually do so by containing an enzymatic activity that destroys the structure of the antibiotic so that it cannot effectively inhibit its target within the cell. Amoxicillin works by destroying the bacterial cell wall, by being incorporated into the growing cell wall, which leads to a cessation of cell wall synthesis. It does not alter ribosome structure. Mutations in RNA polymerase will not lead to resistance to drugs. Azithromycin was effective because the bacteria did not produce an enzyme that destroyed the drug.

A 2-year-old boy with an ear infection was given amoxicillin, but it did not clear up the problem. Switching to azithromycin successfully eradicated the infection, and subsequent laboratory work indicated that the offending bacterium was resistant to amoxicillin. Bacterial resistance to antibiotics is often due to which of the following? (A) Altered ribosome structure (B) Altered cell wall (C) Enzymatic destruction of the antibiotic (D) Inability to transport the drug into the bacteria (E) A mutation in RNA polymerase

*The answer is E.* This patient 's presentation is most consistent with an atypical pneumonia, likely caused by Mycoplasma pneumoniae. Macrolide antibiotics such as azithromycin are the first -line treatment for atypical pneumonias. Macrolides inhibit protein synthesis by blocking the translocation of the peptide chain by binding to the 23S ribosomal RNA (part of the 50S subunit) during translation. The Mycoplasma bacterium has no cell wall, so it is not sensitive to the penicillins or cephalosporins, which act by inhibiting cell wall synthesis,

A 21-year-old college student presents to the student health center complaining of malaise, headaches, fever up to 38.3°C (101°F), chills, and a nonproductive cough for the past week. His temperature is 38.1°C (100.6°F), heart rate is 90/min, respiratory rate is 20/min, and blood pressure is 118/75 mm Hg. On physical examination, the patient appears fatigued. Auscultation of the lungs reveals bilateral diffuse wheezes and rales. An x-ray of the chest reveals fluffy bilateral infiltrates. The physician prescribes ceftriaxone, but the patient fails to show any improvement over the subsequent week. What is the mechanism of action of the antibiotic to which the patient should be switched? A. Blocking the transpeptidase cross-linkage of cell walls B. Forming free radical toxic metabolites that damage bacterial cell DNA C. Inhibiting DNA-dependent RNA polymerase D. Inhibiting fatty acid synthesis E. Inhibiting protein synthesis by blocking the translocation of the peptide chain

*The answer is A.* The woman has lupus, an autoimmune disorder. One class of antibodies developed is against the snurps, small ribonuclear protein complexes, which are involved in mRNA splicing. Autoantibodies are not developed against DNA polymerase (although antibodies against DNA are often found), carbohydrates, tRNA complexes, or peroxisomal proteins.

A 22-year-old woman (see the figure below) sees her physician for a variety of complaints over the past year. These include fevers that come and go, fatigue, joint pain and stiffness, a butterfly rash on the face, sores in her mouth, easy bruising, and increased feelings of anxiety and depression. A diagnostic blood test is likely to show autoimmune antibodies directed against which class of molecules? (A) Ribonuclear protein complexes (B) DNA polymerases (C) Carbohydrates (D) tRNA complexes (E) Peroxisomal proteins

*The answer is A.* The recent weight loss and palpitations are symptoms of hyperthyroidism. This, in combination with exophthalmos, suggests Graves disease. Graves disease is a type II hypersensitivity reaction, meaning that it is antibody mediated. Specifically, the antibodies generated target the thyroid stimulating hormone (TSH) receptor. Such antibodies bind to and stimulate the TSH receptor~ which results in a hyperthyroid state. The only test among the answer choices that involves testing for antibodies is the enzyme-linked immunosorbent assay. In this test, a blood sample is probed to detect either a test antigen or a test antibody (an antibody in this case). Although this is not the test usually performed to screen for Graves disease (usually TSH and triiodothyronine and thyroxine levels are measured), it could detect it.

A 25-year-old woman complains to her primary care physician of recent weight loss and palpitations. Physical examination reveals marked exophthalmos. Which laboratory technique would be most helpful in diagnosing this patient? A. Enzyme-linked immunosorbent assay B. Fluorescence in situ hybridization C. Gel electrophoresis D. Polymerase chain reaction E. Rest riction enzyme analysis F. Southern blot

*The answer is A.* This patient presents with a classic picture of Goodpasture syndrome, an autoimmune disorder in which the body produces immunoglobulins against the α3 chain found in type IV collagen- a structural component of the basement membrane of pulmonary alveoli and renal glomeruli. This is an example of a type II hypersensitivity reaction. Patients are typically young men with tl1e onset of pulmonary symptoms like fatigue, dyspnea, and hemoptysis, in addition to renal symptoms such as dysuria, hematuria, and renal failure. A CT scan is generally done for evaluation of hemoptysis. The CT scan shown here is a good example of the pulmonary manifestations of Goodpasture syndrome, Of all the answer choices, only AIport syndrome is a disease in which type IV collagen is defective. AIport syndrome is a hereditary form of glomerular nephritis with sensorineural hearing loss and sometimes ocular abnormalities. The disease arises from a mutation in one of the α chains of type IV collagen.

A 27-year-old man with no medical history is brought to the hospital after 3 days of hemoptysis. Initial work-up shows elevated levels of creatinine and blood urea nitrogen. On the second day of hospitalization, hematuria is noted. The macromolecule that is affected in this patient is also defective in which of the following diseases? A. Alport syndrome B. Bullous pemphigoid C. Ehlers-Danlos syndrome D. Marfan syndrome E. Osteogenesis imperfecta F. Stickler syndrome

*The answer is C.* Fragile X syndrome is a complex genetic disorder that most closely follows an X-linked dominant pattern of inheritance. It is caused by a trinucleotide repeat expansion (CGG). The disease is characterized by mental retardation and physical features such as macroorchidism (large testicles), long face, large mandible, and large, everted ears. The number of CGG repeats in patients with the full mutation average around 230 4000. In the normal population, this sequence repeats an average of 29 times. Patients with premutations have from 52 to 230 repeats and lack the disease phenotype. Differences in the size of CGG repeats can be shown on a Southern blot. Shorter fragments of DNA migrate more rapidly through a gel toward the positive electrode. Therefore, patients with full mutations (who have longer strands of DNA) will have a "sluggish" band that stays toward the top of the gel, the normal population will show a band at the bottom of the gel, and patients with a premutation will have a band in the middle. Lane C would best represent the Southern blot of a woman who carries a premutation for fragile X and one normal X.

A 28-year-old woman is trying to conceive a child. She has a nephew with fragile X syndrome (a genetic disorder characterized by trinucleotide repeat expansion) and she would like to assess her risk as a carrier for the disease. Blood is drawn, and DNA is extracted and cut by restriction enzymes that flank the CGG repeat region. The DNA is then treated with a labeled probe that binds the affected region of the gene. The woman is found to carry one normal X chromosome and one X chromosome with some expansion of the CGG sequence. However, the number of CGG repeats in this X chromosome is not sufficient to alter phenotype. Which lane on the Southern blot represents this woman's genotype? (A) A (B) B (C) C (D) D (E) E

*The answer is D.* Polymerase chain reaction (PCR) and sequencing can be used to determine if this woman and/or her husband is a carrier of the cystic fibrosis gene, the most common single-gene mutation in white people. This mutation commonly manifests with dysfunction of the lungs, pancreas, and other organs due to the buildup of thick mucus. PCR is used to amplify the region of interest, and sequencing is used to see if the cystic fibrosis mutation is present. A helpful mnemonic is SNoW DRoP for Southern = DNA. Northern = RNA. Western = Protein.

A 31-year-old white woman is trying to get pregnant. She has a niece who suffers from a genetic disease characterized by recurrent respiratory infections and pancreatic failure. She would like to assess her chances of having a child with this disease. Which of the following laboratory techniques could be used to determine if this woman and/or her husband is a carrier of the mutant gene? A. Enzyme-linked immunosorbent assay B. Gel electrophoresis C. Northern blot D. Polymerase chain reaction and sequencing E. Western blot

*The answer is G.* The quest ion describes the technique of a Western blot, which is used to detect proteins through protein-antibody hybridization.

A 32-year-old male that regularly uses IV drugs presents to his primary care provider with fever, chills, muscle aches, fatigue, and a rash. His primary care provider suspects HIV infection and orders several tests. For one of the confirmatory tests, a protein sample is taken and run on an electrophoretic gel, causing it to separate by size and charge. The protein is then transferred to a filter, and labeled antibody is added to allow visualization of the relevant protein. This is an example of which of the following molecular tests? A. Enzyme-linked immunosorbent assay B. Ligase chain reaction C. Northern blot D. Polymerase chain reaction E. Southern blot F. Southwestern blot G. Western blot

*The answer is A.* Deletion of CTT results only in the loss of phe 508; ile 507 and the C-terminal sequence are unaltered because ATC and ATT both code for ile (the coding se- quence is unchanged).

A 4-year-old toddler with cystic fibrosis (CF) is seen by his physician for an upper respiratory infection with Pseudomonas aeruginosa. He is started on oral ciprofloxacin and is referred to a CF center as a potential candidate for gene therapy. Prior genetic testing of the patient identified the mutation causing CF as a 3 base-pair deletion in exon 10 of the CF gene. The nucleotide sequences of codons 506-511 in this region of the normal and mutant alleles are compared below. What effect will this patient's mutation have on the amino acid sequence of the protein encoded by the CF gene? A. Deletion of a phenylalanine residue with no change in C-terminal sequence B. Deletion of a leucine residue causing a change in the C terminal sequence C. Deletion of a phenylalanine residue causing a change in the C-terminal sequence D. Deletion of a leucine residue with no change in C-terminal sequence

*The answer is B.* Red cell hemolysis after drug exposure suggests a red cell enzyme defect, most easily confirmed by enzyme assay to demonstrate deficient activity. A likely diagnosis here is glucose-6-phosphate dehydrogenase (G6PD) deficiency, probably the most common genetic disease (it affects 400 million people worldwide). Tropical African and Mediterranean peoples exhibit the highest prevalence because the disease, like sickle cell trait, confers resistance to malaria. DNA analysis is available to demonstrate particular alleles, but simple enzyme assay is sufficient for diagnosis. More than 400 types of abnormal G6PD alleles have been described, meaning that most affected individuals are compound heterozygotes. The phenotype of jaundice and red blood cell hemolysis with anemia is triggered by a variety of infections and drugs, including a dietary substance in fava beans. Sulfonamide and related antibiotics, as well as antimalarial drugs, are notorious for inducing hemolysis in G6PD-deficient individuals. G6PD deficiency exhibits X-linked recessive inheritance, explaining why male offspring but not the parents become ill when exposed to antimalarials.

A 5-year-old Egyptian boy receives a sulfonamide antibiotic as prophylaxis for recurrent urinary tract infections. Although he was previously healthy and well-nourished, he becomes progressively ill and presents to your office with pallor and irritability. A blood count shows that he is severely anemic with jaundice due to hemolysis of red blood cells. Which of the following would be the simplest test for diagnosis? a. Northern blotting of red blood cell mRNA b. Enzyme assay of red blood cell hemolysate c. Western blotting of red blood cell hemolysates d. Amplification of red blood cell DNA and hybridization with allele specific oligonucleotides (PCR-ASOs) e. Southern blot analysis for gene deletions

*The answer is B.* This vignette describes Huntington disease, which is caused by increased numbers of the sequence CAG on the allele of the Huntingtin gene. Variable number tandem repeats (VNTRs) are a type of DNA polymorphism created by a tandem arrangement of multiple copies of short DNA sequences; it is detected by DNA-DNA hybridization. Therefore, the best technique for assaying for this genetic alteration is a Southern blot. Southern blotting is a standard method for analyzing the structure of DNA cleaved by restriction enzymes. In this process, DNA is digested by restriction enzymes and separated by size using gel electrophoresis, allowing specific DNA molecules to be detected on a filter by their hybridization to labeled probes. Restriction enzymes that cut sequences on either side of a VNTR region to produce fragments that vary in size among individuals. Labeled DNA probes are used to identify and visualize these restriction fragments. In this case, a larger fragment indicates more CAG repeats.

A 50-year-old man is brought to the physician by his wife because he has grown increasingly irritable over the past several months. His wife also notes that sometimes his hand performs a jerky, rhythmic motion but the patient will pretend to have done this on purpose by smoothing his hair or scratching his head after it happens. His father experienced similar symptoms in his mid-fifties. What is the mechanism of the laboratory technique that would most accurately diagnose the patient's underlying genetic condition? A. Attachment of DNA-binding proteins to specific DNA sites B. Hybridization of labeled DNA probes to DNA sequences C. Hybridization of labeled DNA probes to RNA sequences D. Immunoassay to identify the presence of antibody or antigen in a sample E. Use of labeled antibodies to identify mutant protein

*The answer is C.* This patient likely has Ehlers-Danlos syndrome (EDS), a disease of faulty collagen synthesis. Patients with EDS have hyperextensible skin, hypermobile joints, (like that evidenced in these images) and easy bleeding manifested by excessive bruising. It also may be associated with joint dislocation, organ rupture, or berry aneurysm. EDS carl exhibit autosomal dominant or recessive inheritance, and its clinical severity can vary greatly. EDS is classified into several types depending on the specific mutation. Hypermobility type is the most common. Classical type demonstrates skin and joint manifestations resulting from type V collagen defects. However, vascular type can also have milder skin and joint manifestations with a type III collagen defect Vascular type is associated with vascular and organ rupture; this patient's emergency neurosurgery was probably prompted by a ruptured berry aneurysm. Type III collagen (reticulin) makes up the skin, blood vessels, uterus, fetal tissue, and granulation tissue.

A 53-year-old man presents to a clinic for the first time. He had not seen a doctor for years until a few months ago, when he presented to the emergency department with a severe headache. He underwent "life-saving brain surgery" and he now wants to take better care of his health. On examination, you notice he has flexible joints and compliant skin, as well as an abundance of bruises all over his body (resembling those shown in the image). When asked, he says these things " run in the family" but have never caused him any problems. His medical history is otherwise unremarkable. Which of the following is most likely defective in this patient's disease process? A. Type IV collagen B. Type II collagen C. Type III collagen D. Hyaline E. Fibrillin F. Type I collagen

*The answer is D.* The child has adenosine deaminase (ADA) deficiency, or SCID (severe combined immunodeficiency disease). The circulating blood cells pick up deoxyadenosine, which is toxic in the absence of adenosine deaminase, a necessary enzyme for the metabolism of deoxyadenosine. By incorporating functional ADA in bone marrow cells, the stem cells for the production of the circulating blood cells now express ADA, and the mature cells produced from the stem cells will now express ADA, and be resistant to the toxic effects of deoxyadenosine. Placing the gene in other tissues will not address the problem of the death of blood cells.

A 6-month-old has had recurrent bouts of pneumonia, and a chest X-ray shows a greatly reduced thymus. Gene therapy would successfully treat this disorder if the gene is placed in which cell type or organ? (A) Muscle (B) Liver (C) Lung (D) Bone marrow (E) Kidney

*The answer is A.* The child has Menkes disease, in which cellular copper transport is abnormal and produces a functional copper deficiency. Lysyl oxidase in collagen metabolism requires copper. His fragile bones and blood vessels result from weak, poorly crosslinked connective tissue.

A 6-month-old infant is seen in the emergency room with a fractured rib and subdural hematoma. The child's hair is thin, colorless, and tangled. His serum copper level is 5.5 nM (normal for age, 11-12 nM). Developmental delay is prominent. A deficiency of which enzyme activity most closely relates to these symptoms? A. Lysyl oxidase B. Prolyl hydroxylase C. γ-Glutamyl carboxylase D. Phosphotransferase in Golgi E. α-1,4-glucosidase

*The answer is B.* This patient 's history of poor oral intake, hemarthrosis, and signs of microhemorrhage are consistent with vitamin C deficiency, also known as scurvy. Several vitamin deficiencies cause anemia and other hematologic disorders, but scurvy is characteristically associated with gum bleeding and swelling as well as poor wound healing. Vitamin C is vital for proline hydroxylation in collagen synthesis; therefore, deficiency of this vitamin will cause defects in systems requiring collagen.

A 6-year-old boy is brought to the emergency department by his parents after complaining of left knee pain and refusing to bear weight on the limb. The patient also complains of mouth pain. On physical examination, the patient is unkempt and has multiple abrasions on his A elbows and knees that appear to be healing poorly. He has petechiae on his chest and back, and oral examination shows inflamed and bleeding gums. Laboratory results show a hemoglobin level of 10.2 g/dL. The patient's mother reports that they have been homeless and living in a shelter for the past year, and oral intake has been poor for the past several months. Which of the following mechanisms of action is most likely responsible for this patient 's symptoms? A. Impaired carboxylation of glutamic acid B. Impaired collagen synthesis C. Impaired free radical scavenging D. Impaired immune response E. Inhibition of the tricarboxylic acid cycle

*The answer is D.* This child and his uncle appear to have Friedreich ataxia, a trinucleotide repeat disease. Like other trinucleotide repeat diseases, illness occurs because the unstable microsatellite regions on certain chromosomes have triplet codons that expand, typically worsening from generation to generation (and often making the age of onset earlier for each successive generation). These regions of massively expanded triplet repeats (most commonly between 600 and 1200 in Friedreich ataxia) cause a decrease in the product of a gene, frataxin, at the translation stage.

A 6-year-old boy is diagnosed with a worsening ataxic gait and a cardiac dysrhythmia. His uncle also has this condition, but his symptoms did not appear until he was 12 years of age. What is the molecular mechanism of this disease? (A) Unstable repeats affect protein folding (B) Unstable repeats affect protein splicing (C) Unstable repeats cause an amino acid substitution (D) Unstable repeats impede protein translation (E) Unstable repeats result in a truncated protein

*The answer is G.* This patient has xeroderma pigmentosa, an autosomal recessive disease characterized by a defect in excision repair. This disease results in an inability to repair thymidine dimers that can form in the presence of ultraviolet light. This can lead to the development of skin cancer and photosensitivity.

A 6-year-old boy presents to his pediatrician with skin lesions all over his body. For several years he has been very sensitive to sunlight. Neither the boy's parents nor his siblings have the same skin lesions or sun sensitivity. Biopsies of several of the boy's lesions reveal squamous cell carcinoma. Which mutation would one expect to see in this patient's DNA? (A) Methylation of the gene (B) Missense mutation in the gene (C) Nonsense mutation in the middle of the gene (D) Point mutation within the enhancer region (E) Point mutation within the operator region (F) Point mutation within the promoter region (G) Thymidine dimers

*The answer is D.* Messenger RNA (mRNA) is produced from DNA by RNA polymerase II. After processing, which includes removal of noncoding regions of mRNA (introns), the mRNA molecule is transported to the cytoplasm for translation. Remember that the mRNA molecule is composed of groups of three sequential bases known as codes. Since there ere 64 possible combinations of the four bases found in DNA, there are 64 possible codons. Because there are only twenty amino acids, most amino acids have more than one potential codon. For instance. UUU and UUA both code for the amino acid phenylalanine. Some codes call for the termination of synthesis of the polypeptide chain and are referred to as stop codons, including UAA, UAG, and UGA. Changes in the DNA sequence (mutations) can result in alterations of protein structure, affecting enzyme function. In the case described in the question stem, the structure of the protein is significantly shortened secondary to an alteration of the genetic code through a point mutation, which most likely resulted in the introduction of a premature stop codon. Of the choices listed, the single base change of UCA (serine) to UGA results in premature termination of protein synthesis and the formation of a truncated protein molecule. The introduction of a stop codon in the middle of a protein sequence is called a nonsense mutation (Choice D).

A 6-year-old male with hemolytic anemia is found to have an abnormality due to an inactive erythrocyte enzyme. The defective enzyme contains 158 amino acid residues instead of the normal 190 residues. A point mutation in exon 2 of the enzyme gene is identified as the cause for this patients disease. Which of the following mRNA code changes it most likely in this case? A. UAA → UAG B. UUU → UUA C. CUU → AUU D. UCA → UGA E. UAC → CAC

*The answer is D.* Vancomycin is a glycopeptide antibiotic that is effective in fighting only gram-positive bacteria. It binds tightly to a cell wall precursor that contains the terminal amino acid sequence D-ala D-ala and prevents cell wall synthesis. Resistance to vancomycin is transferred via plasmids and encodes enzymes that convert the D-ala D-ala peptide bridge to D-ala D-lac, preventing vancomycin from binding. Vancomycin resistance is much more common with Enterococcus faecium than with Enterococcus faecalis. High-dose ampicillin, often in combination with gentamicin, is generally first-line treatment in urinary tract infections due to vancomycin-resistant Enterococcus.

A 65-year-old woman who has been in the hospital for three weeks receiving cefotaxime to treat Klebsiella pneumonia develops a urinary tract infection. Urine cultures are positive for Enterococcus faecium. Treatment with vancomycin is attempted but is unsuccessful. Which of the following aided in this microorganism's ability to persist despite vancomycin treatment? (A) Alteration of microorganism's gyrase (B) Methylation of microorganism's rRNA at a ribosome-binding site (C) Microorganism's ability to produce β-lactamase (D) Mutation in terminal amino acid of microorganism's cell wall component (E) Mutation in the microorganism's penicillin- binding protein

*The answer is C.* DNA replication to similar in prokaryotes and eukaryotes, with DNA polymerases I and III being the main polymerase enzymes involved in prokaryotic DNA replication. For DNA replication to begin, DNA helical must first unwind the DNA double helix and separate Use parent strands (Choice A). The unwound single-stranded DNA is stabilized by the binding of single-stranded DNA-binding proteins to prevent spontaneous reannealing [Choice B). Synthesis of the daughter strands occur simultaneously from both parent strands. Because DNA synthesis can occur only in the 5' to 3' direction, one daughter strand is synthesized continuously toward the replication fork (leading strand). However, the other strand must be synthesized discontinuously in a direction away from the replication fork (lagging strand), with more and more segments being added as the replication fork moves across the DNA double helix. This results in the formation of Okazaki fragments, short stretches of newly synthesized DNA that are separated by RNA primers. These primers are removed and replaced with DNA, and the Okazaki fragments are subsequently joined together by DNA ligase. Because of the discontinuous nature of DNA synthesis on the lagging strand, DNA ligase acts many more times on the lagging strand than on the leading strand. *Educational Objective:* DNA replication occurs in the 5' to 3' direction on both strands. In contrast to the continuous synthesis of the leading strand, lagging strand synthesis occurs discontinuously and is composed of short stretches of RNA primer plus newly synthesized DNA segments (Okazaki fragments). As a result, lagging strand synthesis requires the repetitive action of DNA primase and DNA ligase

A 65-year-old woman with chronic obstructive pulmonary disease and type II diabetes mellitus comes to the emergency department due to profound fevers and malaise. After initial evaluation, she is hospitalized for septicemia. Blood cultures plated on lactose-containing media grow rapidly dividing gram-negative bacteria. Bidirectional DNA replication in these microbial cells requires synthesis of 2 daughter strands of DNA, each using one of the parent strands as a template Which of the following processes will differ the most between the 2 daughter strands during their synthesis? A. Enzymatic function of DNA helical B. Interaction with single-stranded DNA-binding proteins C. Joining of DNA fragments by ligase D. Proofreading of the nesWy synthesized DNA E. Relief of supercoils by topoisomerase

*The answer is C.* The necessary transcription factor is not binding as well to the DNA, leading to reduced efficiency of transcription. The side chain of aspartic acid can participate in hydrogen bonding with DNA bases, but the side chain of valine (completely carbon-hydrogen bonds) cannot. Transcription factors do not bind to DNA via ionic interactions. An increase in hydrogen bonding between the factor and DNA would lead to enhanced transcription (not reduced transcription). An inability of the transcription factor to bind to DNA would lead to no transcription of the gene; since there is a low level of expression, binding of this factor to the DNA must be occurring, although the affinity for the factor and DNA has been reduced.

A cell is producing a certain transcription factor that contains a single point mutation, such that N is converted to V. Circular dichroism experiments show that this altered factor has the same secondary structure as the nonmutated factor. Under normal conditions, serum stimulation of quiescent cultures shows strong induction of five genes. When quiescent cells harboring the mutant transcription factor are exposed to serum, the level of expression of those five genes stays at basal levels (there is no increase in mRNA production). This finding is most likely due to which of the following? (A) Loss of ionic interactions between the transcription factor and DNA (B) Increase in hydrogen bonding between transcription factor and DNA (C) Decrease in hydrogen bonding between the transcription factor and DNA (D) Increase in ionic interactions between the transcription factors and DNA (E) Inability of the transcription factor to bind to the DNA

*The answer is B.* Before DNA replication can actually begin, unwinding protein must open segments along the DNA double helix. A defective unwinding protein slows the overall rate of DNA synthesis, but does not alter the size of replicated DNA fragments. Defects in DNA synthesis or transcription may produce a phenotype of accelerated aging, as in Cockayne's syndrome (usually defective in a transcription factor). After unwinding, DNA-directed RNA polymerase (primase) catalyzes the synthesis of a complementary RNA primer of approximately 50 to 100 bases on each DNA strand. Then DNA-directed DNA polymerase III adds deoxyribonucleotides to the 3′ end of the primer RNA, which replicates a segment of DNA, the Okazaki fragment. DNA polymerase I then removes the primer RNA and adds deoxyribonucleotides to fill the gaps between adjacent Okazaki fragments. The fragments are finally joined together by DNA ligase to create a continuous DNA chain

A child presents with severe growth failure, accelerated aging that causes adult complications such as diabetes and coronary artery disease, and microcephaly (small head) due to increased nerve cell death. In vitro assay of labeled thymidine incorporation reveals decreased levels of DNA synthesis compared to controls, but normal-sized labeled DNA fragments. The addition of protein extract from normal cells, gently heated to inactivate DNA polymerase, restores DNA synthesis in the child's cell extracts to normal. Which of the enzymes used in DNA replication is likely to be defective in this child? a. DNA-directed DNA polymerase b. Unwinding proteins c. DNA polymerase I d. DNA-directed RNA polymerase e. DNA ligase

*The answer is C.* Polymerase chain reaction (PCR) is used to amplify a particular segment of DNA, including a gene of interest, and then subsequent sequencing or restrictive digestion will enable the detect ion of specific gene mutations or polymorphisms. This approach is commonly used to detect gene mutations leading to cystic fibrosis (CF). CF is an autosomal recessive disease caused by a genetic mutation in the cystic fibrosis transmembrane conductance regulator (CFTR) gene. Many mutations in the CFTR gene can lead to CF, including the Δ-F508 mutation that involves the loss of three nucleotides that normally encode for phenylalanine at position 508 on the gene.

A couple is expecting their first child, and they come to the genetics clinic for counseling regarding cystic fibrosis. Previous testing has shown that the woman and her husband are carriers for the Δ-F508 mutation. They are now requesting test ing of the fetus. Which test is the preferred method for detecting the suspected DNA mutation? A. Enzyme-linked immunosorbent assay B. Northern blot C. Polymerase chain reaction and sequencing D. Western blot

*The answer is C.* Knowing the son is homozygous for the normal allele, one can conclude that the two restriction fragments shown in his pattern derived from chromosomes without the mutation. It is also clear that the upper (larger) fragment came from his mother's chromosome and the lower (smaller) fragment came from his father's chromosome. The fetus has the fragment from his mother's normal chromosome. The other fragment (top one on the blot) must have come from the father's chromosome with the mutation. The fetus therefore is heterozygous for the mutation and the normal allele of the β-globin gene.

A couple seeks genetic counseling because both the man and the woman (unrelated to each other) are carriers of a mutation causing β-thalassemia, an autosomal recessive condition. The couple has one son who is phenotypically normal and has been shown by DNA analysis to be homozygous for the normal allele. They wish to know whether the fetus in the current pregnancy will have β-thalassemia. Using a probe for the β-globin gene that detects a BamHI RFLP, the following results are obtained. What is the best conclusion about the fetus? A. The fetus has inherited the mutation from both parents. B. The fetus has inherited the mutation from the mother but not from the father. C. The fetus has inherited the mutation from the father but not from the mother. D. The fetus has not inherited the mutation from either parent. E. The results are inconclusive.

*The answer is B.* The woman has a Robertsonian translocation between chromosomes 14 and 21 (the two chromosomes are fused together at their stalks.. When she creates her eggs, there is an imbalance in the amount of DNA representing chromosomes 14 and 21 in the eggs, such that fertilization of the eggs will lead to either monosomy or trisomy with these chromosomes, most of which are incompatible with life.mPolyploid outcomes would be three or more times the normal number of chromosomes, which does not occur here; and the Robertsonian translocation will not affect the distribution of the X chromosome. Trisomy 21 will lead to a live birth, Down syndrome, although there is still a risk of miscarriage with trisomy 21 conceptions. The risk is lower, however, than an imbalance of DNA brought about by the segregation of the chromosomes containing the Robertsonian translocation. Euploid cells have a number of chromosomes which are exact multiples of the haploids (in humans haploid is 23, diploid is 46, and polyploid is 69 or 92 chromosomes)

A couple sees an obstetrician due to difficulties of the woman keeping a pregnancy to term. She has had three miscarriages over the past 6 years, and the couple is searching for an answer. Karyotype analysis of the woman gave the result of 45,XX,der(14;21). A likely potential cause of the miscarriages may be which of the following? (A) Imbalance of DNA in polyploid conceptions (B) Imbalance of DNA in euploid conceptions (C) Triple X conceptions (D) Zero X conceptions (E) Trisomy 21 conceptions

*The answer is E.* The process of transduction involves the transfer of a portion of DNA from one bacterium to the chromosome of another bacterium by means of a viral infection. Conjugation is the transfer of a so-called male chromosomal DNA to the DNA of an acceptor, or female, bacterial cell. Colinearity defines the relationship between genes and proteins in that the sequence of amino acids in proteins is a result of the sequence of base triplets in template genes. Recombination is simply the exchange of sequences between two molecules of DNA. Transformation results when exogenous DNA fragments are incorporated into the chromosome of another organism, as in the transformation of pneumococcal bacteria that led Avery and McLeod to recognize the genetic significance of DNA.

A culture of bacteria not resistant to tetracycline develops an infection from a virus that is derived from the lysis of tetracycline-resistant bacteria. Most of the bacterial progeny of the original culture is found to have become resistant to tetracycline. What phenomenon has occurred? a. Conjugation b. Colinearity c. Recombination d. Transformation e. Transduction

*The answer is D.* Expression of cDNA in bacteria does not allow for posttranslational events to occur (bacteria do not carry out such events). Thus, the cleavage of insulin (removal of the C peptide) to form mature insulin would not occur in bacteria. Thus, in making recombinant insulin, the A and B chains are produced separately in bacteria, purified, and then mixed together under appropriate conditions to allow the disulfide bonds to form between the chains to form mature insulin. One cannot express the entire insulin gene in bacteria, as bacteria do not splice mRNA. It has been less expensive to synthesize insulin in bacteria than to try and find a eukaryotic cell line which would process insulin appropriately, and at high levels.

A difficult step in the engineering of recombinant insulin was which of the following? (A) Expressing the entire insulin cDNA in bacteria (B) Expressing the entire insulin gene in bacteria (C) Finding bacteria to properly cleave the insulin preproprotein (D) Allowing the A and B chains to come together appropriately (E) Finding the appropriate eukaryotic cell in which to express the gene

*The answer is C.* After the locus responsible for a genetic disease is mapped to a particular chromosome region, "candidate" genes can be examined for molecular abnormalities in affected individuals. The connective tissue abnormalities in Stickler syndrome make the COL2A1 collagen locus an attractive candidate for disease mutations, prompting analysis of COL2A1 gene structure and expression. Western blotting detects gene alterations that interfere with protein expression, while use of the reverse transcriptase polymerase chain reaction (RT-PCR) detects alterations in mRNA levels. Each analysis should detect one-half the respective amounts of COL2A1 protein or mRNA in the case of a promoter mutation that abolishes transcription of one COL2A1 allele. Southern blotting detects nucleotide changes that alter DNA restriction sites, but this is relatively insensitive unless large portions of the gene are deleted. Fluorescent in situ hybridization (FISH) analysis using DNA probes from the COL2A1 locus is a sensitive method for detecting deletions of the entire locus, and DNA sequencing of the entire gene provides the gold standard for detecting any alteration in the regulatory or coding sequences. Nucleotide sequence changes are still subject to interpretation, since they may represent polymorphisms that do not alter gene function. Population studies and/or in vitro studies of gene expression are often needed to discriminate DNA polymorphisms from mutations that disrupt gene function. For any autosomal locus, the interpretation of molecular analyses is complicated by the presence of two homologous copies of the gene.

A family in which several individuals have arthritis and detached retina is diagnosed with Stickler syndrome. The locus for Stickler syndrome has been mapped near that for type II collagen on chromosome 12, and mutations in the COL2A1 gene have been described in Stickler syndrome. The family became interested in molecular diagnosis to distinguish normal from mildly affected individuals. Which of the results below would be expected in an individual with a promoter mutation at one COL2A1 gene locus? a. Western blotting detects no type II collagen chains b. Southern blotting using intronic restriction sites yields normal restriction fragment sizes c. Reverse transcriptase-polymerase chain reaction (RT-PCR) detects one-half normal amounts of COL2A1 mRNA in affected individuals d. Fluorescent in situ hybridization (FISH) analysis using a COL2A1 probe detects signals on only one chromosome 12 e. DNA sequencing reveals a single nucleotide difference between homologous COL2A1 exons

*The answer is B.* The Ames test is a rapid and relatively inexpensive bacterial assay for determining mutagenicity of potential toxic chemicals. Since many chemical carcinogens are mutagenic, it seems obvious that damage to DNA is a central event in carcinogenesis as well as mutagenesis. Dr. Bruce Ames developed a tester strain of Salmonella that has been modified not to grow in the absence of histidine because of a mutation in one of the genes for the biosynthesis of histidine. Toxic chemicals that are mutagens are placed in the center of the plate and result in reversions of the original mutations, so that histidine is synthesized and the mutated revertants multiply in histidine-free media. Since many carcinogens are converted to active forms by metabolism in the liver, preliminary incubation with liver homogenates may precede the bacterial assay. Essentially all chemicals known as carcinogens in humans cause mutagenesis in the Ames test. The other options—carcinogenicity screening in immunosuppressed (nude) mice, computer modeling, or incubation with mammalian cell cultures—may provide some information, but are less efficient and validated than the Ames test. Contamination of Michigan cattle feed with polychlorinated biphenyls (PCBs) did occur through an industrial mistake.

A farming couple in Northern Michigan consult their physician about severe skin rashes and ulcers noted over the past year. They also have lost many cattle over the past year, and claim that their cattle feed changed in consistency and smell about 1 year ago. Chemical analysis of the feed shows high concentrations of polychlorinated biphenyls, a fertilizer related to known carcinogens. The physician sends the chemical to a laboratory for carcinogen testing, which is performed initially and rapidly by a. Inoculation of the chemical into nude mice b. Incubation of mutant bacteria with the chemical to measure the rate of reverse or "back" mutations c. Incubation with stimulated white blood cells to measure the impact on DNA replication d. Computer modeling based on the structures of related carcinogens e. Incubation with mammalian cell cultures to measure the rates of malignant transformation

*The answer is E.* This is a palindromic sequence and is therefore the correct answer. A palindromic sequence in DNA occurs when reading the nucleic acids in the 5' to 3' direction on one strand matches reading the nucleic acids on the complementary strand in the other direction. In this case, the complementary strand of 5' TGTACA 3' is 3' ACATGT 5 ', Reversing that strand would read 5 ' TGTACA 3', which is identical to the original strand. This is a true DNA palindromic sequence.

A graduate student is running a laboratory experiment on Escherichia coli. For the experiment, she must insert a segment of human DNA into the circular genome of the bacteria. To accomplish this, she utilizes a restriction enzyme, which is known for its ability to cut DNA at palindromic sequences. Which of the following is a palindromic sequence of DNA? A. CAGAGC B. CGTCGC C. TAGGAT D. TATAAG E. TGTACA

*The answer is C.* The girl is expressing HPFH (hereditary persistence of fetal hemoglobin). This can come about by deletions on the locus control region of the β-globin gene cluster (since fetal hemoglobin is α2γ2, and adult hemoglobin is α2β2, mutations in the locus control region of the α-gene cluster will not affect fetal hemoglobin synthesis). A general loss of transcription factors would not lead to increased transcription of the γ-chains, nor would a general increase in all transcription factor expression in the cell. While inappropriate looping may help to lead to γ globin gene expression, the looping needs to be modulated by transcription factors for gene expression to occur.

A healthy teenage girl has come to her pediatrician for a presports physical. Results of hemoglobin electrophoresis indicated an elevation of fetal hemoglobin. This can come about via which of the following mechanisms? (A) Overall increased expression of all transcription factors (B) Overall reduced expression of all transcription factors (C) Deletions in the locus control region of the β-globin gene cluster (D) Deletions in the locus control region of the α-globin gene cluster (E) Inappropriate looping of chromosomal DNA, allowing transcription of previously inaccessible genes to occur

*The answer is C.* The sibling has, in addition to sickle cell disease, hereditary persistence of fetal hemoglobin (HPFH). Individuals with HPFH express the γ-globin chain throughout their life, and it can be at high levels. Since this child is expressing both the HbS protein and the γ-protein, some normal fetal hemoglobin can be formed in this child, with reduced levels of HbS formed. This reduces the level of sickling and allows oxygen delivery to the tissues. Thus, the HPFH protects against the effects of homozygous HbS expression, and the sibling shows few, if any, symptoms of his HbS mutations. Alterations in the expression of the δ-chain have not been observed. Reduced α-chain synthesis would lead to an anemia, and there is no effective way to reduce the synthesis of HbS chain. Increased ζ-synthesis has also not been observed.

A hematologist is studying an African American family as one of the children was recently diagnosed with sickle cell disease. His sibling shows no symptoms of the disease, although genetic tests showed homozygosity for the HbS gene. An analysis of his red blood cells is likely to show which of the following? (A) Reduced alpha chain synthesis (B) Reduced sickle chain synthesis (C) Increased gamma chain synthesis (D) Increased zeta chain synthesis (E) Increased delta chain synthesis

*The answer is E.* This patient is presenting with scurvy. Scurvy is largely found in alcoholics due to poor overall nutrition that results in vitamin C (ascorbic acid) deficiency. As a result, proline and lysine are not hydroxylated in the endoplasmic reticulum during the formation of collagen. This leads to ecchymoses, petechiae, bleeding gums, and the appearance of the characteristic corkscrew hairs.

A homeless man presents to the emergency department complaining of weakness and fatigue. On examination, the physician observes a malnourished, unkempt man with alcohol on his breath. He has diffuse bruising on his arms and legs and corkscrew-shaped hair all over his body. This disorder is due to which of the following? A. Defect of glycosylation of proline and isoleucine in the Golgi apparatus B. Defect of glycosylation of proline and lysine in the endoplasmic reticulum C. Defect of glycosylation of proline and lysine in the Golgi apparatus D. Defect of hydroxylation of proline and isoleucine in the Golgi apparatus E. Defect of hydroxylation of proline and lysine in the endoplasmic reticulum

*The answer is A.* Type A and B blood differ by the presence of one sugar on glycosylated proteins. Individuals with type A blood add one type of sugar (N-acetylgalactosamine), while individuals with type B blood add a different sugar (galactose) due to difference in the specifi city of a glycosyl transferase that recognizes the base carbohydrate structure. The differences in blood group antigens are not due to acylation or proteolytic processing on the cell surface. As these carbohydrates are O-linked, dolichol (which is required for N-linked glycosylation) is not required for their synthesis.

A hospital laboratory made an error and mistyped a patient's blood as AB, instead of B. When given type A blood, the patient had an adverse reaction. The major difference between individuals with AB and B type blood is due to which of the following? (A) The presence of a specific glycosyl transferase (B) The presence of a specific acyl transferase (C) The presence of a specific peptidase (D) The lack of dolichol pyrophosphate (E) Increased levels of dolichol pyrophosphate

*The answer is C.* For the sake of this answer, let us assume that the overexpressed gene (named A) is located on chromosome X and the mutated gene (named B) is located on chromosome Y. The gene on chromosome Y is producing a transcriptional repressor that binds to the promoter region of the gene A on chromosome X. When the repressor is expressed, gene A transcription is reduced. The protein product of gene B is acting in trans in regulating gene A expression. A mutation that inactivates the protein product of gene B, then, would be unable to repress gene A transcription, and lead to overexpression of gene A. If the promoter for gene A had a TATA box mutation, one would expect reduced expression (due to an inability of the basal transcription complex to bind), rather than enhanced expression. Similarly, if the locus control region of the gene is deleted, then reduced expression for gene A would be expected, not enhanced expression. If a transcriptional activator (gene B) suffered a missense mutation that reduced its affinity for DNA, there would be less transcription of gene A, not enhanced expression. And, finally, promoter regions do not undergo splicing.

A human genetic condition in which too much of a gene is routinely expressed has been mapped to a locus on a different chromosome from where the gene in question is located. Which one of the following is a potential explanation for the condition? (A) The activated gene has a TATA box mutation (B) The locus control region for the gene is deleted (C) A gene encoding a transcriptional repressor has been mutated (D) A transcriptional activator sustained a missense mutation, which reduces its affinity for DNA (E) A variant promoter region is formed owing to a splice site mutation

*The answer is A.* Individuals with Li-Fraumeni syndrome inherit a mutated copy of p53, the product of the TP53 gene (on chromosome 17). p53 is a transcription factor whose major job is to monitor the health of DNA; if DNA alterations are found, p53, acting as a transcription factor, will initiate new gene transcription to arrest the cell cycle until the DNA damage is repaired and to also induce genes necessary for DNA repair. If the DNA cannot be repaired, p53 will initiate gene transcription leading to cellular death (apoptosis). In the absence of p53 activity, damaged DNA will be replicated, which increases the probability of errors, eventually causing a mutation that leads to a cancer. Thus, the initial inactivating event is impaired gene transcription by p53, which is the trigger for all other events that follow. Mutations in p53 do not lead, directly, to enhanced gene transcription (this may occur as a result of secondary mutations, but not directly from the mutations in p53) or to alterations in protein synthesis. p53 mutations also do not alter chromosome structure.

A long-standing patient of yours has developed multiple tumor types during his life (41 years old). You have diagnosed him as having a specific syndrome involving p53. The multiple cancers that result from this syndrome are primarily due to which of the following initial direct effects of the inherited mutation? (A) Impaired gene transcription (B) Enhanced gene transcription (C) Impaired protein synthesis (D) Enhanced protein synthesis (E) Altered chromosomal structure

*The answer is A.* Challenges for gene therapy include the construction of recombinant viral genomes that can propagate the replacement gene (gene constructs or vectors), delivery of the altered gene to the appropriate tissues (gene targeting), and recombination at the appropriate locus so that replacement of the defective gene is achieved (site-specific recombination). The latter step positions DNA sequences in the vector so that the replacement gene pairs and recombines precisely with homologous DNA in the native gene. Ex vivo transfection (introduction of vector DNA into patient cells outside the body) is an ideal method for gene targeting if the engineered cells can repopulate the tissue/organ in question. Transfection of bone marrow stem cells with a functional adenosine deaminase gene, followed by bone marrow transplantation back to the patient, has been successful in restoring immunity to children with severe deficiency. Even when tissue targeting and precise gene replacement are feasible, mimicking the appropriate patterns of gene expression can be a substantial barrier to gene therapy. Injection of deficient enzymes into serum (enzyme therapy) has been successful in disorders such as Gaucher's disease (storage of lipids in the spleen and bone), and takes advantage of cellular pathways that target enzymes to lysosomes.

A major obstacle to gene therapy involves the difficulty of homologous gene replacement. Which of the following strategies addresses this issue? a. A recombinant vector contains complementary DNA sequences that will facilitate site-specific recombination b. A recombinant vector expresses antisense nucleotides that will hybridize with the targeted mRNA c. A recombinant vector replaces inessential viral genes with a functional human gene d. A recombinant vector transfects patient cells, which are returned to the patient e. A recombinant vector contains DNA sequences that target its expressed protein to lysosomes

*The answer is B.* Once individual genes have been identified as being either upregulated or downregulated after differentiation, the DNA corresponding to the gene can be used as a probe in Northern blots using mRNA obtained at various times after differentiation is induced. In this manner, the time course of increase, or decrease, in mRNA expression can be determined. None of the other techniques will allow this information to be easily obtained. Western blots require antibodies to the proteins produced from the genes identified, which most likely are not available. A microarray analysis is too complex to easily determine the temporal order of gene expression. RFLP analysis does not address the question of temporal gene expression.

A microarray experiment looking at genes expressed by a cell line both before and after differentiation of the line indicated 15 potential genes which were upregulated. A simple technique to enable the scientist to determine the temporal order of induced gene expression is which of the following? (A) Southern blot (B) Northern blot (C) Western blot (D) RFLP analysis (E) Microarray analysis

*The answer is E* This patient has hemoptysis and hematuria, suggesting either Goodpasture syndrome or granulomatosis with polyangiitis (formerly called Wegener granulomatosis), Goodpasture syndrome is caused by autoantibodies to type IV collagen, whereas in granulomatosis with polyangiitis antineutrophil cytoplasmic antibodies (ANCA) develop within neutrophils. Therefore Goodpasture syndrome is the correct answer. The hematuria in this patient is indicative of rapidly progressive glomerulonephritis. Type IV collagen present in basement membranes within the lung alveoli and renal glomeruli are targeted by autoantibodies, causing the inflammatory reaction and aforementioned symptoms.

A middle-age man schedules an urgent visit with his physician because his urine has been darker and lower in volume than normal. He has also had fevers and headaches. He also reports a cough that does not seem to improve with time and has recently been producing red-tinged sputum. Pulmonary examination reveals diffuse crackles. The patient's urine is positive for blood and protein, and his blood urea nitrogen level is elevated at 24 mg/ dL. What is the target of the pathological process causing this patient's pulmonary symptoms? A. Surfactant B. Type I collagen C. Type I pneumocytes D. Type II pneumocytes E. Type IV collagen

*The answer is C.* This case is an example of Burkitt's lymphoma, which may affect the tonsils or other lymphoid tissues. The translocation places the myc oncogene on chromosome 8 downstream of the very active heavy chain locus on chromosome 14, activating myc gene expression in B cells and their derivatives. The translocation is likely an aberrant form of the normal DNA rearrangements that generate unique heavy chain genes in each B cell. The translocation joins one chromosome 8 to one chromosome 14, leaving their homologues unaffected. The cause for the phenotype must therefore be trans-acting, since cis-acting effects would pertain only to the translocated loci and not affect the homologous untranslocated loci. Activation of a tumor promoting gene (oncogene) on chromosome 8 could produce an enlarged tonsil, while underactivity of immunoglobulin production due to one-half expression could decrease immune function but would not completely ablate the processes in choices a, b, d, and e. At the genetic level, trans-acting events are autosomal dominant in that one of the two homologous loci is abnormal and produces a phenotype. Mutations of cis-acting events must disrupt both homologous loci to produce phenotypes, making them autosomal recessive at the genetic level.

A middle-aged man presents with a markedly enlarged tonsil and recurrent infections with serum immunoglobulin deficiency. Chromosome analysis demonstrates a translocation between the immunuglobulin heavy chain locus on chromosome 14 and an unidentified gene on chromosome 8. Which of the following is the most likely cause of his phenotype? a. The translocation has deleted constant chain exons on chromosome 14 and prevented heavy chain class switching b. The translocation has deleted the interval containing diversity (D) and joining (J) regions c. The translocation has activated a tumor-promoting gene on chromosome 8 d. The translocation has deleted the heavy chain constant chain Cμ so that virgin B cells cannot produce IgM on their membranes e. The translocation has deleted an immunoglobulin transcription factor gene on chromosome 8

*The answer is A.* DNA can be damaged in a number of ways (spontaneous mutations, chemical reactions, or ultraviolet/ionizing radiation) through the following mechanisms: 1. Depurination of DNA and base excision repair (spontaneous or chemical) 2. Formation of thymine dimers (ultraviolet rays) 3. Breaks in DNA chains and oxidative damage (ionizing radiation) 4. Cross-linkage, intercalation alkylation (chemical I pharmacologic agents) The DNA repair mechanism utilized by cells for base alterations is called base excision repair (not to be confused with nucleotide excision repair). Nitrates consumed in the diet lead to the deamination of cytosine, adenine, and guanine to form uracil, xanthine, and hypoxanthine, respectively. These resulting bases are not normally present in DNA and are recognized by specific glycostasis, which cleave these altered DNA bases from the parent DNA molecule. This leaves an empty sugar phosphate site called a basic site, or apurinic-apyrimidinic site (AP). An endonuclease then cleaves the 5' end of the AP site before a lyase enzyme subsequently completes the removal of the AP site from the DNA molecule by removing the sugar phosphate group. DNA polymerase then fills the gap with the correct sugar-phosphate-base, which is finally joined to the strand by ligase (Choice A). *Educational Objective:* Base excision repair is used to correct defects in single bases induced spontaneously or by exogenous chemicals. In this process, glycosylase remove the defective base, and the corresponding sugar-phosphate is cleaved and removed by endonuclease, followed by the action of lyase. DNA polymerase then replaces the missing nucleotides and ligase reconnects the DNA strand.

A mouse is fed with food that is rich in nitrates. As a result, the chromosomal DNA of the intestinal epithelial cells undergoes accelerated cytosine deamination. Which of the following is the correct order of enzymes that repair the damage? A. Glycosylase, endonuclease, lyase, DNA polymerase, ligase B. Endonuclease, DNA polymerase, lyase, glycosylase, ligase C. Lyase, endonuclease, glycosylase, DNA polymerase, ligase D. Endonuclease, DNA polymerase, glycosylase, lyase, ligase E. Glycosylase, ligase, lyase, endonuclease, DNA polymerase

*The answer is A.* The Southern, Western, Northern and Southwestern blot procedures are techniques used to analyze and identify DNA fragments, proteins, mRNA, and DNA-bound proteins, respectively. The best method for determining whether a genie is being expressed is to analyze for the presence of its mRNA using a Northern blot (Choice A). in the case described above, Northern blot analysis of each of the cell culture samples can determine if mRNA corresponding to the gene of interest is being transcribed. All of the "blot" tests rely on the same basic techniques. First, the components of the unknown sample - DNA for Southern blots (Choice C), mRNA for Northern blots (Choice A), protein for Western blots (Choice B), and DNA-bound protein for Southwestern blots (Choice D)- are separated by size and charge via gel electrophoresis. The resultant bands are then blotted onto a nitrocellulose membrane and incubated with a labeled hybridization probe or antibody to identify the specific DNA fragment, mRNA molecule or protein of interest.

A mutation in a non-coding sequence is believed to affect expression of the gene coding for a specific fetal enzyme. Liver and bone marrow cells from the fetus and his parents are obtained. Which of the following is the best method to determine if this gene is being transcribed in cultures of isolated cells? A. Northern blot B. Western blot C. Southern blot D. Southwestern blot E. ELISA

*The answer is B.* Genetic information flows from DNA to RNA to proteins. Most eukaryotic DNA sequences consist of coding exons, non-coding introns, and two promoter regions (the CAAT box and the TATA box). The CAAT box is located 60 to 80 bases upstream of the beginning (5' end) of the coding region and the TATA box is located 25 bases upstream from the beginning of the coding region. Gene transcription begins when RNA polymerase II attaches at one of these promoter sites in a process facilitated by numerous general transcription factors (In eukaryotes, RNA polymerase II alone is unable to recognize the TATA box). Transcription factors and DNA enhancer regions can associate with these promoter sites and increase the affinity of RNA polymerase II for the promoter site thereby increasing gene expression. Though promoters are not directly translated into protein, promoter mutations can cause abnormal gene expression by altering the ability of RNA polymerase II and transcription factors to bind. *Educational Objective:* The TATA box is a promoter region that binds transcription factors and RNA polymerase II during the initiation of transcription. lt is located approximately 25 bases upstream from the beginning of the coding region.

A mutation in the TATA box of a eukaryotic gene that codes for a transmembrane protein is most likely to affect which of the following functions? A. DNA methylation B. Transcription initiation C. Translation initiation D. RNA elongation E. Posttranscrlptional RNA splicing F. Polypeptide folding following translation

*The answer is E.* DNA polymerases are the main enzymes responsible for DNA replication. In E. coli, there are three major DNA polymerases: I, II, and Ill. Other enzymes that are essential for the replication of DNA include primase, helicase, ligase, and topoisomerase I and II. During DNA replication, new daughter strands are synthesized in the 5' to 3' direction, using the parent strands as complimentary templates. Synthesis of DNA in the 3' to 5' direction does not occur in this process. Before the process of replication begins, the parent DNA double helix is unwound and separated by the enzyme helicase. Helicase binds DNA at the origin of replication with the assistance of DNA protein and acts at the replication fork to separate DNA. This separation and unwinding of the DNA produces positive supercoiling that can lead to DNA fracture, if not relieved. Topoisomerases I and II (II is also known as gyrase) relieve unwinding tension. DNA polymerases can not begin synthesis of daughter strands without a free 3'-hydroxyl group, which is provided by an RNA primer and synthesized by the enzyme primase (DNA dependent RNA polymerase). DNA replication then proceeds with the leading strand being formed continuously in the 5' to 3' direction toward the replication fork, and the lagging strand being formed discontinuously in the 5' to 3' direction away from the replication fork. Replication of the lagging strand results in the formation of numerous short DNA segments, called Okazaki fragments, and each separate DNA segment requires a new RNA primer upon which to initiate synthesis. The fragments of the lagging strand are ultimately bound together by ligase after numerous RNA primers have been removed and replaced with DNA. The removal of RNA primers is accomplished by DNA polymerase I, the only bacterial DNA polymerase with 5' to 3' exonuclease activity. This activity allows DNA polymerase Ito function both as an excision-repair enzyme and as the enzyme that removes RNA primers. *Educational Objective:* Bacterial DNA polymerase I have 5' to 3' exonuclease activity, which is used to excise RNA primers. The gaps created after RNA excision is then replaced with DNA in the 5' to 3' direction by DNA polymerase I.

A mutation that leaves prokaryotes unable to replicate their DNA is induced in an experimental seeing. The ability to remove RNA primers during DNA replication is affected by this experimental mutation. Which of the following enzymes is most likely nonfunctional? A. Helicase B. Primase C. Gyrase D. DNA polymerase III E. DNA polymerase I F. Ligase

*The answer is C.* β-Thalassemia is an autosomal recessive disorder of defective β-globin synthesis, resulting in anemia of varying degrees depending on the nature of the mutation and the degree of functionality of the expressed gene product (if it is expressed at all). In humans, the globin locus is located on = chromosome 11. Similarly to other proteins, the synthesis of β-globin includes transcription (DNA to RNA), posttranscriptional modification (such as 5' capping, 3' polyadenylation, and alternative RNA splicing), translation (mRNA to protein), and post-translational modification (eg, O-glycosylation, N glycosylation, and sulfation). In this case, the mutation responsible for this patient's condition is one that occurs in a noncoding intervening region, also known as an intron (as shown by gene sequencing), which results in the production of a truncated protein (as shown by protein electrophoresis). This suggests that the mutation does not affect transcription or translation. Of the answer choices, post-transcriptional modification such as RNA splicing is most likely to be altered by a single nucleotide substitution in a noncoding region. Approximately 15% of all genetic diseases are caused by mutations that affect RNA splicing. Mutation of a splice donor site would result in the inclusion of an intron, whereas mutation of an acceptor site would result in the exclusion of an exon. A reduced-molecular weight protein would result in both cases. Attempted translation of an intron typically results in a premature stop and protein truncation.

A new study attempts to identify the genetic causes of β-thalassemia in 100 patients. On molecular testing, the following results are obtained from one of the patients: 1. Karyotype: 46,XY 2. Fluorescence in situ hybridization: no duplication or deletion in the β-globin locus on chromosome 11. 3. Gene sequencing of β-globin locus: single nucleotide substitution in a large noncoding intervening sequence This patient's β -globin gene was cloned and further purified by electrophoresis, which shows a significantly truncated β globin protein. The mutation responsible for this patient's β-thalassemia most likely leads to defects during which of the following processes? A. Meiosis I B. Meiosis II C. Posttranscriptional modification D. Posttranslational modification E. Transcription

*The answer is A.* Sufficient DNA for analysis can be obtained by amplification of leukocyte DNA using the polymerase chain reaction (PCR). Short segments of DNA (oligonucleotide primers) are designed to be complementary to areas flanking the DNA region of interest—in this case, the portion of the β-globin gene that may harbor the sickle cell anemia mutation. Some 20 to 30 cycles of cooling (to anneal the primers), synthesis (with heat stable DNA polymerase), and heating (to melt the DNA and allow the next cycle) can amplify the targeted DNA segment over 1 million-fold. Hybridization in duplicate to allele-specific oligonucleotides (ASOs; one ASO for the hemoglobin A mutation, one ASO for the S mutation) can establish the diagnosis of normal (AA alleles), sickle trait (AS alleles), or sickle cell anemia (SS alleles). Newborns have fetal hemoglobin (α- and γ globin) with little expression of hemoglobin A (α- and β-globin genes) until 3 to 6 months of life, so testing for anemia or for abnormal hemoglobin with antibodies would not be helpful. DNA polymorphisms (nucleotide sequence variations) occur approximately once per 200 to 500 base pairs (bp) of human DNA. If the sequence variation affects the recognition site for a restriction endonuclease, the altered segment sizes produced by endonuclease digestion allow detection of the sequence change [restriction fragment length polymorphism (RFLP)]. If the nucleotide change causing the RFLP is adjacent to (linked with) or coincident with a disease mutation, then one size variant of the RFLP may be diagnostic. However, mutations of known sequence (such as that for sickle cell anemia) are better detected by PCR and ASOs. The use of several highly variable RFLPs produces a pattern of restriction fragments that is highly distinctive for each individual (DNA fingerprinting) but not diagnostic for a particular disease.

A newborn baby has a sibling with sickle cell anemia and is at risk for the disease. The appropriate diagnostic test for sickle cell anemia in this baby will include a. DNA amplification b. Hemoglobin antibodies c. DNA restriction d. Red cell counting e. DNA fingerprinting

*The answer is D.* Collagen peptides assemble into helical tertiary structures that form quaternary triple helices. The triple helices in turn assemble end to end to form collagen fibrils that are essential for connective tissue strength. Over 15 types of collagen contribute to the connective tissue of various organs, including the contribution of type I collagen to eyes, bones, and skin. The fact that only one of two α2 alleles is normal in this case implies that a mutant α2 allele could be responsible for the disease (even if the α2 locus is on the X chromosome, since the baby is female with two X chromosomes). The mutant α2 collagen peptide would be incorporated into half of the type I collagen triple helices, causing a 50% reduction in normal type I collagen. (A mutant α1 collagen peptide would distort 75% of the molecules since two α1 peptides go into each triple helix). The ability of one abnormal collagen peptide allele to alter triple helix structure with subsequent degradation is well documented and colorfully named protein suicide or, more properly, a dominant-negative mutation.

A newborn female has a large and distorted cranium, short and deformed limbs, and very blue scleras (whites of the eyes). Radiographs demonstrate multiple limb fractures and suggest a diagnosis of osteogenesis imperfecta (brittle bone disease). Analysis of type I collagen protein, a triple helix formed from two α1 and one α2 collagen chains, shows a 50% reduction in the amount of type I collagen in the baby's skin. DNA analysis demonstrates the presence of two normal α1 alleles and one normal α2 allele. These results are best explained by a. Deficiency of α1 collagen peptide synthesis b. Inability of α1 chains to incorporate into triple helix c. Defective α1 chains that interrupt triple helix formation d. Incorporation of defective α2 chains that cause instability and degradation of the triple helix e. A missense mutation that alters the synthesis of α1 chains

*The answer is C.* Sequencing is a laboratory technique that uses fluorescently labeled dideoxynucleotides to randomly terminate growing strands of DNA. A capillary is used to separate the varying lengths of DNA. The DNA sequence can then be read using a laser that detects the fluorescently labeled fragments that elute from the capillary, based on the length of the fragment. In older Sanger sequencing techniques, radiolabeled dideoxynucleotides were used to randomly terminate the growing strands of DNA. Gel electrophoresis was then used to separate the fragments of DNA, and the sequence was read based on the position of the bands on the gel.

A nucleic acid fragment is added to a tube along with a polymerase, a primer, and deoxynucleotides. The tube also contains the four bases as fluorescently labeled dideoxynucleotides. The products in the tube are then run through a capillary, and a laser detects the fluorescence of the products that come out of the capillary. For which of the following purposes would the described laboratory technique be utilized? A. To amplify DNA fragments B. To create an allele-specific oligonucleotide probe C. To decipher the order of nitrogenous bases in a strand of DNA D. To determine the base pairing of a segment of DNA with a DNA probe E. To determine the base pairing of a segment of RNA with a DNA probe F. To establish the presence of a given protein G. To show the presence of a specific antibody in plasma

*The answer is C.* Sequencing is a laboratory technique that uses dideoxynucleotides to randomly terminate growing strands of DNA. Gel electrophoresis is used to separate the varying lengths of DNA. The DNA sequence can then be read based on the position of the bands on the gel.

A nucleic acid fragment is added to four different tubes along with a polymerase, a radiolabeled primer, and deoxynucleotides. Each tube also contains one of the four bases as dideoxynucleotides. The four tubes are then run on electrophoresis gel and visualized by autoradiography. For which of the following purposes would the described laboratory technique be utilized? (A) To amplify DNA fragments (B) To create an allele-specific oligonucleotide probe (C) To decipher the order of nitrogenous bases in the human genome (D) To determine the base pairing of a segment of DNA with a DNA probe (E) To determine the base pairing of a segment of RNA with a DNA probe (F) To establish the presence of a given protein (G) To show the presence of a specific antibody in plasma

*The answer is C.* The patient has the symptoms of myotonic dystrophy, an autosomal dominant disorder brought about by a triplet repeat expansion in the myotonic dystrophy gene. The presence of this expansion increases the size of DNA observed in Southern blots, and since it is autosomal dominant, only one of the two chromosomes has to contain this expansion for the patient to have the disease. Myotonic dystrophy is due to a CTG triplet repeat expansion in the 3′ untranslated region of the DM1 gene. The DM1 gene codes for a serine/threonine kinase and is located on chromosome 19, band q13. Myotonic dystrophy is not caused by a gene duplication, gene deletion (as would by muscular dystrophy), a change in SNP patterns, or differences in restriction length polymorphisms (although one may be created by the triplet repeat expansion).

A patient displays increasing muscle weakness and delayed muscle relaxation. He shows a slack jaw and drooping eyelids. The physician has noted a loss of muscle mass in the calves over the years. A molecular analysis by Southern blot using a probe against the suspected disease gene shows the following. The molecular reason for the difference between the normal and disease gene is which of the following? (A) Gene duplication (B) Gene deletion (C) Triplet repeat expansion (D) Increased SNPs in the disease state (E) Differences in restriction fragment length polymorphisms (RFLPs)

*The answer is B.* Glucocorticoids bind to a cytoplasmic receptor and translocate to the nucleus, where they bind to glucocorticoid response elements on the DNA, leading to a complex of proteins at this site. A number of transactivating factors recruited to the DNA contain histone acetyl transferase (HAT) activity, which leads to further unwinding of the DNA from the nucleosome, enabling gene transcription. Patients who become resistant to glucocorticoid treatment show a reduction in histone acetylation in response to the drug for a variety of reasons. The resistance does not appear to be due to inability of the drug to enter target cells. Once inside the cell, the drug will bind to the receptor, and dimerization of the receptor is normal. In some resistant individuals, the dimerized receptor has trouble translocating to the nucleus, which leads to the reduction in HAT activity. In other patients, however, translocation is normal, and the reduction in HAT activity appears to be due to an inability to recruit transactivating factors to the complex, which contain the HAT activity. The levels of transactivating factors are normal, but it has been hypothesized that the receptor is phosphorylated under resistance conditions, leading to an inability to attract the transactivating factors.

A patient has asthma, but has become resistant to glucocorticoid inhalation. A potential mechanism for this resistance is which of the following? (A) Inability of glucocorticoids to enter target cells (B) Inability to induce histone acetylation (C) Reduction of levels of transactivating factors in the nucleus (D) Cytokine induction of protein kinases (E) Increased dimerization of the glucocorticoid receptor

*The answer is D.* Microarray experiments allow for rapid screening of many genes in one experiment. The technique is based on hybridizing cDNA samples from cells to an array of known DNA sequences, which correspond to a battery of genes. Thus, in a single experiment, one can determine the expression levels of 1,000 genes in both the disease and normal state. Southern blots cannot do this, as one would have to probe the blot with 1,000 different probes to try to obtain similar results. Southwestern blot, in which protein binding to DNA is measured, is not relevant to this situation. MicroRNAs regulate gene transcription, but will not tell you about the expression of the bulk of genes in the cell (and microarray genes can also be analyzed in a microarray experiment). ELISAs examine one protein through

A patient has been diagnosed with a particular form of cancer. Appropriate treatment of this cancer, however, requires knowledge of which molecular markers are being expressed by the tumor as compared to normal cells of the same tissue. This is most easily accomplished by which of the following techniques? (A) Southern blot (B) Southwestern blot (C) MicroRNA analysis (D) Microarray analysis (E) ELISAs

*The answer is A.* Abetalipoproteinemia results from the absence of apolipoprotein B in chylomicrons and very low density lipoprotein (VLDL), caused by a mutation in MTTP, the microsomal triglyceride transfer protein. If one isolates plasma (removing the cells), and performs a Western blot for the presence of apolipoprotein B, one would see greatly reduced levels as compared to normal individuals. As apoB is made either in the intestine (apoB48) or the liver (apoB100), apoB mRNA would not be present in blood cells, so a Northern blot would not be informative. Southern blot of DNA would only work if there was a specific probe for MTTP which would distinguish a disease gene from a normal gene. Microarray, which measures mRNA levels, would not show any difference between normal and disease states, as the apoB protein is still made in the liver and intestine; however, it cannot be incorporated into chylomicrons or VLDL due to the lack of functional MTTP activity. PCR analysis suffers from the same problem as Southern blotting.

A patient has been diagnosed with abetalipoproteinemia. Such a disorder can be determined by which of the following techniques using samples obtained from the blood? (A) Western blot (B) Northern blot (C) Southern blot (D) PCR analysis (E) Microarray

*The answer is A.* The patient has lupus, which is an autoimmune disorder. To test for the presence of antibodies in the patient's sera, one can use a Western blot, and run typical antigens (such as splicesome proteins) on the gel, and then determine if the patient's sera contain antibodies against the proteins. ELISAs will also work for this assay. All of the other techniques mentioned require examining nucleic acids, which would not help in determining if autoimmune antibodies were being generated in the patient.

A patient has seen her physician for a variety of complaints over the past several years. These include fevers which come and go, fatigue, joint pain and stiffness, a rash on the face, sores in her mouth, easy bruising, and increased feelings of anxiety and depression. A molecular technique to provide a confirmation of the suspected diagnosis is which of the following? (A) Western blot (B) Southern blot (C) Northern blot (D) DNA sequencing (E) RFLP analysis

*The answer is D.* Cyclosporin A binds to the protein cyclophilin in immunocompetent lymphocytes, and this protein complex leads to the inactivation of calcineurin. Calcineurin, in response to increases in cytoplasmic calcium, will activate its phosphatase activity and dephosphorylate cytoplasmic nuclear factor of activated T-lymphocytes (NF-AT), a transcription factor. When dephosphorylated NF-AT will translocate to the nucleus, it will interact with nuclear factors and bind to DNA, initiating new gene transcription. When calcineurin is inactivated via cyclosporin-cyclophilin binding, NF-AT cannot translocate to the nucleus, and cytokine synthesis by the cell (second messengers) is compromised. Immunocompetent cells have very low levels of calcineurin, which make them susceptible to cyclosporin treatment. Drug treatment does not directly affect the translation of cytokine genes (it is indirect because the mRNA for these genes is never made), nor does it lead to the activation of transcription of cytokine receptors. Cyclosporin does not lead to the phosphorylation of transcription factors, nor does it stimulate translation of cytokine genes.

A patient is taking cyclosporin A after receiving a kidney transplant. Cyclosporin A protects against organ rejection by which of the following mechanisms? (A) Blocking translation of cytokine genes (B) Activating transcription of cytokine receptors (C) Stimulating the phosphorylation of transcription factors (D) Blocking the dephosphorylation of specific transcription factors (E) Stimulating translation of cytokine genes

*The answer is E.* Gene therapy refers to a group of techniques by which gene structure or expression is altered to ameliorate a disease. Because of ethical and practical difficulties, germ-line therapy involving alterations of genes in primordial germ cells is not being explored in humans. Although germ-line genetic engineering is being performed in animals with the goals of improved breeding or agricultural yield, it alters the characteristics of offspring rather than the treated individuals. Somatic cell gene therapy is targeted to an affected tissue or group of tissues in the individual, and is most effective if stem cells such as bone marrow can be treated. Somatic cell gene therapy offers the hope of replacing damaged tissue without the rejection problems of transplantation. For autosomal recessive disorders, only one of the two defective alleles must be replaced or supplemented.

A patient suffers from adenosine deaminase (ADA) deficiency, an autosomal recessive immune deficiency in which bone marrow lymphoblasts cannot replicate to generate immunocompetent lymphocytes. The treatment option that would permanently cure the patient is a. Germ-line gene therapy to replace one ADA gene copy b. Germ-line gene therapy to replace both ADA gene copies c. Somatic cell gene therapy to replace one ADA gene copy in circulating lymphocytes d. Somatic cell gene therapy to replace both ADA gene copies in circulating lymphocytes e. Somatic cell gene therapy to replace one ADA gene copy in bone marrow lymphoblasts

*The answer is C.* The biosynthesis of both coenzyme Q and dolichol is dependent on isoprene units, specifically isopentenylpyrophosphate, which is derived from mevalonic acid in the de novo pathway of cholesterol biosynthesis. Lovastatin inhibits HMGCoA reductase, which reduces mevalonate production. This can have, then, the unintended consequence of a reduction in dolichol production, thereby leading to underglycosylation of processed proteins. A reduction of coenzyme Q synthesis can lead to muscle weakness (a side effect of the statin class of drugs), but coenzyme Q is not involved in protein glycosylation. Cholesterol, HMG-CoA, and ketone bodies are not required for protein glycosylation as well.

A patient taking lovastatin and Zetia® for elevated cholesterol was found to produce lower levels of glycosylated proteins. This is most likely due to the unintended consequence of blocking the synthesis of which of the following compounds? (A) Coenzyme Q (B) Cholesterol (C) Dolichol (D) HMG-CoA (E) Ketone bodies

*The answer is D.* The HIV tests are looking for the presence of antibodies in the patient's sera against HIV proteins. So, in the confirming Western blot, HIV proteins are separated by size on a polyacrylamide gel, the proteins transferred to filter paper, and the filter paper treated with the sera sample. If the sera contain antibodies to HIV proteins, they will bind to the proteins on the filter paper, and the presence of antibodies on the filter paper is then detected using second antibodies. None of the other answers are appropriate for this type of test.

A patient was given an initial ELISA screening test for HIV, which was positive. A second, more specific test was then undertaken, which consisted of a Western blot. The samples that were run through the polyacrylamide gel for this test consisted of which of the following? (A) Patient sera (B) Patient plasma (C) HIV RNA (D) HIV proteins (E) Extracts of patient white blood cells

*The answer is B.* Lane B represents the Southern blot of a heterozygous carrier of sickle cell anemia. The β-A-globin gene results in a 1.15-kb fragment of DNA cut by the MstII restriction enzyme. The β-S-globin gene results in a 1.35-kb band because the single base-pair mutation responsible for sickle cell anemia eliminates an MstII restriction site. A heterozygote will have two bands indicating one normal allele with an intact MstII site (two fragments), and a mutant allele with a missing MstII site (one fragment).

A patient who is a carrier of sickle cell trait presents to the clinic. The single base-pair mutation for sickle cell anemia destroys the MstII restriction enzyme recognition site represented by an asterisk in the image. The restriction enzyme-binding sites are shown as arrows on the map. DNA from this patient is treated with MstII and run on an electrophoresis gel. The DNA is then hybridized with a labeled probe that binds to the normal gene in the position shown on the map. In the Southern blot shown in the image, which lane represents the patient? (A) A (B) B (C) C (D) D (E) E (F) F

*The answer is A.* Insulin is synthesized as a preproinsulin. The presequence is the signal sequence, which is cleaved when the protein enters the endoplasmic reticulum. Proinsulin contains the C-peptide, which is removed to form mature insulin, and disulfide bonds that hold the A and B chains together. The modification is thus proteolytic processing, and not glycosylation, modification of side chains, acylation, or altered quaternary structure. The test is a C peptide level. If a person is producing insulin, a C-peptide level can show this. This is also a way to see if a person has type 1 diabetes mellitus (no C-peptide, since they cannot produce endogenous insulin) or type 2 (normal or high C-peptide levels, since this disease is a reduced response to normally produced insulin).

A patient with type 2 diabetes has been prescribed recombinant insulin to help control his disease. Three months after starting this regime, a blood test is done, which indicates that the patient is still producing endogenous insulin, in addition to the recombinant insulin the patient is taking. The blood test has, at its basis, which of the following? (A) Posttranslational proteolytic processing (B) Posttranslational glycosylation (C) Posttranslational modification of amino acid side chains (D) Posttranslational acylation (E) Posttranslational quaternary structure formation

*The answer is C.* Incorporation of a Shine-Dalgarno sequence into the expression vector will promote ribosome binding to the translation start site on the mRNA produced by transcription of the cDNA insert.

A pharmaceutical firm is interested in the bacterial production of thymidylate synthase in large quantities for drug- targeting studies. An important step in the overall cloning strategy involves the ligation of synthase cDNA into a plasmid vector containing a replication origin, an antibiotic resistance gene, and a promoter sequence. Which additional nucleotide sequence should be included in this vector to ensure optimal production of the thymidylate synthase? A. Operator sequence B. PolyA sequence C. Shine-Dalgarno sequence D. Attenuator sequence E. 3′-splice acceptor sequence

*The answer is A.* Acetylation of nucleosome core histones is strongly associated with transcriptionally active chromatin. Other modifications (choices B, C and E) are associated with down-regulation of gene expression. Deamination of cytosine in DNA (choice D) is not related to chromatin remodeling and increased gene expression.

A pharmacologist employed by a pharmaceutical company is investigating the mechanism of action of a new drug that significantly inhibits the division of tumor cells obtained from patients with acute myelogenous leukemia. He has determined that the drug serves as a potent inactivator of chromatin-modifying activity that up-regulates the expression of a cluster of oncogenes in these tumor cells. Which type of chromatin-modifying activity is most likely stimulated by the enzyme target of this drug? (A) Acetylation of core histones (B) Binding of histone H1 to nucleosomes (C) Deacetylation of core histone H4 (D) Deamination of cytosine bases in DNA (E) Methylation of cytosine bases in DNA

*The answer is E.* Polymorphic loci have multiple alleles because of DNA sequence variation, including one or more with frequencies greater than 1%. This higher frequency and benign connotation differentiate polymorphic loci from those that harbor multiple disease-causing alleles. The DNA sequence changes may alter restriction sites [producing restriction fragment length polymorphisms (RFLPs)], change the numbers of repeated segments [producing variable numbers of tandem repeats (VNTRs)], or alter the genetic code (producing variant proteins, or protein polymorphisms). Polymorphisms may cosegregate (be inherited together) with disease alleles, allowing diagnosis by linkage analysis or estimates of risk through allele association (a.k.a. linkage disequilibrium, as with certain HLA alleles and diabetes mellitus). Different mutant alleles may cause indistinguishable phenotypes (allelic heterogeneity), as may mutations at different loci (genetic heterogeneity).

A polymorphism is best defined as a. Cosegregation of alleles b. One phenotype, multiple genotypes c. Nonrandom allele association d. One locus, multiple abnormal alleles e. One locus, multiple normal alleles

*The answer is E.* Puromycin is virtually identical in structure to the 3′-terminal end of tyrosinyl-tRNA. In both eukaryotic and prokaryotic cells, it is accepted as a tyrosinyl-tRNA analogue. As such, it is incorporated into the carboxy-terminal position of a peptide at the aminoacyl (A) site on ribosomes, causing premature release of the nascent polypeptide. Thus, puromycin inhibits protein synthesis in both human and bacterial cells. Streptomycin, like tetracycline and chloramphenicol, inhibits ribosomal activity. Mitomycin covalently cross-links DNA, which prevents cell replication. Rifampicin is an inhibitor of bacterial DNA-dependent RNA polymerase.

A potent inhibitor of protein synthesis that acts as an analogue of aminoacyl-tRNA is a. Mitomycin C b. Streptomycin c. Nalidixic acid d. Rifampicin e. Puromycin

*The answer is D.* Ehlers-Danlos syndrome (EDS) is a heterogeneous group of inherited connective tissue disorders characterized by joint hypermobility, cutaneous fragility, and hyperextensibility. Eleven different forms of EDS have been identified. This presentation is most consistent with vascular EDS, also known as EDS 4. In addition to the classic characteristics, EDS 4 presents with vascular problems, including renovascular hypertension and arterial or uterine rupture. This form of EDS is characterized by a decreased amount of type III collagen, which is a structural protein in blood vessels. EDS 4 is a rare and severe form; patients often have a shortened lifespan because of spontaneous rupture of large arteries including berry aneurysms (red arrow in the left-hand image), leading to diffuse subarachnoid hemorrhage (black arrow shown in the right-hand image).

A previously healthy 23-year-old man presents with a headache that started 2 hours ago. He describes it as being the "worst headache of his life." His blood pressure is 138/95 mm Hg and his heart rate is 56/min. The physician notes highly stretchable skin and bruising over his bony prominences. The examination is otherwise unremarkable. What is the most likely defect associated with this patient's disease? A. Fibrillin-1 gene mutation B. Keratin gene mutation C. Type I procollagen gene mutation D. Type III collagen gene mutation E. Type IV collagen gene mutat ion

*The answer is B.* Promoter sites are initiation sites for transcription. Transcription starts when RNA polymerase binds to the promoter. It then unwinds the closed promoter complex, where DNA is in the form of a double helix, to form the open promoter complex in which about 17 base pairs of template DNA are unwound. RNA synthesis then begins with either a pppA or a pppG inserted at the beginning 5′-terminus of the new RNA chain, which is synthesized in the 5′ to 3′ direction.

A promoter site on DNA a. Transcribes repressor b. Initiates transcription c. Codes for RNA polymerase d. Regulates termination e. Translates specific proteins

*The answer is A.* Since a Northern blot is being run, RNA from the various tissues is run through the gel to be separated based on size. Seeing different sizes of hybridizing bands, depending on the tissue of origin, with the same probe, and from the same animal, indicates that alternative splicing of the mRNA has occurred between the tissues. mRNA instability would lead to reduced levels of signal and not differences in sizes. RNA editing would not alter the overall size of the mRNA; it would just alter the sequence of the mRNA. Start codons are within the mRNA; utilizing different start codons would not alter the size of the mRNA. Differential polyadenylation is only correct if it is a part of differential splicing, which is the best answer.

A researcher has isolated the cDNA for gene Y. He performs the following Northern blot experiment using samples prepared from different tissues. A potential explanation for this finding is which of the following? (A) Alternative splicing (B) mRNA instability (C) RNA editing (D) Different AUG start codons in different tissues (E) Differential polyadenylation

*The answer is C.*/

A researcher has obtained an antibody to cytosolic protein X and runs a Western blot using as samples a variety of tissue types. The results of the Western blot are shown below. A potential interpretation of the results is which of the following? (A) Codon degeneracy within the genetic code (B) Tissue-specific posttranslational modifications (C) Tissue-specific alternative splicing of the primary transcript (D) Polyadenylation is lacking in certain tissues (E) Differences in location of 5′-cap formation in the tissues

*The answer is A.* The experiment described above is known as the Northern Blot technique, a procedure used to detect specific mRNA sequences in a sample to assess for gene expression. In this experiment, the Northern Blot identities three different mRNA transcripts containing exon 4, with varying patterns of expression in the different tissues. This is consistent with alternative splicing, e process whereby the exons of the pre-mRNA produced by transcription of s gene are reconnected in multiple ways during post-transcriptional processing. The resulting finalized mRNAs are then translated into different protein isoforms. Thus, a single gene can code for multiple proteins when the same gene is spliced differently in different tissues. Alternative splicing is a normal phenomenon in eukaryotes that greatly increases the biodiversity of proteins that can be encoded by the genome. It is thought that at least 70% of the 30,000 genes in the human genome undergo alternative splicing and that on average, a given gene produces 4 altamativaly spliced variants. Thus, the human genome is able to encode a total of 80,000 to 100,000 proteins which differ in their sequence and function. Abnormal variations in splicing are implicated in many diseases (e.g., beta-thalassemia, cancer). Alternative splicing also plays a prominent role in the lifecycle of many retroviruses. For instance, HIV produces a single primary RNA transcript that is alternatively spliced to produce over 40 different mRNAs.

A researcher is studying the expression pattern of a particular gene. Messenger RNA is isolated from several tissues, subjected to electrophoresis, blotted, and probed with radiolabeled DNA containing sequences from axon 4 from that gene. An x-ray film is then placed over the blotting membrane. with the results of the autoradiogram shown below: Which of the following best explains the autoradiogram findings in the different tissues? A. Alternate RNA splicing B. DNA rearrangement C. DNA mutation D. Enhancer effect E. Transcription factor effect

*The answer is B.* Mature mRNA from eukaryotes has a poly-A tail, which is added posttranscriptionally by poly-A polymerase. The poly A tail will hybridize to the oligo-dT on a column, thereby allowing the mRNA to bind to the column and other types of RNA to pass through the column. Altering the salt conditions (through a reduction of salt concentration) can then elute the mRNA specifically from the column. Phenol extraction is required for nucleic acid isolation, but it is not specific for mRNA. Electrophoresis, either agarose or polyacrylamide, will separate nucleic acids by size but does not by itself lead to mRNA isolation. An oligo-dA column will not hybridize to the poly-A tail of mRNA.

A researcher needs to prepare RNA for Northern blot analysis. Initial experiments using total RNA produce no signal when the experiment is completed. A method to increase the sensitivity of the assay would be to do which of the following to the total RNA sample? (A) Separate by size on agarose gel electrophoresis (B) Run the RNA through an oligo-dT affinity column (C) Run the RNA through an oligo-dA affinity column (D) Separate by size on polyacrylamide gel electrophoresis (E) Perform a phenol extraction of the total RNA

*The answer is B.* The melting temperature Tm of duplex DNA is the temperature at which half the base pairs are denatured. Adenine-thymine (A-T) base pairs have two hydrogen bonds, in contrast to cytosine-guanine (C-G) base pairs, which have three hydrogen bonds. Duplex DNA molecules rich in A-T base pairs have a much lower Tm than those rich in C-G base pairs. As DNA is heated, fractions with a higher A-T content melt or denature before those with a higher C-G content. Most mammals, including humans, have satellite DNA fractions that are highly repetitive and clustered in particular chromosome regions. Satellite DNAs are named for their altered density (satellite band) on centrifugation, caused by higher G-C content. Their function is unknown.

A sample of human DNA is subjected to increasing temperature until the major fraction exhibits optical density changes due t disruption of its helix (melting or denaturation). A smaller fraction is atypical in that it requires a much higher temperature for melting. This smaller, atypical fraction of DNA must contain a higher content of a. Adenine plus cytosine b. Cytosine plus guanine c. Adenine plus thymine d. Cytosine plus thymine e. Adenine plus guanine

*The answer is B.* The given strand of DNA contains 25%T; the complementary strand will contain 20%T (this must be equivalent to the content of A in the given strand, since A and T base pair, and [A] = [T] in duplex DNA). For the entire duplex then, the T content is the average of 25% and 20%, or 22.5% for the duplex. The [A] in the duplex will also be 22.5% (again, since [A] = [T]), and the concentrations of [G] and [C] will each be 27.5% for the duplex.

A single-stranded DNA molecule contains 20%A, 25%T, 30%G, and 25%C. When the complement of this strand is synthesized, the T content of the resulting duplex will be which one of the following? (A) 20% (B) 22.5% (C) 25% (D) 27.5% (E) 30%

*The answer is C.* DNA can be copied rapidly using the polymerase chain reaction (PCR) method. In this procedure, DNA is mixed with two specific primers: deoxynucleotides and a heat-stable polymerase. The solution is heated to denature the double-stranded DNA, producing two fragments of single-stranded DNA that are then cooled to allow synthesis. Because DNA synthesis proceeds from 5' to 3', the primers need to hybridize with each DNA strand so that synthesis can proceed in a 5' to 3' direction. Two primers are required, one that binds to the strand shown in the vignette image (ie, the coding strand) and a second that binds to its complementary strand (ie, the noncoding strand). The primer of the coding strand is complementary to its 3' end. The primer to the noncoding strand is identical to the coding strand's 5' end. To summarize, in order to copy a segment of DNA via PCR, a complementary primer to the 3' end is required along with a reversed copy of the 5' end. To design the primers for this strand, the following steps are performed : 1) Determine the complementary strand to the one displayed, which will be: 3' TATCATGGC---CTGATCGTG 5' 2) Determine the primer that will hybridize to the strand shown in the question, which will be starting from the 3' end of the original strand since the polymerase move in 5'-3' direction. The primer will be 3 ' CTGATCGTG, which is conventionally written as 5' GTGCTAGTC 3', which is the first primer. 3) Determine the primer that will hybridize to the complementary strand in (step 1), and the primer will be 5' ATAGTACCG 3', which is the second primer.

A team of researchers is studying the mechanisms of genetic shift and genetic drift in HIV. To further the investigation, the scientists need to use a technique where they amplify the sequence of DNA shown below: 5' ATAGTACCG------------GACTAGCAC 3' 3' TATCATGGC ------------ CTGATCGTG 5' They decide to use polymerase chain reaction (PCR) in order to do so. Which of the following would be the most appropriate DNA primer to use for this cycle of PCR? A. 5'-CGGTACTAT; 5'-CTGATCGTG B. 5'-ATAGTACCG; 5'-GACTAGCAC C. 5'-ATAGTACCG; 5'-GTGCTAGTC D. 5'-CGGTACTAT; 5'-CACGATCAG E. 5'-CGGTACTAT; 5'-GTGCTAGTA

*The answer is A.* The woman has a form of porphyria, an inborn error in the biosynthesis of heme. The buildup of heme intermediates is what generates the symptoms observed. Barbiturates are degraded through the cytochrome P450 series of enzymes, which are induced by the drug. The enzymes require heme, so heme synthesis is also induced, and whenever heme synthesis is induced, the toxic intermediates get accumulated. The end product of heme degradation is bilirubin, and if excessive bilirubin accumulates, jaundice is the result. Both dolichol and ubiquinone synthesis are from isoprenoids, and not related to the heme pathway.

A woman developed the following symptoms after taking certain drugs such as barbiturates. The symptoms included severe pain in the abdomen, hallucinations, disorientation, and a reddish tint to the urine. These symptoms appeared due to the induction of genes involved in which of the following pathways? (A) Cytochrome synthesis (B) Cytochrome degradation (C) Ubiquinone biosynthesis (D) Ubiquinone degradation (E) Dolichol synthesis

*The answer is C.* CREB (cyclic AMP response element binding protein) is a transcription factor that is activated by protein kinase A and that regulates, in part, the expression of phosphoenolpyruvate carboxykinase (PEPCK), a necessary protein for gluconeogenesis. Since the patient is anorexic, her blood glucose levels are being maintained primarily by gluconeogenesis, and the enzymes for that pathway need to be upregulated. The release of glucagon and epinephrine, both of which would be elevated in this patient, leads to the activation of protein kinase A and an increase in gene transcription for those genes regulated by CREB. Under these conditions, protein synthesis will be limited, so factors necessary for protein synthesis would not be generally activated (eIF4, eEF2, and ribosomal protein S6). Neither glucagon nor epinephrine works through steroid hormone receptors (they both utilize serpentine receptors on the cell membrane).

A woman with a BMI of 16.5 visits her family physician because she always feels tired. The history indicates that the woman is always on a diet, exercises over 3 h/day, and perceives herself as fat. Blood work indicates that her glucose levels are only slightly below normal under fasting conditions. The patient's ability to maintain her blood glucose levels near normal results, in part, from activation of which of the following proteins? (A) eIF4 (B) eEF2 (C) CREB (D) Ribosomal subunit S6 (E) Steroid hormone receptor

*The answer is E.* This patient has injuries associated with minor trauma and dentinogenesis imperfecta (manifested by the discolored teeth), which point toward a structural defect in collagen synthesis. Osteogenesis imperfecta is an autosomal disorder caused by genetic defects in type I collagen synthesis. It is characterized by a generalized decrease in bone mass (osteopenia) that leads to brittle bones and in its most severe form can cause death in utero. Teeth will be discolored blue-gray to yellow-brown which is secondary to the abnormally colored dentin that shines through the translucent enamel. The disorder is also frequently associated with blue sclerae, short stature, and progressive hearing loss. Patients are often diagnosed after multiple emergency department visits because of traumatic injuries and may be mistaken for victims of child abuse.

A young boy presents to the emergency department for the fifth time because of injuries associated with minor trauma. His mother states that these have generally been sustained from falling from standing and during normal play with other children. On several occasions, there were no external signs of injury. On physical exam, he is noted to have discolored teeth. His mother says that his diet seems to be well balanced, but she has a tough time keeping him away from candy. He showers with assistance every day and brushes his teeth twice daily. What other symptoms is this patient most likely to experience? A. Aortic aneurysm B. Chiari malfo rmation C. Cleft palate D. Future fractures from abuse E. Progressive hearing loss

*The answer is A* Thalassemias are the result of an imbalance in the synthesis of α- and β-globin genes. If this were to occur, a Western blot analysis of the α and β chains would show a difference in the amount of each in the red blood cells, suggesting either an α- or β-thalassemia (in an α-thalassemia, one would see less α chains or variants of α chains being produced, as compared to just one, normal β chain. The opposite would be true for a β-thalassemia). As many α-thalassemias are caused by gene deletions, FISH might be another way to determine this condition, using a probe specific for the α-globin gene. Most β-thalassemias are not due to gene deletions. PCR for γ-globin (fetal Hb) or RNA polymerase will not address an imbalance in α- and/or β-globin chain synthesis. Similarly, Western blots of snurps or TFIID will not address an imbalance in synthesis (if there was a problem with snurps, all RNA splicing would be affected, not just the α or β-globin gene; similarly, if TFIID were altered, all mRNA synthesis would be altered). Clinical labs will also use hemoglobin electrophoresis to quantitate the levels of globin chains in a patient. The illustration used in the question was obtained from a patient with β-thalassemia.

A young child of Mediterranean parents was brought to the pediatrician due to lethargy, tiredness, and pallor. Blood analysis revealed a microcytic anemia, although iron levels were normal (see the figure below). What test should be run to determine that the child has a variant of thalassemia? (A) Western blotting of the peptide chains in hemoglobin (B) PCR of the gene for RNA polymerase (C) Western blot of snurps in the child (D) Western blot of TFIID (E) PCR of the gene for γ-globin in the child

1. *The answer is A.* The sequence now contains TAA which will be transcribed to UAA in the mRNA. 2. *The answer is C.* The transcription promoter TATA has been changed to TCTA. Don't choose the distractor B. The question is not about translation.

A. ATGCAA...→ ATGTAA B. ATGAAA...→ GTGAAA C. TATAAG...→ TCTAAG D. CTTAAG...→ GTTAAG E. ATGAAT...→ ATGCAT The options above represent mutations in the DNA with base changes indicated in boldface type. For each mutation described in the questions below, choose the most closely related sequence change in the options above. 1. Nonsense mutation 2. Mutation decreasing the initiation of transcription

*The answer is A.* This patient has Fanconi anemia, an autosomal recessive disorder that makes cells more susceptible to chromosomal breakage in response to DNA cross-linking agents. Over 85% of cases are diagnosed between the ages 4 and 15 years. Patients commonly present with thrombocytopenia or leukopenia, which gradually progresses to pancytopenia and eventually to bone marrow failure. Bone marrow biopsy will show fatty infiltration. In this pat ent, early fatigue, mucosal bleeding, and recurrent infections are secondary to pancytopenia caused by progressive marrow failure. Two-thirds of patients with Fanconi anemia show some form of congenital abnormality, especially thumb abnormalities, hypogonadism, and microcephaly. Patients can be treated with corticosteroids or growth factors in the short term, but bone marrow transplant is needed to address the pancytopenia and its sequelae.

An 8-year-old boy comes to the clinic for a routine visit with his pediatrician. The child has short stature, polydactyly, radial hypoplasia, small testicles, and kidney malformations. The patient's mother reports that her son is easily fatigued and suffers from frequent nosebleeds. CBC is shown below: WBC 2900/mm3 RBC 3. 78 million/mm3 Hgb 9 g/dl Hct 34.7% MCV 91.6 µm3 Plt 142,000/mm3 Which of the following is the correct treatment for this patient's underlying disease? A. Bone marrow transplant B. Dietary modification C. Folate supplementation D. Vitamin B12 supplementation E. Vitamin C supplementation

*The answer is A.* The patient's blue sclerae (likely what the mother noticed to be different' about her child's eyes) and history of fractures with minor trauma (like the fracture described in the vignette) are characteristic of osteogenesis imperfecta (OI), a condition most commonly caused by decreased production of otherwise normal type I collagen, a major component of bone. This genetic disorder leads to diffuse bone weakness and brittleness. The classic sign of the disease is blue sclerae, which is due to the translucency of the abnormal connective tissue over the choroid. Many patients with type I OI also will develop hearing loss by age 30. Bone biopsy will reveal a disorganized, woven pattern. More than 200 gene mutations have been associated with OI. At least 8 subtypes of OI have been described, with type I being the most common. Most forms of OI have an autosomal-dominant inheritance pat tern.

An 8-year-old boy is brought to his physician because of severe pain in his left foot He was running on the playground when he tripped and fell. He was unable to bear weight on his left foot immediately after his fall. His mother adds that her son also has had a radial fracture in the past 2 years and that his eyes appear different from those of his siblings. Results of a complete blood cell count and erythrocyte sedimentation rate are normal. An X-ray of the foot is shown here. What is the most likely cause of this patient's disease? A. Abnormal type I collagen synthesis B. Abnormal type III collagen C. Abnormal type IV collagen synthesis D. Defect of fibrillin synthesis E. Malignant pro liferation of neuroectodermal cells

*The answer is C.* The patient has sickle cell anemia, and hydroxyurea treatment is designed to activate transcription of the γ-globin chain, which is normally only expressed during development (fetal hemoglobin). When expressed, the γ-globin gene will form functional hemoglobin tetramers with the α-globin chains, thereby reducing the effects of the mutated β-globin chain. Hydroxyurea does not bind to hemoglobin and denature it; it does not reduce the synthesis of the β-chains, nor does it alter oxygen levels in the blood or 2,3-bisphosphoglycerate levels in the erythrocyte. The γ-chain is normally turned off at birth as part of the hemoglobin-switching pathway. The challenge to scientists at this time is to understand how to reactivate γ-chain synthesis in patients with both sickle cell disease and β-thalassemias.

An African American patient has displayed vasoocclusive episodes for most of his life. The incidents are more prevalent under conditions in which blood oxygen levels are low, such as during exercise or taking trips to locations at high altitudes. The patient has been placed on hydroxyurea. The rationale behind this treatment is which of the following? (A) To prevent vaso-occlusive episodes through hydroxyurea induced protein degradation (B) To reduce synthesis of a defective protein (C) To induce synthesis of a functional protein (D) To enhance oxygen levels in the blood (E) To activate the enzyme that produces 2,3-bisphosphoglycerate

*The answer is A.* Imbalance of globin chain synthesis occurs in the thalassemias. Deficiency of α-globin chains (α thalassemia) is common in Asian populations and may be associated with abnormal hemoglobins composed of four β globin chains (hemoglobin H) or (in fetuses and newborns) of four γ-globin chains (hemoglobin Bart's). Mutation in a transcription factor necessary for expression of α-globin could ablate α-globin expression, since the same factor could act in trans on all four copies of the α-globin genes (two α-globin loci). Mutation of a regulatory sequence element that acts in cis would inactivate only one α-globin gene, leaving others to produce α-globin in reduced amounts (mild α thalassemia). Deletions of one α-globin would produce a similar mild phenotype, and deficiencies of transcription factors regulating α- and β-globin genes would not produce chain imbalance.

An Asian child has severe anemia with prominence of the forehead (frontal bossing) and cheeks. The red cell hemoglobin concentration is dramatically decreased, and it contains only β-globin chains with virtual deficiency of α-globin chains. Which of the following mechanisms is a potential explanation? a. A transcription factor regulating the α-globin gene is mutated b. A regulatory sequence element has been mutated adjacent to an α-globin gene c. A transcription factor regulating the β-globin gene is mutated d. A transcription factor regulating the α-globin and β-globin genes is deficient e. A deletion has occurred surrounding an α-globin gene

*The answer is B.* As part of the life cycle of the virus, the RNA genome of the virus is converted to DNA, which integrates randomly into the host chromosome. Host cell RNA polymerase II then transcribes the viral DNA, producing viral RNA, which is translated to produce viral proteins, and which is also utilized as the genome for new viral particles. RNA polymerase does not contain 3′-5′ exonuclease activity (which DNA polymerase does), so RNA polymerase cannot check its work and cannot fix errors when a mismatch is made. The accumulated effect of these errors increases the mutation rate of the virus much more than organisms containing DNA genomes. Since the enzyme that creates the viral DNA is reverse transcriptase, which also has no error-checking capability, the risk for mutations is greatly enhanced. DNA polymerase does check its work but is not used in the viral life cycle. The DNA repair enzymes are not altered by HIV infection. Uracil is a normal component of the viral RNA genome, whereas thymine is not, but neither of these facts results in an increase in mutation rate.

An IV drug user is tested, and found positive, for infection by HIV. If the patient is only placed on one antiviral medication, viral loads will initially be reduced, but will then rapidly increase. The resistance to the drug occurs due to which of the following? (A) Lack of error checking in DNA polymerase (B) Lack of error checking in RNA polymerase (C) Lack of DNA repair enzyme systems in HIV-infected cells (D) Incorporation of uracil in the RNA genome of HIV (E) Incorporation of thymine in the RNA genome of HIV

*The answer is C.* Proteins can be separated on the basis of their overall charge at a given pH by ion exchange chromatography. At low pH all proteins have an overall positive charge because carboxyl groups are protonated. Thus, proteins tend to bind to a cation exchange column that has immobilized the negative charges. Usually, negatively charged sulfonic polystyrene resin is used, and Na charges are exchanged for the positively charged protein groups. Once binding has occurred, the pH and NaCl concentration of the eluting medium are increased, and proteins that have a low density of net negative charge emerge first, with those having a higher density of negative charge following. The only information that can be obtained from the information given in the question is that the enzyme has been purified over 100-fold. The turnover rate of the enzyme cannot be deduced. Likewise, the yield, which is the amount of original enzyme protein recovered, cannot be determined. The structure of the enzyme is not revealed by the information given.

An adolescent presents with shortness of breath during exercise and is found to be anemic. A hemoglobin electrophoresis is performed that is depicted in the figure below. The adolescent's sample is run with controls including normal, sickle trait, and sickle cell anemia, and serum. The adolescent is determined to have an unknown hemoglobinopathy. Which one of the lanes contains the adolescent's sample? a. Lane A b. Lane B c. Lane C d. Lane D e. Lane E

*The answer is D.* The gene that produces the deadly toxin of Corynebacterium diphtheriae comes from a lysogenic phage that grows in the bacteria. Prior to immunization, diphtheria was the primary cause of death in children. The protein toxin produced by this bacterium inhibits protein synthesis by inactivating elongation factor 2 (EF-2, or translocase). Diphtheria toxin is a single protein composed of two portions (A and B). The B portion enables the A portion to translocate across a cell membrane into the cytoplasm. The A portion catalyzes the transfer of the adenosine diphosphate ribose unit of NAD1 to a nitrogen atom of the diphthamide ring of EF-2, thereby blocking translocation. Diphthamide is an unusual amino acid residue of EF-2.

An immigrant family from rural Mexico brings their 3-month-old child to the emergency room because of whistling inspiration (stridor) and high fever. The child's physician is perplexed because the throat examination shows a gray membrane almost occluding the larynx. A senior physician recognizes diphtheria, now rare in immunized populations. The child is intubated, antitoxin is administered, and antibiotic therapy is initiated. Diphtheria toxin is often lethal in unimmunized persons because it a. Inhibits initiation of protein synthesis by preventing the binding of GTP to the 40S ribosomal subunit b. Binds to the signal recognition particle receptor on the cytoplasmic face of the endoplasmic reticulum receptor c. Shuts off signal peptidase d. Blocks elongation of proteins by inactivating elongation factor 2 (EF-2, or translocase) e. Causes deletions of amino acid by speeding up the movement of peptidyl-tRNA from the A site to the P site

*The answer is D.* The deadly mushroom A. phalloides has several toxins. A major toxin is α-amanitin, an octapeptide that inhibits mRNA synthesis by very tightly binding RNA polymerase II (DNA-dependent RNA polymerase). As little as one of the mushrooms (know as the deathcap, death-cup, or avenging angel) delivers a lethal dose of about 10 mg α-amanitin. Severe, irreversible liver damage occurs quickly, leading to death. At higher concentrations, the toxin can inhibit RNA polymerase III and tRNA synthesis. Polymerase I is unaffected. Since α-amanitin is effective at concentrations of 10−9 to 10−8 M, it has been useful as a research tool for studying RNA polymerase function.

An immigrant from eastern Europe is rushed into the emergency room with nausea, vomiting, diarrhea, and abdominal pain. His family indicates he has eaten wild mushrooms. They have brought a bag of fresh, uncooked mushrooms from a batch he had not yet prepared. You note the presence of Amanita phalloides, the death-cap mushroom. A liver biopsy indicates massive hepatic necrosis. Care is supportive. A major toxin of the death-cap mushroom is the hepatotoxic octapeptide α amanitin, which inhibits a. DNA primase b. RNA nuclease c. DNA ligase d. RNA polymerase e. RNA/DNA endonuclease

*The answer is B.* The patient has developed a mutation in an intron which acts, only a small percentage of the time, as a splice donor site instead of the normal site at the intron/exon boundary. Thus, when this site is utilized by the splicesome, a piece of the intron is incorporated into the mRNA product, producing a longer than normal mRNA. This is an infrequent event, however, as judged by the finding that the density of the normal sized mRNA band on the gel is darker than this abnormal band. A nonsense mutation in the DNA will not affect transcription (although it does affect the protein product made from the mRNA). The lack of a cap would result in an unstable mRNA that perhaps would not be translated but would not significantly change the size of the mRNA. Poly-A polymerase adds the poly-A tail and would add the same size tail to both species of mRNA. If the polyadenylation signal were mutated, then the overall mRNA size would be larger, but there would not be two different proteins produced. Since the patient has a β-thalassemia, defective β-globin protein is being produced from the larger mRNA. Loss of methionine codons will affect translation, but not transcription.

An individual having β-thalassemia minor exhibits two bands on a Northern blot using a probe against exon 1 of β-globin. The smaller band is of normal size and "heavier" than the other larger band, which consists of approximately 247 additional nucleotides. One explanation for this finding is which of the following? (A) The presence of a nonsense mutation in the DNA (B) A mutation which creates an alternative splice site (C) A lack of capping of the mRNA (D) An extended poly-A tail (E) A loss of AUG codons

*The answer is A.* The decreased amount of AAT protein, its abnormal mobility, and the engorgement of liver ER suggest a mutant AAT that is inefficiently transported from the ER to serum. Since other serum protein abnormalities were not mentioned, general deficiencies of protein synthesis arising from defective energy metabolism or defective signal recognition particles are unlikely. A mutation affecting the N-terminal methionine of AAT or its signal sequence should drastically decrease its synthesis and import to the ER lumen. This would not explain the engorgement of liver ER. The usual binding of the signal recognition particle to the signal sequence of AAT, followed by import into the ER lumen, seems intact. An altered amino acid necessary for signal peptidase cleavage of the signal sequence of AAT might be invoked, but a general deficiency of the signal peptidase should disrupt many secreted proteins and be an embryonic lethal mutation. AAT deficiency is a well characterized autosomal dominant disease with common ZZ, SZ, and SS genotypes that can cause childhood liver disease and adult emphysema. The Z and S mutations alter AAT conformation and interfere with its secretion from ER to serum. Lack of AAT protection from proteases in lung is thought to cause the thinning of alveolar walls and dysfunctional "air sacs" of emphysema.

An older man with severe emphysema is found to have decreased amounts and abnormal mobility of α1 antitrypsin (AAT) protein in his serum when analyzed by serum protein electrophoresis. Liver biopsy discloses mild scarring (cirrhosis) and demonstrates microscopic inclusions due to an engorged endoplasmic reticulum (ER). The most likely explanation for these findings is a. Defective transport from hepatic ER to the serum b. A mutation affecting the N-terminal methionine and blocking initiation of protein synthesis c. A mutation affecting the signal sequence d. Defective structure of the signal recognition particles e. Defective energy metabolism causing deficiency of GTP

*The answer is B.* Chloramphenicol inhibits 50S peptidyltransferase. It is a bacteriostatic drug used in severe meningitis due to Haemophilus influenzae, Neisseria, and Streptococcus pneumoniae. Its use is limited to severe infect ions because of its serious toxicities, including aplastic anemia, bone marrow suppression, and gray baby syndrome.

Antibiotics are widely used in clinical practice to fight infection. Antibiotics work in many different ways to inhibit bacterial growth, including blocking cell wall synthesis, disrupting cell membranes, and blocking protein synthesis. Which of the following antibiotics inhibits bacterial 50S peptidyltransferase, an enzyme used in protein synthesis? A. Amikacin B. Chloramphenicol C. Clindamycin D. Gentamicin E. Tetracycline

*The answer is C.* Ribonucleases (RNAses) are enzymes that specifically digest RNA. There are many different types of RNAses, each target ing different classes of RNA. Some RNAses cleave at specific nucleotide sequences. Some work on single-stranded or double-stranded RNAs. There are three large categories of RNA: mRNA, tRNA, and rRNA. RNA is a cent ral component of ribosomes, the protein synthesis units in the cell. Prokaryotic ribosomes have two subunits aptly named large (50S) and small (30S). Within the large subunit, there are two different molecules of rRNA, measuring 5S and 23S. The small subunit contains a 16S-rRNA molecule. In this particular experiment, the RNAse used is specific for free single-stranded RNAs. Bacterial mRNA, tRNA, and rRNA are all generated as a single polycistronic strand, which is cleaved at specific sites. mRNA remains linear making it the most susceptible to RNAse; rRNA and tRNA each form base pairs with themselves to assume their 3-dimensional configurations; making them more resistant to this RNAase as it only targets single-stranded RNA. Only rRNA is further modified by the addition of ribonucleoproteins, making it the most resistant to degradation by RNAase. Furthermore, rRNA is the most abundant form of RNA (80%). Therefore, following RNAse digestion, one would only expect to see rRNA in the greatest quantity, representing the 5S, 16S, or 23S subunit.

Bacteria are lysed and spun to separate out cellular debris. The supernatant is incubated with a potent, sequence-nonspecific ribonuclease that targets free single-stranded RNA for 4 hours. Any remaining RNA is further isolated and purified on a denaturing agarose gel. Which of t he following types of RNA would one expect to see in the greatest quantity on the gel? A. Messenger RNA B. MicroRNA C. Ribosomal RNA D. Small interfering RNA E. Transfer RNA

*The answer is C.* Hydroxyproline is found uniquely in collagen. Although collagen is also rich in glycine, many other proteins contain significant amounts of glycine.

Collagen, the most abundant protein in the human body, is present in varying amounts in many tissues. If one wished to compare the collagen con- tent of several tissues, one could measure their content of A. Glycine B. Proline C. Hydroxyproline D. Cysteine E. Lysine

*The answer is A.* This would be a noninformative result. Enzyme X would not cut at site Z, so there would be no difference between the normal and disease gene when the DNA is cut with enzyme X, so both genes would give rise to a 5 kb piece of DNA. In order to distinguish between the normal and the disease gene, one needs to cut with two enzymes, X and Z; using the probe indicated in a Southern blot, the disease gene would then show a piece of DNA that is 1.2 kb in size, while the normal gene would show a piece of DNA that is 5.0 kb in size.

Consider the following map of a genomic region of DNA, showing restriction endonuclease sites for enzymes X and Y. You have a probe to this region (as indicated on the figure). A certain disease maps to this region of DNA and creates a new restriction site, Z, which is cut with restriction enzyme Z. As a diagnostic tool, a carrier would exhibit which bands, when DNA that has been completely restricted with enzyme X is run through a gel and a Southern blot, is performed utilizing the indicated probe? (A) 5.0 kb (B) 1.0 and 5.0 kb (C) 1.5 and 5.0 kb (D) 1.5 and 3.5 kb (E) 0.5 and 1.0 kb

*The answer is B.* An individual with the genotype of patient X is producing 50% normal α chain (two normal genes, two deleted) and 100% β-globin for a ratio of 1:2 of α to β (each chromosome 11 contains two α-genes, for a total of four α-genes per cell). Such a ratio would lead to little, if any, clinical symptoms. Patient Y is producing 100% α chain and 20% β chain for a 5:1 ratio of α to β. This ratio is big enough to lead to clinical symptoms. Recall, the major biochemical problem in thalassemia is the imbalance in synthesis of α and β chains, leading to nonfunctional α4 and β4 tetramers forming from the excess chains produced. The other ratios presented as answers do not represent the mutations present in the patients.

Consider two individuals, each with some form of thalassemia. Patient X has a deletion of the α genes on one chromosome but normal expression of all other α and β genes. This person has a mild form of the disease. Patient Y has a normal complement of α genes but has a homozygous mutation in the β genes in which an abnormal splice site is used 80% of the time, producing a transcript with a premature stop codon. Patient Y has a more severe disease than patient X. Why is patient Y's disease more severe than patient X's? (A) The ratio of α/β in patient X is 1:2, whereas in patient Y it is 1:5 (B) The ratio of α/β in patient X is 1:2, whereas in patient Y it is 5:1 (C) The ratio of α/β in patient X is 2:1, whereas in patient Y it is 1.2:1 (D) The ratio of α/β in patient X is 2:1, whereas in patient Y it is 1:1.2 (E) The ratio of α/β in patient X is 1:2, whereas in patient Y it is 1.2:1

*The answer is D.* Specific transcription factors (e.g., any steroid receptor) bind to specific DNA sequences (enhancers) and to RNA polymerase at a single promoter sequence and enable the RNA polymerase to transcribe the gene more efficiently.

Enhancers are transcriptional regulatory sequences that function by enhancing the activity of (A) general transcriptional factors (B) RNA polymerase to enable the enzyme to transcribe through the terminating region of a gene (C) transcription factors that bind to the promoter but not to RNA polymerase (D) RNA polymerase at a single promoter site (E) spliceosomes

*The answer is A.* Virulent strains of bacteria that cause severe, life-threatening respiratory tract infections can often be successfully treated with erythromycin. These include Mycoplasma pneumoniae, various Legionella species, and Bordetella pertussis. The mechanism of action of erythromycin is to specifically bind the 50S subunit of bacterial ribosomes. Under normal conditions, after mRNA attaches to the initiation site of the 30S subunit, the 50S subunit binds to the 30S complex and forms the 70S complex that allows protein chain elongation to go forward. Elongation is prevented in the presence of erythromycin.

Erythromycin is the antibiotic of choice when treating respiratory tract infections in legionnaire's disease, whooping cough, and Mycoplasma-based pneumonia because of its ability to inhibit protein synthesis in certain bacteria by a. Inhibiting translocation by binding to 50S ribosomal subunits b. Acting as an analogue of mRNA c. Causing premature chain termination d. Inhibiting initiation e. Mimicking mRNA binding

*The answer is C.* One codon AUG signals the initiation of protein synthesis and 3 codes (UAA, UAG, and UGA) stop protein synthesis. Stop codons function only to terminate translation they do not add amino acids to the polypeptide chain. Transfer RNA molecules (RNA) transport amino acids to the ribosome end ensure placement of the proper amino acid. One end of the molecule always has an adenine residue that serves as the amino acid binding site, the opposite end contains a specific nucleic acid sequence known as the anticodon that is complementary to one or more mRNA codons (due to base wobbling). The type of amino acid that is bound to each RNA molecule is determined by its anticodon sequence this ensures that the proper amino acid is added at each mRNA codon during protein synthesis. Translation of the mRNA template proceeds in the 5' to 3' direction. The mRNA of the dysfunctional protein in the question stem contains an abnormal stop codon at position 30. This stop codon will bind a release factor, halting protein synthesis and preventing downstream codons from adding further amino acids to the polypeptide chain. Therefore, the 29th codon (5'-UUG-3') is the last codon to add an amino acid to the truncated protein. which will be carried by the 5'-CAA-3' anticodon (codon-anticodon binding occurs in opposite directions [ie, 5' to 3' binds 3' to 5']).

Geneticists are studying a malfunctioning protein that causes impaired endothelial cell migration and angiogenesis. They have noticed that the amino acid sequence of the protein is truncated compared to normal controls. The mutated mRNA strand is isolated, and analysis shows a nonsense mutation located near the 3' end of the coding region. The 3' terminal coding sequence of the abnormal mRNA strand is shown below, beginning with the codon for the 26th amino acid. Which of the following RNA anticodons is responsible for adding the last amino acid to the truncated polypeptide during protein translation? A. 5'-AAC-3' B. 5'-AUC-3' C. 5'-CAA-3' D. 5'-GCU-3' E. 5'-UCG-3'

*The answer is C.* Two molecules of GTP are used in the formation of each peptide bond on the ribosome. In the elongation cycle, binding of aminoacyl-tRNA delivered by EF-Tu to the A site requires hydrolysis of one GTP. Peptide bond formation then occurs. Translocation of the nascent peptide chain on tRNA to the P site requires hydrolysis of a second GTP. The activation of amino acids with aminoacyl-tRNA synthetase requires hydrolysis of ATP to AMP plus PPi.

Guanosine triphosphate (GTP) is required by which of the following steps in protein synthesis? a. Aminoacyl-tRNA synthetase activation of amino acids b. Attachment of ribosomes to endoplasmic reticulum c. Translocation of tRNA-nascent protein complex from A to P sites d. Attachment of mRNA to ribosomes e. Attachment of signal recognition protein to ribosomes

*The answer is E.* Missense mutations, which cause the substitution of one amino acid for another, may significantly alter the function of the resultant protein without altering the size of DNA restriction fragments detected by Southern blotting. In this case, northern blot results would most likely also be normal. Single-base changes may also result in nonsense mutations. Large insertions or deletions in the exon or coding regions of the gene alter the Southern blot pattern and usually ablate the activity of one gene copy. In the case of an autosomal locus like that for ornithine aminotransferase, the homologous allele remains active and gives 50% enzyme activity (heterozygote or carrier range with a normal phenotype). Similar effects on enzyme activity would be predicted from complete gene deletions at one locus, while a duplication might produce 150% or 50% of normal enzyme activity depending on the status of promoter sites.

Gyrate atrophy is a rare autosomal recessive genetic disorder caused by a deficiency of ornithine aminotransferase. Affected individuals experience progressive chorioretinal degeneration. The gene for ornithine aminotransferase has been cloned, its structure has been determined, and mutations in affected individuals have been extensively studied. Which of the mutations listed below best fits with test results showing normal Southern blots with probes from all ornithine aminotransferase exons but absent enzymatic activity? a. Duplication of entire gene b. Two-kb deletion in coding region of gene c. Two-kb insertion in coding region of gene d. Deletion of entire gene e. Missense mutation

*The answer is D.* Since the probe is from mouse DNA, and the sample is human DNA, low stringency conditions need to be used to allow slight mismatches in hybridization to be tolerated. Thus, low temperature reduces thermal motion, and allows for mismatches, and high salt will reduce the potential disruptive ionic interactions that might result from mismatches and alterations in backbone structure. The combination of low temperature and high salt will provide the highest probability of hybridization between mouse and human DNA.

Having cloned the cDNA for mouse gene X, one wants to use this cDNA as a probe to screen a human cDNA library and isolate the human homolog of this gene. Hybridization conditions should be which of the following for the initial screening of the library? (A) High temperature, low salt (B) Low temperature, low salt (C) High temperature, high salt (D) Low temperature, high salt (E) Low temperature, no salt

*The answer is C.* ATP is required for the esterification of amino acids to their corresponding tRNAs. This reaction is catalyzed by the class of enzymes known as aminoacyl-tRNA synthetases. Each one of these enzymes is specific for one tRNA and its corresponding amino acid. amino acid + tRNA + ATP → aminoacyl-tRNA + AMP + PPi As with most ATP hydrolysis reactions that release pyrophosphate, pyrophosphatase quickly hydrolyzes the product to Pi, which makes the reaction essentially irreversible. Since ATP is hydrolyzed to AMP and PPi during the reaction, by convention the equivalent of two high-energy phosphate bonds is utilized

How many high-energy phosphate-bond equivalents are utilized in the process of activation of amino acids for protein synthesis? a. Zero b. One c. Two d. Three e. Four

*The answer is B.* All of the modes of therapy are theoretically possible, and enzyme therapy (i.e., injection of purified enzyme) has been successful in several lysosomal deficiencies, particularly those in which the central nervous system is not affected (i.e., Gaucher's disease). Unfortunately, antibodies frequently develop to the injected enzyme and limit the term of successful enzyme delivery. Heterologous bone marrow transplant, preferably from a related donor, offers the most realistic and effective therapy since the graft provides a permanent source of enzyme. Bone marrow transplants do have a 10% mortality, however, and the enzyme diffuses poorly into the central nervous system. Somatic gene therapy (i.e., delivery of enzyme to somatic cells via viral vectors or transfected tissue) is now possible; however, targeting of the gene product to appropriate tissues and organelles is still a problem. Transfected autologous bone marrow transplant (i.e., marrow from the patient) has been used in a few cases of adenosine deaminase deficiency, an immune disorder affecting lymphocytes. Germ-line gene therapy requires the insertion of functional genes into gametes or blastomeres of early embryos prior to birth. The potential for embryonic damage, lack of knowledge regarding developmental gene control, and ethical controversies regarding selective breeding or embryo experimentation make germ-line therapy unrealistic at present.

Hurler's syndrome is caused by a deficiency of L-iduronidase, an enzyme normally expressed in most human cell types. It was demonstrated by Neufeld that exogenous L-iduronidase could be taken up by deficient cells via a targeting signal that directed the enzyme to its normal lysosomal location. Which of the therapeutic strategies below would be the most realistic and efficient mode of therapy? a. Germ-line gene therapy b. Heterologous bone marrow transplant c. Infection with a disabled adenovirus vector that carries the L iduronidase gene d. Injection with L-iduronidase purified from human liver e. Autologous bone marrow transplant after transfection with a virus carrying the L-iduronidase gene

*The answer is A.* The replication of double-stranded DNA is semiconservative, meaning that each strand separates and serves as a template for synthesis of a new complementary strand. The first round of replication of a labeled DNA helix in a cold (unlabeled) solution will yield two daughter double-stranded molecules, each with one labeled and one unlabeled strand. The second round of replication will yield four double-stranded DNA molecules. Two of these will have one original labeled strand and one unlabeled strand; the other two will have two unlabeled strands and contain no radioactivity.

If a completely radioactive double-stranded DNA molecule undergoes two rounds of replication in a solution free of radioactive label, what is the radioactivity status of the resulting four double-stranded DNA molecules? a. Half should contain no radioactivity b. All should contain radioactivity c. Half should contain radioactivity in both strands d. One should contain radioactivity in both strands e. None should contain radioactivity

*The answer is B.* The cholesterol tag on the dsRNA allowed cells to take up the dsRNA, which was processed by intracellular ribonucleases to make a specific silencing RNA for the apolipoprotein B mRNA. Binding of the processed dsRNA to the apoB mRNA will lead to either the destruction of the mRNA or the blockage of translation of the mRNA. In either event, there will be a reduction in apoB translation such that cells can no longer produce apoB100 or apoB48. The dsRNA does not affect the transcription of the apoB gene, nor does it interfere with apoB folding once it becomes transcribed and translated. The dsRNA does not affect the turnover of the apoB protein, nor does it edit the apoB mRNA (other systems in the cell will do that)

In a study with mice exhibiting hypercholesterolemia, cholesterol was affixed to double-stranded RNA, which targeted the dsRNA to enter cells through cholesterol diffusion through the plasma membrane. The dsRNA was targeted to bind to mRNA that encoded the apolipoprotein B gene and resulted in a lowering of circulating cholesterol levels. This result occurs due to which of the following? (A) Inhibition of apolipoprotein B transcription (B) Inhibition of apolipoprotein B translation (C) Inhibition of apolipoprotein B folding (D) Enhanced degradation of apolipoprotein B (E) RNA editing of the apolipoprotein B mRNA

*The answer is D.* Bacterial DNA contains stop signals, some of which require ρ protein. This has been demonstrated by examining the synthesis of mRNA in the presence and absence of ρ protein. In the absence of ρ protein, longer RNA molecules are often synthesized. This would seem to indicate that mRNA length can be controlled by the cell. In addition, anti-terminator proteins are needed to allow certain genes to be properly expressed. Mammalian mechanisms for transcription termination, and the likely presence of factors regulating termination, are not yet characterized.

In bacterial RNA synthesis, the function of factor ρ is to a. Bind catabolite repressor to the promoter region b. Increase the rate of RNA synthesis c. Eliminate the binding of RNA polymerase to the promoter d. Participate in the proper termination of transcription e. Allow proper initiation of transcription

*The answer is D.* DNA synthesis cannot occur until an RNA primer is made. A specific type of RNA polymerase called primase synthesizes a short stretch of RNA of about five nucleotides that is complementary to the template DNA strand in duplex DNA near the replication fork. This function cannot be carried out by DNA polymerase. In contrast, both DNA polymerase and RNA polymerase work in the 5′ to 3′ direction and add nucleoside monophosphates from nucleotide triphosphates to the growing polynucleotide chains of DNA or RNA. Only DNA polymerase edits as it synthesizes DNA and fills the gap between Okazaki fragments.

In contrast to DNA polymerase, RNA polymerase a. Fills in the gap between Okazaki fragments b. Works only in a 5′ to 3′ direction c. Edits as it synthesizes d. Synthesizes RNA primer to initiate DNA synthesis e. Adds nucleoside monophosphates to the growing polynucleotides

*The answer is B.* When region B is deleted from the promoter region, there is an increase in overall expression of the reporter gene (the second and last lines of the figure indicate this). Removal of any other region leads to a decrease in expression, indicating that positive- acting transcription factors bind to those regions of DNA.

In order to better analyze the promoter region of a particular gene, this cloned region of the gene was placed in front of a reporter gene and the resultant vectors placed in eukaryotic cells to measure the expression of the reporter, using various deleted constructs. The results obtained were as follows: Indicate which region (as designated by the letters A, B, C, D, and E) binds an inhibitory transcription factor.

*The answer is D.* Thymidine can only be found in DNA so it would eb the most specific in labeling DNA.

In order to determine the rate of DNA synthesis in various mammalian tissues, the administration of which of the following labeled substances would be most specific in labeling DNA? (A) Adenosine (B) Cytosine (C) Guanosine (D) Thymidine (E) Uridine

*The answer is B.* The replacement of the codon UAG with UAA would be a silent mutation since both codons are "stop" signals. Thus, transcription would cease when either triplet was reached. There are three termination codons in mRNA: UAG, UAA, and UGA. These are the only codons that do not specify an amino acid. A missense or a substitution mutation is the converting of a codon specifying one amino acid to another codon specifying a different amino acid. A nonsense mutation converts an amino acid codon to a termination codon. A suppression counteracts the effects of another mutation at another codon. The addition or deletion of nucleotides results in a frame-shift mutation.

In the following partial sequence of mRNA, a mutation of the template DNA results in a change in codon 91 to UAA. What type of mutation is it? a. Missense b. Silent c. Nonsense d. Suppressor e. Frame shift

*The answer is E.* Because of wobble codons for the same amino acid often differ in the third base. Option B would be acceptable, except that it is a stop codon.

In the genetic code of human nuclear DNA, one of the codons specifying the amino acid tyrosine is UAC. Another codon specifying this same amino acid is A. AAC B. UAG C. UCC D. AUG E. UAU

*The answer is C.* Transcription factors can be either positive or negative acting. In either event, to exert an effect on transcription, the factor must bind to the DNA and then either promote RNA polymerase binding to DNA (in which case, it is a positive-acting factor) or inhibit RNA polymerase binding to DNA (in which case it is a negative-acting factor). Negative-acting factors can do so by blocking the binding of positively acting factors to the DNA or by blocking necessary transactivating factors from binding to other factors already bound to the DNA. In either event, the net result is a reduction in the rate at which RNA polymerase binds to the promoter region to initiate transcription. Once initiated, the rate of phosphodiester bond formation is constant. RNA editing is a rare event, and is not used to regulate gene transcription. If HAT activity were stimulated, gene transcription would be increased, as the acetylated histones would have a reduced affinity for DNA, and it would be easier for RNA polymerase to bind to the promoter region. Increasing enhancer binding to DNA would also increase the rate at which RNA polymerase would bind to the promoter, as enhancers increase the association of positive-acting transactivation factors, which would promote RNA polymerase binding to the promoter and transcription.

Induction of certain transcription factors leads to a decrease in the expression of certain genes. This occurs through which of the following mechanisms? (A) Decreasing the rate of RNA polymerase-catalyzed phosphodiester bond formation (B) Inducing the synthesis of a protein that posttranscriptionally edits mRNA such that translation initiation is blocked (C) By decreasing the rate of RNA polymerase binding to the promoter (D) Through stimulation of proteins with HAT activity (E) Increasing enhancer binding to DNA

*The answer is C.* Since both strands of parental DNA serve as templates for the synthesis of new DNA, it appears that DNA synthesis must be 5′ to 3′ for one daughter strand and 3′ to 5′ for the other daughter strand at the replication fork. Despite the apparent need for 3′ to 5′ synthesis, all DNA polymerases and repair enzymes can only synthesize DNA in the 5′ to 3′ direction. The apparent contradiction is solved by understanding that one strand of DNA is synthesized continuously in the 5′ to 3′ direction while the other strand is made up of small fragments known as Okazaki fragments. The small Okazaki fragments are, in fact, synthesized in a 5′ to 3′ direction and then joined together by DNA ligase. Each Okazaki fragment is about 1000 nucleotides long. Thus, while the overall direction of growth of the lagging strand that is made up of small fragments is in fact in the 3′ to 5′ direction, the actual polymerization of individual nucleotides is in the 5′ to 3′ direction. Crossing over of the DNA strands does not occur during replication.

It is well known that DNA polymerases synthesize DNA only in the 5′ to 3′ direction. Yet, at the replication fork, both strands of parental DNA are being replicated with the synthesis of new DNA. How is it possible that while one strand is being synthesized in the 5′ to 3′ direction, the other strand appears to be synthesized in the 3′ to 5′ direction? This apparent paradox is explained by a. 3′ to 5′ DNA repair enzymes b. 3′ to 5′ DNA polymerase c. Okazaki fragments d. Replication and immediate crossover of the leading strand e. Lack of RNA primer on one of the strands

*The answer is D.* Homeotic genes are genes which regulate the development of anatomical structures in various organisms such as echinoderms, insects, mammals, and plants. Multiple developmental abnormalities due to mutation in a single gene.

Klein-Waardenburg syndrome is a single-gene disorder that includes dystopia canthorum (lateral displacement of the inner corner of the eye), impaired hearing, and pigmentary abnormalities. The gene involved is most likely to be a (A) pseudogene (B) proto-oncogene (C) transgene (D) homeotic gene (E) tumor suppressor gene

*The answer is C.* Methionyl-tRNA is the special tRNA used in eukaryotes for initiation. Initiation of protein synthesis by bacterial, mitochondrial, and chloroplast ribosomes requires N formylmethionyl-tRNA. The mitochondria of eukaryotic cells are similar to bacteria in the size of their ribosomal RNAs (23S and 16S) and their mechanisms for protein synthesis. Mitochondrial and prokaryotic ribosomes (including those of chloroplasts and bacteria) use formylmethionyl-tRNA for initiation of protein synthesis and are sensitive to inhibitors like streptomycin, tetracycline, and chloramphenicol that have little effect on eukaryotic cells. The latter drugs are useful as antibiotics in animals and humans since they inhibit bacteria but do not gain entry into mitochondria.

Methionyl-transfer (t) RNA is used for initiation of protein synthesis by which of the following? a. Chloroplast ribosomes b. Eukaryotic mitochondrial ribosomes c. Eukaryotic cytoplasmic ribosomes d. Bacterial ribosomes e. Bacterial cytoplasm

*The answer is C.* By using recombinant DNA techniques, mRNAs can be produced that yield chimeric proteins. By forming mRNAs that produce otherwise cytosolic proteins, as when α-globin is engineered with a cleavable amino terminal signal sequence, this otherwise cytosolic protein becomes a secretory protein and is translocated into the lumen of endoplasmic reticulum. The signal sequence thus contains all the information needed to direct the translocation of protein across endoplasmic reticulum. These experiments were performed by adding chimeric mRNA to an in vitro system of protein synthesis composed of endoplasmic reticulum vesicles, ribosomes, tRNAs, and other factors required for protein synthesis. Without the modified amino terminal signal sequence, the α-globin is released into the experimental solution, and with the signal sequence it is synthesized into the lumen of the endoplasmic reticulum vesicles.

Modification of mRNA so that a signal sequence is added to the amino terminus of the cytosolic protein, α-globin, results in a. No change in physiology of the protein b. Proteolytic cleavage within the cytosol c. Translocation across the endoplasmic reticulum d. Cytosolic localization of the protein e. Signal recognition particle synthesis

*The answer is C.* Thalassemias cause anemia due to genetically defective synthesis of either the α or β chains of hemoglobin. While many different types of mutations can cause thalassemias, mRNA production or processing is commonly affected. Premature termination of mRNA transcription leading to shorter-than-normal chains, frame-shift mutations leading to abnormal amino acid sequences in chains after the mutation (addition or deletion), and aberrant splicing are some of the RNA problems that can lead to dysfunctional or nonfunctional chains. In some thalassemias, one of the chains (α or β) may be missing completely. However, the absence of both chains is a lethal mutation that would not be seen in the population.

Most thalassemias are the result of mutations causing a. Increased α chain synthesis b. "Sticky" hemoglobin c. RNA processing or production defects d. Protein folding problems e. Absence of both A and B hemoglobin

*The answer is E.* Protein synthesis occurs in the cytoplasm, on groups of free ribosomes called polysomes, and on ribosomes associated with membranes, termed the rough endoplasmic reticulum. However, proteins destined for secretion are only synthesized on ribosomes of the endoplasmic reticulum and are synthesized in such a manner that they end up inside the lumen of the endoplasmic reticulum. From there the secretory proteins are packaged in vesicles. The Golgi apparatus is involved in the glycosylation and packaging of macromolecules into membranes for secretion.

New proteins destined for secretion are synthesized in the a. Golgi apparatus b. Smooth endoplasmic reticulum c. Free polysomes d. Nucleus e. Rough endoplasmic reticulum

*The answer is E.* A homeobox is a highly conserved DNA sequence, usually about 180 nucleotides in length. A gene containing a homeobox sequence is called a homeobox gene. Homeobox genes typically code for DNA-binding transcription factors which alter the expression of genes involved in morphogenesis (Choice E). Morphogenesis is the proper formation and placement of tissues, organs and structural elements of the body. Homeobox genes were initially described in Drosophila, where mutations of these genes caused limbs and appendages to develop in the incorrect locations. *Educational Objective:* Homeobox genes code for DNA-binding transcription factors that play an important role in morphogenesis.

Numerous human genes contain a highly conserved 180-nucleotide motif called a homeobox. Most homeobox-containing genes code for which of the following protein types? A. Cytoplasmic enzymes B. Transport proteins C. Structural proteins D. DNA replication enzymes E. Transcription regulators F. Cell surface receptors

*The answer is D.* A carrier would have one normal gene, and one disease gene, which contains the new restriction enzyme site. After amplifying this area of DNA by PCR, and then cutting with the appropriate restriction enzyme, this individual would show a 1.5 kb band from the normal gene, and both 0.6 and 0.9 kb bands from the disease gene.

One studies a disease caused by a single point mutation in which a restriction enzyme site is gained by the mutation. To determine if someone is a carrier of the disease, PCR primers were generated which allowed a 1.8 kb fragment to be amplified across both X restriction sites in the normal gene. After treatment of the amplified DNA by enzyme X, a carrier of the disease would be expected to exhibit which size fragments, as determined by ethidium bromide staining of an agarose gel? (A) 1.5 kb only (B) 0.6 and 0.9 kb (C) 0.6 and 1.5 kb (D) 0.6, 0.9, and 1.5 kb (E) 0.9 and 1.5 kb

*The answer is E.* Normal parents having two affected children, male and female, is suggestive of autosomal recessive inheritance. This interpretation fits with the usual inheritance of oculocutaneous albinism, implying a 1/4 risk for a newborn in whom signs and symptoms of albinism are not yet evident. The defect in melanin synthesis in albinism decreases the amount of this protective pigment in skin and increases the exposure of DNA in skin cells to sunlight. Ultraviolet rays from sunlight cause DNA cross-linkage between at least two bases in the same or opposite strands of DNA. Cross-linking occurs through the formation of thymine-thymine dimers. The DNA cross-links cause higher rates of mutation and skin cancer in albinism, mandating the wearing of protective clothing, sunglasses, and sunscreens by affected individuals. DNA deletions and point mutations are less common than DNA cross-links after sunlight exposure.

Parents bring their newborn daughter to you for consultation about diagnosis and management. Their first two children, a boy and a girl, have a complete form of albinism with pink irides, blond hair, and pale skin. Which of the following represents your correct advice concerning the newborn child? a. A 1/8 risk for albinism and skin cancer from DNA deletions b. A 1/8 risk for albinism and skin cancer from DNA cross-linkage c. A 1/4 risk for albinism and skin cancer from DNA point mutations d. A 1/4 risk for albinism and skin cancer from DNA deletions e. A 1/4 risk for albinism and skin cancer from DNA cross-linkage

*The answer is B.* The following mutations were shown in the DNA changes: a. Missense (cys to leu) b. Nonsense (tyr to stop) c. Missense (ser to phe) d. Harmless (tyr to tyr) e. Missense (leu to phe) f. No mutation g. Missense (phe to tyr) Most of the mutations shown result in a missense effect, with a different amino acid being incorporated into the same site in a protein. This may or may not have an effect depending upon its location. Some single-base mutations are harmless because of the degeneracy of the genetic code, whereby more than one triplet code exists for all amino acids except tryptophan and methionine. Choice a contains two mutations, one degenerate and the other missense. DNA template: 3′-ACGACGACG-5′ to 3′-ACAAACACG-5′ mRNA: 5′-UGCUGCUGC-3′ to 5′-UGUUUGUGC-3′ Protein: cys-cys-cys to cys-leu-cys Nonsense mutations occur when the reading of the normal termination signal is changed. This can occur by mutation to a stop signal as in choice b, by deletions near a stop codon, or by insertions.

Part of the triplet genetic code involving mRNA codon triplets that start with U is shown below. Using the portion of the genetic code shown, which of the following mutations in the 3′ to 5′ DNA template strand corresponds to a nonsense mutation? a. ACGACGACG to ACAAACACG b. AGGAATATG to AGGAATATT c. AGAATAACA to AAAATAACA d. AAAATGAGC to AAAATAAGC e. AACAACAAC to AACAAGAAC f. AGAATCAAA to AGAATCAAA g. AAAAAGAGG to ATAAAGAGG

*The answer is B.* The most efficient DNA diagnosis would involve analysis of the two L-iduronidase genes in an individual, looking for any change in DNA sequence (nucleotide substitutions, deletions, duplications) that would alter function of the L-iduronidase protein. Alternative forms of a gene are called alleles, and an allele that has changed during transmission from parent to child may be called a mutant allele. Many human characteristics are encoded by genes, each occupying a particular address or locus on a chromosome. Genes consist of DNA segments that encode RNA along with flanking DNA sequences that regulate gene expression. Within the coding regions are DNA segments that are transcribed and then translated into protein (exons), and those that are transcribed but removed by RNA splicing (introns). If a DNA mutation produces a disease through its altered protein, as with L-iduronidase gene mutations in Hurler's syndrome, the mutant gene (strictly, the mutant allele) may be called abnormal. Human autosomes (chromosomes 1 through 22) and X chromosomes in females have two homologous loci in each individual, harboring two identical (homozygous) or different (heterozygous) alleles. Males have only one X chromosome and often only one allele per sex chromosome locus because the Y has minimal coding material. The complete set of genetic material (all the genes and loci) in each biological species is called the genome.

Patients with Hurler's syndrome are known to have mutations at the L-iduronidase locus. The diagnosis of Hurler's syndrome is most efficiently made by analyzing a patient's DNA for a. A region of DNA that does not encode RNA b. Alternative forms of the L-iduronidase gene c. The entire set of genes in one leukocyte d. A nucleotide substitution in the L-iduronidase gene e. The position of the L-iduronidase gene on a chromosome

*The answer is A.* One of the most common types of inherited cancers is nonpolyposis colon cancer. Most cases are associated with mutations of either of two genes that encode proteins critical in the surveillance of mismatches. Mismatches are due to copying errors leading to one- to five-base unmatched pieces of DNA. Two- to five-base-long unmatched bases form miniloops. Normally, specific proteins survey newly formed DNA between adenine methylated bases within a GATC sequence. Mismatches are removed and replaced. First, a GATC endonuclease nicks the faulty strand at a site complementary to GATC, then an exonuclease digests the strand from the GATC site beyond the mutation. Finally, the excised faulty DNA is replaced. In HNPCC, the unrecognized mismatches accumulate, leading to malignant growth of colon epithelium. The other forms of DNA repair are important for rectifying damage from ultraviolet light.

Patients with hereditary nonpolyposis colon cancer [HNPCC] have genes with microsatellite instability, that is, many regions containing abnormal, small loops of unpaired DNA. This is a result of a mutation affecting a. Mismatch repair b. Chain break repair c. Base excision repair d. Depurination repair e. Nucleotide excision repair

*The answer is E.* RNA polymerase II forms precursor mRNA (pre~mRNA) from the DNA template. Posttranscriptional processing of pre-mRNA fonts mature mRNA, which is ready for export from the nucleus and translation. Posttranscriptional processing involves the following: 1. *5' capping:* A 7-methyl-guanosine cap is added to the 5' and of the mRNA. 2. *Polyadenylation:* A poly-A tail is added to most eukaryotic mRNA molecules by poly-A polymerase. Poly-A tails are not transcribed from the DNA template. Instead, a consensus sequence (usually "AAUAAA") found within the 3' end of the gene being transcribed directs the addition of the poly-A tail onto the mRNA. This tail protects the mRNA from degradation within the cytoplasm after it exits the nucleus. Segments of mRNA downstream from the consensus sequence are likely part of the poly-A tail (Choice E) 3.*Splicing:* The initial mRNA, called pre-mRNA or heterogeneous nuclear RNA (hnRNA), contains sequences from coding and non-coding regions of DNA, known as exons and introns, respectively. Removal of introns (non coding mRNA segments) occurs during splicing. *Educational Objective:* The polyadenylation signal sequence at the 3' end of the mRNA transcript is responsible for addition of the poly-A tail. The poly-A tail is not transcribed from DNA, but rather added as a posttranscriptional modification downstream of a consensus sequence (usually "AAUAAA"). This tail protects the mRNA from degradation within the cytoplasm after it exits the nucleus .

Precursor mRNA undergoes substantial post-transcriptional processing before it becomes the finalized mRNA template. Because of this extensive processing, the final mRNA sequence can be very different from that of its encoding DNA. An mRNA molecule transcribed from a eukaryotic gene is shown schematically below. Which of the following portions is not transcribed from the DNA template?

*The answer is E.* The major effects of radiation are to damage cellular DNA by opening purine rings and rupturing phosphodiester bonds. Chemical agents such as formaldehyde can cross-link DNA, and inhibitors of DNA methylation, such as methotrexate (an inhibitor of folic acid), were the first anticancer drugs. Experimental gene therapies for cancer include the inhibition of oncogene expression and the enhancement of tumor suppressor gene activity. These therapies target particular DNA-RNA transcription complexes or signal transduction cascades that are active in cancer cells.

Radiation therapy is employed for many cancers, including irradiation of the central nervous system to destroy lymphoblasts in leukemia. Which of the following accounts for the destruction of rapidly growing cells? a. Cross-linking of DNA b. Demethylation of DNA c. Cleavage of DNA double strands d. Disruption of DNA-RNA transcription complexes e. Disruption of purine rings in DNA

*The answer is D.* Nuclear chromosomes contain most of the DNA found in human cells. However, mitochondria also contain their own DNA called mitochondrial DNA (mtDNA). This DNA exists as a small circular chromosome with a slightly different genetic code than that of nuclear DNA, consistent with the endosymbiotic theory that mitochondria originated as prokaryotic cells that were later engulfed by ancient eukaryotes. Over time, most of the genes coding for mitochondrial proteins have migrated to nuclear DNA. However, mtDNA still codes for about 14 proteins (some involved in oxidative metabolic pathways) and the ribosomal and transfer RNA needed for mitochondrial protein synthesis. Each mitochondrion contains 1-10 copies of maternally derived mtDNA. As a result. diseases arising from mutations in mtDNA are transmitted from the mother to all of her offspring. Mitochondria can be identified on electron microscopy by their characteristic double membrane and wavy cristae. (Cholce A) The rough endopasmic reticulum has a stippled appearance secondary to the presence of numerous ribosomes bound to its membranes. These ribosomes are involved In the synthesis of integral membrane proteins and proteins destined for export or packaging into granules of organelles. (Choice B) The dark region identified within the nucleus is the nucleolus, the site of synthesis and assembly of ribosomal components. There is no lipid membrane separating the nucleolus from the rest of the nucleus (Choice C) The lighter "electron-Iucent" regions within the nucleus signify euchromatin (unpackeged DNA actively being transcribed). (Choice E) This electron-dense membrane-bound spherical structure represents an exocrine grande containing enzymes and other proteins packaged for secretion.

Researchers analyzing eukaryotic genome structure and function perform an experiment to extract DNA from exocrine pancreatic cells, During the purification process, they isolate small circular DNA molecules that resemble a bacterial chromosome. Further analysis shows that these molecules code for proteins transfer RNA, and ribosomal RNA. From which of the following cellular structures did these DNA molecules most likely originate?

*The answer is B.* A variety of genetic diseases, such as sickle cell anemia, Huntington's chorea, and cystic fibrosis, can be detected by restriction fragment length polymorphism (RFLP) analysis. In order for RFLP to be able to detect and follow the inheritance of these genes, the detected mutation must be at or closely linked to an altered restriction site. Mutations within the restriction sites change the size of restriction fragments. The different-sized fragments migrate in different positions during electrophoresis of bands visualized by Southern blot analysis, which utilizes fluorescent or radiolabeled DNA probes.

Restriction fragment length polymorphism (RFLP) analysis can only be used to follow the inheritance of a genetic disease if a. mRNA probes are used in combination with antibodies b. The disease-causing mutation is at or closely linked to an altered restriction site c. Proteins of mutated and normal genes migrate differently upon gel electrophoresis d. Mutations are outside of restriction sites so that cleaving still occurs e. Restriction fragments remain the same size but their charge changes

*The answer is D.* All options represent single-base changes in the mutant sequence in the stem, but only choice D reestablishes a palindrome.

Restriction fragment length polymorphisms may be produced by mutations in the sites for restriction endonucleases. For instance, a single base change in the site for the nuclear SalI produces the sequence GTGGAC, which can no longer be recognized by the enzyme. What was the original sequence recognized by SalI? A. GTAGAC B. GCGGAC C. CTGGAC D. GTCGAC E. GTGTAC

*The answer is C.* Prokaryotic ribosomes have a sedimentation coefficient of 70S and are composed of 50S and 30S subunits. Eukaryotic cytoplasmic ribosomes, either free or bound to the endoplasmic reticulum, are larger—60S and 40S subunits that associate to an 80S ribosome. Nuclear ribosomes are attached to the endoplasmic reticulum of the nuclear membrane. Ribosomes in chloroplasts and mitochondria of eukaryotic cells are more similar to prokaryotic ribosomes than to eukaryotic cytosolic ribosomes. Like bacterial ribosomes, chloroplast and mitochondrial ribosomes use a formylated tRNA. In addition, they are sensitive to many of the inhibitors of protein synthesis in bacteria.

Ribosomes similar to those of bacteria are found in a. Plant nuclei b. Cardiac muscle cytoplasm c. Pancreatic mitochondria d. Liver endoplasmic reticulum e. Neuronal cytoplasm

*The answer is D.* Chemotherapeutic agents derived from alpha-amanitin, found in Amenita phalloides (death cap mushrooms), aim to inhibit RNA polymerase II. RNA polymerase II transcribes messenger RNA (mRNA). Messenger RNA are the largest form of RNA that code for functioning proteins in eukaryotic cells.

Scientists develop a new chemotherapeutic agent derived from alpha-amanitin. It is found to decrease levels of mRNA. Levels of rRNA and tRNA are not affected. What is the most likely mechanism of action of this new drug? A. Inhibition of RNA polymerase III B. Blockage of the ribosomal P site C. Inhibition of RNA polymerase I D. Inhibition of RNA polymerase II E. Blockage of the ribosomal A site

*The answer is A.* The recognition of overlapping sequences on DNA by different factors allows either positive-acting or negative-acting effects, depending on which transcription factor is present in higher concentration. The γ-globin gene is turned off after birth, and this has to do, in part, with factors bound to the promoter. When stage selector protein (SSP) is bound, γ-globin synthesis is favored, but when SP1 binds to this area instead, γ-globin synthesis is repressed. A similar story occurs at the CAAT sites; when CP1 is bound, γ-globin synthesis is favored, but when CAAT displacement protein binds, CP1 can no longer bind, and gene transcription is inhibited. Thus, γ-globin expression is regulated by the concentrations of both positive-acting and negative-acting factors available in the cell. The overlapping binding sites were not designed for redundancy in case of mutation, to provide ribosome binding sites (the ribosomes bind to RNA produced from exons, not from the promoter region), or to provide a target for interfering RNAs (the siRNAs are targeted to mRNA). The multiple binding sites, by themselves, do not promote looping of DNA, but once transcription factors have bound to the DNA, looping may occur as the transactivating factors bind to the proteins bound to DNA.

Shown below is a partial map of the promoter region, and promoter-proximal region, for the γ-globin gene. The overlapping binding sites for transcription factors allow for which of the following to occur? (A) The ability to modulate the binding of positive, or negative, transacting factors to the DNA (B) The ability to reduce the risk of losing transcriptional control via mutation in this region (C) Promoting looping of this DNA region (D) Providing a target for interfering RNAs (E) Providing ribosome binding sites for translation initiation

*The answer is E.* Starting with the S site at the "top" of the figure, the next restriction cut would be at the X site in the insert, generating a DNA fragment of 1,000 bp (700 + 300). The next site is the S site in the insert, which is 800 bp long. Moving along the DNA the next cut site is the X site in the vector, 900 bp from the previous S site. This leaves, then, a 1,500 bp piece of the vector. The total size of the vector plus insert is 4,200 bp, and the pieces generated, 800 bp, 900 bp, 1,000 bp, and 1,500 bp, add up to the total, 4,200 bp.

Shown below is a vector plus insert, with restriction endonuclease sites also shown. The insert was obtained from a gene, and exon 1 of the gene is indicated by the black box. If this vector plus insert were to be cleaved to completion with enzymes S and X simultaneously, what size bands would be observed? (A) 1,600 bp, 2,600 bp (B) 400 bp, 1,500 bp, 2,300 bp (C) 400 bp, 1,000 bp, 1,300 bp, 1,500 bp (D) 800 bp, 1,500 bp, 1,900 bp (E) 800 bp, 900 bp, 1,000 bp, 1,500 bp

*The answer is C.* Since the probe is going to be used in a Northern blot, the DNA for the probe must be obtained from an exon, as mRNA will only contain sequences which are complementary to exonic DNA. The exonic DNA is represented by the black box in the figure, which is within the 500 bp SH restriction fragment of the insert. The other parts of the insert represent introns and would not hybridize to mRNA in a Northern blot

Shown below is a vector plus insert, with restriction endonuclease sites also shown. The insert was obtained from a gene, and exon 1 of the gene is indicated by the black box. Which labeled DNA fragment would work best for a Northern blot analysis? (A) The 300 bp HX restriction fragment (B) The 1,100 bp HS restriction fragment (C) The 500 bp HS restriction fragment (D) The 1,800 bp SS restriction fragment (E) The 800 bp XS restriction fragment

*The answer is D.* The complementary probe will be antiparallel to the coding strand of the mutant allele, with all sequences written 5′ → 3′.

Sickle cell anemia is caused by a missense mutation in codon 6 of the β-globin gene. A man with sickle cell disease and his phenotypically normal wife request genetic testing because they are concerned about the risk for their unborn child. DNA samples from the man and the woman and from fetal cells obtained by amniocentesis are analyzed using the PCR to amplify exon 1 of the β-globin gene. Which 12-base nucleotide sequence was most likely used as a specific probe complementary to the coding strand of the sickle cell allele? A. CCTCACCTCAGG B. CCTGTGGAGAAG C. GGACACCTCTTC D. CTTCTCCACAGG E. CTTCTCCTCAGG

*The answer is D.* Sickle cell anemia is an autosomal recessive hemoglobinopathy with an incidence of 1 in 500 African American births. It is caused by a single-nucleotide substitution in codon 6 of the β-globin gene. This mutation abolishes an enzyme site so that a larger DNA fragment is obtained after Southern blot analysis with the appropriate enzyme. Single-nucleotide substitutions do not change the length of coding regions (exons). The amplification of DNA segments using the polymerase chain reaction (PCR) allows more sensitive detection of restriction enzyme differences, and can be followed by allele-specific oligonucleotide (ASO) hybridization to determine the presence of normal versus sickle alleles. The equivalence of DNA in most tissues (with the exception of red blood cells that extrude their nucleus) makes DNA diagnosis a powerful technique that is independent of gene or protein expression. Western blotting is a technique that uses antibodies to highlight the size and amount of mutant protein in cell extracts. Since single-nucleotide changes in the gene may not affect protein size or conformation, western blotting is generally less sensitive and specific than DNA diagnosis.

Sickle cell anemia is caused by a point mutation in the hemoglobin gene, resulting in the substitution of a single amino acid in the β-globin peptides of hemoglobin. This mutation is best detected by which of the following? a. Isolation of DNA from red blood cells followed by polymerase chain reaction (PCR) amplification and restriction enzyme digestion b. Isolation of DNA from blood leukocytes followed by Southern blot analysis to detect globin gene exon sizes c. Isolation of DNA from blood leukocytes followed by DNA sequencing of globin gene introns d. Isolation of DNA from blood leukocytes followed by polymerase chain reaction (PCR) amplification and allele-specific oligonucleotide (ASO) hybridization e. Western blot analysis of red blood cell extracts

*The answer is E.* In the β-globin chain of hemoglobin S, a valine residue replaces a glutamic acid at the sixth amino acid position from the N-terminus. The amino acid substitution is the result of a single base change (point mutation) from thymine to adenine at the second position of the sixth codon. Crossing over among homologous β-globin genes might exchange alleles, if equal, or generate mutant alleles with duplicated/ deficient nucleotides, if unequal. Two-base insertions would change the reading frame of the genetic code (frame-shift mutation) and produce a nonsense peptide after the point of insertion. Three-base deletions could also cause frame shifts or, if one codon were removed, delete one amino acid. Nondisjunction involves abnormal segregation of chromosomes at meiosis or mitosis, and would produce nonviable individuals or somatic cells with additional or missing copies of chromosome 11 and its β-globin locus.

Sickle cell anemia is the clinical manifestation of homozygous genes for an abnormal hemoglobin molecule. The mutation in the β chain is known to produce a single amino acid change. The most likely mechanism for this mutation is a. Crossing over b. Two-base insertion c. Three-base deletion d. Nondisjunction e. Single-base substitution (point mutation)

*The answer is C.* In the mRNAs of bacteria and the exonic mRNA regions of mammals, the triplet nucleotides comprising codons are continuous, without "spacers" to mark the end of one codon and the beginning of another. These RNA regions and their product peptides are also colinear with the gene (DNA) sequence. Bacterial mRNA molecules are polycistronic and code for more than one polypeptide chain or enzyme, allowing their coordinate regulation in response to metabolic or environmental signals. There are only three "nonsense" or chain terminating codons and 61 "sense" codons that encode for 20 amino acids. Redundancy of the code (several codons code for the same amino acid) is compensated for by the "wobble" hypothesis of Crick. The complementary anticodons of charged transfer RNAs hybridize stringently at the first two positions of the codon but weakly ("wobbly") at the third position. One aminoacyl-transfer RNA can thus recognize several different codons, each identical at the first two positions but different at the third.

Studies of the genetic code in bacteria have revealed that a. Messenger RNA (mRNA) molecules specify only one polypeptide chain b. Many triplets can be "nonsense" triplets c. No signal exists to indicate the end of one codon and the beginning of another d. The nucleotide on the 5′ end of a triplet has the least specificity for an amino acid e. Gene sequence and encoded proteins are not colinear

*The answer is D.* Oncogenes are cancer-producing genes. They are closely related to normal cellular genes and are often tyrosine kinases, growth factors, or receptors for growth factors. The expression of oncogenes leads to the translation and eventual transcription of the protein product of the oncogene. Thus, template-directed RNA synthesis but not DNA synthesis occurs during the expression of oncogenes. In contrast, template-directed DNA synthesis rather than RNA synthesis occurs during the repair of thymine dimers, the polymerase chain reaction, the functioning of the replication fork, and the growth of RNA tumor viruses. In the final stages of the repair of thymine dimers, once the dimer has been excised, DNA polymerase I enters the gap to carry out template-directed synthesis. In functioning of the replication fork, DNA polymerase III holoenzyme carries out synthesis of DNA during replication. Template-directed DNA synthesis is required for the growth of RNA tumor viruses (retroviruses). Once released into the host cytoplasm, retroviral RNA synthesizes both the positive and minus strands of DNA, using reverse transcriptase. This unique enzyme catalyzes the initial RNA-directed DNA synthesis, hydrolysis of RNA, and then DNA-directed DNA synthesis. The newly formed viral DNA duplex integrates into the host cell DNA prior to transcription. In this form, the retrovirus is inherited by daughter host cells. The polymerase chain reaction is a method of amplifying the amount of DNA in a sample or of enriching particular DNA sequences in a population of DNA molecules. In the polymerase chain reaction, oligonucleotides complementary to the ends of the desired DNA sequence are used as primer for multiple rounds of template-directed DNA synthesis.

Template-directed RNA synthesis occurs in which of the following? a. Point mutation b. Triplet repeat expansion c. Initiation of the polymerase chain reaction d. Expression of oncogenes e. Repair of thymine dimers

*The answer is D.* The collagen molecule consists of three polypeptide alpha chains held together by hydrogen bonds to form a rope-like triple helix structure. The variation in amino acid sequences in the collagen alpha chains gives rise to collagen diversity in different tissues. For example, type 1 collagen is present in skin, bones, tendons, blood vessels, and the cornea. In these tissues, collagen provides tensile strength, and it (along with keratin) is responsible for skin strength end elasticity. Ehlers-Danlos syndrome (EDS) is a group of rare hereditary disorders involving a defect in the synthesis of collagen found in skin, tendons, ligaments, and muscles. EDS usually manifests clinically as over-flexible (hypermobile) joints, over-elastic (hyperelastic) skin, and fragile tissue susceptible to bruising, wounding, and hemarthrosis. There are several different types of EDS defined by the extent of involvement of skin, joints, and other tissues. Common mutations leading to EDS phenotypes include deficiencies of the lysyl-hydroxylase and pro-collagen peptidase enzymes responsible for collagen synthesis. *Educational Objective:* Ehlers-Danlos syndrome is a heritable connective tissue disease associated with abnormal collagen formation. EDS usually manifests clinically as over flexible (hypermobi1e) joints, over-elastic (hyperelastic) skin, and fragile tissue susceptible to bruising, wounding, and hemarthrosis.

The 14-year-old daughter of a circus contortionist says that her skin scars and bruises easily. Physical examination reveals the findings seen below. A. Fibrillin B. Laminin C. Fibronectin D. Collagen E. Kerstin F. Elastin

*The answer is B.* The process of DNA replication is similar In eukaryotes and prokaryotes. The key steps involved in DNA replication are: 1. Unwinding of double stranded DNA (dsDNA) by helical to produce single stranded DNA (ssDNA) 2. Formation of a replication fork 3. Formation of an RNA primer by the action of the enzyme primase 4. Synthesis and concurrent proofreading of daughter DNA strands by DNA polymerase 5. Ligation of Okazaki fragments on lagging strands by lipase and removal and replacement of RNA primers with DNA by DNA polymerase I 6. Reconstitution of chromatin and ligation of daughter strands In E.coli, the three major types of DNA polymerase are DNA polymerase I, II and Ill. In eukaryotes, there are five major DNA polymerases: alpha, beta. gamma, delta and epsilon. Though the eukaryotic genome is much larger and more complex than the prokaryotic genome, interestingly, the size of the eukaryotic genome is not the source of its complexity. Its complexity results from the presence of a large number of non-coding DNA regions between coding regions. Within genes there are introns (Non-coding regions - Think "IN" between) separating exons (Coding regions - Think "EX"pressed). Prokaryotes rarely have introns within their genes. In contrast to prokaryotes, which typically have a single origin of replication, eukaryotes have multiple origins of replication. With multiple origins of replication, the genome can be copied much more quickly= because multiple regions are being replicated at once. *Educational Objective:* Multiple origins of replication make eukaryotic DNA synthesis quick and effective despite the large size of the genome compared to that of prokaryotic organisms.

The DNA replication process in eukaryotic cells closely mimics that in prokaryotic cells, but the volume of genetic material to be replicated is typically much greater in eukaryotic cells. Which of the following ensures fast DNA replication in eukaryotic cells? A. Energy-independent DNA unwinding B. Multiple origins of replication C. No RNA primers synthesized during replication D. Continuous synthesis of the lagging strand E. No proofreading by DNA polymerase (pol δ)

*The answer is C.* Insertion (choice c) or deletion (choice d) of nucleotides shifts the reading frame unless the change is a multiple of 3 (choice e). Frame shifts may create unintended stop codons as in choice c. Point mutations resulting in nucleotide or amino acid substitutions are conveniently named by their position in the protein, i.e., P21A (choice b). The protein change P21A could also be denoted by the corresponding change in the DNA reading frame, i.e., C63A. Deletions may be prefixed by the letter delta, as with ΔF25 (choice e).

The DNA sequence M, shown below, is the sense strand from a coding region known to be a mutational "hot spot" for a gene. It encodes amino acids 21 to 25. Given the genetic and amino acid codes CCC = proline (P), GCC = alanine (A), TTC = phenylalanine (F), and TAG = stop codon, Which of the following sequences is a frame-shift mutation that causes termination of the encoded protein? M 5′-CCC-CCT-AGG-TTC-AGG-3′ A. -CCA-CCT-AGG-TTC-AGG B. -GCC-CCT-AGG-TTC-AGG C. -CCA-CCC-TAG-GTT-CAG D -CCC-CTA-GGT-TCA-GG— E. -CCC-CCT-AGG-AGG——

*The answer is B.* In the Sanger technique, the sequence is read from the bottom of the gel to the top of the gel (5′ to 3′), going from one size of DNA to the next largest size. This is due to the way in which the fragments are generated.

The DNA sequence represented by the gel shown below, which was generated using the Sanger dideoxy sequencing technique, is which of the following? (A) ACTAACGCTTA (B) ATTCGCAATCA (C) ATCTATCGATC (D) GCCCTTTAAAA (E) AAAATTTCCCG

*The answer is D.* Since the sequence in the stem represents the coding strand, the mRNA sequence must be identical (except U for T). No T in the DNA means no U in the mRNA.

The base sequence of codons 57 -58 in the cytochrome β5 reductase gene is CAGCGC. The mRNA produced upon transcription of this gene will contain which sequence? A. GCGCTG B. CUGCGC C. GCGCUG D. CAGCGC E. GUCGCG

*The answer is E.* The first event that occurs in mRNA synthesis is the binding of transcription factor TFIID to the TATA box. This consensus sequence portion of virtually all eukaryotic genes coding for mRNA is centered at about −25 and is similar to a 10-sequence promoter box found in prokaryotes. TFIID contains a TATA box binding protein. The following sequence occurs in the initiation of mRNA synthesis: 1. TFIID binding to the TATA box 2. TFIIA binding 3. TFIIB binding 4. RNA polymerase II binding 5. TFIIE binding When all these elements are bound to DNA, the basal transcription apparatus complex is formed and can transcribe DNA slowly. Other factors are required for fast, efficient mRNA synthesis.

The consensus sequence 5′ TATAAAA 3′ found in eukaryotic genes is quite similar to a consensus sequence observed in prokaryotes. It is important as the a. Only site of binding of RNA polymerase III b. Promoter for all RNA polymerases c. Termination site for RNA polymerase II d. Major binding site of RNA polymerase I e. First site of binding of a transcription factor for RNA polymerase II

*The answer is B.* A Northern blot analysis detects RNA using a DNA probe to bind to the sample RNA (DNA-RNA hybridization). In this example, detection of glucokinase messenger RNA from the liver sample would demonstrate that the liver cells are expressing glucokinase.

The enzyme hexokinase catalyzes the first step of glycolysis, the phosphorylation of glucose in the 6-carbon position. In the liver, a different kinase called glucokinase catalyzes this step. A researcher wishes to determine what her other tissues in addition to the liver express glucokinase. Which of the following techniques would be most useful for this researcher? A. Northern blot analysis to detect DNA B. Northern blot analysis to detect RNA C. Southern blot analysis to detect DNA D. Southern blot analysis to detect RNA E. Western blot analysis to detect RNA

*The answer is E.* The AIDS treatment drug azidothymidine (AZT) exerts its effect by inhibiting viral reverse transcriptase. Thus, it prevents replication of the human immunodeficiency virus. Reverse transcriptase is an RNA-directed DNA polymerase. The RNA of retroviruses utilizes reverse transcriptase to synthesize DNA provirus, which in turn synthesizes new viral RNA. AZT inhibits DNA provirus production, but does not directly inhibit synthesis of new viral RNA.

The first drug to be effective against AIDS, including the reduction of maternal-to-child AIDS transmission by 30%, was AIDS drug azidothymidine (AZT). Which of the following describes its mechanism of action? a. It inhibits viral protein synthesis b. It inhibits RNA synthesis c. It inhibits viral DNA polymerase d. It stimulates DNA provirus production e. It inhibits viral reverse transcriptase

*The answer D.* The directing of nascent polypeptide chains to the endoplasmic reticulum is regulated by signal recognition particles (SRPs). The signal sequence of a nascent protein is recognized by an SRP, which complexes with the ribosome, mRNA, and the nascent protein. The complexed SRP then binds to an SRP receptor on the surface of the endoplasmic reticulum. After the ribosome is transferred to ribophorins and the translocation begins, SRP is released back into the cytosol. Ribosomes with nascent protein without a signal sequence do not participate in this process and instead synthesize proteins that are released into the cytosol.

The function of signal recognition particles is to a. Cleave signal sequences b. Detect cytosolic proteins c. Direct the signal sequences to ribosomes d. Bind ribosomes to endoplasmic reticulum e. Bind mRNA to ribosomes

*The answer is C.* This particular genetic defect is a known cause of thalassemia intermedia, which is a form of beta-thalassemia that is clinically less severe than beta-thelassemia major. Hypochromic, microcytic anemia is the classic laboratory finding in patients with thelassemia. Red blood cell morphology is quite variable depending on the type of thalassemia, and can include marked anisopoikilocytosis, target cell formation, tear drop cells, ardor Heinz bodies.

The genetic evaluation of a family with hereditary anemia reveals a point mutation in the β-globin gene, which results in the replacement of guanine (G) by cytosine (C) in the β-globin mRNA molecules three bases upstream from the AUG codon (position 6). The mutation results in a 30% decrease in the efficiency of protein synthesis. Homozygous patients with the mutation are most likely to have which of the following? A. Red blood cell sickling B. Spherocytosis C. Microcytosis D. Hyperchromia E. Iron deficiency

*The answer is A.* The Kozak sequence occurs on eukaryotic mRNA and is defined by the following sequence: GCCGCCRCCAUGG, where R is either adenine or guanine. When the methionine codon (AUG) is positioned near the beginning of a mRNA molecule and is surrounded by the Kozak sequence, it serves as the initiator for translation (i.e. mRNA binding to ribosomes). Among other factors, a purine (G or A) positioned three bases upstream from the AUG appears to be a key factor in this initiation process. A mutation in which guanine (G) is replaced by cytosine (C) in this particular position of the β-globin gene has been associated with thalassemia intermedia. (Choice B) Translocation is catalyzed by the elongation factor eEF2 and requires GTP hydrolysis. (Choice C) Peptide bond formation is catalyzed by peptide transferase on eukaryotic ribosomes. A defect in the mRNA coding for ribosomes would be required to interfere with this function. (Choice D) Termination of polypeptide synthesis occurs at stop codons. A mutation producing a premature stop codon would have this effect. (Choice E) Protein targeting is achieved by the amino acid sequence of the N-terminal section of a formed protein. This sequence is often removed from the final protein product once it reaches its destination, *Educational Objective:* The Kozak sequence plays a role in the initiation of translation. A mutation three bases upstream from the start codon (AUG) in this sequence is associated with thelassemia intermedia.

The genetic evaluation of a family with hereditary anemia reveals a point mutation in the β-globin gene, which results in the replacement of guanine (G) by cytosine (C) in the β-globin mRNA molecules three bases upstream from the AUG codon (position 6). Which of the following is most likely impaired in these patients? A. mRNA binding to ribosomes B. Translocation during translation C. Peptide bond formation D. Termination of polypeptide synthesis E. Protein targeting

*The answer is B.* DNA which was produced from the genomic RNA integrates randomly into the host chromosome. Cellular RNA polymerase then transcribes the viral DNA to produce viral RNA, which is used in the translation of viral proteins, and as the genomic material for new virions. RNA polymerase lacks 3′-5′ exonuclease activity, thus the enzyme cannot correct any errors it may make while transcribing the viral DNA. The RNA produced, which carries errors in transcription, is then packaged into a new virus particle, and this mutation may lead to a change that confers a growth advantage to this strain of virus. The lack of proofreading by RNA polymerase is not usually a problem in eukaryotic cells, as many messages are produced from a single gene, and if 1% of those messages produce a mutated protein it will be compensated by the 99% of the messages which produce a normal protein. In the viral case, however, the mRNA turns into the genomic material, which will lead to mutations in all future descendants of that virus. This is why HIV is treated with multiple, different antivirals simultaneously, to destroy any virus which mutates to be resistant to the antiviral agents. DNA polymerase has error checking capabilities, and will not significantly increase the mutation rate of the integrated viral DNA. DNA primase may make errors, but they are corrected when the RNA primer is removed from the DNA. Telomerase only works on the ends of chromosomes, and the viral DNA does not usually integrate at those positions. DNA ligase activity is not required for viral RNA production.

The high mutation rate of the human immunodeficiency virus (HIV) is due in part to a property of which of the following host cell enzymes? (A) DNA polymerase (B) RNA polymerase (C) DNA primase (D) Telomerase (E) DNA ligase

*The answer is B.* During the course of protein synthesis on a ribosome, peptidyl transferase catalyzes the formation of peptide bonds. However, when a stop codon such as UAA, UGA, or UAG is reached, aminoacyl-tRNA does not bind to the A site of a ribosome. One of the proteins, known as a release factor, binds to the specific trinucleotide sequence present. This binding of the release factor activates peptidyl transferase to hydrolyze the bond between the polypeptide and the tRNA occupying the P site. Thus, instead of forming a peptide bond, peptidyl transferase catalyzes the hydrolytic step that leads to the release of newly synthesized proteins. Following release of the polypeptide, the ribosome dissociates into its major subunits.

The hydrolytic step leading to the release of a polypeptide chain from a ribosome is catalyzed by a. Stop codons b. Peptidyl transferase c. Release factors d. Dissociation of ribosomes e. UAA

*The answer is B.* Exons are the coding portions of genes and consist of trinucleotide codons that guide the placement of specific amino acids into protein. Introns are the noncoding portions of genes that may function in evolution to provide "shuffling" of exons to produce new proteins. The primary RNA transcript contains both exons and introns, but the latter are removed by RNA splicing. The 5′ (upstream) and 3′ (downstream) untranslated RNA regions remain in the mature RNA and are thought to regulate RNA transport or translation. A poly(A) tail is added to the primary transcript after transcription, which facilitates transport and processing from the nucleus. The discovery of introns complicated Mendel's idea of the gene as the smallest hereditary unit; a modern definition might be the colinear sequence of exons, introns, and adjacent regulatory sequences that accomplish protein expression. Using these principles, one can determine the size of the stimulin gene. It contains a coding region of 300 bp (100 amino acids × 3 bp per amino acid), plus 100 bp in the intron, plus 70 + 30 = 100 bp in the untranslated regions (total = 500 bp). The mature RNA contains the same number of bp except for the 100 bp in the intron (500 − 100 = 400 bp). Transcription begins at the start of the 5′ untranslated region (70 bp) and the splice site occurs 30 bp (10 × 3) into the coding region at the beginning of the intron.

The hypothetical "stimulin" gene contains two exons that encode a protein of 100 amino acids. They are separated by an intron of 100 bp beginning after the codon for amino acid 10. Stimulin messenger RNA (mRNA) has 5′ and 3′ untranslated regions of 70 and 30 nucleotides, respectively. A complementary DNA (cDNA) made from mature stimulin RNA would have which of the following sizes? a. 500 bp b. 400 bp c. 300 bp d. 100 bp e. 70 bp

*The answer is A.* Splice junction mutations will theoretically produce a larger mRNA unless the mRNA is unstable; the larger protein may have abnormal function but retain peptide regions that react with antibody to the authentic protein. Nucleotide insertions or deletions other than multiples of 3 alter the reading frame of the code and scramble the amino acid sequence distal to the frame-shift mutation. Such altered mRNAs may be of increased or smaller size, depending on their stability, as may the translated protein, depending on the presence of stop codons within the shifted reading frame. Only the protein upstream of the frame-shift mutation retains immune cross-reactivity and normal function. Point mutations (nucleotide substitutions) may have substantial functional impact if the altered codon results in an amino acid substitution. If no amino acid substitution occurs, they are called silent mutations.

The hypothetical "stimulin" gene with two exons encoding a protein of 100 amino acids is found to have abnormal expression in cell culture. Which of the following mutations would produce a 500-bp stimulin mRNA and a 133-amino acid peptide that reacts with antibodies to stimulin protein? a. Splice junction mutation preventing RNA splicing b. Frame-shift mutation in codon #2 c. Silent point mutation in the third nucleotide of codon #50 d. Nonsense mutation at codon #2 e. Deletion of exon 1

*The answer is B.* The lactose (lac) operon is a classic model for understanding gene regulation. It is negatively controlled through two regulatory genes—the lac I gene that constitutively (always) expresses a repressor protein and the operator (o) region to which the repressor binds. The lac operon is inducible by lactose and lactose analogues, inactivating the repressor and uncovering the operon and its neighboring promoter (p) sequence. RNA polymerase then transcribes the inducible, structural genes β-galactosidase (z), permease (y), and transacetylase (a). The RNA transcript is polycistronic, so that one regulatory site allows transcription of all three genes needed for the metabolism of lactose. The bacterial ribosomes immediately attach to the nascent RNA transcript, allowing for simultaneous transcription and translation. When all the lactose is metabolized, the repressor returns to its native conformation, binds to the operator, and shuts down lac operon transcription.

The lactose operon is negatively controlled by the lactose repressor and positively controlled by which of the following? a. Increased concentrations of glucose and cyclic AMP (cAMP) b. Decreased concentrations of glucose and cAMP c. Increased concentrations of glucose, decreased concentration of cAMP d. Decreased concentrations of glucose, increased concentration of cAMP e. Increased concentrations of glucose and adenosine triphosphate (ATP)

*The answer is C.* RNA is susceptible to alkaline hydrolysis, whereas DNA is not. The major difference between the two polynucleotides is the presence of a 2′-hydroxyl group on the sugar ribose in RNA, versus its absence in deoxyribose, a component of DNA. Under alkaline conditions, the hydroxyl group can act as a nucleophile and attack the phosphodiester linkage between adjacent nucleotides, breaking the linkage and leading to the transient formation of a cyclic nucleotide. As this can occur at every phosphodiester linkage in RNA, hydrolysis of the RNA will occur due to these reactions. As DNA lacks the 2′-hydroxyl group, this reaction cannot occur, and DNA is very stable under alkaline conditions. The fact that DNA contains thymine, and RNA uracil (both true statements) does not address the base stability of DNA as compared to RNA. Both DNA and RNA contain 3′-hydroxyl groups, which are usually in 3′-5′ phosphodiester bonds in the DNA backbone. The procedure of Southern blotting is used in the diagnosis of various disorders, including some instances of hemoglobinopathies and diseases induced by triplet-repeat expansions of DNA (such as myotonic dystrophy).

The procedure of Southern blotting involves treatment of the solid support (nitrocellulose) containing the DNA with NaOH to denature the double helix. Treatment of a Northern blot with NaOH, however, will lead to the hydrolysis of the nucleic acid on the filter paper. This is due to which major chemical feature of the nucleic acids involved in a Northern blot? (A) The presence of thymine (B) The presence of uracil (C) The presence of a 2′-hydroxyl group (D) The presence of a 3′-hydroxyl group (E) The presence of a 3′-5′ phosphodiester linkage

*The answer is E.* DNA methylation occurs mainly at CpG dinucleotides that often cluster in at the upstream promoter regions of genes (CpG islands). While these are generally correlated with gene inactivation, there are many exceptions. Double crossovers at meiosis can substitute a normal allele for a mutant allele (conversion), and reverse transcriptases can copy intronless mRNA into complementary DNAs (cDNAs) that integrate into the genome as pseudogenes. Immunoglobulin genes undergo gene rearrangement to unite variable, joining, and constant regions for expression of a unique antibody. Unequal crossing over between sister chromatids is thought to be an important mechanism for variation in copy number within gene clusters.

The process that occurs at the 5 position of cytidine and often correlates with gene inactivation is a. Gene conversion b. Sister chromatid exchange c. Pseudogene d. Gene rearrangement e. DNA methylation

*The answer is D.* The POMC gene provides a mammalian example in which several proteins are derived from the same RNA transcript. Unlike the polycistronic mRNA of the bacterial lactose operon, mammalian cells generate several mRNAs or proteins from the same gene by variable protein processing or by alternative splicing. Variable protein processing preserves the peptide products of some gene regions but degrades those from others. Alternative splicing would often produce proteins composed of different exon combinations with the same terminal exon and carboxy-terminal peptide, but could remove the terminal exon in some proteins and produce different C-terminal peptides. Different transcription factors or enhancers in different brain regions could regulate the total amounts of POMC gene transcript but not the types of protein produced. Elongation of protein synthesis involves GTP cleavage but is not differentially regulated in mammalian tissues.

The proopiomelanocortin (POMC) gene encodes several regulatory proteins that affect pituitary function. In different brain regions, proteins encoded by this gene have different carboxy-terminal peptides. Which of the following best explains the regulatory mechanism? a. POMC transcription is regulated by different factors in different brain regions b. POMC translation elongation is regulated by different factors in different brain regions c. POMC transcription has different enhancers in different brain regions d. POMC protein undergoes different protein processing in different brain regions e. POMC protein forms different allosteric complexes in different brain regions

*The answer is A.* For a PCR experiment, one requires primers that will allow DNA synthesis to occur across the region of DNA to be amplified. Since DNA polymerase always synthesizes DNA in the 5′ to 3′ direction, the 3′ ends of the primers must face each other and bracket the region of DNA to be amplified. Primers A and D will only allow DNA synthesis away in the opposite direction from the region of interest, due to their orientation on the DNA template strand.

The region of DNA shown below is to be amplified by PCR. The appropriate pair of primers to use is which of the following? (A) B and C (B) A and D (C) A and C (D) B and D (E) C and D

*The answer is C.* Self-splicing of the introns of some primary ribosomal RNA transcripts occurs because of the presence of catalytic RNAs (ribozymes) generated from the introns. This occurs in the absence of protein catalysis. In contrast, the splicing of messenger RNA is carried out by spliceosomes. Spliceosomes are large complexes of three kinds of small ribonucleoprotein particles (snRNPs) and the messenger RNA precursor. The snRNPs are involved in recognizing the 5′ splice site and the 3′ splice site and then binding to these sites. Once the spliceosome is bound, it mediates excision of the intron and splicing of the two adjacent exons.

The removal of introns and subsequent self-splicing of adjacent exons occurs in some portions of primary ribosomal RNA transcripts. The splicing of introns in messenger RNA precursors is a. RNA-catalyzed in the absence of protein b. Self-splicing c. Carried out by spliceosomes d. Controlled by RNA polymerase e. Regulated by RNA helicase

*The answer is D.* The primary transcripts of all eukaryotic mRNAs are capped at the 5′ end. Prokaryotic RNAs and eukaryotic tRNA and rRNA are not capped. The cap is composed of 7-methylguanylate attached by a pyrophosphate linkage to the 5′ end. This is known as cap 0. One of the adjacent riboses is methylated in cap 1, and both of the adjacent riboses are methylated in cap 2. The cap protects the 5′ ends of mRNAs from nucleases and phosphatases and is essential for the recognition of eukaryotic mRNAs in the protein-synthesizing system. When prokaryotic monocistronic mRNAs are artificially capped, translation occurs in a eukaryotic, in vitro translation system.

The so-called caps of RNA molecules a. Allow tRNA to be processed b. Occur at the 3′ end of tRNA c. Are composed of poly A d. Are unique to eukaryotic mRNA e. Allow correct translation of prokaryotic mRNA

*The answer is A.* In an expression library, cDNA clones are screened on the basis of their ability to direct bacterial synthesis of a foreign protein of interest. Radioactive antibodies specific to this protein can be used to identify the colonies of bacteria that contain the cDNA vector. As was the case for probing genomic libraries, bacteria grown on a master plate are blotted onto a nitrocellulose replica plate and then lysed. The released proteins may then be labeled with 125I antibodies. In contrast, northern blotting can be used to identify RNA molecules separated by gel electrophoresis. In northern blotting, RNA molecules separated by gel electrophoresis can be identified by hybridization with probe DNA following transfer to nitrocellulose. Mutant and wild-type oligonucleotides can be used as probes to analyze polymerase chain reaction products. Conversely, the products of polymerase chain reaction can be used to analyze cDNA libraries.

The western blot use what type of probe? a. Antibody b. mRNA c. Products of polymerase chain reaction (PCR) d. tRNA e. Mutant and normal oligonucleotides f. rRNA g. cDNA clone

*The answer is A.* σ factor is a bacterial protein that can associate with and become a subunit of bacterial RNA polymerase. σ factor confers specificity of initiation on the core enzyme. In the presence of σ factor, RNA polymerase chooses the correct strand of duplex DNA for transcription and initiates transcription at the appropriate promoter region. In some bacteria, such as Bacillus subtilis, a specific σ factor is synthesized to change transcriptional selectivity and effect cellular changes like sporulation.

The σ factor found in many bacteria is best described as a a. Subunit of RNA polymerase responsible for the specificity of the initiation of transcription of RNA from DNA b. Subunit of DNA polymerase that allows for synthesis in both 5' to 3' and 3' to 5' directions c. Subunit of the 50S ribosome that catalyzes peptide bond synthesis d. Subunit of the 30S ribosome to which mRNA binds e. Factor that forms the bridge between the 30S and 50S particles constituting the 70S ribosome

*The answer is B.* A single nucleotide substitution in a promoter-proximal region has the capability of increasing hydrogen bonding to a transacting factor, thereby increasing the interaction between the factor and DNA and enhancing recruitment of RNA polymerase to the promoter. Since the mutation is in the promoter region, it is not involved in splicing (which occurs at exon-intron borders), it does not code for an amino acid (since exons are not in the promoter region), it is not related to sigma (which is prokaryotic specific), and the DNA helix does not have to be significantly melted in this area to allow transcription factor binding.

Theoretically, a disease could result from an increased expression of a particular gene. This can occur in eukaryotes through a single-nucleotide mutation in a promoter-proximal element. This is best explained by which one of the following? (A) More efficient splice site recognition (B) Increased opportunity for hydrogen bonding to a transacting factor (C) Beneficial amino acid replacement derived from the missense mutation (D) Increased amount of sigma factor binding (E) Reduced energy need to melt the DNA helix at this position

*The answer is E.* This is a Western blot . In HIV testing, the Western blot is performed to confirm HIV after an enzyme-linked immunosorbent assay result is positive. In a Western blot, HIV proteins from known HIV-infected cells are separated by electrophoresis. The separated proteins are blotted on to a filter membrane. This membrane is incubated with the patient 's serum. f the patient is HIV-positive, the patient's serum will contain anti-HIV antibodies. These antibodies will bind to the HIV proteins on the membrane. Then enzyme-linked anti-human IgG antibodies are added; these recognize and bind to the patient's anti-HIV antibodies. An enzymatic reaction causes bands to form on the membrane at the position of specific HIV proteins. The Western blot is positive in the presence of two or more bands corresponding to the HIV proteins p24, gp41, and gp120/160, indicating that the patient's serum contains specific antibodies to these HIV proteins and is therefore infected with HIV. False-positive and false-negative results are very rare with the Western blot, so it is an excellent confirmatory test.

To confirm a diagnosis of HIV, a laboratory technician runs a set of proteins on an electrophoretic gel, transfers them to a filter membrane, and then incubates the filter membrane with the patient's serum. The technician adds labeled anti-human IgG antibodies. Which of the following tests is being performed? A. Enzyme-linked immunosorbent assay B. Northern blot C. Polymerase chain reaction D. Southern blot E. Western blot

*The answer is B.* RFLPs can be detected with a straightforward Southern blot. Chromosomal translocations are too rare in the population to be used for general DNA fingerprinting. DNA methylation patterns are epigenetic and can be variable amongst individuals, so they are not sufficiently reliable to be used for DNA fingerprinting. Deletions and triplet repeat expansions are too rare and variable in the population to be used as a general screen for DNA identification purposes.

Two babies were born at the same time in the hospital, but their nametags may have been mixed up DNA fingerprinting was done to determine the parentage of the children, using Southern blots of restricted DNA and a specific probe. The basis for seeing differences between individuals using this technique is which of the following? (A) Chromosomal deletions (B) RFLPs (C) Chromosomal translocations (D) Triplet repeat expansions (E) DNA methylation

*The answer is B.* The boys have an X-linked recessive condition called α thalassemia/mental retardation or ATR-X syndrome. The X-encoded gene has an unknown function in brain as well as being a factor that regulates α-globin gene transcription. In order to affect all four α-globin genes, the X-encoded gene must produce a trans-acting factor; second mutations altering enhancers or promoters would be cis-acting and affect only one α-globin gene. Pseudogenes are functionless gene copies, so altered expression would not influence α-globin chain synthesis.

Two boys with mental disability are found to have mutations in a gene on the X chromosome that has no homology with globin genes. Both are also noted to have deficiency of α-globin synthesis with α thalassemia. Which of the following is the best explanation for their phenotype? a. The mutation disrupted an enhancer for an α-globin pseudogene b. The mutation disrupted an X-encoded transcription factor that regulates the α-globin loci c. There is a second mutation that disrupts an enhancer near the α-globin gene d. There is a DNA rearrangement that joins the mutated X chromosome gene with an α-globin gene e. There is a second mutation that disrupts the promoter of an α-globin gene

*The answer is A.* Among the conclusions offered, only A is consistent with the results on the blot. Infant A's pattern shows a PCR product (lower on the blot) matching F1 and another PCR product (higher on the blot) matching M1. Neither of infant A's PCR products match F2 (choices B and E). The upper PCR product in infant B's pattern does not match with either F1 or M1 (choices C and D). Although unlikely given the situation, another possibility is consistent with the blot. Infant A could be the child of M2 and F1, although this is not offered as an option.

Two sets of parents were friends in a small town and had babies on the same day. The wristbands of the two similar-looking infants (A and B) were inadvertently mixed at the pediatric care unit. In order to accurately identify the parents of the respective infants, PCR analysis was performed on samples of blood taken from the two infants and both sets of parents (Father 1 and Mother 1 versus Father 2 and Mother 2). Shown below is the analysis of the PCR products by gel electrophoresis. What is the best conclusion from the analysis? A. A is the child of Parents 1. B. A is the child of Parents 2. C. B is the child of Parents 1. D. Father 1 (F1) could be the father of both infants. E. (Father 2 (F2) could be the father of both infants.

*The answer is C.* In two-dimensional electrophoresis, proteins are separated first by charge (they are run through a pH gradient, and when they reach their isoelectric point, they stop migrating), and then by size. As indicated in the figure to the question, larger sizes migrate more slowly and would be closer to the "top" of the figure. Thus, spot A is of a larger molecular weight than spot B, as spot B migrates farther in the sizing portion of the two dimensional electrophoresis. The horizontal positions are related to the isoelectric points of the proteins, and spot A reaches its isoelectric point at a higher pH than does spot B. This means that protein B is more acidic than protein A. Overall, then, the protein expressed by spot A is larger and more basic than the protein represented by spot B.

Two-dimensional PAGE gels were run on cells from normal tissue, and also on the same tissue which has been affected by a certain disease. Which of the following best describes spot A in the disease state, as compared to spot B in the normal state? (A) Lower molecular weight, same isoelectric points (B) Higher molecular weight, more acidic protein (C) Higher molecular weight, more basic protein (D) Same molecular weight, more basic protein (E) Lower molecular weight, more acidic protein

*The answer is D.* The acquisition of a new viral surface protein is often all that is necessary for a vines to infect a new type of host cell. In the illustration above, the virus A genome obtains some of the surface components of virus B (and vice versa) while both viruses are present simultaneously in the same human cell. This exchange is defined as phenotypic mixing, which generally occurs when a host cell is co-infected by two viral strains and progeny virions are produced that contain nucleocapsid or envelope proteins from the first strain and the genome of the second strain (or vice versa). Since virus A acquired some virus B surface proteins but no genetic exchange occurred, the progeny are considered phenotypically mixed. The next generation will revert to having only virus A type surface proteins (virus A phenotype) and will again be non-infectious for human epithelium.

Virus A cannot normally infect human epithelial cells. After exposure to virus B in non-human cells, virus A acquires the ability to infect human epithelial cells; however, the viral particles generated as a result of this infection still cannot infect human epithelial cells. The phenomenon described is best categorized as which of the following? A. Reassortment B. Recombination C. Transformation D. Phenotypic mixing E. Interference

*The answer is E.* Despite some differences, protein synthesis in prokaryotes and eukaryotes is quite similar. The small ribosomal subunit is 30S in prokaryotes and 40S in eukaryotes. The large ribosomal subunit is 50S in prokaryotes and 60S in eukaryotes. The intact ribosome is consequently larger in eukaryotes (80S) and smaller in prokaryotes (70S). At the start of translation, initiation factors, mRNA, and initiation aminoacyl-tRNA bind to the dissociated small ribosomal subunit. The initiation tRNA in prokaryotes is N-formyl methionine in prokaryotes and simply methionine in eukaryotes. Only after the small ribosomal subunit is primed with mRNA and initiation aminoacyl-tRNA does the large ribosomal subunit bind to it. Once this happens, elongation factors bring the first aminoacyl-tRNA of the nascent protein to the A site. Then peptidyl transferase forges a peptide bond between the initiation amino acid and the first amino acid of the forming peptide. The now uncharged initiation tRNA leaves the P site and the peptidyl-tRNA from the A site moves to the now vacant P site with the two amino acids attached. The ribosome advances three bases to read the next codon and the process repeats. When the stop signal is reached after the complete polypeptide has been synthesized, releasing factors bind to the stop signal, causing peptidyl transferase to hydrolyze the bond that joins the polypeptide at the A site to the tRNA. Factors prevent the reassociation of ribosomal subunits in the absence of new initiation complex.

What is the correct order of the following steps in protein synthesis? 1. A peptide bond is formed. 2. The small ribosomal subunit is loaded with initiation factors, messenger RNA, and initiation aminoacyl-transfer RNA. 3. The intact ribosome slides forward three bases to read a new codon. 4. The primed small ribosomal subunit binds with the large ribosomal subunit. 5. Elongation factors deliver aminoacyl-tRNA to bind to the A site. a. 1, 2, 5, 4, 3 b. 2, 3, 4, 5, 1 c. 4, 5, 1, 2, 3 d. 3, 2, 4, 5, 1 e. 2, 4, 5, 1, 3

*The answer is E* Introns have been shown to contain genes for microRNAs, which are processed to small, interfering RNAs, which can regulate gene expression either by binding to and initiating degradation of a particular mRNA or by binding to a particular mRNA and blocking translation of the mRNA. These small RNA molecules do not affect the transcription of the target mRNA, nor posttranscriptional processing (capping and polyadenylation). They also do not affect the export of mRNA into the cytoplasm, nor do they alter ribosome biogenesis. As an example, the miR-17-92 cluster encodes seven microRNAs and resides within an intron of the C13 or F25 gene on chromosome 13. These miRNAs are upregulated in lung cancer, and may contribute to the progression of the disease by downregulating their target genes.

When first discovered, introns were not thought to code for a functional product. Recently, however, introns have been found to code for products that can regulate the expression of a large number of genes. This regulation occurs at which stage of gene expression? (A) Transcription of mRNA (B) Posttranscriptional processing (C) Export of mRNA into the cytoplasm (D) Ribosome biogenesis (E) Degradation of the mRNA

*The answer is B.* This scenario describes the exchange of genetic information between two virus strains that have non-fragmented. double-stranded DNA genomes. Such an exchange amounts to recombination, which may be defined as the exchange of genes between two chromosomes via crossing over within homologous regions. The resulting progeny will have traits not present simultaneously in either parent virus.

When introduced individually, two different mutant strains of adenovirus do not cause cytopathic effects in a human cell culture. But when the cultured cells are simultaneously exposed to both mutant strains, a new viral progeny strain is produced that causes cellular enlargement and aggregation. Which of the following most likely contributed to the formation of the progeny viral strain? A. Reassortment B. Recombination C. Transformation D. Phenotypic mixing E. Interference

*The answer is C.* The drug is 3′-deoxyadenosine. This nucleoside will be activated when it enters cells (to the triphosphate level), and will be recognized by RNA polymerase and incorporated into growing RNA chains. Since the compound lacks a 3′-hydroxy group, RNA synthesis is terminated, as the next phosphodiester bond cannot be created. This drug is not recognized by DNA polymerase because it contains a 2′-hydroxy group, and DNA polymerase will only recognize nucleotides which lack 2′-hydroxy groups. This drug does not resemble tRNA, or mRNA, or rRNA, so it will not have a direct effect on protein synthesis. Protein synthesis may be decreased, however, if no mRNA is present to translate. However, the drug does not directly inhibit the mechanisms of translation.

Which answer below best predicts the effect of the following drug on the pathways indicated?

*The answer is C.* In all of the forms of DNA repair in normal cells, a common sequence of events occurs. 1. The single or multiple base abnormality is surveyed and detected by a specific protein or proteins. 2. The DNA is nicked on one side of the damaged DNA. 3. A specific enzyme excises the damaged portion (steps 2 and 3 can be combined if an excinuclease cuts on both sides of the damaged DNA). 4. The damaged portion of the strand is replaced by resynthesis catalyzed by DNA polymerase I. 5. A ligase seals the final gap. With some variability, these general principles apply in nucleotide excision repair (segments of about 30 nucleotides), base excision repair of single bases, and mismatch repair of copying errors (one to five bases).

Which is the most correct sequence of events in gene repair mechanisms in patients without a mutated repair process? a. Nicking, excision, replacement, sealing, recognition b. Sealing, recognition, nicking, excision, replacement c. Recognition, nicking, excision, replacement, sealing d. Nicking, sealing, recognition, excision, replacement e. Nicking, recognition, excision, sealing, replacement

*The answer is D.* Several operons in E. coli, including the lac operon, are subject to catabolite repression. In the presence of glucose, there is decreased manufacture of cyclic AMP (cAMP) by adenylate cyclase. Low glucose levels increase production of cAMP, which binds to the catabolite activator protein (CAP). The cAMP-CAP complex binds to the promoters of several responsive operons at catabolite activator protein (CAP) binding sites, greatly enhancing transcription of operon RNA. This positive control stimulates use of more exotic metabolites when glucose is not available and conserves energy when glucose is plentiful. High levels of glucose lower cAMP levels and direct metabolism toward constitutive glucose pathways such as glycolysis.

Which of the following best describes the negatively controlled lactose operon in Escherichia coli? a. An inducer (lactose) binds to the operator, enhancing simultaneous transcription and translation of β-galactosidase (z), permease (y), and transacetylase (a) genes b. An inducer (lactose) alters the repressor protein and uncovers the operator and promoter, allowing simultaneous transcription and translation of β-galactosidase (z), permease (y), and transacetylase (a) genes c. The repressor (lactose) alters the operator protein and uncovers the promoter, allowing simultaneous transcription and translation of β-galactosidase (z), permease (y), and transacetylase (a) genes d. The repressor (lactose) alters the catabolite repression protein and uncovers the operator and promoter, allowing simultaneous transcription and translation of β-galactosidase (z), permease (y), and transacetylase (a) genes e. An inducer (lactose) alters the repressor protein, uncovers the β-galactosidase (z) operator, and allows transcription. The inducer also uncovers separate operators for the permease (y) and transacetylase (a) genes

*The answer is E.* In the leading strand, DNA is synthesized continuously in the 5′ to 3′ direction by DNA polymerase. In contrast, in the lagging strand, which is in the 3′ to 5′ direction, DNA polymerase III synthesizes small (approximately 1000 nucleotides) Okazaki fragments. For the synthesis of these small fragments, all the same roles and steps apply except that additional enzymes are needed to fill the gap between the fragments and join the fragments. Consequently, DNA ligase is repeatedly needed to join the ends of the DNA fragments along the growing lagging strand. DNA ligase catalyzes the formation of a phosphodiester bond between the 3′ hydroxyl group at the end of one DNA chain and the 5′ DNA phosphate group at the end of the other. DNA ligase is only functional when double helical DNA molecules are the substrate. It does not work on single-stranded DNA. DNA ligase effects the joining of strands of DNA not only during the normal synthesis of DNA, but during the splicing of DNA chains in genetic recombination as well as the repair of damaged DNA.

Which of the following descriptions of DNA replication is not common to the synthesis of both leading and lagging strands? a. RNA primer is synthesized b. DNA polymerase III synthesizes DNA c. Helicase (rep protein) continuously unwinds duplex DNA at the replication fork during synthesis d. Nucleoside monophosphates are added in a 5′ to 3′ direction along the growing DNA chain e. DNA ligase repeatedly joins the ends of DNA along the growing strand

*The answer is B.* Primase is a DNA-dependent RNA polymerase located in the primosome at the replication fork of DNA. Primase initiates DNA synthesis by synthesizing a 10-base RNA primer. The DNA-RNA helix formed binds DNA polymerase III, which synthesizes a DNA fragment (the Okazaki fragment) in a 5′ to 3′ direction. When the RNA primer of the previous Okazaki fragment is met, DNA polymerase I replaces III and digests the RNA primer, replacing it with appropriate DNA bases. When the RNA primer is completely removed, DNA ligase synthesizes the last phosphodiester bond, thereby sealing the space. What is left is a new lagging strand extended by the new Okazaki fragment with the 10-base RNA primer at its 5′ end. Reverse transcriptase is a DNA polymerase that uses RNA as a template found in retroviruses as well as normal eukaryotic cells. Unlike DNA polymerase I and III, which proofread for errors during normal synthesis, reverse transcriptase has no proofreading capabilities. Hence, it has an exceedingly high error rate that contributes to the high rate of mutation in retroviruses like HIV.

Which of the following enzymes can be described as a DNA-dependent RNA polymerase? a. DNA ligase b. Primase c. DNA polymerase III d. DNA polymerase I e. Reverse transcriptase

*The answer is E.* Reverse transcriptase is an RNA-dependent DNA polymerase that can synthesize first a single strand and then a double-stranded DNA from a single-strand RNA template. It was originally found in animal retroviruses. Primase is a DNA-dependent RNA polymerase enzyme that synthesizes an RNA molecule 10 to 200 nucleotides in length that initiates or "primes" DNA synthesis. DNA ligase joins DNA fragments and DNA gyrase winds or unwinds DNA. Transfer RNA, 5SRNA, and other small RNAs are synthesized by RNA polymerase III (RNA polymerase I synthesizes ribosomal RNA and RNA polymerase II synthesizes messenger RNA).

Which of the following enzymes can polymerize deoxyribonucleotides into DNA? a. Primase b. DNA ligase c. DNA gyrase d. RNA polymerase III e. Reverse transcriptase

*The answer is A.* The addition of the 7-methyl G cap immediately after transcription has started.

Which of the following is a co transcriptional event in RNA synthesis? A. Addition of a 7-methyl G cap B. Splicing C. Poly A tailing

*The answer is E.* Chain termination is determined by three codons: UAA, UAG, and UGA. Aside from chain termination codons, each group of three bases in a sequence codes for an amino acid. The next three bases specify another amino acid. Thus, the genetic code is nonoverlapping. The triplet genetic code is degenerate, which is to say that for most amino acids there is more than one code word. The triplets of bases (codons) that specify the same amino acid usually differ only in the last base of the triplet.

Which of the following is a true statement about translation? a. The genetic code can be overlapping b. The first nucleotide in a codon has less specificity than the others c. Only one group of nucleotides codes for each single amino acid d. Every codon (three nucleotide bases) specifies an amino acid e. Specific nucleotide sequences signal termination of peptide chains

*The answer is A.* Antibody classes, called isotypes, are determined by the constant region of heavy chains. There are five isotypes that include IgM, IgD, IgG, IgE, and IgA. During B cell maturation, DNA rearrangements produce light chains with unique V-J segments and heavy chains with unique V-D-J segments. After activation, B cells can change their preferred DNA recombination to join the V-D-J segment to a different constant (C) segment. Different constant region exons are clustered downstream of the Cμ exon used for initial IgM synthesis by the activated B cell. Recombination at designated switch sites places another constant chain exon (e.g., Cγ for IgG) next to the V-D-J segment and allows the activated B cell to secrete a different antibody isotype (heavy chain class switching). Alternative splicing allows activated B cells to switch from membranebound to secreted IgM.

Which of the following is involved in determining antibody class? a. Different rearrangements of V-D-J and C segments in heavy chains b. Different V-D-J rearrangements in heavy chains c. Different V-J rearrangements in light chains d. Alternative splicing to switch from κ to λ light chains e. Alternative splicing to switch from membrane-bound to secreted antibody

*The answer is C.* Signal recognition particles (SRPs) recognize the signal sequence on the N-terminal end of proteins destined for the lumen of the endoplasmic reticulum (ER). SRP binding arrests translation and an SRP receptor facilitates import of the SRP, ribosome, and nascent protein into the ER lumen. A signal peptidase removes the signal sequence from the protein, which may remain in the membrane or be routed for secretion. Common to both eukaryotic and prokaryotic protein synthesis is the requirement for ATP to activate amino acids. The activated aminoacyltRNAs then interact with ribosomes carrying mRNA. Peptidyl transferase catalyzes the formation of peptide bonds between the free amino group of activated aminoacyl-tRNA on the A site of the ribosome and the esterified carboxyl group of the peptidyl-rRNA on the P site; the liberated rRNA remains on the P site.

Which of the following is required for certain types of eukaryotic protein synthesis but not for prokaryotic protein synthesis? a. Ribosomal RNA b. Messenger RNA c. Signal recognition particle d. Peptidyl transferase e. GTP

*The answer is C.* A phosphodiester bond links nucleotides in nucleic acids. Answer A is an amide bond (the type found linking amino acids together in proteins). Answer B is an ester linkage (the type found in triacylglycerol, in which fatty acids are attached to a glycerol backbone). Answer D is a phosphoanhydride bond (similar to that found at the 1 position of 1,3 bisphosphoglycerate), and answer E is an ether linkage (found in ether lipids, for example).

Which of the following is the type of bond that allows nucleotides to form long polymers?

*The answer is B.* A nucleoside consists of a purine or pyrimidine base linked to a pentose sugar. The 1′ carbon of the pentose is linked to the nitrogen of the base. In DNA, 2′-deoxyribose sugars are used; in RNA, ribose sugars are used. Nucleotides are phosphate esters of nucleosides with one to three phosphate groups, such as adenosine monophosphate (AMP), adenosine diphosphate (ADP), or adenosine triphosphate (ATP). The nitrogenous bases are adenine, thymine, guanine, and cytosine in DNA, with thymine replaced by uridine in RNA. Nucleotide polymers are chains of nucleotides with single phosphate groups, joined by bonds between the 3′-hydroxyl of the preceding pentose and the 5′-phosphate of the next pentose. Polymerization requires high-energy nucleotide triphosphate precursors that liberate pyrophosphate (broken down to phosphate) during joining. The polymerization reaction is given specificity by complementary RNA or DNA templates and rapidity by enzyme catalysts called polymerases.

Which of the following molecules is found in a nucleoside? a. A pyrophosphate group b. A 1′ base linked to a pentose sugar c. A 5′-phosphate group linked to a pentose sugar d. A 3′-phosphate group linked to a pentose sugar e. A terminal triphosphate

*The answer is D.* At a physiologic pH of 7.4, mRNAs (like DNA) are polyanionic owing to the negatively charged phosphate hydroxyl groups. Mammalian mRNAs are synthesized from DNA as single-stranded linear molecules. Because they are not double-stranded, the concentrations of the different bases in mRNA are variable rather than exhibiting the A = T and G = C pattern of double-stranded DNA (A does not equal U in mRNA). The hybridization of RNA with its complementary template DNA is antiparallel. In both DNA and RNA, sugar units equal base units equal phosphate units. However, their bases consist of pyrimidines as well as purines.

Which of the following most correctly describes mammalian messenger RNAs? a. They are usually transcribed from both DNA strands b. They are normally double-stranded c. Their content of uridine equals their content of adenine d. They have an overall negative charge at neutral pH e. Their ratio of ribose to purine bases equals 1

*The answer is E.* Insertion of one extra nucleotide causes a frame-shift mutation and mistranslation of all the mRNA transcribed from beyond that point in the DNA. All the other mutations cited in the question usually cause an error in the identity of only one amino acid (choice a or b), removal of one amino acid from the sequence (choice d), or no error at all in the amino acid sequence (choice c). There is a chance that the mutations in choices a or b will give a "nonsense," or chain-terminator, mutation, and this is about as likely to be lethal as is a frame shift.

Which of the following mutations is most likely to be lethal? a. Substitution of adenine for cytosine b. Substitution of cytosine for guanine c. Substitution of methylcytosine for cytosine d. Deletion of three nucleotides e. Insertion of one nucleotide

*The answer is D.* Zinc fingers are related to steroid receptors.,

Which of the following protein classes is related to the transcription factor for steroid receptors? A. Helix-turn-helix B. Leucine zipper C. Rel domains D. Zinc fingers

*The answer is C.* Certain regulatory elements act on genes on the same chromosome ("cis"), while others can regulate genes on the opposite chromosome ("trans"). The terminology makes analogy to carbon-carbon double bonds where two modifying groups may both be above or below the bond (cis) or opposite it (trans). Cis regulatory elements like the lac operator and promoter or mammalian enhancers are usually DNA sequences (regulatory sequences) adjoining or within the regulated gene. Trans regulatory elements like the lac repressor protein or mammalian transcription factors are usually diffusible proteins (regulatory factors) that can interact with adjoining target genes or with target genes on other chromosomes. Classification of bacterial elements as cis or trans requires mating experiments where portions of a second chromosome are introduced by transfection (with bacteriophage) or conjugation (with other bacteria). The distinction between cis and trans is fundamental for understanding how regulators work.

Which of the following regulators are said to act in "cis"? a. The lac repressor and mammalian transcription factors b. The lac repressor and the lac operator c. The lac operator and mammalian enhancers d. The lac operator and mammalian transcription factors e. Mammalian transcription factors and enhancers

*The answer is C.* Northern blotting is analogous to Southern blotting, a technique that was first described by Edward Southern. DNA fragments are separated on agarose gels by electrophoresis and then transferred to nitrocellulose filters. The filters are then exposed to labeled probes, which hybridize to the DNA fragments. Northern blotting is an analogous procedure that uses more powerful denaturing substances to extend the RNA molecules and ensure that their electrophoretic migration is inversely proportional to their length. Labeled RNA or DNA segments (probes) are used to identify particular RNA species within the size-separated array. Western blotting is a technique for detecting proteins. It uses a different type of denaturing gel and labeled antibodies as probes to detect specific proteins.

Which of the following results is provided by northern blot analysis? a. Detects specific base pairs b. Detects DNA molecules c. Detects RNA molecules d. Detects proteins e. Determines chromosome structure

*The answer is B.* The two subunits of ribosomes are composed of proteins and rRNA. Ribosomes are found in the cytoplasm, in mitochondria, and bound to the endoplasmic reticulum. Transcription refers to the synthesis of RNA complementary to a DNA template and has nothing immediately to do with ribosomes.

Which of the following statements about ribosomes is true? a. They are an integral part of transcription b. They are found both free in the cytoplasm and bound to membranes c. They are bound together so tightly they cannot dissociate under physiologic conditions d. They are composed of RNA, DNA, and protein e. They are composed of three subunits of unequal size

*The answer is D.* Like bacterial DNA, eukaryotic DNA is replicated in a semiconservative manner. However, in contrast to most bacterial DNA, which is circular in structure, nuclear chromosomal DNA is a single, uninterrupted molecule that is linear and unbranched. A eukaryotic chromosome contains a strand of DNA at least 100 times as large as the DNA molecules found in prokaryotes. Eukaryotic, but not prokaryotic, DNA molecules are bound to small basic proteins called histones. The histone-DNA complex formed is referred to as chromatin.

Which of the following statements correctly describes eukaryotic nuclear chromosomal DNA? a. Each discontinuous piece making up the chromosomes of eukaryotes is about the same size as each prokaryotic chromosome b. Unlike bacterial DNA, no histones are associated with it c. It is not replicated semiconservatively d. It is a linear and unbranched molecule e. It is not associated with a specific membranous organelle

*The answer is B.* The nucleolus is an organelle unique to eukaryotic cells. It is the site where hundreds of copies of genes repeated in tandem for three of the four ribosomal RNAs are transcribed by RNA polymerase I to give a 45S primary transcript. Enzymatic modification and cleavage remove spacer regions to yield 28S, 18S, and 5.8S ribosomal RNA. The 5S subunit is synthesized by RNA polymerase III in the nucleoplasm rather than in the nucleolus. Ribosomal proteins combine with the ribosomal subunits to assemble into a 60S subunit containing the 5S, 5.8S, and 28S RNAs and a 40S subunit containing the 18S RNA. Combined, the two subunits produce a functional eukaryotic ribosome with a sedimentation coefficient of 80S.

Which of the following statements correctly describes the nucleolus of a mammalian cell? a. It differs from that found in bacterial cells in that histones are present b. It may contain hundreds of copies of genes for different types of ribosomal RNAs c. It synthesizes 5S ribosomal RNA d. It synthesizes 60S and 40S ribosomal subunits e. It synthesizes all ribosomal RNA primary transcripts

*The answer is D.* Plasmids are duplex DNA circles that may carry genes determining antibiotic resistance (R factors), sex (F factors), or toxin production (colicinogenic factors) in their bacterial hosts. They can replicate independently of the host chromosome or insert into the host chromosome. Plasmids are one class of mobile genetic elements (transposons) that are normally found in bacteria. Restriction of plasmid vector DNA and mammalian DNA with the same endonuclease produces cohesive ends that may be joined together with DNA ligase. The ligated molecules, which consist of one or more mammalian DNA segments inserted between plasmid DNA ends, are recombinant DNA molecules that can be replicated in the host bacteria. Isolation of the recombinant plasmid DNA by centrifugation and excision of the inserted mammalian DNA segment(s) then provides a pure and abundant sample of the mammalian gene segment that is separated from all other DNA segments in the mammalian genome. Plasmid vectors are useful for gene segments under about 10 kilobases (kb) in size, but bacteriophage and recently yeast artifical chromosomes (YACs) can incorporate DNA segments up to 1000 kb or 1 megabase (Mb) in size. These larger vectors allow genomes like those of humans or mice (3000 Mb in size) to be entirely represented in a collection (library) of about 3000 recombinant molecules.

Which of the following statements correctly describes the recombinant DNA tool known as plasmids? a. They are found more commonly in viruses than in bacteria b. They are single-stranded circles c. They sometimes enhance bacterial susceptibility to antibiotics d. They sometimes enhance bacterial resistance to antibiotics e. They are too small to be useful as vectors for the cloning of mammalian DNA segments

*The answer is E.* Double-stranded DNA is arranged in a double helix as originally deduced by Watson and Crick. The double helical structure of duplex DNA is different than the α-helical structure of portions of proteins. The α-helical structure of proteins is formed of one chain of proteins stabilized by individual hydrogen bonds between components of the amide bonds, that is, between the carbonyl oxygens and the amide nitrogens. In contrast, the hydrogen bonding in double-stranded DNA is important to allow each strand to act as a template for the other complementary strand, with adenine bonding to thiamine and cytosine bonding to guanine. Hydrophobic stacking between bases in the hydrophobic interior of the double strand actually makes a greater contribution to the stability of the DNA double helix than does hydrogen bonding. DNA and protein helices are both composed of polymers of subunits (amino acids and nucleotides) held together by a covalently linked backbone. As pointed out above, hydrogen bonding is important to both the double helix of DNA and the α-helix of proteins. Finally, both are, in fact, spiral structures, although only the helix of proteins is an α helix.

Which of the following statements describes both the spiral structure of double-stranded DNA and the spiral structure found in certain segments of protein? a. They are repeating spiral structures with intervals of pleated sheets b. They have four alternative units arranged in polymeric chains c. They are held together by hydrogen bonding d. They are α-helical e. They have covalently linked backbones

*The answer is D.* Restriction endonucleases are produced by prokaryotes for cleaving both strands of foreign DNA. The host cell's DNA is not degraded because the recognition sites are specifically methylated. The endonucleases recognize specific short symmetrical sequences known as palindromes. These cleavage sites contain twofold rotational symmetry in that the sequence is identical but antiparallel in the complementary strands. In some cases, single-stranded cohesive ends on each of the complementary strands are produced, while in other cases double-stranded blunt ends are formed. Modern analysis of DNA structure is highly dependent upon the use of different restriction endonucleases that permit the specific hydrolysis of DNA into large polynucleotides.

Which of the following statements describing restriction endonucleases is true? a. They always yield overhanging single-stranded ends b. They recognize methylated DNA sequences c. They recognize triplet repeats d. They cleave both strands in duplex DNA e. They always yield blunt ends

*The answer is E.* For transfer RNAs, the 5′ end is often guanosine and is always phosphorylated, while the 3′ end is CCA. Although transfer (t) RNA molecules have many features in common, the primary feature that sets them apart is their specificity for different amino acids and the corresponding specific differences of their anticodons. Each tRNA is an L shaped single chain composed of up to 93 ribonucleotides. Each contains up to 15 methylated bases, and about half of the nucleotides are base-paired into double helices. Activated amino acids attach to the terminal 3′-hydroxyl group of the adenosine.

Which of the following statements is true of all transfer (t) RNAs? a. The 3′ end is phosphorylated b. They are duplex chains c. No methylated bases are found d. The anticodon loop is identical e. The 3′ end base sequence is CCA

*The answer B.* In the classic double-helical model of DNA proposed by Watson and Crick, the purine (adenosine and guanine) and pyrimidine (cytosine and thymine) bases attached to the sugar backbone are perpendicular to the axis and parallel to each other. They are paired (A to G or T to C) and held together by hydrogen bonds. The DNA strands (nucleotide polymers) are joined by linkages between the 3′-hydroxyl of each pentose (deoxyribose) and the 5′-phosphate of its deoxyribose neighbor. Each strand composing the double helix is different and antiparallel. The 3′ end of one strand is opposite the 5′ end of its complement and vice versa (see the figure above). It is this complementary nature of DNA that allows the strands to be templates for one another during DNA replication.

Which of the following statements regarding a double-helical molecule of DNA is true? a. All hydroxyl groups of pentoses are involved in linkages b. Bases are perpendicular to the axis c. Each strand is identical d. Each strand has parallel, 5′ to 3′ direction e. Each strand replicates itself

*The answer is B.* The structures of glycine and alanine are quite similar, with the −H side group of glycine being replaced with the −CH3 side group of alanine. Consequently, a mutation causing such a change is unlikely to produce a dysfunctional protein. In contrast, a mutation changing a splice site could result in either the abnormal exclusion of an exon or the inclusion of an intron in a protein, which would drastically change its properties. Likewise, frame-shift mutations caused by deletion or addition of one or two bases completely distort the remainder of the protein. The closer a frame shift is to the amino-terminal that is synthesized first, the more garbled the protein. All nonsense mutations involve inserting a stop codon in place of whatever amino acid would have been coded. In the case given, only half the protein would be synthesized.

Which of the following would not be expected to result in a dysfunctional protein? a. Mutation affecting the splice site of an intron b. Substitution of glycine for alanine at the carboxyl terminus c. Insertion of two bases in the code for the amino end d. Nonsense mutation affecting the middle of a potential protein product e. Deletion of a single base of a codon near the middle of a potential protein

*The answer is D.* Mammalian regulatory factors are much more diverse than those of bacteria, possessing several types of structural domains. Activators of transcription, such as steroid hormones, may enter the cell and bind to regulatory factors at specific sites called ligand-binding domains; these intracellular "receptors" are analogous to G protein-linked membrane receptors that extend into the extracellular space. Response elements are not regulatory factors but DNA sequences near the transcription site for certain types of genes (e.g., steroid-responsive and heat shock-responsive genes). Regulatory factors interact with specific DNA sequences through their DNA-binding domains, and with other regulatory factors through transcription-activating domains. Some regulatory factors have antirepressor domains that counteract the inhibitory effects of chromatin proteins (histones and nonhistones).

Which of the structural domains of mammalian regulatory factors may be called intracellular receptors? a. Response elements b. Antirepressor domains c. Transcription-activating domains d. Ligand-binding domains e. DNA-binding domains

*The answer is A.* In mammals, RNA polymerase binds to promoter sites upstream from the start site. These include the TATA box (TATAAT), the CAAT box, and the GC box. DNA primase and helicase are involved in DNA replication and do not bind specifically to sequences upstream of genes. Restriction endonucleases recognize specific sequences in double-helical DNA and cleave both strands. Histones nonspecifically bind to chromosomal DNA and constitute about half the mass of mammalian chromosomes.

Which one of the following binds to specific nucleotide sequences that are upstream of the start site of transcription? a. RNA polymerase b. Primase c. Helicase d. Histone protein e. Restriction endonuclease

*The answer is C.* Point mutations that are frame-shift mutations put the normal reading frame out of register by one base pair. The insertion of an extra base pair or the deletion of one or more base pairs falls into this category. Transitions and transversions are not frame-shift mutations; they are substitutions of one base pair for another. Substitutions are the most common type of mutation. In transitions, a purine is replaced by a purine or a pyrimidine by a pyrimidine. In transversions, a purine is replaced by a pyrimidine or vice versa. It has been suggested that transitions occur spontaneously owing to the tautomeric changes in base hydrogen bond locations. Transversions can be caused by defective DNA polymerases.

Which one of the following causes a frame-shift mutation? a. Transition b. Transversion c. Deletion d. Substitution of purine for pyrimidine e. Substitution of pyrimidine for purine

*The answer is B.* The "genetic code" uses three-nucleotide "words," or codons, to specify the 20 different amino acids (see the chart below). There are 64 different three-base pair codons (three positions with four possible nucleotides at each). It follows that the genetic code must be degenerate, i.e., different codons can specify the same amino acid. Three codons are reserved as "stop" signals that result in peptide chain termination. The linear correspondence of codons in DNA and of amino acids in protein domains is interrupted by the presence of introns in DNA. Codons differ from the dinucleotide tandem repeats that provide useful DNA polymorphisms, or the trinucleotide repeats that can be responsible for disease. The genetic code is universal in the sense that codon-amino acid relationships are the same in all organisms.

Which statement about the "genetic code" is most accurate? a. Information is stored as sets of dinucleotide repeats called codons b. The code is degenerate (i.e., more than one codon may exist for a single amino acid) c. Information is stored as sets of trinucleotide repeats called codons d. There are 64 codons, all of which code for amino acids e. The sequence of codons that make up a gene exhibits an exact linear correspondence to the sequence of amino acids in the translated protein

*The answer is C.* RNA polymerase II produces heterogeneous nuclear RNA, which is processed into mRNA. Thus, inhibition would lead to a decrease in the production of mRNA.

While hiking in the forest near her home, a 32-year-old woman picks mushrooms for a salad. A few hours after eating a mushroom, she experiences diarrhea and nausea and goes to t he hospital. In the emergency department, the mushrooms are identified as Amanita phalloides, a poisonous mushroom that contains the toxin α-amanitin. This toxin works by inhibiting an enzyme, halting the production of mRNA. Which of the following enzymes does α-amanitin inhibit? A. Helicases B. RNA polymerase I C. RNA polymerase II D. RNA prirnase E. Topoisomerase

*The answer is E.* Xeroderma pigmentosum appears to be due to the inability of an excision-repair system to remove thymine dimers, which are formed on exposure of DNA to ultraviolet radiation. This results in a deficiency in the ability to repair the damaged DNA. Mutagenesis by this mechanism is presumably the basis for the multiple neoplasms that occur in patients who have this disease.

Xeroderma pigmentosum is an inherited human skin disease that causes a variety of phenotypic changes in skin cells exposed to sunlight. The molecular basis of the disease appears to be a. Rapid water loss caused by defects in the cell membrane permeability b. The inactivation of temperature-sensitive transport enzymes in sunlight c. The induction of a virulent provirus on ultraviolet exposure d. The inability of the cells to synthesize carotenoid-type compounds e. A defect in an excision-repair system that removes thymine dimers from DNA

*The answer is B.* Since the disease is autosomal recessive, both chromosomes must carry a mutated gene for the person to have the disease, so only one band should be seen on the gel. Since the mutation which leads to the disease destroys a restriction endonuclease site, the disease-carrying allele will have a larger size on the gel than the normal allele, which will be cleaved by the restriction enzyme. Thus, for individual A, for example, the 1.8 kb band corresponds to the mutant allele, while the 1.2 kb and 0.6 kb bands correspond to the normal allele, which contains the restriction site that enables the DNA to be cut into two pieces. Individual A is a carrier of the disease, as is individual B. Individual D has two normal copies of the gene and is not a carrier of the disease.

You are studying a family that exhibits a rare autosomal recessive disease. The disease is caused by a point mutation which abolishes an EcoR1 restriction site in the genome. To genotype the family, one can amplify a 1.8 kb region of this area of the genome using polymerase chain reaction (PCR), restrict the PCR products with EcoR1, and then separate the restricted DNA on an agarose gel. The results of such an experiment are shown below, using DNA obtained from each of the f ve family members. Which family members have the disease? (A) A and B (B) C and E (C) D (D) A, B, and D (E) C, D, and E

*The answer is A.* All 3 tissues contain the gene (the probe was produced from baboon mRNA, implying the gene is also there).

mRNA encoding glucose 6-phosphatase was isolated from baboon liver and used to make a ³²P-cDNA probe. DNA was then isolated from marmoset and from human tissue, digested with a restriction endonuclease, Southern blotted, and probed with the ³²P-cDNA. Which of the following conclusions can be drawn from the results of this analysis shown below? A. The glucose 6-phosphatase gene is present in baboon, marmomest and human liver. B. Both marmoset and human liver express the glucose 6-phosphatase gene. C. There are two glucose 6-phosphatase genes in the human liver. D. The glucose 6-phosphatase gene is on different chromosomes in the marmoset and in the human. E. The human and marmoset tissue used in this experiment is from liver.

*The answer is B.* The patient has Fragile X syndrome, caused by the expansion of a triplet nucleotide repeat (CGG) within the FMR1 gene on the X chromosome. The expansion interferes with the normal functioning of this gene product in the brain, leading to the symptoms observed. This is an extreme example of an insertion mutation. A point mutation occurs when only one base is substituted for another, and the change in base results in an amino acid change in the protein. Deletion is removal of one or more nucleotides from the gene. Mismatch (answer D) repair is a DNA repair process that is utilized when a mismatch is found in the DNA. A silent mutation is one in which a base change in the DNA leads to no change in the corresponding amino acid in the protein (due to degeneracy in the genetic code).

A 17-year-old male has large, prominent ears, elongated face, large testicles, hand flapping, low muscle tone, and mild mental retardation. Which type of mutation does his diagnosis represent? (A) Point (B) Insertion (C) Deletion (D) Mismatch (E) Silent

*The answer is D.* The patient has HNPCC, which is due to specific mutations in proteins involved in mismatch repair (mutations in at least four different proteins have been identified that lead to HNPCC). Mismatch repair is not involved in thymine dimer removal, nor base excision repair (the removal of uracil from DNA). HNPCC does not involve a defective DNA ligase, nor does the disease result in defective DNA packaging (solenoid formation) in the nucleus.

A 33-year-old man had a screening colonoscopy, and was diagnosed with a right-sided, mucinous colon cancer, with no other lesions or polyps seen. The reason he had a colonoscopy at such an early age is that his father and paternal uncle had colon cancers diagnosed by age 40. His paternal grandmother had ovarian and uterine cancers. A likely defect in the patient is a reduction in the ability to carry out which one of the following processes? (A) Removal of thymine dimers from the DNA (B) Inability to remove the base U from DNA (C) Loss of DNA ligase activity (D) Inability to correct mismatched bases in newly synthesized DNA (E) Inability to form a solenoid structure from individual nucleosomes

*The answer is B.* Two drugs are utilized for latent tuberculosis: isoniazid\ and rifampin. Isoniazid works by blocking the synthesis of mycolic acid, a necessary component of the cell wall of the bacteria that leads to tuberculosis. Rifampin works by inhibiting bacterial RNA polymerase, and blocking the synthesis of new proteins. Neither drug affects DNA polymerase or peptidyl transferase (chloramphenicol is the antibiotic that inhibits bacterial peptidyl transferase activity). Rifampin also has no effect on IF-1.

A 42-year-old man is placed on a two-drug regimen to prevent the activation of the tuberculosis bacteria, as his tuberculin skin test (PPD) was positive, but he shows no clinical signs of tuberculosis, and his chest X-ray is negative. One of the drug's mechanism of action is to inhibit which one of the followingmenzymes? (A) DNA polymerase (B) RNA polymerase (C) Peptidyl transferase (D) Initiation factor 1 of protein synthesis (IF-1) (E) Telomerase

*The answer is C.* Transposons have the ability to move DNA elements from one piece of DNA to another, including antibiotic-resistance genes from R-plasmids to the host chromosome. Thus, over time, as a bacteria obtains plasmids with antibiotic-resistance genes on them, the transposons can move the gene to the bacterial chromosome so it is always expressed by the cell, and no longer requires the plasmid for antibiotic resistance. Alterations in the membrane structure do not occur, nor do large deletions of the bacterial chromosome (antibiotic-resistance genes are not normal components of the bacterial chromosome). Antibiotic resistance is neither due to a loss of energy production, nor to spontaneous mutations in existing genes, as the bacteria do not encode genes that may confer antibiotic resistance to begin with.

A 72-year-old man acquired a bacterial infection in the hospital while recuperating from a hip replacement surgery. The staph infection was resistant to a large number of antibiotics, such as amoxicillin, methicillin, and vancomycin, and was very difficult to treat. The bacteria acquired its antibiotic resistance owing to which one of the following? Choose the one best answer. (A) Spontaneous mutations in existing genes (B) Large deletions of the chromosome (C) Transposon activity (D) Loss of energy production (E) Alterations in the membrane structure

*The answer is E.* The short segments of the newly synthesized DNA that accumulate at 42°C are Okazaki fragments. They are usually joined together by DNA ligase, which most likely exhibits reduced activity at 42°C in this mutant. If the ligase is not functioning, Okazaki fragments would not be joined during replication, so the cells would contain short fragments of the DNA. Endonucleases and exonucleases cleave DNA strands in the middle and at the ends, respectively. They do not join fragments together, nor does DNA polymerase. Unwinding enzymes "unzip" the parental strands, and if these were defective, DNA synthesis most likely would not occur at the nonpermissive temperature, and short DNA fragments would not accumulate.

A bacterial mutant grows normally at 32°C but at 42°C accumulates short segments of newly synthesized DNA. Which one of the following enzymes is most likely to be defective at the nonpermissive temperature (the higher temperature) in this mutant? (A) DNA primase (B) DNA polymerase (C) An exonuclease (D) An unwinding enzyme (helicase) (E) DNA ligase

*The answer is B.* The variant cell line will mischarge a tRNAala with glycine, but only the tRNAala that has, as its anticodon, IGC. IGC will recognize three codons: GCA, GCU, and GCC. Thus, when these codons are present in the mRNA, glycine will replace alanine. Thus, reading the RNA from the 5′ end (as translation reads the mRNA 5′ to 3′) provided in a series of 3, we have AUG (which is methionine),\ GCG (which is alanine—the anticodon for this codon is CGC; IGC will not recognize the GCG codon), GAC (which is aspartic acid), UCG (which is serine), GCU (which, in this cell line, is glycine and not alanine, since the IGC anticodon will recognize the GCU codon), and AUG (methionine). The answer choices utilize the single-letter codes to represent the amino acids.

A cell contains a mutated alanine-tRNAala synthetase that recognizes glycine instead of alanine as its substrate. The anticodon of the tRNA recognized by this enzyme is IGC. When the cell translates the following portion of an mRNA molecule (presented in frame beginning with the 5′ nucleotide), what will be the amino acid sequence of the protein produced from this stretch of mRNA? -AUGGCGGACUCGGCUAUG- (A) M-G-S-D-G-M (B) M-A-D-S-G-M (C) M-A-D-S-A-M (D) M-A-S-D-A-M (E) M-G-D-S-A-M

*The answer is D.* There are 64 codons in the genetic code, the majority of which code for amino acids. Because there are only 20 amino acids, most amino acids have more than one codon. For example, GUU, GUC, GUA, and GUG all code for valine. In addition, there are codons that call for the initiation and termination of protein synthesis. AUG. which codes for methionine, is the universal start codon. UAA, UAG, and UGA are stop codons. The stop codons do not code for amino acids. Instead, when the ribosome encounters a stop codon. releasing factors bind to the ribosome and stimulate release of the formed polypeptide chain and dissolution of the ribosome-mRNA complex.

A fragment of an mRNA molecule isolated from a Gram-negative bacterium is shown below. Which of the following recognizes codon 128? (A) snRNA (B) Uncharged tRNA (C) Charged tRNA (D) Releasing factor 1 (E) Elongation factor 2 (F) Transcription factor II D

*The answer is A.* tRNA is the smallest of the RNA subtypes. The image shows the tRNA secondary structure. The 3' and 5' ends of the molecule form the so-called "acceptor stem." The 3' end is the amino acid attachment site. the anticodon is located on the opposite end of the tRNA molecule (site C). It recognizes and binds the mRNA codon and assures placement of the proper amino acid in the growing polypeptide chain. The TΨC arm, which contains a sequence of three nucleotides present in all tRNA molecules (ribothymidine, pseudouridine, and cytosine) and the D loop has unpaired bases. The TΨC arm, D loop, and anticodon are responsible for the "cloverleaf" secondary structure of the molecule. The D and TΨC loops help to determine the tRNA tertiary structure as well. The 5' end of tRNA generally is composed of a terminal guanosine and does not participate in amino acid or mRNA binding.

A schematic structure of tRNA is shown on the slide below. Which of the following sites is responsible for amino acid binding?

*The answer is C.* A tRNA with the anticodon IGA will bind to UCU, UCC, and UCA (with wobbling at the third position of the codon). A tRNA with the anticodon CGA will bind to UCG. And the third tRNA, with an anticodon of ICU, can base-pair with both AGU and AGC. Recall that I will base-pair with A, C, or U in the wobble position.

Given the serine codons shown below, what is the minimum number of tRNAs which is required to bind to them? UCU, UCC, UCA, UCG, AGU, AGC (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

*The answer is D.* The product of DNA replication will be complementary to the template, and antiparallel. Reading from the 5′ end of the template, the product will be 3′-TAAGCTAACGGGTGCA-. When written 5′-3′ (standard notation) one has -ACGTGGGCAATCGAAT-. Recall that uracil (U) is not placed into DNA by DNA polymerase.

The sequence of part of a DNA strand is the following: -ATTCGATTGCCCACGT-. When this strand is used as a template for DNA synthesis, the product will be which one of the following? (A) TAAGCTAACGGGTGCA (B) UAAGCUAACGGGUGCA (C) ACGUGGGCAAUCGAAU (D) ACGTGGGCAATCGAAT (E) TGCACCCGTTAGCTTA

*The answer is A.* On a molar basis, DNA contains equal amounts of adenine and thymine and of guanine and cytosine. Uracil is not found in DNA. There are 10 base pairs per turn of the helix, and the overall charge on the molecule is negative, due to the phosphates in the backbone (each phosphodiester bond contains one negative charge).

Which one of the following is true for a double- stranded DNA molecule?

*The answer is B.* Diphtheria toxin, after entering cells, is cleaved by a protease to form an active enzyme, which, utilizing NAD+ as a substrate, ADP-ribosylates eEF2, thereby inhibiting protein translation. ATP, FAD, acetyl-CoA, and UDP-glucose are not required for the ADP-ribosylation reaction. The final modified product is an arginine with an ADP-ribose attached.

A 3-year-old boy, whose parents did not immunize him due to fears of postimmunization side effects, exhibited fever, chills, severe sore throat, lethargy, trouble breathing, and a husky voice. Physical exam indicated greatly enlarged lymph nodes, an increased heart rate, and swelling of the palate. A picture of the boy's throat is shown below. A necessary cofactor for allowing these symptoms to appear in the child is which of the following? (A) ATP (B) NAD+ (C) FAD (D) Acetyl-CoA (E) UDP-glucose

*The answer is A.* Lynch syndrome (hereditary nonpolyposis colon cancer) is an autosomal dominant disease caused by defective DNA mismatch repair. DNA replication occurs with a high degree of fidelity because mismatched nucleotides are repaired through the proofreading activity of DNA polymerases delta and epsilon. However, this proofreading functionality is not infallible, base substitutions and smell insertion-deletion mismatches occur due to errors in base pairing every 10^6 bases on average. It is the function of the DNA mismatch repair system to fix these errors is shortly after the daughter strands are synthesized. The mismatch repair system involves several genes, including MSH2 and MLH1, which code for components of the human Muts and MutL homologs. Mutations in these 2 genes account for around 90% of cases of Lynch syndrome. Mismatch repair begins with MutS homolog detecting a mismatch on the newly created daughter strand, which is distinguished from the parent strand by occasional nicks in the phosphodiester bonds. MutL homolog is then recruited, and the resulting complex slides along the DNA molecule until 1 of the daughter strand nicks is encountered. At this point, exonuclease 1 is loaded onto the end activated by the repair complex. The daughter strand is then degraded backward past the initial mismatch point, leaving a variable gap of single-stranded DNA that is stabilized by ssDNA-binding protein. The complex then dissociates while DNA polymerase delta loads at the 3' end of the discontinuity and begins synthesizing a new daughter strand segment. Finally, DNA ligase I seals the remaining nick to complete the is repair process.

A 34-year-old woman comes to the physician with abdominal pain and melena. She also complains of progressive fatigue and a 5 kg (11 lb) weight loss over the last 2 months. She has a strong family history of colon, endometrial, and ovarian cancer. Colonoscopy shows a protuberant, friable mass in the ascending colon, and biopsy is diagnostic for colon adenocarcinoma. Genetic analysis confirms a mutation consistent with Lynch syndrome (hereditary nonpolyposis colon cancer). Which of the following is most likely responsible for the development of colon cancer in this patient? (A) Nucleotide mismatches that escape repair (B) Covalent bonds between adjacent pyrimidines (C) Insertion of abnormal bases (Eg, uracil) into DNA (D) Empty sugar-phosphate residues in the DNA molecule (E) Double-strand breaks in DNA

*The answer is E.* The quinolone family of antibiotics (which includes ciprofloxacin) inhibits DNA gyrase, a prokaryotic-specific topoisomerase involved in unwinding the DNA strands for replication to occur. In the absence of gyrase activity, there would be no DNA replication, and the bacteria would not be able to proliferate. The quinolones do not affect eukaryotic topoisomerases. Splicing of hnRNA only occurs in eukaryotic cells. The gyrase is neither involved in nucleotide excision repair, nor in any aspect of protein synthesis.

A woman has been complaining of a burning sensation when urinating, and a urine culture demonstrated a bacterial infection. The physician placed the woman on ciprofloxacin. Ciprofloxacin will be effective in eliminating the bacteria because it interferes with which one of the following processes? (A) mRNA splicing (B) Initiation of protein synthesis (C) Elongation of protein synthesis (D) Nucleotide excision repair (E) DNA replication

*The answer is E.* The damage to the DNA caused by UV light (pyrimidine dimers) can be repaired by the nucleotide excision repair pathway. In some cases, the missing enzyme is a repair endonuclease. The boy has XP, as determined by the increase in thymine dimers in his DNA after exposure to UV light. Since the dimers cannot be repaired, the DNA polymerase will "guess" when replication occurs across the dimers, increasing the mutation rate of the cells. Eventually, a mutation occurs in a gene that regulates cell proliferation, and a cancer results. An increase or decrease in mutation rate is not related to the rate of recombination during mitosis, nor to a lack of DNA primase activity (which would lead to reduced DNA synthesis, not inaccurate DNA synthesis). Base excision repair is normal in patients with XP.

A 15-year-old boy was diagnosed with skin cancer. He had always been sensitive to sunlight, and had remained indoors for most of his life. An analysis of his DNA, from isolated fibroblasts, indicated an increased level of thymine dimers when the cells were exposed to UV light. The boy developed a skin tumor owing to an increased mutation rate, which was caused by which one of the following? (A) A lack of DNA primase activity (B) Decreased recombination during mitosis (C) Increased recombination during mitosis (D) Loss of base excision repair activity (E) Loss of nucleotide excision repair activity

*The answer is E.* The child is exhibiting the symptoms of Cockayne syndrome, which is due to a defect in transcription-coupled DNA repair. During transcription of genes, if the RNA polymerase notices DNA damage, transcription will stop while the transcription-coupled DNA repair mechanism will correct the DNA damage. This syndrome can be due to mutations in either the ERCC6 or ERCC8 gene, and the protein products of both the genes are involved in repairing the DNA of actively transcribed genes. The key to answering the question is the amount of DNA damage in growing cells (which are transcriptionally active) versus the damage in quiescent cells (which express fewer genes). The symptoms described are also unique to individuals with this disorder. The repair of thymine dimers and the processes of base excision repair, nucleotide excision repair, and mismatch repair are all functional in individuals with this disorder.

A 4-year-old boy displays a failure to thrive, extreme sensitivity to the sun, hearing loss, severe tooth decay, pigmentary retinopathy, and premature aging. An analysis of fibroblasts from the boy demonstrated extensive DNA damage in cells trying to grow, but minimal damage in quiescent cells, which have a greatly reduced rate of transcription as compared to the growing cells. This child most likely has a defect in which one of the following processes? (A) Repair of thymine dimers (B) Base excision repair (C) Nucleotide excision repair (D) Mismatch repair (E) Transcription-coupled DNA repair

*The answer is B.* Rifampin binds to prokaryotic RNA polymerase but cannot bind (due to the different structures of the RNA polymerase between prokaryotic and eukaryotic cells) to eukaryotic RNA polymerase. The drug does not bind to ribosomes, DNA, snurps, or transcription factors. Snurps are not present in prokaryotic cells.

A man with a bacterial infection was prescribed rifampin to resolve the infection. Rifampin does not affect eukaryotic cells due to which of the following? (A) Differences in ribosome structure between eukaryotes and prokaryotes (B) Structural differences in RNA polymerase between eukaryotes and prokaryotes (C) Differences in transcription factors between eukaryotes and prokaryotes (D) Inability of the drug to bind to DNA containing nucleosomes (E) Differences in snurp structure between eukaryotes and prokaryotes

*The answer is C.* The patient has Kaposi's sarcoma and AIDS. The causative agent is HIV, an RNA virus. The Western blot technique is used to identify whether a specific blood sample contains antibodies that will bind to HIV-specific proteins. The HIV proteins are run through a gel, transferred to filter paper, and probed using the sera from the patient. If the patient has antibodies to the HIV proteins, then a positive result will be obtained. A Southern blot is used to identify the DNA, and in this case it is easier to check for the presence of anti-HIV antibodies in the patient's sera. A Northern blot would check for viral RNA, but it is more efficient, and reliable, due to the low levels of viral RNA, to check for anti-HIV proteins instead. The Sanger technique identifies a portion of the DNA chain through sequencing the bases in the DNA, and is not used for determining the HIV status. A Southwestern blot is used to detect DNA binding to proteins, and would not be applicable for AIDS testing.

A thin, emaciated 25-year-old male presents with purple plaques and nodules on his face and arms, coughing, and shortness of breath. In order to diagnose the cause of his problems most efficiently, you would order which one of the following types of tests? (A) Southern blot (B) Northern blot (C) Western blot (D) Sanger technique (E) Southwestern blot

*The answer is D.* The nontemplate strand in the DNA is the same as the coding strand, and will have the same sequence of the RNA that is produced, except that T is in place of U. Both the nontemplate strand and RNA produced will be a complementary sequence to the template strand. As all sequences are written in the 5′ to 3′ direction, unless otherwise specified, an RNA sequence of AAUUGGCU would correspond to a DNA sequence, on the nontemplate strand, of AATTGGCT. The template strand would be the complement of the nontemplate strand, or AGCCAATT (written 5′ to 3′). Note that this sequence is also the complement of the RNA that has been produced. The base U is not found in DNA (so answer choices B and E cannot be correct).

An RNA produced from a fragment of DNA has the sequence of AAUUGGCU. The sequence of the nontemplate strand in the DNA that gave rise to this sequence is which one of the following? (A) AGCCAATT (B) AAUUGGCU (C) AATTGGCT (D) TTAACCGA (E) UUAACCGA

*The answer is C.* RNA polymerase III requires transcription factors to bind to promoters, which are labeled as TFIIIx, where x is a variable letter. Polymerase III will synthesize tRNA molecules and 5S rRNA. RNA polymerase II synthesizes mRNA while RNA polymerase I synthesizes primarily rRNA. Accessory factors for the polymerases are labeled TFIIx or TFIx, respectively. MicroRNAs are synthesized by polymerase II. hnRNA is the precursor for mature mRNA, also synthesized by RNA polymerase II.

TFIIIA is a necessary transcription factor for the synthesis of which class of molecules? (A) mRNA (B) rRNA (C) tRNA (D) hnRNA (E) microRNAs

*The answer is C.* Levofloxacin is a member of the quinolone family of antibiotics that inhibits bacterial topoisomerases, primarily DNA gyrase (etoposide is the drug that inhibits eukaryotic topoisomerases). Without gyrase activity, the DNA of the bacterial chromosome cannot be unwound properly, and DNA replication would cease, leading to the death of the bacteria. The quinolone family of antibiotics does not directly affect DNA polymerases, DNA ligase, or DNA primase.

A 50-year-old female has shortness of breath, cough, and fever for 3 days. She lives with her husband and has no medical problems. Her pulse ox in the office is 89 and her pulse rate is 110. She is admitted for treatment of community-acquired pneumonia, and her intravenous (IV) antibiotic treatment includes levofloxacin. A mutation in which bacterial enzyme would be required for levofloxacin resistance to be observed? (A) DNA primase (B) DNA polymerase III (C) DNA gyrase (D) DNA ligase (E) DNA polymerase I

*The answer is E.* DNA polymerases are the main enzymes responsible for DNA replication. In E. coli, there are three major DNA polymerases: I, II, and III. Other enzymes that are essential for the replication of DNA include primase, helicase, ligase, and topoisomerase I and II. During DNA replication, new daughter strands are to 3' direction, using the parent strands as complimentary templates. Synthesis of DNA in the 3' to 5' direction does not occur in this process. Before the process of replication begins, the parent DNA double helix is unwound and separated by the enzyme helicase. Helicase binds DNA at the origin of replication with the assistance of DNA protein and acts at the replication fork to separate DNA. This separation and unwinding of the DNA produces positive supercoiling that can lead to DNA fracture, if not relieved. Topoisomerases I and II (II is also known as gyrase) relieve unwinding tension. DNA polymerases can not begin synthesis of daughter strands without a free 3'-hydroxyl group, which is provided by an RNA primer and synthesized by the enzyme primase (DNA dependent RNA polymerase). DNA replication then proceeds with the leading strand being formed continuously in the 5' to 3' direction toward the replication fork. and the lagging strand being formed discontinuously in the 5' to 3' direction, away from the replication fork. Replication of the lagging strand results in the formation of numerous short DNA segments called Okazaki fragments, and each separate DNA segment requires a new RNA primer upon which to initiate synthesis. The fragments of the lagging strand are ultimately bound together by ligase after numerous RNA primers have been removed and replaced with DNA.The removal of RNA primers is accomplished by DNA polymerase I, the only bacterial DNA polymerase with 5' to 3' exonuclease activity. This activity allows DNA polymerase I to function both as an excision repair enzyme and as the enzyme that removes RNA primers.

A mutation that leaves prokaryotes unable to replicate their DNA is induced in an experimental setting. The ability to remove RNA primers during DNA replication is affected by this experimental mutation. Which of the following enzymes is most likely nonfunctional? (A) Helicase (B) Primase (C) Gyrase (D) DNA polymerase III (E) DNA polymerase I (F) Ligase

*The answer is A.* The patient is concerned about HNPCC, hereditary nonpolyposis colon cancer, which is due to mutations in genes which are involved in DNA mismatch repair. This colon tumor does not form large numbers of polyps within the intestine, as does the other form of inherited colon cancer, adenomatous polyposis coli (APC). HNPCC is also a right-sided colon cancer. Defects in base excision repair do not lead to HNPCC. A defect in the Wnt signaling pathway, which controls the action of β-catenin, an important transcription factor, may play a role in APC. Defects in repairing double-strand breaks in DNA are linked to breast cancer. Mutations in telomerase would lead to earlier cell senescence and death due to an inability to maintain the proper length of the chromosomes.

A new patient visits your practice due to his concern of developing colon cancer. A large number of relatives have had premature (less than the age of 45) colon cancer, and all cases were right-sided, with the only visible polyps being found on that side. The molecular basis for this form of colon cancer is which of the following? (A) A defect in DNA mismatch repair (B) A defect in base excision repair (C) A defect in the Wnt signaling pathway (D) A defect in repairing double-strand DNA breaks (E) A defect in telomerase

*The answer is C.* The codons for proline are CCU, CCA, CCG, and CCC. The codons for lysine are AAA and AAG. The normal sequence of these two amino acids in the B-chain of insulin is pro-lys, so examples of this are CCGAAG, or CCCAAA. However, in the genetically engineered lispro variant of insulin, the lysine codon comes first, followed by the options of valine codons. The only answer that does this correctly is answer C. Answer A converts a pro-lys to a lys-asn. Answer B converts a pro-asn to lys-pro. Answer D converts a lys-pro to a pro-lys, and answer E converts a lys-pro to lys-pro.

For the synthesis of lispro insulin (Lispro is a synthetic insulin formed by reversing the lysine and proline residues on the C-terminal end of the B-chain.), which one of the following changes in the coding for the B-chain would be required? (A) CAAAAA to AAAAAC (B) CCTAAT to AAACTC (C) CCGAAG to AAACCA (D) AAACCA to CCGAAG (E) AAGCCT to AAACCC

*The answer is F.* A gene that is methylated is less readily transcribed than the one that is not methylated. Polycistronic mRNAs are only produced in prokaryotes, not in eukaryotic cells.

Processes that, in part, can lead to the activation of gene expression in eukaryotes can be best described as which one of the following?

*The answer is D.* Proteins destined for secretion contain a signal sequence at the N-terminal end that causes the ribosomes on which they are being synthesized to bind to the SRP, which transfers the mRNA-ribosome complex to the RER. As they are being produced, they enter the cisternae of the RER, where the signal sequence, including the initial methionine, is removed. It is thus unlikely that the mature protein will contain an N-terminal methionine. Carbohydrate groups can be attached in the RER or the Golgi. Secretory vesicles bud from the Golgi, and the proteins are secreted from the cell by the process of exocytosis. If the proteins have a hydrophobic sequence that embeds in the membrane, they remain attached to the membrane and are not secreted, and become membrane-bound proteins.

Proteins destined for secretion from eukaryotic cells have which of the following in common?

*The answer is B.* Every chromosome has a homolog. Therefore, there will be two copies of every DNA sequence in the genome. Child C2 could have obtained the 9-kb restriction fragment from this mother, and the 8.5-kb fragment from this father. According to this test, child C1 is not genetically related to either this mother or this father, as neither the mother nor the father contain the 9.5- and 7.0-kb fragments expressed by child C1.

Use the following figure to answer this question. Two male infants were born on the same day in the same hospital. Because of concern that the infants had been switched in the hospital nursery, genetic tests based on a DNA restriction fragment that exhibits polymorphism (RFLP) were performed. Blood was drawn from the parents and the infants, the DNA extracted, and PCR performed. The DNA was then treated with the restriction enzyme BanI, and the fragments were separated by gel electrophoresis. The results of a Southern blot test are shown in the figure. A radioactive probe was used that bound to a sequence within the BanI fragments that exhibited polymorphism. Which of the two infants, C1 or C2, is the genetic offspring of this mother (M) and father (F)? (A) C1 could be the offspring of these parents. (B) C2 could be the offspring of these parents. (C) Both infants could be the offspring of these parents (i.e., this test cannot discriminate). (D) Either of these infants could be related to the mother, but neither could be related to the father. (E) Neither infant could be related to thismother or this father.

*The answer is B.* Although D contains the sequence 3′-TAC-5′, which produces a start codon (5′-AUG-3′) in the mRNA, there is a sequence (3′-ATT-5′ in the DNA) that would produce a stop codon (5′-UAA-3′) in the mRNA in frame with this start codon. Sequence B, read 3′ to 5′ (from right to left), would produce a start codon in the mRNA transcribed from it. There are no stop codons in this sequence, so it could produce a protein 300 amino acids in length. Sequences A and C do not contain triplets corresponding to the start codon in mRNA.

Which region (A to D) of the DNA strands shown could serve as the template for transcription of the region of an mRNA that contains the initial codon for translation of a protein 300 amino acids in length?

*The answer is A.* Dactinomycin binds to DNA and blocks the ability of RNA polymerase to transcribe genes, thereby blocking transcription. The drug does not bind specifically to ribosomes, transcription factors, or RNA polymerase II. It also does not interfere with= DNA synthesis.

A 2-year-old has been diagnosed with a rhabdomyosarcoma and is placed on chemotherapy, including the drug dactinomycin. Dactinomycin exerts its effects by which of the following mechanisms? (A) Binding of the drug to DNA, thereby blocking RNA synthesis (B) Binding of the drug to ribosomes, thereby blocking translation (C) Binding of the drug to transcription factors, thereby blocking RNA synthesis (D) Binding of the drug to RNA polymerase II, thereby blocking RNA synthesis (E) Binding of the drug to DNA, thereby blocking DNA synthesis

*The answer is E.* This patient most likely has heterozygous familial hypercholesterolemia. an autosomal dominant LDL receptor defect that causes high LDL levels and increases the risk of premature atherosclerosis. Homozygous familial hypercholesterolemia (a rarer and more severe form of the disease due to the inheritance of 2 defective LDL receptor alleles) often presents with coronary heart disease in childhood adolescence. Southern blotting is a technique that can be used to detect DNA mutations. The process involves the following steps: 1. DNA extraction from the individual's calls 2. Restriction endonuclease digestion of the DNA sample into fragments 3. Gel electrophoresis to separate the various sizes of DNA fragments. Larger fragments move slowly and shorter fragments move faster 4. DNA probe (a single-stranded segment of labeled DNA complementary to the gene of interest) to identify the target gene Once the gene of interest is identified by the DNA probe, various family members' Southern blots can be compared. Because both the patient and his father are affected, the common DNA segment between them (8 kb segment) most likely represents the mutated gene. The patient's son also has the 8 kb segment, meaning that he is probably affected as well.

A 34-year-old man is found to have an LDL level of 310 mg/dL and a normal serum triglyceride level. His father suffered a myocardial infarction at age 39, and his paternal grandfather died of a heart attack at age 40. The patient's wife has a normal lipid profile. DNA samples are obtained from several family members for genetic analysis. Southern blotting of restriction fragments from a region containing the LDL receptor gene shows the following pattern: Which of the following statements best describes the DNA analysis results? (A) The disease is transmitted 'm an X-linked recessive fashion (B) The mutation is probably located in the 10 kb band (C) The mutation is probably located in the 12 kb band (D) The patient's brother most likely inherited t:he mutation (E) The patient's son most likely inherited the mutation

*The answer is C.* An adenine nucleotide in the middle of the intron is a required component for splicing to occur, and the sugar residue attached to this adenine is involved in three phosphodiester linkages; the normal 3′ and 5′ and also 2′ to the splice site. The resulting structure resembles a lariat. Such a structure does not form during capping, polyadenylation, or the normal transcription of genes. It is unique to the splicing mechanism.

A careful analysis of cellular components discovers short-lived RNA species in which an adenine nucleotide is found with three phosphodiester bonds (linked to the 2′, 3′, and 5′ carbons). This transient structure is formed during which of the following processes? (A) mRNA cap formation (B) mRNA polyadenylation (C) Splicing of hnRNA (D) Transcription of microRNAs (E) Transcription of rRNA

*The answer is C.* The child is exhibiting the symptoms of Cockayne syndrome (CS), a defect in transcription-coupled DNA repair. Transcription-coupled DNA repair occurs only on actively transcribed genes; if RNA polymerase is halted due to damage to DNA on an actively transcribed gene, this repair system fixes the DNA such that transcription can continue. Cells derived from patients with CS have a reduction in RNA synthesis in response to UV irradiation, as transcription-coupled DNA repair is reduced, thereby reducing the rate of RNA produced from genes which did contain thymine dimers. There are at least two forms of CS: CS 1 (or A), the form present at birth, and CS 2 (or B), one that occurs later in life, during early childhood. The two forms are due to mutations in two different genes (ERCC8, on chromosome 5, is responsible for CS-A, and ERCC6, on chromosome 10, is responsible for CS-B). The child's symptoms are not due to defects in base excision repair or in DNA ligase (sealing nicks in DNA). DNA replication is normal in these children, as is the proofreading capability of DNA polymerase.

Concerned parents are referred to a specialty clinic by their family physician due to abnormalities in their 18-month-old child's development. The child displays delayed psychomotor development, and is mentally retarded. The child is photosensitive, and also appears to be aging prematurely, with a stooped posture and sunken eyes. The altered process in this autosomal recessive disorder is which of the following? (A) Base excision repair (B) DNA replication (C) Transcription-coupled DNA repair (D) Proofreading by DNA polymerase (E) Sealing nicks in DNA

*The answer is A.* The most efficient way to destroy an α-helix is to insert a proline into the middle of it, as proline cannot form an α helix due to the restriction of rotation of certain bond angles. Choice (A) has a leucine going to proline; all of the other substitutions (val to ala, leu to leu, ile to val, and val to leu) would still allow α-helix formation after the substitution took effect as these are conservative substitutions.

While investigating structure-function studies in a membrane transport protein, a researcher discovered a single nucleotide mutation that led to the loss of a key α-helical segment of the protein. The mutation that led to this finding is most likely which of the following? (A) CUC to CCC (B) GUU to GCU (C) UUG to CUG (D) AUC to GUC (E) GUG to UUG

*The answer is B.* Amino acid activation and attachment to the 3' end of RNA is catalyzed by aminoacyl-tRNA synthetases (AA-tRNA synthetases). Each amino acid/tRNA pair has a specific AA-tRNA synthetase that links them together. Those enzymes are responsible for ensuring that each amino acid binds to the RNA with the proper anticodon. Aminoacyl-tRNA synthetase activation and binding sites are highly specific for their correct amino acids and RNA molecules. Additionally, some AA-tRNA synthetases can "proofread" their specific RNA molecules and hydrolyze the amino acid bond when their tRNAs are incorrectly charged. The error rate for AA-tRNA synthetases is thus very low at less than 1 error per 10^4 charges. During protein synthesis, tRNA acts as an adaptor molecule between the codes found on mRNA and the amino acids being incorporated into the polypeptide chain. The sequence of amino acids in a polypeptide chain is dictated by binding of the RNA anticodon to its complementary codon on the mRNA molecule being translated. Erroneous amino acid tRNA coupling by the AA-tRNA synthetase causes the wrong amino acid to be incorporated into the growing polypeptide chain. In the scenario described in the question, when the ribosome encounters the praline codon on mRNA (e.g., CCC), the complementary RNA (GGG) binds. If this tRNA is improperly charged with leucine, leucine gets incorrectly incorporated into the growing polypeptide chain in proline's place

A 10-year-old boy suffers a laceration to his left knee while playing outside. He is brought to the emergency department where his wound is cleansed with sterile saline solution and closed with sutures. In response to the injury, the patient's fibroblasts begin to increase protein synthesis locally. During this process, an aminoacyl~tRNA synthetase erroneously charges a praline-carrier tRNA with leucine. Which of the following is most likely to happen with the leucine residue? (A) It is properly incorporated into the polypeptide chain at a site requiring leucine (B) It is incorrectly incorporated into the polypeptide chain in praline's place (C) It is randomly incorporated into the polypeptide chain, halting chain elongation (D) It is never incorporated into the polypeptide chain and remains attached to RNA (E) lt is rapidly cleaved by glycosylase

*The answer is D.* The act of splicing requires the breakage of internal phosphodiester bonds, which is the job of an endonuclease. Splicing does not require new RNA synthesis, DNA synthesis, error-checking (the 3′-5′ exonuclease activity), or DNA repair.

A cell line was derived, which was temperature sensitive for splicing hnRNA. At the nonpermissive temperature, splicing was unable to occur. A potential activity that is mutated in the splicesome is which of the following? (A) Ability to carry out RNA synthesis (B) Ability to carry out DNA synthesis (C) Loss of 3′-5′ exonuclease activity (D) Loss of endonuclease activity (E) Loss of ability for transcription-coupled DNA repair

*The answer is C.* The intestine contains an RNA editing complex that alters one base in the apo B100 mRNA, which creates a stop codon, such that when the mRNA is translated, protein synthesis stops after 48% of the codons have been translated. This is a unique type of posttranscriptional modification. The initial transcripts for both apo B48 and apo B100 are the same. Mutations that alter splicing, cap formation, or polyadenylation would not produce the full-size protein in place of apo B48. Mutations in the promoter would alter initiation of transcription, but not the end product formed.

An intestinal cell line was being studied for its ability to produce lipid-containing particles. Surprisingly, a mutated variant of this cell line was unable to do so. Western blot analysis yielded a protein with the same size as apolipoprotein B100. A potential mutation in this cell line, which would lead to this result, is which of the following? (A) Splicing defect (B) Cap formation altered (C) RNA editing defect (D) Inefficient poly-A tail addition (E) Promoter alteration

*The answer is C.* The 5′ cap of mRNA is recognized by initiation factors (specifically eIF4E) to allow ribosome assembly on the mRNA. The absence of a cap would not allow a translation initiation complex to form (Factor eIF4e is required for the mRNA binding to the small ribosomal subunit through cap recognition). Introns are not found in mature mRNA (they are removed by splicing in the nucleus); thus, intron-exon secondary structure would not be present. Pseudouridine is found only in tRNA, not in mature mRNA. Thymine, while found in tRNA by posttranscriptional processing, is not found in mRNA (uracil is placed into the mRNA when an adenine is in the template strand), and the poly-A tail, at the 3′ end of the mRNA, adds stability to the mRNA, but it does not play a role in translation initiation.

A researcher has discovered a temperature-sensitive cell line that displays an overall reduction in protein synthesis. Analysis of the mRNA produced at the nonpermissive temperature indicated that a key structural feature, normally present on mRNA, was missing. Such a structure is most likely which one of the following? (A) Intron-exon secondary structure (B) Pseudouridine (C) The 5′ cap (D) Thymine (E) The poly-A tail

*The answer is C.* During the process of cell division, DNA replication occurs secondary to the coordinated effects of multiple enzymes and proteins. DNA polymerases are the primary enzymes responsible for DNA replication, but they can not function without the assistance of other enzymes such as primase, helicase, ligase. and topoisomerese I and II. In E. coli, there are three primary types of DNA polymerases: DNA polymerase I. II. and Ill. Primase forms the 3' OH group primer to initiate replication of daughter strands. while helicase promotes unwinding and dissociation or the parent strands. On the other hand, topoisomerases reduce positive and negative supercoiling in order to relieve the strain produced by DNA unwinding. DNA replication requires a high degree of fidelity, therefore, as the synthesis of the daughter strands proceeds. DNA polymerases proofread to ensure that the daughter DNA is the exact complement of the parent DNA. All three prokaryotic DNA polymerase: have proofreading activity and remove mismatched nucleotides via a 3' to 5' exonuclease activity

Some proteins that participate in bacterial DNA synthesis have specify exonuclease activity. Which of the flowing is the best statement about the 3' to 5' exonuclease activity of DNA polymerase Ill? (A) It cuts DNA at specific DNA sequences (B) It nicks the DNA strands that have formed thymidine dimers (C) It removes an improper base-pair nucleotide during replication (D) It cleaves DNA strands to relax positive supercoils (E) It can remove groups of nucleotides (up to ten) at a time

*The answer is A*. The poison in poisonous mushrooms is α-amanitin, an inhibitor of eukaryotic RNA polymerases, primarily RNA polymerase II. As the family ate the mushrooms containing the poison, RNA polymerase II stopped functioning, and mRNA was no longer produced. This led to a lack of protein synthesis. There is no direct effect on the synthesis of lipids, carbohydrate, or DNA, other than replacing the required enzymes due to protein turnover. However, the net effect of α-amanitin poisoning would be to stop protein synthesis, which may then lead to a cessation of lipid or DNA synthesis. α-Amanitin has no direct effect on DNA repair.

A family, while on a picnic, picked some wild mushrooms to add to their picnic salad. Shortly thereafter, all the members of the family became ill, with the youngest child showing the most severe symptoms. The family is suffering these effects owing to a primary inability to accomplish which one of the following in their cells and tissues? (A) Synthesize proteins (B) Synthesize lipids (C) Synthesize DNA (D) Synthesize carbohydrates (E) Repair damage in DNA

*The answer is C.* The patient has sickle cell anemia. Sickle cell anemia arises owing to a substitution of valine (GTG) for glutamate (GAG). This is the definition of a missense mutation (one amino acid replaced by another), and since only one amino acid is replaced, it is also a point mutation. A nonsense mutation is a point mutation that converts a codon to a stop codon and premature termination of the growing peptide chain. A silent mutation is the result of a DNA change that does not change the amino acid sequence of the protein. Since this is a single nucleotide substitution, it is not due to an insertion or deletion of genetic material.

A young black man was brought to the emergency room (ER) due to severe pain throughout his body. He had been exercising vigorously when the pain started. He has had such episodes about twice a year for the past 10 years. An analysis of the blood shows a reduced blood cell count (anemia), and odd-looking red blood cells that were no longer concave and looked like an elongated sausage. The type of mutation leading to this disorder is best described as which one of the following? (A) Insertion (B) Deletion (C) Missense (D) Nonsense (E) Silent

*The answer is A.* Telomerese is a reverse transcriptase enzyme (RNA-dependent DNA polymerase) that adds TTAGGG repeats to the 3' end of DNA strands at the terminal end of chromosomes, the telomere region. Telomerase is similar to other reverse transcriptase enzymes in that it synthesizes single-stranded DNA using single-stranded RNA as a template. Telomerase is composed of two main subunits the reverse transcriptase TERT and the RNA template TERC. The TERC template is a "built-in" part of the telomerase enzyme. The TERC RNA template is repeatedly read by the reverse transcriptase to add TTAGGG DNA sequence repeats to the ends of chromosomes (telomeres). Stem cell have very long telomeres and active telomerase, but with every cell division, the length of telomeres progressively shortens. Terminally differentiated adult somatic cells have very short telomeres. Critical shortening in telomere length is thought to be one signal for programmed cell death. On the other hand, cancer cells up-regulate their telomerase activity, preventing cellular death by maintaining the length of their telomeres. Cancer cells are considered immortal because these cells continue to divide without aging or shortening of their telomeres, thus, telomerase is a potential target for the treatment of cancers. Syndromes of premature aging such as Bloom syndrome are associated with shortened telomeres. Stem cells are undifferentiated cells that have the potential to differentiate into other cell types, Embryonic and adult stem cells are two major types of stem cells. While embryonic stem cells are present in the very early stages of embryogenesis and give rise to every cell type in adult humans, adult stem cells are thought to be present in most tissues where they are responsible for replacement of dead cells. For instance, the epidermis is continuously replaced from stem cells present in the basal cell layers. Similarly, bone marrow stem cells replace peripheral red and white blood cells. Stem cells have very long telomeres they are able to proliferate indefinitely in a controlled manner, giving rise to replacement stem cells and daughter cells that differentiate into the desired tissue.

Cells obtained from a 73-year-old male demonstrate a high activity of an enzyme that has reverse transcriptase activity. It adds TTAGGG repeats to the 3'-ends of chromosomes. Which of the following cells were most likely obtained from the patient? (A) Epidermal base calls (B) Pancreatic islet B-calls (C) Neurons (D) Erythrocytes (E) Myocardial ceils

*The answer is E.* DNA polymerases are the primary enzymes responsible for DNA replication. In E. Coli, there are three major DNA polymerases: DNA polymerase I, Il, and III. DNA replication requires a high degree of fidelity in order to preserve the genetic code in daughter cells and prevent potentially lethal mutations. This high fidelity replication is accomplished by the 3' to 5' "proofreading" exonuclease activity of DNA polymerase. (Only DNA polymerase I has 5' to 3' exonuclease activity). This 5' to 3 exonuclease activity of DNA polymerase I functions to remove the RNA primer (3-hydroxyl group), which is used by DNA polymerase III for the initiation of DNA replication. The 5' to 3' exonuclease activity of DNA polymerase I also performs exonuclease excision and repair of damage to parent DNA. *(Choice A)* Before the process of DNA replication begins, the parent strand of DNA unwinds and dissociates secondary to the action of the enzyme helicase. *(Choice C)* The enzymes topoisomerase I and Il release tension caused during DNA strand unwinding by relieving both negative and positive supercoils in eukaryotic cells. The enzyme topoisomerase lI, also known as DNA gyrase in prokaryotic cells, has a slightly different function than the eukaryotic topoisomerase ll and does not have 5 to 3' exonuclease activity *(Choice B)* Once a strand is unwound, DNA polymerases begin to synthesize complementary strands in the 5 to 3' direction, but they require a free 3-hydroxyl group primer (RNA primer). This free 3-hydroxyl group is placed on the strand to be duplicated by an RNA polymerase called primase. *(Choice F)* The Okazaki fragments of the lagging strand are bound together by the enzyme ligase. *Educational Objective:* DNA polymerase I has 5' to 3' exonuclease activity in addition to its 5 to 3 polymerase and 3' to 5' exonuclease activities. This 5' to 3' exonuclease activity is used to remove the RNA primer (which initiates DNA polymerization) and to remove damaged DNA.

DNA exonucleases hydrolytically remove one nucleotide at a time from the end of a DNA chain. Which of the following enzymes has 5' to 3' exonuclease activity? (A) Helicase (B) Primase (C) Gyrase (D) DNA polymerase III (E) DNA polymerase I (F) Ligase

*The answer is C.* Sequences are read from the bottom to the top of the gel. In this region, the sequences of the CF and normal genes are identical for the first eight bases. Positions 12 to 15 of the normal gene and 9 to 12 of the CF gene are identical. Therefore, there is a 3-base deletion in the CF gene corresponding to bases 9 to 11 of the normal gene. Since the deletion is a multiple of three (3 bases), there is no frameshift involved. This DNA pattern is indicative of the δF508 mutation, in which the codon for a phenylalanine residue is deleted from the gene. This is the most common mutation in CF patients, occurring in about 70% of the individuals with CF.

Use the following figure to answer this question. The gene for CF has been isolated and sequenced. The gel pattern for the DNA sequence of the region that differs from the normal gene in the most common form of CF is shown in the figure. What do the results of this gel indicate about the disease-causing mutation in the altered gene? (A) It is due to a single nucleotide change. (B) It is due to an insertion of a small number of bases. (C) It is due to a deletion of a small number of bases. (D) It is due to a cytosine deamination. (E) It is due to a frameshift mutation in the DNA.

*The answer is C.* This is a prime example of recombinant DNA technology to manufacture a very useful treatment for human diseases, by using site-directed mutagenesis. As the amino acid and DNA sequences of mature insulin is known, the engineering of lispro required that a proline codon be converted to a lysine codon, and the adjacent lysine codon converted to a proline codon. A polymorphism refers to the differences in DNA sequences amongst individuals in a population at a particular location within the genome. A polymorphism would not result in the synthesis of lispro insulin. DNA fingerprinting is used to identify unknown DNA samples by comparing the polymorphisms present in the sample DNA to a particular individual's DNA. Other than identical twins, everyone's DNA is different, and can be distinguished by fingerprinting using genetic polymorphisms. Repressor binding to a promoter is part of gene regulation in prokaryotypes, and would not be useful in generating the genetic changes necessary to produce lispro insulin. The PCR can amplify a particular DNA segment between two known segments of DNA, but the technique does not lead to the alteration of amino acid sequence between lispro and normal insulin.

Your diabetic patient is using the short-acting insulin lispro to control his blood glucose levels. Lispro is a synthetic insulin formed by reversing the lysine and proline residues on the C-terminal end of the B-chain. This allows for more rapid absorption of insulin from the injection site. The engineering of this drug is an example of which of the following technologies? (A) Polymorphism (B) DNA fingerprinting (C) Site-directed mutagenesis (D) Repressor binding to a promoter (E) PCR

*The answer is C.* The man has contracted diphtheria, and needs the diphtheria antitoxin and then antibiotics to remove the offending organism, C. diphtheriae. As the patient has not received medical care over the past 20 years, he has also missed his diphtheria vaccine, which should be received every 10 years. Diphtheria toxin blocks eukaryotic protein synthesis by phosphorylating an initiation factor, which inhibits protein synthesis in the cells. The toxin does not directly affect DNA or RNA synthesis, nor does it, as a primary target, reduce ATP production by the mitochondria or allow the plasma membrane to become leaky.

A 38-year-old homeless man who has not received any medical care in the last 20 years presents with 2 days of shortness of breath, chills, fever, drooling, painful swallowing, and a "croupy" cough. A physical examination reveals a bluish discoloration of his skin and a tough, gray membrane adhered to his pharynx. The underlying mechanism through which this disease affects normal cells is which one of the following? (A) DNA synthesis is inhibited in the target cells. (B) RNA synthesis is inhibited in the target cells. (C) The process of protein synthesis is inhibited in the target cells. (D) The plasma membrane becomes leaky in the target cells. (E) ATP generation is reduced in the target cells.

*The answer is D.* The child is suffering from a form of xeroderma pigmentosum, a disorder in which thymine dimers (created by exposure to UV light) cannot be appropriately repaired in DNA. Nucleotide excision repair enzymes recognize bulky distortions in the helix, whereas base excision repair recognizes only specific lesions of a small, single, damaged base. The mechanism whereby thymine dimers are removed from DNA is nucleotide excision repair in which entire nucleotides are removed from the damaged DNA. In base excision repair, only a single base is removed; the sugar phosphate backbone is initially left intact. This disorder is not due to alterations in transcription (synthesizing RNA from DNA), DNA replication, or translation (synthesizing proteins from mRNA). Another example of a disease resulting from a defect in nucleotide excision repair is Cockayne syndrome. Neurological diseases (such as Alzheimer's) may also have a deficiency in nucleotide excision repair.

An 8-month-old child is brought to the pediatrician's office due to excessive sensitivity to the sun. Skin areas exposed to the sun for only a brief period of time were reddened with scaling. Irregular dark spots have also appeared. The pediatrician suspects a genetic disorder in which of the following processes? (A) DNA replication (B) Transcription (C) Base excision repair (D) Nucleotide excision repair (E) Translation

*The answer is C.* RNA molecules that carry out functions without first being translated into proteins are referred to as non-coding RNA. Some important forms of non-coding RNA include small nuclear RNA (smyRNA), ribosomal RNA (mRNA), and transfer RNA (tRNA). Small nuclear RNA molecules are transcribed by RNA polymerase II or III and are typically associated with specific proteins forming small nuclear ribonucleoproteins (snRNPs, or °'snurps"). A collection of snRNPs on pre-mRNA is referred to as a spliceosome. These spliceosomes remove intron sequences from pre-mRNA by cleaving the 5' end of the intros and joining that end to the branch point. The 3' end is subsequently cleaved with the free ends of the remaining exon mRNA and ligated with a phosphodiester linkage. Anti~snRNP antibodies are present in mixed connective tissue disease.

A 23-year-old Caucasian female with history of joint pains develops a facial rash. Her serum tests are positive for antibodies against small nuclear ribonucleoprotein particles (snRNP). These snRNP particles participate in the function of which of the following? (A) Peroxisomes (B) Proteasomes (C) Spliceosomes (D) Nucleosomes (E) Ribosomes

*The answer is B.* RNA polymerase is looking for substrates that contain a 2′-hydroxyl group (recall, DNA polymerase utilized dNTPs, which normally lack a hydroxyl group at the 2′ position). As this substrate lacks a 2′-hydroxyl group, therefore the binding affinity of this drug for RNA polymerase is very low, such that the likelihood that this chain terminator will be incorporated into a growing RNA chain is minimal. DNA polymerase, however, utilizes substrates lacking a 2′-hydroxyl group, and can bind and utilize this substrate.

Dideoxynucleotides are effective agents against DNA synthesis, but appear to have little, or no, effect on RNA synthesis. This is most likely due to which of the following? (A) Lack of a 3′-OH group (B) Lack of a 2′-OH group (C) Presence of a 5′-phosphodiester bond (D) Presence of an N-glycosidic bond at carbon 1 (E) Factor TFIID does not recognize deoxyribonucleotides

*The answer is A.* Methotrexate resistance most often occurs due to amplification of the gene for DHFR, the target for methotrexate treatment. Through overproduction of DHFR, there is sufficient enzyme available to overcome the effects of the drug given to the patient. Resistance does not come about by altering the rate of entry of the drug into the cell, by inactivating DHFR, or by inducing an enzyme that can degrade methotrexate.

A patient has been prescribed methotrexate for chemotherapy. After an initial success in reducing tumor growth, the tumor resumes its rapid growth. One potential mechanism for this is which of the following? (A) Amplification of the dihydrofolate reductase (DHFR) gene (B) Amplifi cation of the gene for degrading metho trexate (C) Reduced transcription of the gene allowing methotrexate entry into the cell (D) Increased transcription of the gene allowing methotrexate efflux from the cell (E) An inactivating mutation in the gene for DHFR

*The answer is D.* A mutation is defined as any change in the DNA sequence. When such changes occur within exons or important non coding regions, altered protein quantity or function may result. Deletion of a single nucleotide causes a frameshift mutation, as does deletion or insertion of any number of nucleotides that are not multiples of three. With a frameshift mutation, a shift in the reading frame by one or two base pairs causes the production of an entirely different protein (often with a premature stop codon). in the above vignette, a nucleotide deletion alters lactose metabolism by decreasing the formation of a required enzyme.

A 34-year-old woman with a history of chronic urinary tract infections comes to the office complaining of dysuria. A urine sample is obtained and sent for culture. Gram-negative bacteria isolated from the urine are found to form pink colonies on lactose-containing media. Several days later, bacterial isolates from a new urine sample do not ferment lactose due to the deletion of a single nucleotide from the DNA sequence. This genomic change is most consistent with which of the following: (A) Conservative mutation (B) Missense mutation (C) Nonsense mutation (D) Silent mutation (D) Frameshift mutation

*The answer is A.* DNA replication begins at multiple sites within eukaryotic chromosomes called origins of replication. At these sites, the parent DNA double helix is separated and unwound in a process facilitated by the helicase enzyme and single-stranded DNA-binding proteins. The locations where unwound DNA meets the non-separated double helix are known as replication forks. Replication forks travel bidirectionally away from the origin of replication as DNA polymerase synthesizes complementary daughter DNA strands. Synthesis of the daughter strands occurs simultaneously from both parent strands. Because DNA synthesis can occur only in the 5' to 3' direction, one daughter strand is synthesized continuously toward the replication fork (leading strand). However, the other strand must be synthesized discontinuously in a direction away from the replication fork (lagging strand), with more and more segments being added as the replication fork moves across the DNA double helix. This results in the formation of Okazaki fragments, short stretches of newly synthesized DNA that are separated by RNA primers. These primers are removed and replaced with DNA, and the Okazaki fragments are subsequently joined together by DNA ligase.

A 4-year-old is brought to the physician for fatigue and persistent bone pain. Physical examination shows diffuse lymphadenopathy and multiple purpora over his arms and legs. Laboratory analysis reveals anemia and thrombocytopenia, and a peripheral blood smear shows lymphoblasts. After further workshop, he is diagnosed with acute lymphoblastic leukemia and started on a chemotherapy regiment that includes doxorubicin. This agent intercalates between DNA base pair and inhibits DNA replication, a process that normally occurs at sites known as replication forks. As the replication forks move across the DNA molecule, 2 distinct daughter strands are formed. Which of the following is unique to the daughter strand that is synthesized in the opposite direction of the growing replication fork? (A) Synthesis of multiple short DNA fragments (B) 5' to 3' exonuclease activity of DNA polymerase (C) 3' to 5' exonuclease activity of DNA polymerase (D) 3' to 5' polymerase activity of DNA polymerase (E) RNA primer synthesis before DNA strand synthesis

*The answer is C.* The patient has an acute version of hepatitis C infection, which primarily affects the liver and its function. The two-drug treatment for hepatitis C is ribavirin and modified interferon (it is modified so it is more stable). The interferon works by activating a kinase (protein kinase R) that phosphorylates a key initiation factor for protein synthesis, thereby inhibiting the factor from participating in protein synthesis. This leads to a reduction in protein synthesis, and reduced replication of the virus infecting the cells. Interferon does not inhibit DNA repair, enhance the elongation phase of protein synthesis, or affect ribosome formation.

A 47-year-old woman, who has been on kidney dialysis for the past 7 years, has developed jaundice, fatigue, nausea, a lowgrade fever, and abdominal pain. A physical examination indicates a larger-than-normal liver, and blood work demonstrates elevated levels of aspartate aminotransferase (AST) and alanine aminotransferase (ALT). The physician places the patient on two drugs, one of which is a nucleoside analog, geared to inhibit DNA and RNA syntheses. The primary function of the other drug is to do which one of the following? (A) Inhibit DNA repair in infected cells (B) Enhance the rate of the elongation phase of protein synthesis (C) Reduce the rate of initiation of protein synthesis (D) Inhibit ribosome formation (E) Promote ribosome formation

*The answer is B.* The drug given to prevent Neisseria infection (used prophylaxically), which is common in crowded conditions such a freshman dormitory rooms or military barracks, is rifampin. Rifampin inhibits RNA polymerase, and also exhibits a red color. Loss of rifampin in the urine or tears would give a reddish tint to those fluids. Rifampin does not interfere with DNA synthesis, the bacterial membrane, the process of protein synthesis, or ATP generation by the bacteria.

An 18-year-old college freshman shares a dorm room with three roommates. One of his roommates has been diagnosed with meningococcal meningitis, caused by the bacteria Neisseria meningitidis. The other three roommates are isolated, and treated twice a day with an antibiotic as prophylaxis against this organism, as none of them had received the meningococcal vaccine prior to enrollment. They are told that this antibiotic can give a reddish discoloration of their urine or tears. The reason this drug is effective in killing the bacteria is which one of the following? (A) DNA synthesis is inhibited. (B) RNA synthesis is inhibited. (C) The process of protein synthesis is inhibited. (D) The bacterial membrane becomes leaky. (E) ATP generation is reduced.

*The answer is E.* The lac operon is the sequence of the E. coli genome which is required for the metabolism of lactose. The lac operon consists of a regulatory gene (i) , promoter region (p), operator region (o), and three structural genes (z, y, and a). The z gene codes for β-galactosidase (β-gal), which is primarily responsible for the hydrolysis of lactose to glucose and galactose. The y gene codes for permeable, a transmembrane enzyme that increases the permeability of the cell to lactose. The a gene encodes a β-galactoside transacetylase, which transfers acetyl groups to β galactosides and is unnecessary for lactose metabolism by E. coli. In prokaryotes, one mRNA transcript contains the sequences for many proteins, and a single mRNA molecule can be translated into multiple proteins or polypeptides. For instance, all three proteins of the lac operon (β-galactosidase, permeable, and transacetylase) are synthesized from a single mRNA molecule containing the z, y, and a gene sequences, respectively. Transcription and translation of the genes of the lac operon is typically synchronous. Remember that a single mRNA molecule which codes for more than one protein is referred to as a polycistronic mRNA, and while most prokaryotic mRNA molecules are polycistronic, eukaryotic mRNA is rarely polycistronic.

E. coli colonies grown on a lactose-containing medium up-regulate the production of the enzymes β-galactosidase and galactoside permeate. Which of the following best explains the synchronous production of both enzymes in response to lactose? (A) There are two activator binding sites for one activator protein (B) There are two operators for one repressor protein (C) There are two repressors for one inducer (D) There are two promoters in close proximity to each other (E) There is one mRNA coding for both enzymes

*The answer is E.* DNA polymerase requires a primer in order to synthesize DNA. The primer is provided by a small piece of RNA, synthesized by DNA primase (an RNA-polymerase). RNA synthesis does not require a primer. Once a small piece of RNA is synthesized, DNA polymerase will begin to add deoxyribonucleotides to the end of the RNA. Later, during DNA replication, another DNA polymerase will come along and remove the RNA, replacing the RNA bases with deoxyribonucleotides. However, as initially synthesized, Okazaki fragments will contain uracil. While the deamination of cytosine can produce uracil, this is much more frequent in the more stable DNA than RNA. Thymine cannot be easily converted to uracil in DNA (it would require a demethylation) and does not contribute to uracil content in Okazaki fragments. Mismatch repair does not operate on DNA: RNA hybrids (which form when the primer is synthesized), and DNA polymerase does not recognize uracil, so it would not make the type of mistake in which uracil was placed into DNA.

The isolation of nascent Okazaki fragments during DNA replication led to the surprising discovery of uracil in the fragment. The uracil is present due to which of the following? (A) Deamination of cytosine (B) Chemical modification of thymine (C) An error in DNA polymerase (D) Failure of mismatch repair (E) The need for a primer

*The answer is C.* The child has Bloom's syndrome, a DNA synthesis defect due to a defective DNA helicase. The defective helicase leads to an increased mutation rate in the cells, through an unknown mechanism. Cells derived from patients with Bloom's syndrome display a significant increase in recombination events between homologous chromosomes as compared to normal cells (increased sister chromatid exchange rate). Mutations in the helicase increase genomic instability; the normal Bloom's protein suppresses sister chromatid exchange, and helps to maintain genomic stability. Bloom's syndrome is not due to a mutation in either DNA or RNA polymerase, an exonuclease, or an endonuclease.

A 10-year-old boy, small for his age in both height and weight with a calculated, projected adult height of less than 5 feet, is photophobic, and develops a "butterfly" rash over his nose and cheeks if exposed to the sun. He has a high-pitched voice, large nose, prominent ears, and has had multiple pneumonias in his childhood. An examination of fibroblasts from this patient demonstrated an increased sister chromatid exchange rate during mitosis as compared to cells from a normal child. The defective enzymatic activity in this child can be traced to which one of the following activities? (A) A DNA polymerase (B) An RNA polymerase (C) A helicase (D) An exonuclease (E) An endonuclease

*The answer is D.* Since the patient has an iron deficiency, leading to a microcytic anemia, cells will upregulate their mechanism for acquiring iron, which is through the transferrin receptor. Ferritin is the iron storage protein within cells, and if intracellular iron levels are low, there is no need to upregulate the synthesis of ferritin (its synthesis is actually downregulated under these conditions). The iron travels in the circulation bound to transferrin, so increasing the number of transferrin receptors on the cell surface will enable a more efficient transport of iron and transferrin into the cells. The regulation of transferrin receptor synthesis is at the level of translation, as is the regulation of ferritin synthesis. Thus, cells under these conditions will increase their translation of the transferrin receptor mRNA.

A 16-year-old girl has been losing weight and feeling lethargic over the past 4 months and is taken to the physician by her parents. During the history, the parents expressed concern that their daughter had seemed to eat very little during the day, a claim denied by the patient. Laboratory results indicated an iron deficiency and a microcytic anemia. The cells of the patient have adapted to the iron deficiency in which one of the following ways? (A) Increased transcription of ferritin mRNA (B) Reduced transcription of the transferrin receptor mRNA (C) Increased translation of the ferritin mRNA (D) Increased translation of the transferrin receptor mRNA (E) Increased degradation of the transferrin receptor mRNA

*The answer is D.* Nucleosomes are structural subunits present inside the nucleus composed of nuclear proteins called histones. There are five major subtypes of histories: H1, H2A, H2B, H3, and H4. The nucleosome core is composed of two molecules each of H2A, H2B, H3, end H4, making eight total histone proteins in each nucleosome core. During the initial steps of DNA packaging into chromatin, the DNA double helix wraps around the nucleosome core twice, but in contrast to the other history proteins, H1 histones are not part of the nucleosome. H1 histones participate in DNA packaging by binding the segment of DNA that lies between nucleosomes and as facilitating the packaging of nucleosomes into more compact structures. The association of DNA with histones gives the appearance of a 'beaded chain," as this structure undergoes further rounds of coiling and association with other structural proteins, such as nuclear scaffold proteins, before ultimately forming chromosomes.

A 17-year-old female is being evaluated for amenorrhea and short stature. Karyotype analysis reveals 46 chromosomes that contain DNA material tightly packed with additional proteins. Which of the following proteins outside the nucleosome core facilitates nucleosome packing into a more compact structure? (A) Topoisomerase I (B) snRNP (C) Ubiquitin (D) Histone H1 (E) Histone H3 (F) Histone H4

*The answer is C.* Chloramphenicol blocks peptide bond formation in prokaryotic ribosomes (with no effect on eukaryotic ribosomes). This concept is the basis of certain antibiotic therapy; the differences in ribosome structure between eukaryotic and prokaryotic cells allow selective drug inhibition. The child has meningococcal meningitis, and chloramphenicol, despite its side effects of inhibiting mitochondrial protein synthesis, is a very effective agent for this disorder. Cycloheximide has the same effect as chloramphenicol in eukaryotic cells but has no effect on prokaryotes. Rapamycin leads to the blockage of translation initiation, not peptide bond formation. Puromycin is a chain terminator, stopping protein synthesis but not directly inhibiting peptide bond formation. Rifampin inhibits prokaryotic mRNA synthesis and has no direct effect on translation. Rifampin might be used as prophylaxis for household contacts of meningococcal meningitis but is not as effective a treatment for the actual disease.

A 2-year-old girl exhibits a very high fever of sudden onset and complains of a stiff neck. Physical exam reveals a positive Brudzinski and Kernig sign and petechiae on the extremities. The pediatrician, in addition to rushing the child to the hospital, prescribes a drug that blocks prokaryotic peptide bond formation, even though it can have serious side effects. That drug is which of the following? (A) Rifampin (B) Rapamycin (C) Chloramphenicol (D) Cycloheximide (E) Puromycin

*The answer is B.* Amatoxins are found in a variety of poisonous mushrooms (eg, Amanita phalloides, known as death cap) and are responsible for the majority of mushroom poisoning fatalities worldwide. Ingestion of 1 or more amatoxin-containing mushrooms is a life-threatening emergency. After absorption by the gastrointestinal tract, amatoxins are transported to the liver via the portal circulation where active transport by organic anion transporting polypeptide (OATP) and sodium taurocholate co-transporter (NTCP) concentrates the toxin within the liver cells. There, amatoxins bind to DNA-dependent RNA polymerase type II and halt mRNA synthesis, ultimately resulting in apoptosis. Other organ systems with rapid cellular turnover can also be affected in amatoxin poisoning, including the gastrointestinal tract and proximal convoluted renal tubules. Symptoms typically start 6-24 hours after ingestion and include abdominal pain, vomiting, and severe, cholera-like diarrhea that may contain blood and mucus. Severe poisoning can lead to acute hepatic and renal failure. Urine testing for α-amanitin can confirm suspected ametoxin poisoning.

A 24-year-old man comes to emergency department complaining of abdominal pain, vomiting, and severe watery diarrhea. He recently returned from a camping trip and admits to eating wild mushrooms that he collected in the woods. His past medical history is insignificant and he takes no medications. He does not use illicit drugs. On physical examination, ha is ill-appearing and jaundiced. His liver edge is soft. tender, and palpable 4 cm below the right costal margin. Laboratory tests are significant for elevated levels of alanine aminotransferase, aspartate aminotransferase, and bilirubin. Synthesis of which of the following is most likely to be directly inhibited by the responsible toxin? (A) DNA (B) Messenger RNA (C) Protein (D) Ribososmal RNA (E) Transfer RNA

*The answer is E.* The child has diphtheria, which is caused by a bacterium (Corynebacterium diphtheriae), which produces a toxin that leads to the inhibition of eEF2 (eukaryotic elongation factor 2), which is required for the movement of tRNA from the "A" site to the "P" site. The toxin catalyzes the ADP-ribosylation (using NAD+ as a substrate) of eEF2 to bring about this inhibition. If one treats such a child with nicotinamide (the reaction product resulting from the loss of ADP-ribose from NAD+), one can reverse and block the ADP-ribosylation reaction catalyzed by the toxin. The toxin has no effect on protein kinase A, nor does it glycosylate a G protein. Diphtheria causes sore throat, fever, swollen nodes (bull neck), weakness, hoarseness, painful swallowing, and chills. The hallmark of the disease is a thick, gray membrane covering the pharynx.

A 3-year-old boy, whose parents did not immunize him due to fears of postimmunization side effects, exhibited fever, chills, severe sore throat, lethargy, trouble breathing, and a husky voice. Physical exam indicated greatly enlarged lymph nodes, an increased heart rate, and swelling of the palate. A picture of the boy's throat is shown below. The molecular mechanism responsible for these physical observations in the boy is which of the following? (A) Activation of protein kinase A (B) Activation of an elongation factor for translation (C) Glycosylation of a G protein (D) Inhibition of protein kinase A (E) Inhibition of an elongation factor for translation

*The answer is F.* After transcription, the preliminary, unprocessed mRNA is known as precursor mRNA, or heterogeneous nuclear RNA (hnRNA). Eukaryotic pre-mRNA undergoes significant posttranscriptional processing before leaving the nucleus, including 5'-capping, poly A tail addition, and intron splicing. Once mRNA is finalized, it leaves the nucleus bound to specific packaging proteins. Upon entering the cytoplasm, these mRNA complexes often associate with ribosomes to undergo translation. However, certain mRNA sequences instead associate with proteins that are found in P bodies. P bodies are distinct foci found within eukaryotic cells that are involved in mRNA regulation and turnover. They play a fundamental role in translation repression and mRNA decay, and contain numerous proteins including RNA exonucleases. mRNA decapping enzymes. and constituents involved in mRNA quality control and microRNA-induced mRNA silencing. P bodies also seem to function as a form of mRNA storage, as certain mRNAs are incorporated into P bodies only to be later released and utilized for protein translation.

A 32-year-old man is recovering from extensive burns. Fi-broblasts near the site of injury actively synthesize precursor mRNA to be used as templates for protein synthesis. After transcription, extensive processing of the precursor RNA occurs to form the finalized mRNA sequence. The finalized mRNA then exits the nucleus a undergoes translation by ribosome complexes before being degraded. Which of the following steps involving the processing end handling of mRNA occurs only within the cytoplasm of cells? (A) 5'-terminal guanosine triphosphate addition (B) Methylation of the 5'-terminal guanine (C) Multiple adenine nucleotide attachment to the 3'-end (D) Interaction with snRNP (E) Removal of intervening sequences (F) Interaction with P bodies

*The answer is F.* This question describes a scenario in which uracil is found in association with bacterial DNA during prokaryotic DNA replication. In general, uracil is found only in RNA, so the question essentially asks which enzyme involved in DNA synthesis catalyzes the formation of RNA strands. In prokaryotic DNA replication, primase (an RNA polymerase) is responsible for synthesizing a short RNA primer using the separated strands of DNA at the replication fork as templates. DNA replication then proceeds, with DNA polymerase using the 3' hydroxyl group of the RNA primer as a starting point for synthesis. Primase is a crucial enzyme for bacterial replication as DNA polymerase cannot initiate DNA synthesis without this short nucleic acid sequence primer.

A 34-year old women with a history of recurrent urinary tract infections comes to the physician with dysuria and increased urinary frequency. Her urine culture grows colonies of Gram-negative bacteria. The bacteria is isolated and placed in a growth-enhancing nutrient solution, where they undergo rapid cellular division. As they are actively dividing, the bacterial cells are lysed and their DNA is extracted and purified. Analysis of the partially replicated DNA fragments shows the presence of uracil. This finding is most likely mediated by which of the following enzymes? (A) DNA ligase (B) DNA polymerase I (C) DNA polymerase III (D) Gyrase (E) Helicase (F) Primase

*The answer is A.* Warfarin is metabolized by a specific subset of induced p450 isozymes. The p450 system is used by cells to modify the xenobiotic (in this case the warfarin) such that it can be more easily excreted. Erythromycin, along with other macrolide antibiotics, inhibits the p450 oxidizing system, which in this case would lead to a higher blood level of warfarin and, therefore, the balance of clotting and bleeding is shifted toward excessive bleeding. A stimulation of p450 production by erythromycin would lead to a lower level of warfarin (due to increased metabolism and loss of warfarin by p450) and the potential of excessive clotting. This effect of p450 is a common drug-drug interaction. The causative agents of community-acquired pneumonia do not affect vitamin K absorption in the small intestine, or distribution throughout the body. Erythromycin does not affect mitochondrial transcription, although it may affect mitochondrial translation. Inhibition of mitochondrial protein synthesis, however, will not alter the inhibition of cytochrome p450 activity, and the increased levels of warfarin present, which may lead to increased bleeding.

A 40-year-old male is well controlled on warfarin for a factor V leiden deficiency and recurrent deep vein thrombosis. He presents today with a community-acquired pneumonia, and is placed on erythromycin. Three days later, he develops bleeding and his INR is 8.0 (indicating an increased time for blood clotting to occur, where INR is international normalized ratio). Which of the following best explains why this bleeding occurred? (A) The erythromycin inhibited cytochrome P450 (B) The erythromycin stimulated cytochrome P450 (C) The causative agent of the pneumonia inhibited vitamin K utilization (D) The causative agent of the pneumonia stimulated vitamin K utilization (E) The erythromycin inhibited mitochondrial translation (F) The erythromycin inhibited mitochondrial transcription

*The answer is B.* DNA can be damaged by ultraviolet rays, which leads to the formation of thymine dimers from two adjacent thymine residues. These thymine dimers are repaired by several mechanisms. The most common defect that causes xeroderma pigmentosum is the absence of UV-specific endonuclease. This UV-specific endonuclease recognizes distortions in the structure of DNA caused by thymine dimers, and subsequently excises stretches of single stranded DNA which contain these defects. The gap created following this excision is then filled in by DNA polymerase, which uses the opposite DNA strand as a template. The new strand of DNA is then joined on both ends to the existing DNA by the enzyme ligase. Remember that xeroderma pigmentosum (XP), like most enzymatic disorders, is an autosomal recessive disease. Patients suffering from xeroderma pigmentosum (XP) exhibit photosensitivity, poikiloderma, and hyperpigmentation in sun-exposed areas and also possess a markedly increased risk of developing skin cancers. There are at least eight different forms of XP, and all involve defects in nucleotide excision repair *(Choice A)* 3' to 5' exonuclease activity describes the "proofreading" ability of DNA polymerase. This proofreading ability allows for the recognition and repair of mismatched bases during DNA replication; defective repair of mismatched bases is associated with hereditary nonpolyposis colon cancer. *(Choice C)* Topoisomerase enzymes relieve DNA supercoiling produced during unwinding and separation by helicase. Topoisomerase, also known as gyrase in prokaryotes, is the target of the anticancer drug etoposide and the fluoroquinolone group of antibiotics. *(Choice D)* DNA ligase is responsible for creating a phosphodiester linkage between the phosphate group of the 5' end of a DNA fragment and the hydroxyl group of the 3' end. DNA ligase is particularly active in joining the numerous fragments of DNA that result from discontinuous replication of the lagging strand. *(Choice E)* Helicases are responsible for unwinding and separating the double-stranded DNA into single-stranded DNA in preparation for DNA replication. *Educational Objective:* UV-specific endonuclease deficiency is the most common cause of the autosomal recessive disorder xeroderma pigmentosum.

A 6-year-old Caucasian male experiences recurrent skin lesions on his face and upper extremities that rapidly progress to cancer. You suspect xeroderma pigmentosum as a diagnosis. Which of the following enzymes is most likely nonfunctional if this is the correct diagnosis? (A) 3- 5' exonuclease (B) Endonuclease (C) Topoisomerase (D) DNA ligase (E) Helicase

*The answer is A.* Ototoxicity (hearing loss) occurs with a subset of antibiotics because, in addition to affecting prokaryotic ribosomes, the drugs also have an effect on mitochondrial ribosomes. Mitochondria contain their own DNA, RNA polymerase, and protein synthesizing apparatus (recall that it is thought that during evolution, bacteria invaded eukaryotic cells and formed a symbiotic relationship, with the bacteria eventually becoming the mitochondria), which is very similar to the prokaryotic apparatus. Thus, certain drugs will affect mitochondrial protein synthesis, and the effects seem to be greatest on those organs that have high energy needs (such as neuronal tissue). Erythromycin does not affect mitochondrial RNA synthesis or DNA replication. It does not affect the ear drum, nor does it increase neuronal signaling in the inner ear.

A Russian youngster was prescribed erythromycin for a bacterial infection, but he developed hearing loss due to use of this drug. This occurred due to which of the following? (A) Inhibition of mitochondrial protein synthesis (B) Inhibition of mitochondrial RNA synthesis (C) Inhibition of mitochondrial DNA replication (D) Weakening and tearing of the eardrum (E) Increased neuronal signaling in the inner ear

*The answer is D.* The iron response element binding protein (IRE-BP) binds to loops in RNA (RNA secondary structure), and in the case of ferritin mRNA, the IRE-BP blocks translation when iron levels are low. In the case of the transferrin receptor mRNA, the IRE-BP stabilizes the mRNA when iron levels are low, allowing efficient translation of the mRNA such that transferrin receptors can be synthesized and placed in the membrane. If a patient had a mutated IRE-BP, which could no longer bind to its target sequence in mRNA, then the transferrin receptor mRNA would be unstable and degraded, and cells would not be able to take up iron from the circulation, leading to higher than normal iron levels in the blood yet still would result in cellular depletion of iron. However, the lack of IRE-BP binding would allow ferritin translation and synthesis, leading to high levels of ferritin inside the cell despite the low levels of intracellular iron. Ceruloplasmin is involved in copper transport, not iron. A mutation in transferrin would lead to lower circulating iron levels in the blood as the iron carrier would be mutated. Mutations in the transferrin receptor would not lead to high intracellular levels of ferritin (intracellular iron levels would be low, the IRE-BP would remain bound to the ferritin mRNA, and translation would be blocked). Mutations in transcobalamin would affect vitamin B12 transport, not iron.

A hypothetical patient was suffering from excessive free iron in the blood, yet a cellular analysis indicated low intracellular levels of iron, despite high intracellular levels of ferritin, and normal transferrin levels in the blood. The disease was shown to be caused by a single base change in the DNA that led to a dysfunctional protein. The mutation is likely to be in which of the following proteins? (A) Transferrin (B) Transferrin receptor (C) Transcobalamin (D) Iron response element binding protein (E) Ceruloplasmin

*The answer is B.* Transfer RNA (RNA) is one form of non-coding RNA composed of between 74-93 nucleotides. Specific molecules of tRNA transfer certain amino acid residues to the growing polypeptide chain during translation. The tRNA molecule functions by recognizing the three base codon on the mRNA molecule being translated through its anticodon region, which contains complementary bases. The secondary structure of RNA resembles a cloverleaf and contains the following regions: The acceptor stem is created through the base pairing of the 5'-terminal nucleotides with the 3'-terminal nucleotides. The CCA tail hangs off the 3' end, is with the amino acid bound to the 3' terminal hydroxyl group. tRNA is 'loaded' is with the appropriate amino acid through the process of aminoacylation. which is catalyzed by aminoacyl RNA synthetase. The acceptor stem helps to mediate correct RNA recognition by the proper aminoacyl RNA synthetase. A 3' CCA tail is added to the 3' end of RNA as a posttranscriptional modification in eukaryotes and in most prokaryotes. In some prokaryotic tRNAs, the tail region is directly transcribed from the genome. Several enzymes utilize this tail to help recognize tRNA molecules, and its presence is as necessary for protein translation. The D arm contains numerous dihydrouracil residues, which are modified bases that are often present in tRNA. The D arm (along with the acceptor stem end anticodon arm) facilitates correct RNA recognition by the proper as aminoacyl tRNA synthetase. The anticodon arm contains sequences which are complementary to the as mRNA codon and is read in the 3' to 5' direction. During translation, the ribosome complex selects tile proper tRNA molecule based solely upon its anticodon sequence. The T arm contains the TΨC sequence that is necessary for binding of RNA to ribosomes. The TΨC sequence refers to the presence of thymidine pseudouridine, and cytidine residues in this arm of tRNA. tRNA is the only RNA species that contains the nucleoside thymidine. A 5' terminal phosphate.

A laboratory Is performing experiments to determine the structure and function of several different types of RNA molecules. An RNA molecule is isolated from a culture of gram-positive bacteria that consists of 90 nucleotides. It is found to contain dihydrouracil, thymidine. and acetylcytosine residues. Which of the following is the most likely composition of the 3'-end of this molecule? (A) TATA (B) CCA (C) Poly-A (D) Methylguanosine triphosphate (E) AUG (F) UAG

*The answer is E.* The child has XP, a defect in nucleotide excision repair such that thymine dimers, created by exposure to UV light, cannot be removed from the DNA. XP will not affect the repair of apurinic or apyrimidinic sites (sites missing just the base from DNA, which requires the AP endonuclease for repair). An increase in sister chromatid exchange rates is a finding in Bloom's syndrome, which is a defect in a helicase required for both DNA and RNA syntheses. Patients with Bloom's syndrome are small for their age, unlike those with XP who follow normal developmental milestones. XP does not result in unusual base pairs in DNA, rather the formation of thymine dimers between the adjacent T residues in one strand of DNA. These T residues are still complementary to the A residues in the other strand. XP does not affect the ability of telomerase to extend the ends of the linear chromosomes in the cell.

A newborn has found to be very photophobic, and his skin burns even with minimal exposure to sunlight, eventually forming skin blisters. Neither parent exhibits this trait, although both are prone to burning when in the sun for a short period of time. As the child grows, he is found to be at average height and weight for his age, and is progressing normally along the developmental guidelines. He is, however, kept inside at all times, and is carefully wrapped if he has to leave the house. Fibroblasts isolated from this child are grown in culture, and in an experiment, exposed to UV light. An analysis of the fibroblast DNA will demonstrate which one of the following? (A) A preponderance of apurinic sites and apyrimidinic sites (B) An increase in sister chromatid exchange rate (C) A preponderance of abnormal base pairs in the DNA (D) Loss of telomeres within the DNA (E) An increase in cross-linked bases within the strands of DNA

*The answer is E.* Changes in the genetic code can result in the formation of altered proteins. For instance, the protein formed in the cell culture described in the question stem is a larger, nonfunctional protein. The normal protein is shorter and functional. This is explained by mutations at the splice site (Choice E). After transcription, mRNA contains sequences from both introns and exons; this type of RNA is called pre-RNA or heteronuclear RNA (hnRNA). The pre-RNA must be processed to mature mRNA by posttranscriptional modifications including 5' methylguanosine capping. addition of a 3' polyadenine (Poly A) tail, and splicing. Only exons contain the proper base pairs in the correct order that will result in the formation of an appropriate functional protein, therefore, intrans are excised before translation by a process known as splicing. Mutation of splice sites result in the formation of larger proteins that are usually nonfunctional. but often retain the immunoreactivity of the normal protein (binding to antibodies).

A nonfunctional protein expressed in a cell culture contains 158 amino acid residues rather than the 130 amino acid residues normally seen in the functional protein. The nonfunctional protein is still detected by specific antibodies against the functional protein. Which of the following point mutations best explains the observed finding? (A) Silent mutation in intron 2 (B) Missense mutation in exon 1 (C) Frameshift mutation in codon 3 of axon 1 (D) Nonsense mutation In axon 2 (E) Splice site mutation

*The answer is B.* If an intronic mutation created an alternative splice site, the splicesome would utilize this site for splicing a certain percentage of the time, forming an mRNA that would not code for functional β-globin protein. This would lead to a reduction in β globin synthesis relative to α-globin synthesis, thereby creating a β-thalassemia. If a microRNA were created which targeted the β-globin mRNA, then there would be a drastic reduction in β-globin synthesis as all β-globin mRNA would be targeted for destruction, which is not observed. Since the introns would be normally spliced from the mature mRNA, the creation of a transcription initiation site would have no effect on the mature mRNA. Similarly, the creation of a stop codon in an intron would have no effect on the mature mRNA. The polyadenylation signal is not found in introns, so the mutation could not be at this location within the gene.

A patient displays tiredness and lethargy, and blood work demonstrates an anemia. Western blot analysis indicates significantly greater levels of α-globin than β-globin. Molecular analysis indicates a single nucleotide change in an intron of the β-globin gene. How does such a mutation lead to this clinical finding? (A) A microRNA is produced, which is targeted against the β-globin mRNA, thereby reducing β-globin production (B) Creation of an alternative splice site, such that β-globin levels are decreased (C) Creation of a new transcription initiation site, such that the mRNA for β-globin is now out of frame (D) Creation of a stop codon in the β-globin mRNA (E) Elimination of the polyadenylation signal, thereby reducing β globin production

*The answer is B.* The patient has ingested α-amanitin, a toxin that, at very low concentrations, inhibits RNA polymerase II and blocks the transcription of single-copy genes. RNA polymerase I and III are more resistant to the effects of amanitin, and this toxin has no effect on telomerase or any type of DNA polymerase. The inability to synthesize new proteins in all cells leads to the symptoms observed. Amanitin poisoning initially causes gastrointestinal disturbances, then electrolyte imbalance and fever, followed by liver and kidney dysfunction. Death can follow 2 to 3 days after ingestion.

A patient suffering from chills, vomiting, and cramping was rushed to the emergency department. He had eaten wild mushrooms for dinner that he had picked earlier in the day. His symptoms are due to an inhibition of which of the following enzymes? (A) RNA polymerase I (B) RNA polymerase II (C) RNA polymerase III (D) Telomerase (E) DNA primase

*The answer is D.* Rapamycin inhibits the mammalian target of rapamycin (mTOR), which is a protein kinase. One of the many targets of mTOR is eIF4E binding protein (eIF4E is a required initiation factor for protein synthesis). When not phosphorylated, the binding protein binds tightly to eIF4E and prevents it from participating in the formation of the translational initiation complex, thereby blocking protein synthesis. When phosphorylated at multiple locations by mTOR, the binding protein falls off the initiation factor and allows translational initiation complexes to form. In the presence of rapamycin, mTOR has no kinase activity, and the binding protein remains bound to eIF4E, thereby inhibiting protein synthesis. The drug does not affect ribosome assembly. The drug also has no effect on transcription or RNA processing

A patient underwent a kidney transplant and among the many drugs she received posttransplant was rapamycin. Rapamycin aids in preventing an immune response to the transplant via which of the following mechanisms? (A) The drug inhibits ribosome subunit assembly (B) The drug inhibits cap formation (C) The drug specifi cally inhibits RNA polymerase III (D) The drug inhibits initiation of protein synthesis (E) The drug inhibits antibody-specific transcription factors from binding to DNA

*The answer is C.* RNA interference is an important mechanism by which short (20-30 base pair) non-coding RNA sequences induce posttranscriptional gene silencing. Types of silencing RNA include small interfering RNA (siRNA) and microRNA (miRNA). The human genome encodes over 1000 miRNA genes, each one capable of repressing hundreds of target genes. Altered expression of even a few miRNA genes can lead to cellular dysregulation and has been implicated in the development of many diseases, including hematologic and solid malignancies. In addition, synthetic siRNA sequences can be introduced into cells to silence specific pathogenic genes (eg. c-Myc oncogene) and are being explored as possible therapeutic agents After being transcribed, miRNA undergoes processing in the nucleus to form a double-stranded precursor that is then exported into the cytoplasm. There, the precursor is cleaved into a short RNA helix by a ribonuclease protein called dicer individual strands are then separated and incorporated into RNA-induced silencing complex (RISC). This multiprotein complex uses its associated miRNA as a template to bind to complementary sequences found on target mRNAS. An exact match generally results in mRNA degradation, but a partial match also causes translational repression by preventing ribosome and transcription factor binding

A pharmaceutical corporation is investigating new therapeutic agents for the treatment of Burkitt lymphoma. A double-stranded RNA molecule consisting of 21 base pairs is created that is complementary to a region of mRNA encoding c-Myc. Introduction of this molecule into tumor cells results in a significant reduction in cell growth. Western blot analysis of equivalent numbers of treated and untreated cells is shown. (A) DNA replication (B) DNA transcription (C) mRNA translation (D) Proteasome activity (E) Splicing

*The answer is D. One of the genes that encoded protein X had a mutation at the transcription termination site, which enabled the mRNA to be transcribed into a longer form (237 nucleotides longer). The reading frame of the mRNA was intact, as was the start and stop codons, so the protein produced from this lengthened mRNA was normal. If a nonsense mutation had been created in gene X, a truncated, likely inactive, protein would have been produced. The loss of an intron/exon junction would alter the splicing pattern of the mRNA, and would most likely alter the reading frame of the protein and create a nonfunctional protein. Another possibility is that the loss of an intron/exon junction would produce an elongated protein (due to intronic DNA being transcribed as part of the mRNA), with a concomitant loss of activity. Gaining an alternative splice site would potentially lead to two forms of the final protein being produced, yet only one is seen by Western blot. Inefficient transcription initiation would not produce two distinct mRNAs.

A researcher, while studying a liver cell line, found the following anomalous result. He was studying protein X production within the liver cell. Western blot analysis using a polyclonal antibody showed a normal size, and amount, for protein X. Enzyme assays demonstrated normal levels of activity for protein X. Northern blot analysis, however, yielded two bands of equal intensity: one the expected size and the other 237 nucleotides longer. One possible explanation for this finding is which of the following? (A) A nonsense mutation in the DNA (B) A loss of an intron/exon junction (C) Inefficient transcription initiation (D) Loss of a transcription termination site (E) Gain of an alternative splice site

*The answer is B.* DNA polymerase requires a primer in order to synthesize DNA. The primer is provided by a small piece of RNA, synthesized by DNA primase (an RNA-polymerase). RNA synthesis does not require a primer. Once a small piece of RNA is synthesized, DNA polymerase will begin to add deoxyribonucleotides to the end of the RNA. Later, during DNA replication, another DNA polymerase will come along and remove the RNA, replacing the RNA bases with deoxyribonucleotides. However, as initially synthesized, Okazaki fragments will contain uracil. While the deamination of cytosine can produce uracil, this is much more frequent in the more stable DNA than RNA. Thymine cannot be easily converted to uracil in DNA (it would require a demethylation), and does not contribute to uracil content in Okazaki fragments. Mismatch repair does not operate on DNA:RNA hybrids (which form when the primer is synthesized), and DNA polymerase does not recognize uracil, so it would not make the type of mistake in which uracil were placed into DNA.

A scientist is replicating human DNA in a test tube and has added intact DNA, the replisome complex, and the four deoxyribonucleoside triphosphates. To the surprise of the scientist, there was no DNA synthesized, as determined by the incorporation of radio-labeled precursors into acid-precipitable material. The scientist's failure to synthesize DNA is most likely due to a lack of which of the following in his reaction mixture? (A) Reverse transcriptase (B) Ribonucleoside triphosphates (C) Templates (D) Dideoxynucleoside triphosphates (E) Sigma factor

*The answer is B.* The nucleolus is the site within the nucleus at which rRNA is produced, and ribosomal subunits assembled. In the absence of nucleoli, mature ribosome content within the cell will decrease, which will lead to an overall reduction in protein synthesis. One of the functions of the mature ribosomal complex is to catalyze the formation of peptide bonds, using the enzymatic activity within the large ribosomal subunit rRNA (23S in prokaryotes, and 28S in eukaryotes). In nucleoli, rRNA genes are transcribed to produce the 45S rRNA precursor, which is trimmed, modified, and complexed with proteins to form ribosomal subunits. The synthesis of tRNA does not require the nucleolus, and charging reactions occur in the cytoplasm, so it is unlikely that the levels of charged tRNAs will be reduced. The capping, splicing, and polyadenylation of mRNA does not require the nucleolus (or ribosomes), and would proceed normally at the nonpermissive temperature. The only reason for initiation factors to be reduced is if their turnover number is high, and new proteins need to be synthesized to replenish the eIF pool. The lack of nucleoli, for 96 hours, should not affect the ability of the cell to produce energy in the form of ATP and GTP.

A scientist is studying a novel hepatocyte cell line that cannot produce a nucleolus when the cells are grown at 42°C. When examining cells that have been at 42°C for 96 hours, the scientist finds that the incorporation of 14C-leucine into proteins is greatly reduced as compared to cells grown at 35°C. This is most likely due to which one of the following at the nonpermissive temperature? (A) Lack of charged tRNA molecules (B) Inability to form peptide bonds during protein synthesis (C) Lack of initiation factors (D) Inability to form mature mRNA (E) Lack of GTP needed for protein synthesis

*The answer is A.* Telomerase is defective at the nonpermissive temperature. Owing to the dual requirements of DNA polymerases that they synthesize DNA in the 5′ to 3′ direction, and their need for a template, replicating the ends of linear chromosomes leads to a 3′ overhang after the replication is complete. The overhang is created when the DNA-RNA primer is removed from the template strand, and there is no primer for the DNA polymerase to extend to fill in the gap. Telomerase solves this problem by carrying its own RNA template, and extending the 3′ overhang. After the gap is filled in as best it can, telomere-binding proteins cap the end of the chromosome to protect it from degradation. Lack of DNA ligase would result in a large number of gaps in the phosphodiester backbone (particularly on the lagging strand), but would not affect the telomeres. Lack of DNA polymerase activity would lead to an overall inhibition of DNA replication, and not just affect the telomeres. Lack of helicase activity would also affect the global DNA replication, not just the replication at the telomeres. Lack of repair polymerase would increase the amount of damage in the DNA, but would not specifically target the telomeres.

A temperature-sensitive cell line would show early senescence when grown at the nonpermissive temperature, and examination of the chromosomes demonstrated many 3′ overhangs at the ends of the DNA fragments. The defective enzyme, at the nonpermissive temperature, is which one of the following? (A) Telomerase (B) DNA ligase (C) DNA polymerase (D) A repair DNA polymerase (E) A helicase

*The answer is B.* The quinolone family of antibiotics is targeted toward the bacterial enzyme DNA gyrase, which is the bacterial counterpart of the mammalian topoisomerases. These are the enzymes which break the phosphodiester backbone to allow relief of torsional strain as the DNA helix is unwinding to allow replication to proceed. Through an inhibition of gyrase, DNA replication in the bacteria is inhibited, which leads to bacterial cell death. Since the topoisomerases are not affected by these drugs, there is no effect on eukaryotic DNA synthesis. DNA polymerase α is unique to eukaryotic cells (the bacteria have DNA polymerases I, II, or III). DNA ligase, primase, and DNA helicase are not targets of this class of drugs. The helicase is the enzyme which allows the DNA strands to unwind; however, it needs to work with gyrase (or topoisomerase) such that the tension created by unwinding can be relieved.

A woman visits her physician due to fever and pain upon urination. Urinary analysis shows bacteria, leukocytes, and leukocyte esterase in the urine, and the physician places the woman on a quinolone antibiotic (ciprofloxacin). The mammalian counterpart to the bacterial enzyme inhibited by this drug is which of the following? (A) DNA polymerase α (B) Topoisomerase (C) Ligase (D) Primase (E) Helicase

*The answer is B.* The boy is displaying kwashiorkor due to a calorie deficient diet low in protein. Since the boy is taking in less protein than he needs, he is becoming deficient in the essential amino acids. Thus, to synthesize new proteins, existing proteins need to be degraded such that a pool of essential amino acids is available for the new protein synthesis. This protein degradation occurs in the muscles, leaving the boy with very thin arms and legs. The liver, despite the increased muscle protein turnover, is still deficient in essential amino acids and reduces its level of protein synthesis, including those proteins normally found in the blood, such as serum albumin. The reduced protein content in the blood reduces the osmotic strength of the blood such that when the blood flows through the body, the osmotic strength of the tissues is higher, and fluid leaves the blood and enters the interstitial space around the tissues. This leads to the protruding abdomen seen in these starving individuals. The problem is not related to reduced muscle protein synthesis (which does occur, but does not affect the osmotic strength of the blood), or reduced intestinal protein synthesis, for the same reason as the muscle. Both protein hydrolysis and triglyceride hydrolysis require water; water is not produced when these molecules are broken down.

A young boy has edema, a protruding abdomen, and very thin arms and legs. The edema has at its origins which of the following? (A) Lack of muscle protein synthesis (B) Lack of liver protein synthesis (C) Lack of intestinal protein synthesis (D) Excessive water production due to protein hydrolysis (E) Excessive water production due to triglyceride hydrolysis

*The answer is B.* The child has I-cell disease (mucolipidosis type II), which is a deficiency in protein sorting, particularly of sending lysosomal enzymes to the lysosome (a lysosomal storage disease). The I of I-cell disease stands for inclusion bodies. If the child develops these clinical and radiologic symptoms later in life, one would consider the diagnosis of Hurler syndrome (mucopolysaccharidosis). Lysosomal enzymes are tagged with mannose-6-phosphate (M6P) during posttranslational modification. Enzymes containing M6P then bind to a M6P receptor, which transports the enzymes to the lysosomes. Lacking such a signal, patients with I-cell disease secrete their lysosomal contents into the plasma and interstitial fluids. This leads to lysosomal dysfunction and cellular and tissue destruction. This disease is not peroxisomal, nor does it enhance protein secretion.

A young child exhibits the following symptoms: Coarse facial features, congenital hip dislocation, inguinal hernias, and severe developmental delay. These symptoms are fully evident at the child's age of 1. Cellular analysis demonstrated the presence of inclusion bodies within the cytoplasm of liver cells. The inclusion bodies are the result of which of the following? (A) Enhanced lysosomal enzyme activity (B) Reduced lysosomal enzyme activity (C) Enhanced peroxisomal enzyme activity (D) Reduced peroxisomal enzyme activity (E) Enhanced protein secretion

*The answer is D.* The child is expressing the symptoms of Hutchinson Gilford progeria, a premature aging disease, which is due to a mutation in the LMNA gene, which encodes lamin A, a nuclear protein. The most common mutation is C1824T, in which the normal cytosine at position 1,824 of the gene is replaced by a thymine. This is a silent mutation as far as the protein is concerned-G608G. However, the introduction of the T creates a cryptic splice site in the gene, such that as the hnRNA is processed, a lamin A mRNA is created that is missing 150 nucleotides, corresponding to a loss of 50 amino acids near the carboxy terminal of the protein. Under normal conditions, lamin A is farnesylated, which allows the protein to be attached to the endoplasmic reticulum membrane. During processing, the enzyme AMPSTE24 cleaves part of the carboxy terminal, releasing the farnesylated portion of the protein such that lamin A can be transferred to the nucleus, where it is involved in providing a scaffold for the nuclear membrane. In the mutant protein (progerin), the site of cleavage is lost owing to the loss of the C-terminal amino acids, although the site of farnesylation still remains. Thus, the progerin that reaches the nucleus is still bound to the nuclear membrane, distorting the nuclear membrane and contributing to nuclear instability. Chromatin binding to the nuclear membrane is also altered, as are the phosphorylation sites in progerin, which makes it more difficult for the nuclear membrane to dissolve during mitosis. Since this is a silent mutation in the mature protein, the tertiary structure of the protein is not altered, and a premature stop codon has not been introduced into the protein (that would be a nonsense mutation, not a silent mutation). Since the protein amino acid sequence is initially the same, an alternative start site for transcription has not been created, nor does a simple base change lead to an inhibition of DNA replication.

An 8-year-old boy has failure to thrive, alopecia totalis, localized scleroderma, a small face and jaw, a "beak" nose, wrinkled skin, and stiff joints. He is determined to have a singlepoint mutation in a nuclear protein, which is a silent mutation in terms of the primary structure of the protein. How could such a mutation lead to a disease? (A) Through altering the tertiary structure of the protein (B) Inhibiting DNA replication (C) By introducing a premature stop codon into the protein (D) By creating an alternative splice site in the gene (E) By creating an alternative start site for transcription in the gene

*The answer is C.* Clarithromycin (an antibiotic in the macrolide family with erythromycin and azithromycin) is specific for the large ribosomal subunit of prokaryotes (it will not bind to eukaryotic ribosomes). When this drug binds to the large ribosomal subunit, translocation of the ribosome (movement along the mRNA) is blocked, which blocks overall protein synthesis. tRNA binding is not affected by clarithromycin, nor is there a blockage of the formation of an initiation complex. It is the large subunit (50S) that contains the peptidyl transferase activity, which is also not blocked by this agent.

An adult male is diagnosed with a typical pneumonia. His physician prescribes clarithromycin, which is specific for prokaryotic cells. Which of the following best explains the mechanism of prokaryotic specificity? (A) The drug binds to the 50S ribosomal subunit of bacteria and inhibits f-met-tRNAi binding (B) The drug binds to the 30S ribosomal subunit of bacteria and blocks initiation of protein synthesis (C) The drug binds to the 50S ribosomal subunit of bacteria and blocks translocation (D) The drug binds to the 30S ribosomal subunit of bacteria and blocks peptide bond formation (E) The drug binds to both ribosomal subunits and prevents bacterial ribosome assembly

*The answer is B.* DNA polymerase rarely makes mistakes when inserting bases into a newly synthesized strand and base-pairing with the template strand. However, mistakes do occur at a frequency of about one in a million bases synthesized, but DNA polymerase has an error checking capability which enables it to remove the mispaired base before proceeding with the next base insertion. This is due to the 3′-5′ exonuclease activity of DNA polymerase by which, prior to adding the next nucleotide to the growing DNA chain, the base put into place in the previous step is examined for correct base-pairing properties. If it is incorrect, the enzyme goes "backwards" and removes the incorrect base, then replaces it with the correct base. The 5′-3′ exonuclease activity of DNA polymerase moves ahead, and is used to remove RNA primers from newly synthesized DNA. If the enzyme could no longer synthesize phosphodiester bonds (the primary responsibility of the enzyme), DNA synthesis would halt. A loss of uracil-DNA glycosylase activity is not a property of DNA polymerase, but that of a separate enzyme system which repairs spontaneous deamination of cytosine bases to uracil within DNA strands. If these were left intact, mutations would increase in DNA. A loss of ligase activity would lead to unstable DNA, as the Okazaki fragments would not be able to be sealed together to form one continuous piece of DNA, and this would most likely lead to cell death, not an increased mutation rate.

Analysis of a cell line that rapidly transforms into a tumor cell line demonstrated an increased mutation rate within the cell. Further analysis indicated that there was a mutation in the DNA polymerase enzyme that synthesizes the leading strand. This inactivating mutation is likely to be in which of the following activities of this DNA polymerase? (A) 5′-3′ exonuclease activity (B) 3′-5′ exonuclease activity (C) Phosphodiester bond making capability (D) Uracil-DNA glycosylase activity (E) Ligase activity

*The answer is A.* Amplifying the DNA region between the primers yields a 3.0-kb piece. The DNA from normal alleles will show a 3.0-kb band on an agarose gel after cutting with the appropriate restriction enzyme, as the normal allele does not contain this site within the amplified region. The mutant allele, however, does have this restriction site, such that after amplifying the DNA, and treating with the restriction enzyme, both a 1.1 and 1.9-kb piece will be generated, both of which would be seen on an ethidium bromide-treated gel. Carriers will have one of each allele (normal and mutant), so that a carrier would show three bands on the agarose gel-1.1, 1.9, and 3.0 kb in size. A person with two normal alleles would show only the 3.0-kb fragments, whereas a person with two mutated alleles would show both the 1.1- and 1.9-kb fragments.

Disease X has been linked to the creation of a new restriction enzyme site, as indicated in the associated figure. A rapid PCR test to determine whether the amplified DNA carries the risk for the disease has been developed. After amplifying this region of the genome with the indicated PCR primers, and treatment of the amplified DNA with the appropriate restriction enzyme, an individual who is a carrier for this disease would express which of the following bands on an ethidium bromide-treated agarose gel? (A) 1.1 kb (B) 1.9 kb (C) 3.0 kb (D) 1.1 and 1.9 kb (E) 1.1 and 3.0 kb (F) 1.9 and 3.0 kb (G) 1.1, 1.9, and 3.0 kb

*The answer is G. A carrier would have one normal allele (which would generate a 3.0-kb EcoR1 fragment), and one mutated allele (which generates fragments of 1.1 and 1.9 kb). The key to answering this question is that the probe is located within the 1.1-kb fragment that is generated in the mutant allele (it is also within the normal 3.0-kb fragment generated from the normal allele). So when the Southern blot is probed with the given probe, only the 1.1-kb piece of DNA will anneal to the probe and be visible in the mutated allele. Since this section of DNA is also present in the 3.0-kb piece of DNA, a 3.0-kb piece of DNA will also be visible on the Southern blot. If the person had two normal alleles, then only the 3.0 kb piece would be visible-if the person had the disease, then only the 1.1 kb piece would be visible on the Southern blot.

Disease X has been linked to the creation of a new restriction enzyme site, as indicated in the associated figure. The relevant area of the DNA is shown, along with EcoR1 restriction sites, sizes of fragments obtained, and an area to which a probe is available (the small red box). A family has had DNA obtained to determine if they are carriers for this disorder, and their DNA was digested with EcoR1, and a Southern blot was done using the available probe. A carrier for this disorder would display which bands on the Southern blot? (A) 1.1 kb (B) 1.9 kb (C) 3.0 kb (D) 1.1 and 1.9 kb (E) 1.1 and 3.0 kb (F) 1.9 and 3.0 kb (G) 1.1, 1.9, and 3.0 kb

*The answer is E.* There are 61 codons that code for amino acids, but only 20 amino acids used in protein synthesis. The genetic code Is thus considered "degenerate" because more than one codon can code for a particular amino acid. For instance. the codons GGU. GGC, GGA, and GGG all correspond to the amino acid glycine. Individual RNA molecules are specific for certain amino acids and recognize the mRNA codons corresponding to that amino acid. Because of the degeneracy of the code, certain RNA molecules can recognize multiple different codons coding for the same amino acid, a phenomenon explained by the "wobble" hypothesis. According to this hypothesis, the 5' base in the RNA anticodon often has a different spatial orientation than the other two bases. The 5' nucleotide of the anticodon may be inosine, nucleotide not found in mRNA. Inosine can form hydrogen bonds with three bases: uracil, adenine, and cytosine. In the case of glycine, for example, one tRNA molecule recognizes 3 codons (GGU, GGC and GGG) because only two base pairs (corresponding to the GG of the codon) form "traditional" bonds. Educational Objective: The genetic code is "degenerate" meaning that there are more codes (61 ) than amino so acids (20). Each tRNA molecule is specific for a given amino acid. Many tRNA anticodons can bind to a few different codes coding for the same amino acid. This is called the wobble" phenomenon.

Experiments have shown that the tRNA molecule with the UCU anticodon can effectively bind to both AGA and AGG mRNA codons. The finding described above is best referred to as: (A) Transition (B) Ambiguity (C) Universality (D) No punctuation (E) Wobble

*The answer is C.* If the rate of degradation of the mRNA is not altered by glucocorticoids, then the increase in mRNA levels should reflect the increase in transcription rate. Because the increase in mRNA level is greater than the increase in transcription rate, the glucocorticoids must also be increasing the mRNA stability (i.e., decreasing the rate of degradation by nucleases). The activity of RNA polymerase II is increased (transcription is increased, so transcription initiation is also increased), and the rate of translation (the binding of ribosomes to mRNA) is increased (the enzyme activity is increased), owing to the increased amount of mRNA available.

Gene transcription rates and mRNA levels were determined for an enzyme that is induced by glucocorticoids. Compared with untreated levels, glucocorticoid treatment caused a 10-fold increase in the gene transcription rate and a 20-fold increase in both mRNA levels and enzyme activity. These data indicate that a primary effect of glucocorticoid treatment is to decrease which one of the following? (A) The activity of RNA polymerase II (B) The rate of mRNA translation (C) The ability of nucleases to act on mRNA (D) The rate of binding of ribosomes to mRNA (E) The rate of transcription initiation by RNA polymerase II

*The answer is B.* The lac operon consists of a regulatory gene (lac I), a promoter region (lac p), an operator region (lac o), and three structural genes (lac Z, lac Y, and lac A). The lac Z gene codes for β-galactosidase, which is responsible for the hydrolysis of lactose to glucose and galactose. The lac Yvgene codes for permeable, which allows lactose to enter the bacterium. The lac p region is the binding site for RNA polymerase during the initiation of transcription. The Lac I repressor protein is the product of the lac I gene and is constitutively expressed. Repressor proteins, when bound to the operator region, prevent binding of RNA polymerase to the promoter region, thus decreasing transcription of the lac Z, lac Y, and lac A genes. Culture of E coli in lactose-containing media causes s conformational change in the repressor protein, preventing its attachment to the operator region and increasing transcription of the lac operon structural genes. Culturing E coli in media containing glucose results in reduced expression of the lac operon, even when the media contains lactose as well. This occurs because the lac operon is positively regulated by the binding of catabolite activator protein (CAP) to a site slightly upstream from the promoter region. This only occurs when cAMP concentrations are high. Since glucose decreases the activity of adenylyl cyclase (reducing intracellular CAMP), the lac operon is repressed in high-glucose conditions. in summary, the lac operon is regulated by 2 distinct mechanisms: 1. Negatively by binding of the repressor protein to the operator locus 2. Positively by cAMP-CAP binding upstream from the promoter region Mutations impairing the binding of the repressor protein to its binding site of the operator region will prevent repression of the genes of the lac operon in the absence of lactose. This results in increased transcription of the genes of the lac operon in lactase-deficient media, although the presence of glucose will prevent maximal transcriptional activity.

Microbiologists are investigating sugar metabolism In wild-type and mutant strains of Escherichia coli. Both strains are found to grow viable colonies on lactose containing media. Each strain is then cultured on a new growth medium containing only glucose. Representative colonies of each strain from the new media undergo Western blot processing using a fluorescently labeled probe specific for β-galactosidase. Wild-type bacterial colonies are found to contain only trace quantities of β-galactosidase. However, the mutant colonies express significant amounts of β-galactosidase. Further analysis reveals that the variant strain contains a mutation that inhibits the binding of a certain protein to its regulatory sequence. in which of the following locations did this mutation most likely occur? (A) Activator protein (CAP) binding site (B) Operator locus (C) Promoter region (D) Activator protein (CAP) gene (E) RNA polymerase astron

*The answer is B.* Cytosine spontaneously deaminates to form uracil while in DNA. This error is repaired by the uracil-DNA glycosylase system, which recognizes this abnormal base in DNA and initiates the process of base excision repair to correct the mistake. Neither thymine nor uracil contains an amino group to deaminate (thus, answers A and E are incorrect). When adenine deaminates, the base hypoxanthine is formed (inosine as part of a nucleoside), and guanine deamination will lead to xanthine production. The deamination of cytosine and conversion to uridine is shown below.

Spontaneous deamination of certain bases in DNA occurs at a constant rate under all conditions. Such deamination can lead to mutations if not repaired. Which deamination indicated below would lead to a mutation in a resulting protein if not repaired? (A) T to U (B) C to U (C) G to A (D) A to G (E) U to C

*The answer is C.* UCA is a codon for serine. Of the answer choices given, only the codon UCU is also a codon for serine, which would result in a silent mutation (serine would be placed in the protein even though the DNA had been mutated from TGA to TGT). Conversion of UCA to UAA will generate a termination codon, and a truncated protein would be produced. The conversion of UCA to CCA would replace the serine with a proline in the amino acid. Conversion of UCA to ACA results in a threonine being placed in the protein in place of serine, and generation of GCA from UCA would result in alanine being incorporated into the protein in place of the serine. Only the change of UCA to UCU results in the exact same amino acid sequence being produced by the protein.

Which one of the following changes in the coding region of an mRNA (caused by a point mutation) would result in translation of a protein identical to the normal protein? (A) UCA → UAA (B) UCA → CCA (C) UCA → UCU (D) UCA → ACA (E) UCA → GCA

*The answer is A.* Constitutive synthesis can occur by either of the two mechanisms. The first is an inability to synthesize lac repressor; the second is to have a mutation in the operator region that renders repressor binding impossible. An inability to synthesize lac repressor can be repaired in trans; if the partial diploid contains a functional lac repressor gene, functional protein will be synthesized from the gene, which can bind to the chromosomal operator region, and regulate lac gene expression. If, however, an oc mutation occurred, the operator region is in cis with the operon and can only regulate regions of DNA that are adjacent to the operator. Thus, if a partial diploid contains a normal operator region on the extrachromosomal region of DNA, that operator region cannot regulate the operon on the chromosomal DNA. Thus, the constitutive mutant that was not rescued in a partial diploid is most likely an oc mutation. If the inducer could no longer bind to the repressor, then no expression would occur, as the repressor would not leave the operator region. In addition, if this were the case, then adding normal repressor to the cell (via the partial diploid) should allow expression of the operon. Similarly, if inducer bound too tightly to the repressor, the introduction of normal repressor should reverse the effects of the mutated repressor. The lac repressor does not contain a transactivation domain.

While studying the lac operon in bacteria, a scientist isolates mutants of Escherichia coli, which always express the genes of the lac operon (constitutive synthesis). The scientist creates partial diploids of the regulatory elements of the lac operon in these mutants of E. coli. In one partial diploid, expression of the lac operon is still constitutive (synthesis of the genes is observed even in the absence of an inducer). A likely explanation for this result is which of the following? (A) There is a mutation in cis with the operon (B) There is a mutation in trans with the operon (C) Inducer can no longer bind to the repressor (D) Inducer binds too tightly to the repressor (E) The transactivation domain of the repressor is mutated

*The answer is D.* The child has the symptoms of Bloom syndrome, a disease in which DNA helicase is defective, and DNA replication is compromised. The DNA helicase is necessary to help stabilize the unwinding of the DNA as the replication fork passes through a stretch of DNA. With reduced helicase activity, genomic instability occurs, with increased risk of mutagenic effects and chromosome damage, including chromosome breaks and translocations. These secondary effects lead to the symptoms observed in the patients. The patients also have a higher than normal risk for various malignancies, due to the increased genomic instability. The mutation is in the BLM gene, which is on chromosome 15. This mutation does not alter, in a direct fashion, DNA polymerase or ligase activity nor RNA polymerase activity. Reverse transcriptase is not a normal component of eukaryotic cells (it is introduced to cells when they are infected by a retrovirus).

You see a 2-year-old child of Ashkenazi Jewish descent who is very small for her age. The patient exhibits a long, narrow face, small lower jaw, and prominent eyes and ears. The child is very sensitive to being outdoors in the sun, often burning easily, with butterfly-shaped patches of redness on her skin. Upon testing, the child is also slightly developmentally delayed. The defective protein in this child is which of the following? (A) DNA polymerase (B) DNA ligase (C) RNA polymerase (D) DNA helicase (E) Reverse transcriptase

*The answer is C.* Ricin, a toxin found in castor oil beans, specifically cleaves an N-glycosidic bond in the 28S rRNA of the large ribosomal subunit (an adenine base is removed, but the phosphodiester backbone remains intact). The sequence of the rRNA that is altered is required for binding elongation factors during protein synthesis. As ribosomes become inactivated by ricin, protein synthesis in cells stops, leading to cell death. Ricin does not inhibit RNA polymerases or aminoacyl tRNA synthetases. Ricin does not initially affect ribosome assembly. The LD50 for ricin is 30 mg/kg body weight.

You see a very sick patient (vomiting and bloody diarrhea, dehydration, and mental status changes) in the emergency department, who, you are told, was an amateur chef trying out a new creation in which he wanted to experiment with the extracts of castor beans. This person's symptoms are all due to which of the following? (A) Inhibition of RNA polymerase I (B) Inhibition of RNA polymerase II (C) Ribosomal inactivation by covalent modifi cation (D) Ribosomal disassembly due to covalent modifi cation (E) Inhibition of amino-acyl tRNA synthetases