Biochemsistry, Histology, Clinical Skills Questions

Nuclear pores A. Tend to be found in areas where the DNA is particularly active B. Rely on diffusion to move molecules between the nucleus and the cytoplasm C. Apparently are not associated with any specialized structures D.Are associated only with ATP-mediated transport E. Do not have any known ability to regulate the movement of material between the nucleus and cytoplasm

A.

The importance of the mordant in hematoxyl in and eosin (H & E) staining is A.It forms a complex with hematoxylin that binds tightly to the tissue being stained B.It forms a complex with eosin that binds preferentially to cytoplasmic elements C.It gives cytoplasmic elements a net positive charge, allowing them to bind to eosin D.It prevents eosin from being washed out by the organic solutions used late in the staining process E.It gives cytoplasmic elements a net negative charge, allowing th em to bind to eosin

A.

A key protein (bcl-2), involved in apoptosis, is found predominantly in which subcellular compartment? A. Mitochondria B. Nucleus C. Golgi apparatus D. Lysosomes E. Plasma membrane

A. BCL-2 is localized to the outer membrane of mitochondria, where it plays an important role in promoting cellular survival and inhibiting the actions of pro-apoptotic proteins.

During prophase of mitosis, the chromosomes are found A. Condensed, and the mitotic spindle is forming B. Aligned at the center of the cell C. Scattered through out the nucleus D. Decondensed in the reformed nuclei E. Separated and moving to opposite poles of the spindle

A. In prophase, chromosomes become visible, the nucleolus disappears, the mitotic spindle forms, and the nuclear envelope disappears.

Identify the structure that controls movement of proteins in and out of the nucleus. (A) Nuclear pore complex (B) Nucleolus (C) Heterochromatin (D) Outer nuclear membrane (E) Euchromatin

A. The NPC selectively controls movements of water-soluble molecules and proteins in and out ofthe nucleus.

A solution containing a mixture of several proteins is processed via gel electrophoresis and transfer to filter paper. Antibodies to a specific protein are then added to the filter. A radioactive protein that combines with the antibody-protein complex is used to determine whether the test is positive. This analysis is called a: A. Northern blot B. Western blot C. Southern blot D. Microarray E. Southwestern blot

B.

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 consist of approximately 247 additional nucleotides. One explanation for this finding is which of the following? A. The presence of nonsense mutation in the DNA B. A mutation which creates an alternative splice site C. Alack of capping of the mRNA D. An extended poly-A tail E. A loss of AUG codon

B.

The dinucleotide CpG in DNA at which: is the site A.Transcription is initiated B.Specific methylation occurs C.Recombination occurs D.The small ribosomal subunit E. DNA transcription terminates

B.

The genes for tryptophan biosynthesis in E.coli are switched off in the presence of high levels of tryptophan because A. Tryptophan interacts with DNA through its ring structure B. Tryptophan binding to the tryptophan repressor leads to its binding to the operator C. Tryptophan binds directly to RNA polymerase, preventing its binding to the operator D. Tryptophan decreases the rate of transcription by RNA polymerase E. Tryptophan binds to the repressor protein, causing it to dissociate from the operator

B.

The purpose of staining a histologic or cytologic slide is to: A.Increase the illumination of the specimen. B.Increase the contrast between the elements of the specimen. C.Produce a uniform light absorbance throughout the specimen. D.Give histologists and pathologists a uniform frame of reference. E.Make studying histology and cytology easier for those starting out.

B.

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 transcription 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

B.

A 23-year-old metal worker presents to your emergency room in a comatose state. Pulse oximetry indicates that his hemoglobin is 100%. When you draw blood for labs, you notice that the venous sample is bright red. You suspect cyanide poisoning. As cyanide blocks the cytochrome oxidase system, you understand that shortly after oxidative phosphorylation stops, glycolysis will stop too. Which of the following organelles is the major site for anaerobic metabolism? A. Mitochondria B. Cytoplasm C. Golgi apparatus D. Nucleolus E. Centrioles

B. Glycolysis occurs in the cytoplasm.

In cytology the nucleus is studied to determine A. The function of the cell B. The health of the cell C. The origin of the cell D. How differentiated an abnormal cell is

B. Not function or differentiated because each cell of a same organism has the same DNA, so the nucleus wouldn't show this. You would check differential expression of protein for this. Nucleus determines the health of a cell (think: pyknosis, karyolysis, karyohexis). State of nucleus can show you whether cell is undergoing apoptosis or not.

The major function of the nucleolus is to A. Synthesize DNA B. Synthesize ribosomal RNA C. Repair DNA D. Transport RNA into the nucleus E. Synthesize mRNA

B. Nucleolus assembles ribosomal units.

DNA is duplicated in the cell cycle during the (A) G2 phase. (B) S phase. (C) Mphase. (D) G, phase. (E) G0 phase

B. The S (synthesis) phase of the cell cycle is the period during which DNA replication and histone synthesis occur, resulting in duplication of the chromosomes. At the end of the S phase, each chromosome consists of two identical chromatids attached to one another at the centromere.

Cholesterol functions in the plasmalemma to (A) increase fluidity of the lipid bilayer. (B) decrease fluidity of the lipid bilayer. (C) facilitate the diffusion of ions through the lipid bilayer. (D) assist in the transport of hormones across the lipid bilayer. (E) bind extracellular matrix molecules.

B. The fluidity of the lipid bilayer is decreased in three ways: (1) by lowering the temperature, (2) by increasing the saturation of the fatty acyl tails of the phospholipid molecules, and (3) by increasing the membrane's cholesterol content.

Which one of the following is an inclusion not bounded by a membrane that is observable only during interphase? (A) Nuclear pore complex (B) Nucleolus (C) Heterochromatin (D) Outer nuclear membrane (E) Euchromatin

B. The nucleolus is an inclusion, not bounded by a membrane, within the nucleus. It is observable during interphase but disappears during mitosis.

A34-year-old man of Italian descent is seen for a yearly physical examination. He has no complains and is in good health. However, he does relay a family history of anemia, and a complete blood count demonstrates a mild anemia; the physician suspects thalassemia minor in the patient. Thalassemia is often due to an alteration in RNA splicing, which is an essential part of mRNA processing in eukaryotes. Which of the following is a correct statement concerning mRNA processing? A. Poly(A) RNA is the initial transcript produced, which is subsequently spliced to mRNA B. The coding region of the gene is found within introns C. The coding region of the genes are found within exons D. All human genes required splicing of introns

C.

Gene transcription rates and mRNA levels were determine 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 in assay is to decrease which of the following? A. The activity of RNA Pol 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 activity of RNA Pol III

C.

Ribosomal proteins and histones synthesized in the cytoplasm have been demonstrated to be incorporated into ribosomes in the nucleolus, or utilized within the nucleus, but without nuclear protein synthesis. Transport of large molecular size materials, such as these proteins through the nuclear pore complex occurs by what process A. Simple diffusion B. Selective transport C. Active transport D. Passive diffusion E. Microvesicular transport

C.

Which of the following organelles divides by fission? (A) Golgi complex (B) RER (C) Peroxisome (D) SER (E) Centriole

C. A peroxisome originates from preexisting peroxisomes. It imports specific cytosolic proteins and then undergoes fission. The other organelle that divides by fission is the mitochondrion

Which of the following consists of globular actin monomers linked into a double helix? (A) Keratin (B) LaminA (C) Microfilament (D) Microtubule (E) Neurofilament

C. Globular actin monomers (G actin} polymerize into a double helix of filamentous actin (F actin}, also called a microfilament, in response to the regulatory influence of a number of actinbinding proteins.

Clumps of nucleoprotein concentrated near the periphery of the nucleus are called (A) nuclear pore complex. (B) nucleolus. (C) heterochromatin. (D) outer nuclear membrane. (E) euchromatin.

C. Heterochromatin is the dark-staining nucleoprotein near the periphery of the nucleus. It is transcriptionally inactive, but may be responsible for proper chromosome segregation during meiosis.

A 3-year-old Caucasian boy has grown along the third percentile for age, despite his parents being tall. He also has required treatment for chronic asthma and his pediatrician notes unusual curvature of his knees and wrists typical of rickets. Testing confirms rickets by x-ray and deficiency of vitamin D. This suggests a diagnosis of cystic fibrosis. The DNA sequence shown below is the sense strand from a coding region known to be a mutational "hot spot" for a gene. It encodes amino acid 21 to 25. Given the genetic and amino acid codes CCC=proline(P) GCC=alanine(A)TTC=phenylalanine(F) TAG=stop codon, which of the following sequence is a frameshift mutation that causes termination of the encoded protein 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-

C. Look for TAG

Which one of the following cytoplasmic organelles divides by fission? A.Centriole B.Golgi complex C.Peroxisome D.Rough endoplasmic reticulum E.Smooth endoplasmic reticulum

C. Peroxisomes come from older peroxisomes via fission.

Large nuclear proteins, such as the DNA and RNA polymerases, end up in the nucleus because they A. Are synthesized on ribosomes found in the nucleus B. Diffuse into the nucleus C. Contain a short peptide sequence which directs them to the nucleus D. Contain a signal sequence which attaches polysomes to the nucleus E. Contain covalently attached lipids

C. Protein that must be imported to the nucleus from the cytoplasm carry nuclear localization signals (NLS) that are bound by importins. A NLS is a sequence of amino acids that acts as a tag.

When the space between the two mitochondrial membranes increases to the point that it occupies more than 1/2 the volume of the organelle, the cristae are no longer distinct and the mitochondrial matrix is greatly reduced in volume, what can be said about the organelle? A. It has been damaged, but may recover B. It has been damaged irreparably C. It is engaging in high level oxidative phosphorylation D. The patient has muscular dystrophy

C. So many cristae, more surface area, more oxidative phosphorylation.

A 6-year-old male with hemolytic anemia is found to have an abnormality due to an inactive erythrocyte enzyme. The defective enzyme contains 156 aa instead of the normal 190 aa. A point mutation in exon 2 of the enzyme gene is identified as the cause for this patient's disease. Which of the following mRNA code changes is most likely in this case? A. UAA → UAG B. UUU → UUA C. CUU → AUU D. UCA → UGA E. UAC → CAC

D.

Skin fibroblasts incubated with radioactive aa synthetize polypeptide chains that assemble to form a triple helix. Which of the following aa is most avidly consumed by the fibroblasts? A.Lysine B.Proline C.Alanine D.Glycine E.Leucine F.Cysteine

D.

The periodic acid-Schiff (PAS) stain is positive with A.Glycogen, neurosecretory granules, and mucus B.Glycogen, muscle fibers, and nuclear chromatin C.Glycogen, mucus, flagella, and cilia D.Glycogen, mucus, and basement membrane material E.Mucous, nuclear chromatin, and basement membrane material

D.

What conditions are necessary for the activation of the lac operon?A.The presence of lactose only B.The presence of both glucose and lactose C.The absence of glucose D.The presence of lactose and absence of glucose E. The absence of glucose and presence of arabinose

D.

The primary advantage of phase contrast microscopy is that it A.Can see through many layers of cells B.Can provide a three dimensional view of a specimen C.Can enhance the staining properties of a specimen D.Can generate enough contrast in a specimen that staining is not necessary E.Is much less expensive than regular brightfield microscopy

D. For phase contrast microscopy, don't need to stain.

Degradation of hydrogen peroxide in the cell cytoplasm formed by the metabolism of fatty acids or amino acids occurs through the use an enzyme, catalase. Catalase is found in large amounts in which cellular organelle? A. Lysosome B. Nucleus C. Mitochondria D. Peroxisome E. Golgi body

D. Hydrogen peroxide degradation is done by peroxisomes.

Which of the following cytoskeletal components is associated with kinesin? (A) Keratin (B) LaminA (C) Micro filament (D) Microtubule (E) Neurofilament

D. Kinesin is a force-generating protein associated with microtubules. It serves as a molecular motor for the transport of organelles and vesicles outward, away from the centrosome.

A male child at puberty is determined to have Klinefelter syndrome. Although the parents have been informed of the clinical significance, they have asked for an explanation of what happened. Identify the item that should be discussed with the parents. (A) Trisomy of chromosome 21 (B) Loss ofa n autosome during mitosis (C) Loss of theY chromosome during meiosis (D) Nondisjunction of the X chromosome (E) Loss of the X chromosome

D. Klinefelter syndrome occurs only in men. This condition results from nondisjunction of the X chromosome during meiosis, resulting in an extra X chromosome in somatic cells. These cells therefore have a normal complement of autosomal chromosomes (22 pairs), and instead of one pair of sex chromosomes (XY), there is an extra X chromosome. These individuals have an XXY genotype, resulting in 47 total chromosomes rather than the normal complement of 46. This syndrome is an example of trisomy of the sex chromosomes. Down syndrome is an example of an autosomal trisomy, specifically trisomy of chromosome number 21. Both syndromes have profound complications.

The nuclear pore complex (A) permits free communication between the nucleus and the cytoplasm. (B) is bridged by a unit membrane. (C) is located only at specific nuclear pore sites. (D) permits passage of proteins via receptor mediated transport. (E) has a luminal ring that faces the cytoplasm.

D. The NPC contains a central aqueous channel that permits passage of small water-soluble molecules. However, movement of proteins in and out of the nucleus is selectively controlled by the NPC via receptor-mediated transport.

A structure that is continuous with RERis the (A) nuclear pore complex. (B) nucleolus. (C) heterochromatin. (D) outer nuclear membrane. (E) euchromatin.

D. The outer nuclear membrane is continuous with the RER.

All cells progress through a cell cycle which can vary between 30 minutes for bacteria to 24 hours for the typical mammalian cell. As part of a research project you want to investigate the molecular mechanisms governing progression through the mammalian cell cycle in tumor cells. You compare the cell cycle of a cell line derived from a colon carcinoma with that of a normal epithelial cell. One of the key covalent modifications which occurs in cells to regulate the cell cycle is A. glycosylation of membrane proteins B. cleavage of localization signals allowing replication inhibitors to leave the nucleus C. methylation of DNA D. phosphorylation on tyrosine residues on cytoplasmic proteins E. phosphorylation of serine residues on downstream target proteins

D. Transitions in the phases of the cell cycle are also dependent on tyrosine phosphorylation.

During anaphase of mitosis, the chromosomes are found A. Condensed, and the mitotic spindle is forming B. Aligned at the center of the cell C. Scattered through out the nucleus D. Decondensed in the reformed nuclei E. Separated and moving to opposite poles of the spindle

E.

Which of the following statements concerning plasma membrane components is TRUE? (A) All G proteins are composed of three subunits. (B) The glycocalyx is usually composed of phospholipids. (C) Ion channel proteins are energy dependent (require adenosine triphosphate). (D) Gated channels are always open. (E) Ankyrin binds to band 3 of the red blood cell plasma membrane.

E. Ankyrin is linked both to band 3 proteins and to spectrin tetramer, thus attaching the spectrin-actin complex to transmembrane proteins of the erythrocyte. There are two types of G proteins: trimeric and monomeric; glycocalyx (the sugar coat on the membrane surface) is composed mostly of polar carbohydrate residues; only carrier proteins can be energy requiring; gated channels are open only transiently.

A medical student goes to the emergency department and is diagnosed with a ruptured bowel, the result of a genetic condition called Ehlers-Danlos type IV syndrome. Which one of the following statements about this patient's condition is true? (A) He has a defect in the synthesis of mRNA encoding type I collagen. (B) He has a defect in the genes encoding type IV collagen. (C) He has defective type II collagen. (D) He has an increased risk of breaking his bones. (E) He has a defect in the translation of mRNA for type III collagen. 72

E. Ehlers-Danlos type IV syndrome is associated with a defect in the synthesis and translation of mRNA for type III reticular collagen

Which one of the following events in collagen synthesis occurs outside of the cell? (A) Synthesis of preprocollagen (B) Hydroxylation of lysine residues (C) Triple-helix formation (D) Carbohydrate addition to procollagen (E) Cleavage of pro collagen by procollagen peptidases

E. In the extracellular space, peptidases cleave off end sequences of pro collagen to yield tropocollagen, which self-assembles to form collagen fibrils.

Which of the following statements about scurvy is true? (A) One of its symptoms is bowlegs. (B) It is caused by excessive glycosylation of tropocollagen. (C) It is caused by a deficiency of vitamin A. (D) It is associated with structurally defective elastic fibers. (E) It is alleviated by eating citrus fruits.

E. Scurvy is caused by a deficiency of vitamin C, a necessary cofactor in the hydroxylation of preprocollagen. Citrus fruits are rich in vitamin C.

The site of transcriptional activity is the (A) nuclear pore complex. (B) nucleolus. (C) heterochromatin. (D) outer nuclear membrane. (E) euchromatin.

E. The pale-staining euchromatin is the transcriptionally active chromatin in the nucleus.

Assume there is a microRNA that participates in regulating the expression of a particular cyclin kinase inhibitor. How might an alteration in this microRNA lead to uncontrolled cell proliferation? (A) Overexpression of the microRNA, so it acts as an oncogene. (B) Reduced expression of the microRNA, so it acts as a tumor suppressor. (C) A total loss of activity of the microRNA, so it cannot bind to its target mRNA. (D) A loss of specificity of the microRNA for its target, so different mRNAs are not targeted. (E) No change in microRNA activity can lead to uncontrolled cell proliferation.

The answer is A. Cyclin kinase inhibitors act as brakes on the cell cycle. If the cyclin kinase inhibitor can be removed from the cell, then the cell cycle could proceed in an uncontrolled fashion. MicroRNAs reduce the amount of protein product formed from the target mRNA. In order to eliminate the production of the cyclin kinase inhibitor, the microRNA would need to be overexpressed, such that all target mRNAs are bound, and translation of the gene product is halted. Reducing the expression of the microRNA would lead to overexpression of the cyclin kinase inhibitor, and more control of the cell cycle. This is also the case if the microRNA lost all activity (overproduction of its target), or lost its specificity (again, overproduction of the target).

Many growth factors, upon binding to their receptor, exhibit downregulation, in which the receptor number on the cell surface decreases. This occurs due to which one of the following processes? (A) Endocytosis (B) Exocytosis (C) Pinocytosis (D) Potocytosis (E) Phagocytosis

The answer is A. Receptor-mediated endocytosis refers to the clustering of receptors over clathrin-coated pits in the inner membrane, and then invagination of the membrane to form an intracellular vesicle that contains the receptor-growth factor complex. Exocytosis is the opposite effect-an intracellular vesicle fuses with the plasma membrane to release its contents into the extracellular space. Pinocytosis refers to endocytosis without the receptors-small particles can be taken into the cell through vesicle formation on the cell surface. Potocytosis refers to receptor-mediated entry into a cell through caveolae, and not through clathrin-coated pits. Phagocytosis refers to the forming of a membrane around a particle, and then the endocytosis of that membrane containing the particle (or bacteria).

The mechanism whereby the illegal substance is entering the muscle cells of the bodybuilder is most likely which one of the following? (A) Simple diffusion (B) Active transport (C) Endocytosis (D) Facilitative diffusion (E) Pinocytosis

The answer is A. The bodybuilder is injecting (most probably) testosterone, a steroid hormone, which aids in building muscle mass. Steroid hormones are lipid-soluble substances, and cross membranes by simple diffusion. The receptor for steroid hormones is present inside the cell (either the cytoplasm or nucleus), and once the steroid hormone enters the cell, it will bind to the receptor in a saturable manner. Once the concentration of the hormone inside and outside the cells is equal, transport will stop. Active transport refers to using energy to concentrate a solute against its concentration gradient, which is not the case for steroid hormone transport across the membrane. Facilitative diffusion requires a membrane-bound carrier (no energy), but as indicated previously, the carrier (receptor) for these hormones is intracellular. Steroid hormones do not enter cells through either endocytosis or pinocytosis. The fact that the bodybuilder never became hypoglycemic after taking the drug suggests that it was not insulin being injected.

A 12-year-old boy is displaying tiredness and lethargy, and is found to have a hypochromic, microcytic anemia. Microscopic examination of the boy's red blood cells demonstrated a spherical shape, rather than concave. The mutation in this child is most often found in a protein located in which part of the red blood cell? (A) The cytoskeleton (B) The nucleus (C) The mitochondria (D) The endoplasmic reticulum (E) The plasma membrane

The answer is A. The boy has hereditary spherocytosis, which is due to a mutation in a red blood cell cytoskeletal protein. The most common mutation is in spectrin, although mutations in ankyrin, band 3, and protein 4.2 can also lead to this phenotype. Owing to the mutation in the cytoskeletal protein, the membrane shape becomes spherical instead of concave. This leads to the removal of the spherical cells by the spleen, leading to both anemia and splenomegaly. Mutations in the proteins in the plasma membrane or the endoplasmic reticulum will not lead to this disorder. Red blood cells do not have a nucleus or mitochondria.

One method to separate proteins is by charge, through a suitable gel-like substance. If normal HbA and sickle cell hemoglobin (HbS) were placed on such a gel, which molecule would migrate more rapidly to the positive pole of the gel? (A) HbA (B) HbS (C) Both would have the same charge, so there is no difference in migration.

The answer is A. The mutation in sickle cell is E6V of the β-globin chain. The sickle hemoglobin molecules have a valine in place of a glutamate in the two β chains within the tetramer. All other amino acids are the same, and so in comparison to normal hemoglobin, the sickle variant has two fewer negative charges. This means that the normal form of hemoglobin (HbA) will migrate more rapidly toward the positive pole of a gel because it contains more negative charges than does HbS.

You are studying cell migration during embryonic development. Neural tubes are harvested from postimplantation mouse embryos and placed in culture on plastic dishes coated with fibronectin. Time-lapse imaging reveals neural crest cells migrating away from the explanted tissue. The cells are observed to undergo continuous changes in cell shape, including the formation and retraction of lamellipodia. What protein is the principal mediator of membrane ruffling and locomotion in these cultured cells? (A) Actin (B) Desmin (C) Lamin (D) Tubulin (E) Vimentin

The answer is A: Actin. Motility is a remarkable property of cells that is essential for embryonic development, wound healing, and lymphocyte trafficking. Cell locomotion involves the coordinated assembly and disassembly of actin microfilaments. Actin filaments are helical structures, with a growing end that adds globular (G-actin) to filamentous F-actin. Assembly of microfilaments can generate membrane protrusions, such as filopodia and lamellipodia. Changes in the shape of lamellipodia over time are referred to as "membrane ruffling." During cell locomotion, the leading edge of the plasma membrane displays cell-substrate adhesion proteins that bind glycoproteins in the extracellular matrix. Desmin and vimentin (choices B and E) are intermediate filament proteins found in mesenchymal cells. Lamins (choice C) are nuclear matrix proteins that stabilize the nuclear membrane and organize chromatin. Tubulins (choice D) form the spindle apparatus, regulate intracellular transport, and control the movement of cilia and flagella.

Release of cytochrome c from the organelle described in Question 16 activates which of the following cellular processes? (A) Apoptosis (B) Autophagy (C) Cell division (D) Cell motility (E) Exocytosis

The answer is A: Apoptosis. Apoptosis is a programmed pathway of cell death that is triggered by a variety of extracellular and intracellular signals. It is often a selfdefense mechanism, destroying cells that have been infected with pathogens or those in which genomic alterations have occurred. Mitochondria play a key role in regulating apoptosis. In response to cellular stress, mitochondria open an outer membrane "permeability transition pore" that permits the release of cytochrome c from the inner mitochondrial membrane to the cytoplasm. Within the cytoplasm, cytochrome c triggers an apoptotic cascade that leads to the activation of effector enzymes (caspases) that degrade chromatin and destabilize the cytoskeleton. During development, apoptosis deletes unwanted cells in limb buds to form the digits. None of the other cellular processes are activated by the release of cytochrome c from mitochondria.

You are asked to lead a seminar on intracellular protein trafficking. What organelle provides a microenvironment for the posttranslational modification and sorting of membrane and secretory proteins? (A) Golgi apparatus (B) Lysosome (C) Peroxisome (D) Plasma membrane (E) Smooth endoplasmic reticulum

The answer is A: Golgi apparatus. The Golgi apparatus is an intracellular organelle that regulates posttranslational modification and sorting of membrane and secretory proteins. Like the ER, the Golgi apparatus is composed of flat membrane sacs (vesicles). Newly synthesized proteins leave the ER in small transport vesicles that fuse with the Golgi membrane network. Here, a variety of glycosyltransferase enzymes attach linear and branched oligosaccharide chains to the asparagine residues (N-linked glycans) and serine/threonine residues (O-linked glycans) of membrane and secretory proteins. The ultimate destination of each protein is determined by intrinsic signal peptides and patterns of protein glycosylation. Mature vesicles leave trans-Golgi membranes as secretory vesicles that may be stored in apical cytoplasm or as transport vesicles that deliver proteins/ glycoproteins to various organelles or membrane domains (e.g., apical, basal, or lateral membranes). Lysosomes (choice B) are vesicles filled with acid hydrolases that degrade cellular debris. Peroxisomes (choice C) are small vesicles filled with catalase and other enzymes that remove reactive oxygen species (e.g., hydrogen peroxide). None of the other organelles are involved in the posttranslational modification of membrane and secretory proteins.

An individual is visiting Mexico City, which is at an altitude of 7,350 feet. The person is having trouble breathing due to difficulty in getting sufficient oxygen to the tissues. Which one of the following treatments might the person try to get hemoglobin to release oxygen more readily? (A) Take a drug that initiates a metabolic alkalosis. (B) Take a drug that increases the production of BPG. (C) Hyperventilate, which will lead to decreased levels of carbon dioxide in the blood. (D) Take a drug that induces the synthesis of the γ subunits of hemoglobin. (E) Take a drug that induces the synthesis of the β subunits of hemoglobin.

The answer is B. In order for hemoglobin to release oxygen more readily, the deoxygenated state of hemoglobin needs to be stabilized. This can occur by decreasing the pH (the Bohr effect), increasing the CO2 concentration, or increasing the concentration of BPG. Fetal hemoglobin (HbF = α2γ2) has a greater affinity for O2 than does HbA (α2β2), so inducing the synthesis of the γ genes would have the opposite of the intended effect. Inducing the concentration of the β chains would not decrease oxygen binding to hemoglobin (in fact, if there is not a concurrent increase in α-gene synthesis, this may be quite detrimental to the individual, as an imbalance in the synthesis of the hemoglobin chains leads to a disorder known as thalassemia, and overall oxygen transport to the tissues would be decreased). Increased BPG would cause O2 to be more readily released. A metabolic alkalosis would raise the pH of the blood, which would stabilize the oxygenated form of hemoglobin. Reducing carbon dioxide levels in the blood through hyperventilation will also stabilize the oxygenated form of hemoglobin, and make it more difficult to deliver oxygen to the tissues.

A message transmitted by which example of a chemical messenger would most likely be negatively affected by a mutation that greatly reduced the fluidity of the plasma membrane? (A) Cytokine (B) Steroid hormone (C) A transforming growth factor (D) Insulin (E) Glucagon

The answer is B. Steroid hormones must enter the cell by passive diffusion, and if the membrane is less fluid, it will be more difficult for the hormone to enter the cell to bind to its receptor. Cytokines, transforming growth factors (TGFs), insulin, and glucagon all bind to transmembrane receptors, which transmit their signal to the cytoplasmic portion of the receptor. It is less likely that those conformational signals will be affected by the fluidity of the membrane than the passage of the steroid hormone through the membrane. Decreased membrane fluidity may impair dimerization of the receptors, but the initial binding events should still occur normally.

On a routine newborn exam, it is noted that the red reflex is absent in one eye. An MRI shows a tumor blocking the retina. Regulation at which phase of the cell cycle would be affected by the mutation that leads to this tumor? (A) G0 to G1 (B) G1 to S (C) S to G2 (D) G2 to M (E) M to G1 (F) G1 to G0

The answer is B. The child has hereditary retinoblastoma, which is due to an inherited mutation in the rb gene. As the rb gene is a tumor suppressor gene, once loss of heterozygosity occurs, the function of rb in the cell cycle is lost. Rb helps to regulate the E2F family of transcription factors. Once cyclin D is synthesized, and activates a pair of CDKs, rb protein is phosphorylated, which causes it to leave a complex with the E2F factors. The removal of rb from the protein complex activates the E2F proteins, which initiate new gene transcription to allow the cell to transition to the S phase of the cell cycle. In the absence of any functional rb gene product, the transition to S phase is unregulated, and occurs continuously, leading to tumor growth. The rb gene product is not required for any other checkpoints in the cell cycle.

A family has been using an additional propane heater in their enclosed apartment during the winter months. One morning, a family member is difficult to awake, and when awake, complains of a splitting headache and being very tired. His mucous membranes are also a cherry red color. These symptoms are the result of which one of the following? (A) Increased oxygen delivery to the tissues (B) Decreased oxygen delivery to the tissues (C) Increased blood flow to the brain (D) Decreased blood flow to the brain (E) Decreased oxygen affinity to hemoglobin

The answer is B. The family member is exhibiting the symptoms of carbon monoxide (CO) poisoning. CO will bind to hemoglobin, with a higher affinity than oxygen, and decrease oxygen delivery to the tissues. In addition to competing with oxygen for binding to hemoglobin, CO, once bound to hemoglobin, shifts the oxygen-binding curve to the left, stabilizing the "R" state, or oxygenated state, which makes it more difficult for oxygen to be released from hemoglobin in the tissues. Thus, in the presence of CO, oxygen affinity for hemoglobin is actually increased. CO poisoning does not affect the blood flow to the brain.

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 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 23-year-old male presents to the ER with a fracture of his humerus, sustained in what appeared to be a minor fall. He has a history of multiple fractures after a seemingly minor trauma. He also has "sky blue" sclera and an aortic regurg murmur. His underlying problem is most likely due to a mutation in which one of the following proteins? (A) Fibrillin (B) Type 1 collagen (C) Type IV collagen (D) α1-Antitrypsin (E) β-Myosin heavy chain

The answer is B. The patient is exhibiting the signs of osteogenesis imperfecta, brittle bones, as exemplified by various mutations in type 1 collagen, the building blocks of the bones. The aortic regurgitation murmur is also due to a lack of type 1 collagen in the extracellular matrix of the aorta. Mutations in fibrillin give rise to Marfan syndrome, which does exhibit long bones, but not brittle or easily broken bones. Marfan syndrome would also be associated with lens dislocation, which is not occurring in this patient. Mutations in type IV collagen would lead to Alport syndrome, not brittle bones, and there is no mention of kidney/urine problems with the patient. A defect in α1-antitrypsin would lead to emphysema (not brittle bones), and a mutation in β-myosin heavy chain would lead to hypertrophic cardiomyopathy, not brittle bones.

A 42-year-old man is placed on a twodrug 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 following enzymes? (A) DNA polymerase (B) RNA polymerase (C) Peptidyl transferase (D) Initiation factor 1 of protein synthesis (IF-1) (E) Telomerase

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 55-year-old woman learns that she has high levels of serum cholesterol (greater than 280 mg/dL; normal less than 200 mg/dL) and is at increased risk for development of ischemic heart disease. The patient asks you to explain the normal pathway for serum cholesterol uptake and clearance. You explain to her that low-density lipoprotein (LDL) receptors present in her liver bind LDL cholesterol and internalize it by forming coated vesicles (endosomes). Which of the following structural proteins mediates LDL receptor internalization by organizing small buds of plasma membrane into endosomes? (A) Actin (B) Clathrin (C) Desmin (D) Laminin (E) Vimentin

The answer is B: Clathrin. The LDL receptor is a transmembrane glycoprotein that regulates plasma cholesterol by mediating endocytosis and recycling of apolipoprotein (apo) E. Lacking LDL receptor function, high levels of LDL circulate, are taken up by tissue macrophages, and accumulate to form arterial plaques (atheromas). Receptor-mediated endocytosis is a mechanism for uptake of specific ligands and receptors that is regulated by clathrin. Clathrin stabilizes small invaginations of the plasma membrane, forming coated vesicles (endosomes). Coated vesicles are transported to lysosomes, where ligands and receptors are separated, and receptors are recycled to the plasma membrane. None of the other proteins regulates receptor-mediated endocytosis.

As part of your research, you investigate the role of cyclins and cyclin-dependent kinases in regulating ES cell growth in vitro. These rapidly dividing cells spend most of their time in which phase of the mitotic cell cycle? (A) G0 (B) G1 (C) G2 (D) M (E) S

The answer is B: G1. The cell cycle can be divided into discrete phases that are referred to as G1, S, G2, and M. Cells that have exited the cell cycle are said to reside in G0. Together, G1, S, and G2 constitute interphase. DNA is replicated for cell division during S-phase. Progression of cells through G1 and G2 are regulated by cyclins and cyclin-dependent kinases. These gap phases provide critical checkpoints for cell division. In most rapidly proliferating cells, G1 is the longest and most variable phase of the cell cycle (not choices A, C, D, and E). During G1, cells "evaluate" the integrity of their genome. DNA damage that cannot be repaired typically leads to programmed cell death. M phase is divided into prophase, metaphase, anaphase, and telophase.

You are studying the role of mitochondrial dysfunction in alcoholic liver disease. Genes for an inner mitochondrial membrane protein and a red fluorescent protein are spliced, and the fusion protein is expressed in mouse embryo fibroblasts. The distribution of mitochondria in the transfected cells is visualized by confocal fluorescence microscopy (shown in the image). Inhibition of the electron transport chain in this organelle leads to which of the following reversible changes in cell behavior? (A) Extension of filopodia (B) Hydropic swelling (C) Intracellular lipid storage (D) Membrane ruffling (E) Protooncogene activation

The answer is B: Hydropic swelling. Fusion proteins containing fluorescent protein markers can be used to examine the distribution of organelles in living cells. In this experiment, mitochondria are identified as long, coiled, rope-like structures. These ATP energy-producing organelles are derived from the oocyte at the time of fertilization. They carry their own DNA, synthesize many of their own proteins, and replicate autonomously during interphase. Mitochondria can assume different sizes and shapes, and they often localize to sites within the cell where energy is most needed. When cellular levels of ATP are depleted (e.g., by exposure to toxins or lack of oxygen), cells undergo acute hydropic swelling. This increase in cell volume is caused by an inability of the plasma membrane Na/K ATPase to pump sodium out of the cell. Without adequate levels of ATP to fuel the membrane pump, sodium and water are retained within the cell, and the cell swells. Inhibition of the mitochondrial electron transport chain over an extended period of time will lead to cellular atrophy. ATP depletion does not lead to the other listed changes in cell morphology or behavior.

Motor neurons are labeled by immunocytochemistry using antibodies directed against a neuron-specific protein that helps maintain the shape of dendrites and axons. This structural protein forms which of the following intracellular organelles? (A) Endoplasmic reticulum (B) Intermediate filaments (C) Microfilaments (D) Microtubules (E) Plasma membrane

The answer is B: Intermediate filaments. The cytoskeleton is an intracellular network of filamentous proteins that provides structural support, transports organelles, regulates cell motility, and controls cell division. It includes microtubules composed of tubulin, microfilaments composed of actin, and intermediate filaments composed of tissue-specific fibrous proteins. Unlike microtubules and microfilaments, intermediate filaments are nonpolar structures composed of protein building blocks that vary from one tissue to another. Intermediate filament protein families include keratins, lamins, vimentins, desmins, and neurofilament proteins. Keratins protect the external surface of the skin. Lamins stabilize the inner nuclear membrane, organize chromatin, and regulate gene expression. Neurons express neurofilament proteins that provide flexible, structural support to help maintain complex patterns of axons and dendrites within the central and peripheral nervous system. Microtubules and microfilaments are present in nerve axons and dendrites and contribute to cell structure, but they are not composed of neuron-specific fibrous proteins. None of the other organelles provide structural support to neurons or glial cells.

As part of your research, you examine integral membrane proteins in cleavage-stage mouse embryos using fluorescence microscopy (shown in the image). A pulse of high-intensity UV light is directed at a small patch on the surface of one blastomere, thereby causing an immediate loss of fluorescence emission (photobleaching). Over the next 10 minutes, fluorescence emission from this patch of membrane recovers. Which of the following cellular properties/processes best explains these experimental findings? (A) Lipid raft assembly (B) Membrane fluidity (C) Patching and capping (D) Protein trafficking (E) Receptor-mediated endocytosis

The answer is B: Membrane fluidity. The plasma membrane separates the cytoplasm and intracellular organelles from the external environment. Loss of plasma membrane integrity results in cell death (necrosis). The plasma membrane is a fluid mosaic of lipids and proteins. Integral proteins pass through the lipid bilayer, whereas peripheral proteins do not. Membrane proteins are essential for cell viability and differentiated cell functions. For examples, membrane proteins serve as pumps, enzymes, channels, receptors, structural molecules, and attachment sites. Oligosaccharides and polysaccharides conjugated to membrane proteins and sphingolipids form a cell surface coat (glycocalyx). In polarized epithelial cells, the plasma membrane exhibits distinct apical, basal, and lateral domains. Fluorescence recovery after photobleaching (FRAP) is an experimental technique that can be used to measure the rate at which lipids and proteins move by lateral diffusion within the plane of the membrane. The viscosity of the plasma membrane has been compared to that of thick molasses. Tight junctions provide a barrier to the lateral diffusion of membrane proteins and lipids. In some cells, the plasma membrane forms microdomains (lipid rafts, choice A) that regulate cell signaling. Patching and capping (choice C) describe the clustering of cell surface molecules by specific cross-linking agents, such as antibodies or pollen. Protein trafficking and endocytosis (choices D and E) do not regulate the lateral diffusion of lipids and proteins in the plasma membrane.

Which one of the following types of bonds is covalent? (A) Hydrophobic (B) Hydrogen (C) Disulfide (D) Electrostatic (E) Van der Waals

The answer is C. Disulfide bonds are an example of covalent bonds. Hydrophobic interactions occur between hydrophobic groups as they come together in space to reduce their interactions with water, and to allow water to maximize its entropy. Hydrogen bonds are the sharing of a hydrogen atom between two electronegative atoms. While the hydrogen is covalently bound to one of those atoms, it is also attracted to the other electronegative group (which creates the hydrogen bond) via partial charge interactions, in a noncovalent manner. Electrostatic interactions are the attraction of fully charged groups between each other (one negatively charged, such as a carboxylic acid, and one positively charged, such as a primary amine), due to the opposite charges attracting each other. Van der Waals interactions are nonspecific interactions between two atoms as they approach each other up to a certain distance; once they get too close, repulsion will occur between the two atoms.

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 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.

The liver enzyme glucokinase catalyzes the phosphorylation of glucose to glucose 6-phosphate. The value of Km for glucose is about 7 mM. Blood glucose is 5 mM under fasting conditions, and can rise in the liver to 20 mM after a high-carbohydrate meal. Therefore, if a person who is fasting eats a high-carbohydrate meal, the velocity of the glucokinase reaction will change in which one of the following ways? (A) Remain at <50% Vmax (B) Remain above 80% Vmax (C) Increase from <50% Vmax to >50% Vmax (D) Decrease from >50% Vmax to <50% Vmax (E) Remain at Vmax

The answer is C. This problem is best solved using the Michaelis-Menton equation and comparing the velocity (as a function of maximal velocity) under fasting and nonfasting conditions. During fasting, [S] = 5 mM, and the Km is 7 mM; so v = (5 × Vmax)/(7 + 5) = 42% Vmax. In the fed state, [S] = 20 mM, and the Km is 7 mM; so v = (20 × Vmax)/(7 + 20) = 74% Vmax. Glucokinase is more active in the fed than in the fasting state, and the velocity will increase from <50% Vmax to >50% Vmax.

A soft tissue biopsy is examined in the pathology department. Normal adipocytes are examined at high magnification. The clear space that has pushed the cytoplasm and nucleus to the periphery of these cells is best described by which of the following terms? (A) Endosome (B) Granule (C) Inclusion (D) Vacuole (E) Vesicle

The answer is C: Inclusion. Differentiated cells synthesize a wide variety of proteins, lipids, and carbohydrates that are stored, transported, or secreted. Adipocytes synthesize and store large quantities of triglycerides. Lipid droplets in the cytoplasm coalesce to form a large inclusion that pushes the cytoplasm and nucleus to the periphery of the cell (shown in the image). Glycogen, hemosiderin (denatured ferritin), and lipofuscin (crosslinked lipids and proteins) are also stored as cytoplasmic inclusions. Other metabolic products are packaged within membrane-bound organelles, termed vesicles (choice E). The cytoplasm of most cells is filled with innumerable small vesicles. With the help of microtubules and motor proteins, vesicles transport proteins, lipids, and carbohydrates from one organelle to another (e.g., from ER to Golgi or plasma membrane to lysosome). Large membrane-bound organelles are referred to as vacuoles (choice D). Examples of vacuoles include phagolysosomes and autophagic vacuoles. Endosomes (choice A) are vesicles that internalize ligands and cell surface receptors and transport them to lysosomes for degradation or for recycling back to the plasma membrane. Granules (choice B) are secretory vesicles that are commonly stored in apical cytoplasm. During exocytosis, secretory granules fuse with the plasma membrane, releasing their contents to the extracellular space.

Which of the following proteins contributes to the structural matrix that anchors chromatin to the nuclear membrane during interphase of the cell cycle? (A) Desmin (B) Keratin (C) Lamin (D) Perlecan (E) Vimentin

The answer is C: Lamin. A network of intermediate filament proteins is associated with the inner nuclear membrane. This nuclear (fibrous) lamina stabilizes the nuclear membrane, organizes chromatin, and regulates gene expression. It is composed largely of lamin A and lamin C proteins that form intermediate filaments. Lamin receptors bind these filamentous proteins to the nuclear membrane. During cell division, the nuclear lamina and nuclear membrane disintegrate to facilitate chromosome segregation and separation. Lamin gene mutations are associated with a variety of diseases (laminopathies) including Hutchinson-Gilford progeria. Patients with progeria undergo accelerated aging. Perlecan (choice D) is a basement membrane protein. None of the other intermediate filament proteins (choices A, B, and E) anchors chromatin to the nuclear membrane.

An 85-year-old woman with Alzheimer disease dies in her sleep. At autopsy, hepatocytes are noted to contain golden cytoplasmic granules that do not stain with Prussian blue (shown in the image). This "wearand-tear" pigment of aging (lipofuscin) accumulates primarily within which of the following cellular organelles? (A) Endosomes (B) Golgi apparatus (C) Lysosomes (D) Peroxisomes (E) Vacuoles

The answer is C: Lysosomes. Lysosomes are acidic vesicles that degrade proteins, lipids, and carbohydrates. They are filled with a variety of acid hydrolases that degrade macromolecules to their constituent parts (e.g., amino acids and simple sugars). In some situations, lysosomes are unable to degrade cellular debris. Examples include (1) endogenous substrates that are not catabolized because a key enzyme is missing (lysosomal storage diseases), (2) insoluble endogenous pigments (lipofuscin and melanin), and (3) exogenous particulates (silica and carbon). Examination of this patient's liver at autopsy reveals insoluble "wear-and-tear" pigment of aging. These pigments are composed of cross-linked lipids and proteins (peroxidation products) that accumulate over time. Lipofuscin is stored within the lysosomes of longlived cells in the brain, heart, and liver. None of the other organelles store lipofuscin.

You are studying the differentiation of epithelial cells lining the intestinal mucosa and identify a common stem cell for the secretory lineage that gives rise to Paneth cells, enterocytes, and goblet cells. Which of the following terms describes the developmental potential of these gastrointestinal stem cells? (A) Embryonic (B) Metaplastic (C) Multipotent (D) Nullipotent (E) Pluripoten

The answer is C: Multipotent. Development proceeds from clusters of self-renewing stem cells to beautiful networks of highly differentiated cells. How stem cells acquire instructions for differentiation remains a mystery. When these instructions are revealed, stem cell- based therapies may transform medicine, providing a source of replacement cells and tissues for patients with chronic diseases. The zygote and early cleavage stage blastomeres are totipotent cells, meaning that they have the ability to form all embryonic and extraembryonic tissues. The inner cell mass of the blastocyst is composed of pluripotent embryonic stem cells (choices A and E) that give rise to all embryonic cells and tissues. Pluripotent embryonic stem (ES) cells can be isolated from human blastocysts and cultured in vitro. ES cells that lose the ability to undergo differentiation are said to be nullipotent (choice D). Metaplastic cells (choice B) have undergone a change in differentiation from one pathway to another. Examples of metaplasia include squamous metaplasia in the lungs of smokers and glandular metaplasia in the esophagus of patients with acid reflux. The correct answer for this question is multipotent (choice C). Gastrointestinal stem cells that have the ability to differentiate into a limited number of derivatives are best described as multipotent, adult stem cells.

Chromosomal translocations can lead to uncontrolled cell growth due to which one of the following? (A) Interference with mitosis (B) Interference with DNA synthesis (C) Unequal crossing over during mitosis (D) Inappropriate expression of translocated genes (E) Loss of gene expression

The answer is D. For most translocations that lead to uncontrolled cell growth, a gene is inappropriately expressed because it has been moved adjacent to a constitutive promoter (such as the myc gene next to the immunoglobulin promoter in Burkitt lymphoma). The dysregulation of cell proliferation does not occur owing to problems with mitosis or DNA replication, nor with crossing over. In most cases, the problem is an increased or inappropriate expression of the translocated gene, and not a loss of gene expression.

Proton gradients across membranes are essential for the functions of which of the following organelles? Choose the one best answer. (A) Lysosomes (B) Mitochondria (C) Nucleus (D) Lysosomes and mitochondria (E) Lysosomes and nucleus (F) Nucleus and mitochondria

The answer is D. Lysosomes depend on a proton gradient to acidify their intracellular milieu, such that the lysosomal hydrolases will be at their pH optima (around 5.5). Mitochondria require a proton gradient across their inner membrane in order to synthesize ATP via oxidative phosphorylation. The nucleus does not concentrate protons; the intranuclear space has the same pH as the cytoplasm.

A 4-year-old boy has had multiple episodes of pneumonia, steatorrhea, and has fallen off his normal growth curve. A sweat test was positive for chloride ions. The reason this boy is at risk for repeat episodes of pneumonia is which one of the following? (A) Elastase destruction of lung cells (B) Defective α1-antitrypsin activity (C) Excessive water in the lungs (D) Dried mucus accumulation in the lungs (E) Loss of lung cells due to a defect in DNA repair

The answer is D. The boy is exhibiting the symptoms of cystic fibrosis, which is due to a mutation in the CFTR. The CFTR is required for chloride transport across the membrane, is activated by phosphorylation by the cAMP-activated protein kinase, and when activated allows chloride to flow down its electrochemical gradient. A defective CFTR also alters the ion composition of mucus, reducing its ability to absorb water through osmosis, leading to the drying of mucus in various ducts and tissues, including the lung cells. The lung cells normally secrete a thin, watery mucus designed to trap small particles, which are moved through the lung so they can be swallowed or removed by coughing. When water cannot leave the lung cells, the mucus dries out, leading to pulmonary dysfunction due to clogged bronchi.

A 5-year-old boy begins to regress in terms of developmental milestones, particularly neurologically. Shortly thereafter, the child enters a coma, and dies 2 years into the coma. Upon autopsy, the myelin sheath in the brain was found to be abnormal, as it contained a large quantity of very long-chain fatty acids in its phospholipids. The adrenal glands were also abnormal in appearance. The child, at the molecular level, had inherited a mutation that led to an inability to catalyze reactions that occur in which one of the following intracellular organelles? (A) Lysosomes (B) Nucleus (C) Mitochondria (D) Peroxisomes (E) Golgi apparatus (F) Nucleolus

The answer is D. The child has the symptoms of X-linked adrenoleukodystrophy, which is an X-linked disorder with a mutation in the ABCD1 gene. The ABCD1 gene is required for the transport of very long-chain fatty acids into the peroxisome for catabolism. In the absence of this activity, the very long-chain fatty acids accumulate, become incorporated into phospholipids, and alter the structure of myelin, leading to the neurological disorders observed. The lysosomes, nucleus, and Golgi apparatus are not involved in very long-chain fatty acid oxidation. The nucleolus is found in the nucleus and is the site of ribosome formation. Mitochondria oxidize fatty acids, but not when they are very long-chain fatty acids (greater than 20 carbons). In those cases, the initial steps of oxidation occur in the peroxisome, and when the chain length has been reduced, the partially oxidized fatty acid is transferred to the mitochondria to finish the oxidation of the compound.

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 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.

A small muscular artery is examined in the pathology department. Smooth muscle fibers in the tunica media appear red, whereas collagen bundles in the tunica adventitia appear blue (shown in the image). This slide was most likely colored using which of the following histochemical stains? (A) Aldehyde fuchsin (B) Hematoxylin and eosin (C) Luxol fast blue/cresyl violet (D) Masson trichrome (E) Periodic acid-Schiff

The answer is D: Masson trichrome. This slide specimen reveals key histologic features of a muscular artery. Erythrocytes in the vascular lumen and smooth muscle cells in the tunica media appear red. Collagen fibers in the tunica media appear blue. This striking pattern of tissue staining was obtained using Masson trichrome. Trichrome reagents use two or more acid dyes to stain tissues with contrasting colors (e.g., red and blue). Trichrome staining methods are widely used for differentiating smooth muscle fibers from collagen connective tissue. This is helpful, because these tissues look similar in slides that are stained using H&E.

You are involved in a translational research project to develop small-molecule inhibitors of pepsin secretion by chief cells in the stomach mucosa. Chief cells store precursor enzymes within zymogen granules. By electron microscopy, these "protein factory" cells would most likely show an abundance of which of the following intracellular organelles? (A) Centrosomes (B) Endosomes (C) Phagolysosomes (D) Rough endoplasmic reticulum (E) Smooth endoplasmic reticulum

The answer is D: Rough endoplasmic reticulum. As mentioned above, the endoplasmic reticulum (ER) is composed of parallel membrane sheets and sacs that are specialized for protein and lipid biosynthesis. Smooth ER lacks ribosomes, and its surface appears smooth when examined by electron microscopy. Smooth ER is particularly abundant in cells that synthesize lipids. By contrast, cells that are actively synthesizing proteins feature an abundance of rough ER. Rough ER features bound ribosomes, and its surface appears rough when examined by electron microscopy. Signal sequences, recognition particles, docking proteins, and translocator proteins collaborate to guide proteins destined for secretion through the lipid bilayer. Chief cells store precursor proteins (e.g., pepsinogen) in membrane-bound dense-core granules.

A spinal cord smear preparation is obtained at autopsy and stained with Luxol fast blue/cresyl violet. The large octopus-like cells on this slide are multipolar motor neurons (shown in the image). What protein forms intracellular tracts that deliver organelles and vesicles to distant nerve terminals via anterograde axonal transport? (A) Actin (B) Clathrin (C) Lamin (D) Tubulin (E) Ubiquitin

The answer is D: Tubulin. Axons are cellular processes that convey electrochemical signals away from neuronal cell bodies. These elongated structures are largely dependent on the neuronal cell body for the delivery of organelles and vesicles and for the removal of cellular waste (e.g., abnormal proteins). Axonal transport is an intracellular shuttle/delivery system that uses microtubules and motor proteins (e.g., kinesin and dynein) to transport vesicles to and from the synaptic membrane. Microtubules are rigid hollow tubes composed of repeating units of αβ-tubulin dimers. These polymeric structures grow from nucleation sites within centrosomes. Tubulins comprise a family of proteins that regulate diverse cellular activities, including: (1) chromosome separation during mitosis and meiosis, (2) intracellular vesicle transport, and (3) the whip-like movement of cilia and flagella. Clathrin (choice B) forms coated membrane vesicles during receptor-mediated endocytosis. Ubiquitin (choice E) is a protein that tags other proteins for degradation by proteasomes. None of the other proteins mediate axonal (axoplasmic) transport.

A new antibiotic has been developed that shows a strong affinity for attacking amino acids with a specific orientation in space. In order for it to work well in humans, the antibiotic must be effective against amino acids in which one of the following configurations? (A) R-configuration (B) L-configuration (C) Aromatic ring configuration (D) Polypeptide chain configuration (E) D-configuration

The answer is E. Amino acids in humans are in the L-configuration (except glycine which is neither L nor D), whereas bacterial amino acids can be in either the L- or D-configuration. An antibiotic would need to be effective against bacterial proteins and not human proteins, so developing an antibiotic that recognizes proteins or polypeptides that contain D amino acids would only be effective against bacterial products. All amino acids are in polypeptide chains, and phenylalanine, tyrosine, and tryptophan are amino acids that contain aromatic rings, and are present in both bacteria and humans. The R and S nomenclature is not commonly used in biochemistry to describe the configuration of amino acids.

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. 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 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 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 16-year-old male high school student was playing basketball for his school when he collapsed on court and could not be resuscitated. An autopsy demonstrated increased thickness of the intraventricular septum and left ventricular wall. These findings could be explained by a mutation in which one of the following proteins? (A) Spectrin (B) α1-Antitrypsin (C) Collagen (D) Fibrillin (E) β-Myosin heavy chain

The answer is E. The student has died from FHC, a thickening of the left ventricle of the heart muscle due to a mutation in β-myosin heavy chain. The exact reason for the hypertrophy, which can be caused by mutations in a variety of sarcomeric proteins, is still unknown. None of the other proteins suggested as answers are muscle sarcomeric proteins. Spectrin is a red blood cell protein, and is not found in the heart. α1-Antitrypsin is a circulating protein synthesized by the liver, and in its absence, emphysema will develop. Collagen is the major structural protein of the body, but there are no mutations in collagen that lead to a greatly hypertrophied heart muscle. A lack of fibrillin leads to Marfan syndrome, which can present with defects in heart valves and the aorta, but not a heart muscle greatly increased in size.

What intracellular protein complex links microtubules of the spindle apparatus to sister chromatids during mitosis and meiosis? (A) Astral fibers (B) Centrioles (C) Centromere (D) Centrosome (E) Kinetochore

The answer is E: Kinetochore. The spindle apparatus organizes and separates chromosomes during mitosis and meiosis. Microtubules of the spindle apparatus link chromosomes to microtubule organizing centers and mediate the movement of paired chromosomes to opposite poles of the cell during anaphase. Centromeres (choice C) are repetitive DNA sequences that provide a point of attachment between the sister chromatid and a nucleation site for the assembly of the kinetochore protein complex. Kinetochores are attachment sites for microtubules of the spindle apparatus. Each kinetochore binds 15 to 20 microtubules. Bundles of microtubules (spindle fibers) originate from microtubule-organizing centers (centrosomes, choice D). Centrosomes are composed of two centrioles (choice B) and a zone of pericentriolar proteins that regulate microtubule nucleation. Centrosomes are associated with the nuclear membrane during interphase and replicated during S phase of the cell cycle. They move to opposite poles of the cell during mitotic prophase as the nuclear envelope disintegrates. Astral fibers (choice A) are microtubules that anchor centrosomes to the plasma membrane. Dyneins are molecular motor proteins that move chromosomes along the spindle apparatus. Failure of sister chromatids to separate during anaphase is referred to as nondisjunction. The resulting embryos are said to exhibit genetic mosaicism.

Which of the following cellular processes describes the uptake of extracellular fluids and small particles by the cell described in Question 18? (A) Autophagy (B) Exocytosis (C) Involution (D) Phagocytosis (E) Pinocytosis

The answer is E: Pinocytosis. Uptake of fluid and macromolecules at the cell surface is referred to as endocytosis. This energy-dependent cellular activity provides cells with essential fluids, nutrients, and proteins. It also enables specialized cells to internalize large particles (e.g., cellular debris and bacteria) for degradation within phagolysosomes. Endocytosis involves the formation of vesicles at the plasma membrane by a process of vesicle budding. Three general mechanisms of endocytosis are described: (1) pinocytosis (constitutive uptake of fluid and small particles), (2) phagocytosis (uptake of large particles by macrophages and other phagocytic cells; choice D), and (3) receptor-mediated endocytosis (clathrin-dependent uptake of specific ligands). Pinocytotic vesicles can be identified by electron microscopy. They are particularly abundant in the cytoplasm of vascular endothelial cells. Autophagy (choice A) enables cells to degrade and eliminate unwanted or damaged organelles. Exocytosis (choice B) is an energy-dependent process of secretion that involves fusion of secretory vesicles with the plasma membrane.

A sample of adrenal cortex obtained at autopsy is fixed with formalin, embedded in paraffin, sectioned at 6 μm, stained with H&E, and examined by light microscopy. Cells of the zona fasciculata appear washed out and "spongy" due to an accumulation of cholesterol and other precursors for steroid hormone biosynthesis. Electron microscopic examination of these "steroid factory" cells would be expected to show an abundance of which of the following organelles? (A) Autophagic vacuoles (B) Dense-core secretory granules (C) Golgi apparatus (D) Rough endoplasmic reticulum (E) Smooth endoplasmic reticulum

The answer is E: Smooth endoplasmic reticulum. Intracellular membranes establish compartment boundaries and organelles that serve different cellular functions. Examples of membrane-bound intracellular organelles include the nucleus, endoplasmic reticulum, Golgi apparatus, mitochondria, peroxisomes, lysosomes, endosomes, and secretory vesicles. The endoplasmic reticulum (ER) is composed of parallel membrane sheets and sacs that are specialized for protein and lipid biosynthesis. Smooth ER lacks ribosomes, and its surface appears smooth when examined by electron microscopy. Smooth ER is particularly abundant in cells that synthesize lipids (e.g., fatty acids, phospholipids, cholesterol, and steroid hormones). In skeletal and cardiac muscle, smooth ER sequesters calcium and regulates muscle contraction. In the liver, smooth ER provides a large surface area for oxidative enzymes (e.g., cytochromes) that degrade toxins and carcinogens. The other organelles may be present in steroid-producing cells, but they would not be abundant.

A portion of the small intestine is collected at autopsy, and sections are stained with periodic acid-Schiff (PAS) and counterstained with hematoxylin. The mucosa of the intestine is examined by light microscopy (shown in the image). PAS is particularly useful for identifying which of the following biological materials? (A) Collagens (B) Lipids (C) Nucleic acids (D) Proteins (E) Sugars

The answer is E: Sugars. Periodic acid-Schiff (PAS) reagent is a histochemical stain that is useful for identifying carbohydrates (oligosaccharides and polysaccharides). In this section of the small intestine, PAS stains mucusproducing goblet cells. Mucins are heavily glycosylated glycoproteins that protect the intestinal mucosa and lubricate the luminal contents. Hematoxylin is a basic dye that is commonly used to identify cell nuclei (nucleic acids) in paraffin sections. Cellular structures that retain hematoxylin are said to be basophilic. Cellular structures that retain eosin are said to be eosinophilic. PAS does not stain collagens, lipids, nucleic acids, or proteins (choices A to D).

Hematopoietic stem cells are cultured in vitro at 37°C in the presence of recombinant erythropoietin. A photomicrograph of a typical "burst-forming unit" committed to the erythrocyte pathway of differentiation is shown in the image. Which of the following histochemical stains can be used as a "vital dye" to distinguish viable from nonviable cells in your cell culture? (A) Aldehyde fuchsin (B) Hematoxylin and eosin (C) Luxol fast blue/cresyl violet (D) Periodic acid-Schiff (E) Trypan blue

The answer is E: Trypan blue. Trypan blue is a nontoxic (vital) dye that is retained by dead cells but excluded by viable nonphagocytic cells. When trypan blue is added to an aliquot of cells in suspension, the percentage of viable cells in the sample can be determined rapidly using a benchtop hemocytometer. One simply divides the number of viable cells in an aliquot by the total number of cells examined. Hematoxylin and eosin (H&E, choice B) are the most commonly used dyes in histology and histopathology. Aldehyde fuchsin (choice A) can be used to identify elastic fibers and mast cell secretory granules. Luxol fast blue/cresyl violet (choice C) is commonly used to stain nervous tissue. As mentioned above, PAS (choice D) is commonly used to identify carbohydraterich cellular components and secretions (e.g., mucus). Erythropoietin is a kidney hormone that promotes the survival and growth of hematopoietic cells that are committed to the erythrocyte pathway of differentiation.

A cervical biopsy is obtained from a 42-year-old woman with a history of abnormal Pap smears. The tissue is tested for human papillomavirus (HPV) by in situ hybridization using cDNA probes. Evidence of HPV viral genome is detected in cells in the cervical biopsy (dark blue spots, shown in the image). The patient is told that she is at increased risk for the development of cervical cancer. She asks you to elaborate. You explain that HPV encodes an early gene (E6) that activates a cellular protein that, in turn, accelerates the degradation of the p53 tumor suppressor protein. Name the protein that is activated by HPV E6. (A) β-Catenin (B) Cathepsin (C) Glucuronyl transferase (D) GTP-activating protein (E) Ubiquitin ligase

The answer is E: Ubiquitin ligase. Epithelial cells in this cervical biopsy exhibit distinct perinuclear vacuoles (shown in the image). These sharply demarcated, clear zones surround the nuclei of HPV-infected cells. The vacuoles are filled with actively replicating virus particles (virions). The gene products of oncogenic DNA viruses, like HPV, are known to inactivate tumor suppressor proteins. Recent studies indicate that they do so, by accelerating the degradation of p53 via the ubiquitin-proteasome pathway (see Question 24). Loss of p53 permits cells to escape cellular senescence and proliferate. Mutations in GTP-activating protein (choice D) are associated with neurofibromatosis. None of the other proteins accelerates the degradation of p53 in cervical epithelial cells.

Lysosomal hydrolases are targeted to the lysosome by the addition of a carbohydrate residue to the protein. An inability to add this carbohydrate leads to a disease in which the lysosomal hydrolases are treated as secreted proteins, and are exported from the cell, rather than taken to the lysosomes. The secreted proteins will have which one of the following effects on the cells and proteins in the circulation? (A) The blood cells will have their membrane proteins digested. (B) The blood cells will have their carbohydrates on the cell surface removed. (C) The blood cell membranes will become leaky, leading to the death of the blood cells. (D) Circulating proteins will be degraded, whereas the blood cells will be protected against the enzymes. (E) Circulating proteins will be targeted to the spleen for removal. (F) There will be no effect on the proteins and cells in the circulation.

The answer is F. Most lysosomal hydrolases have their highest activity near an acidic pH of 5.5 (pH optimum) and little activity in a neutral or basic environment. The intralysosomal pH is maintained near 5.5 by vesicular ATPases, which actively pump protons into the lysosome. The cytosol and other cellular components have a pH near 7.2, and are therefore protected from escaped hydrolases. The pH of the blood is maintained between 7.2 and 7.4, so the escaped lysosomal enzymes will have no activity at that pH, and will not affect the proteins and cells in the circulation. I-cell disease results from the inability to appropriately target lysosomal proteins, and it is a lysosomal storage disease.

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 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 40-year-old tobacco farmer is seen in the ER with bradycardia, profuse sweating, vomiting, increased salivation, and blurred vision. He was spraying his field with malathion when the hose ruptured and he was covered with the malathion. Which of the following types of inhibition of enzymes does this poisoning represent? (A) Competitive (B) Noncompetitive (C) Irreversible (D) Reversible (E) The drug does not work by inhibiting enzyme activity.

The answer is c. Malathion is an organophosphate that inhibits the action of acetylcholinesterase in an irreversible manner. It is one of the most common causes of poisoning worldwide. Malathion forms an irreversible covalent bond between the inhibitor and the active site serine side chain of the enzyme. Without acetylcholinesterase, acetylcholine accumulates in the neuromuscular junction and causes the symptoms described in the case. Both competitive and noncompetitive inhibition are reversible forms of inhibition, and their mechanism of action does not apply to malathion.

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 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 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 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 28-year-old man presents to the ER with a large amount of blood and protein in his urine. He has had a sensorineural hearing loss since his teen years and has misshaped lenses (anterior lenticonus). The physician is suspicious of a genetic disorder that may lead to eventual kidney failure. If this is the case, the patient most likely has a mutation in which one of the following proteins? (A) Spectrin (B) α1-Antitrypsin (C) Collagen (D) Fibrillin (E) β-Myosin heavy chain

The answer is c. The patient has Alport syndrome, a mutation in type IV collagen that alters the basement membrane composition of kidney glomeruli. In the absence of a functional basement membrane, the kidneys have difficulty in properly filtering waste products from blood into the urine, and both blood and proteins can enter the urine. Type IV collagen is also important for hearing (it is found in the inner ear) and for the eye. Type IV collagen forms a meshlike structure, which is different from the rodlike structures found in type I collagen, and is found in almost all basement membrane structures. Given sufficient time, the alteration in the basement membrane in the glomeruli will lead to their destruction, and loss of kidney function. A mutation in α1-antitrypsin will lead to emphysema, mutations in spectrin can lead to hereditary spherocytosis, mutations in fibrillin lead to Marfan syndrome, and mutations in β-myosin heavy chain can lead to FHC.

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 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 53-year-old man, who has been smoking for the past 35 years at a two-pack-a-day rate, visits his physician for a cough that will not go away, and for difficulty in breathing. A chest X-ray rules out cancer, but does display an increased anterior-posterior (AP) diameter, flattened diaphragm, and "air trapping." The patient is told that his condition will not improve, and that he needs to stop smoking to stop the progression of the disease. At the molecular level, this disease is due to which one of the following? (A) Enhanced trypsin activity in the lung (B) Decreased trypsin activity in the lung (C) Enhanced α1-antitrypsin activity in the lung (D) Decreased α1-antitrypsin activity in the lung (E) Enhanced reduction of sulfhydryl groups in the lung (F) Decreased reduction of sulfhydryl groups in the lung

The answer is d. The man has the symptoms of emphysema, due to destruction of lung cells by the protease elastase. Neutrophils in the lung accidentally release elastase as they engulf and destroy inhaled bacteria and other particles, and normally α1-antitrypsin would bind to the elastase and inhibit its activity. In a long-term smoker, however, products from the cigarette smoke oxidize an essential methionine side chain in α1-antitrypsin, rendering it inactive. Thus, over time, noninhibited elastase has been destroying lung tissue until the lung no longer functions properly. Even though the inhibitor will block trypsin activity, the lung damage is the result of increased elastase activity, not trypsin activity. Sulfhydryl groups are not being affected, rather a sulfur in methionine is the target of the cigarette smoke.

A 23-year-old female patient presents to the ER with a feeling of being unable to catch her breath, light-headedness, and "tingling" of her fingers, toes, and around her mouth. This happens whenever she drives through a tunnel, and that is what set off this episode. Which of the following arterial blood pHs would be most consistent with her diagnosis? (A) 8.10 (B) 7.55 (C) 7.15 (D) 6.40 (E) 6.10

the answer is B. Acetazolamide is a carbonic anhydrase inhibitor, which is found primarily in red blood cells. The red blood cells contain carbonic anhydrase that catalyzes the reaction that forms carbonic acid from CO2 and H2O. Under high-altitude conditions, the inhibition of carbonic anhydrase will lead to a decrease in blood pH, which stabilizes the deoxygenated form of hemoglobin. This is due to an increased loss of bicarbonate in the urine by the inhibition of carbonic anhydrase within the kidney. The change in pH increases oxygen delivery to the tissues, and can overcome, in part, the symptoms of altitude sickness. However, in the case of the person with Type I diabetes who begins to produce ketone bodies, the body's main compensatory mechanism to overcome the acidosis is blocked. As ketone bodies are formed and protons generated, the H+ will react with bicarbonate to form carbonic acid. Carbonic anhydrase, which catalyzes a reversible reaction, will then convert the carbonic acid to CO2 and H2O, with the CO2 being exhaled. These reactions soak up excess protons, and help to buffer against the acidosis. If, however, carbonic anhydrase has been inhibited by acetazolamide, then the bicarbonate cannot buffer the blood pH and the acidosis could become more severe. White blood cells, muscle cells, liver cells, and the lens of the eye do not contribute to the buffering of the blood, and inhibition of carbonic anhydrase in those cells would not affect the ability to overcome an acidosis.

An environmentalist attempted to live in a desolate forest for 6 months, but had to cut his experiment short when he began to suffer from bleeding gums, some teeth falling out, and red spots on the thighs and legs. This individual is suffering from an inability to properly synthesize which one of the following proteins? (A) Myoglobin (B) Hemoglobin (C) Collagen (D) Insulin (E) Fibrillin

the answer is c. The environmentalist is suffering from scurvy, a deficiency of vitamin C. The hydroxylation of proline and lysine residues in collagen requires vitamin C and oxygen. In the absence of vitamin C, the collagen formed cannot be appropriately stabilized (owing, in part, to reduced hydrogen bonding between subunits due to the lack of hydroxyproline) and is easily degraded, leading to the bleeding gums and loss of teeth. Globin synthesis might be indirectly affected because absorption of iron from the intestine is stimulated by vitamin C, but globin is not modified through a hydroxylation reaction. Iron is involved in heme synthesis, which regulates globin synthesis. Insulin and fibrillin synthesis are not dependent on vitamin C (lack of insulin will lead to diabetes, and mutations in fibrillin lead to Marfan syndrome).

A young black man was brought to the emergency room (ER) owing 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. An underlying cause in the change of shape of these cells is which one of the following? (A) Increased ionic interactions between hemoglobin molecules in the oxygenated state. (B) Increased ionic interactions between hemoglobin molecules in the deoxygenated state. (C) Increased hydrophobic interactions between hemoglobin molecules in the oxygenated state. (D) Increased hydrophobic interactions between hemoglobin molecules in the deoxygenated state. (E) Increased phosphorylation of hemoglobin molecules in the oxygenated state. (F) Increased phosphorylation of hemoglobin molecules in the deoxygenated state.

the answer is d. The man has sickle cell disease, and his hemoglobin consists of mutated β chains, along with normal α chains. The glutamate at position 6 in the β chains of HbA is replaced by valine in HbS. Valine contains a hydrophobic side chain, whereas glutamate contains an acidic side chain. Under low oxygenation conditions (such as vigorous exercise), the HbS molecules will polymerize owing to hydrophobic interactions between the valine on the β chain and a hydrophobic patch on another HbS molecule. Under well-oxygenated conditions, the valine in the β chain is not exposed on the surface of the molecule, and it cannot form an interaction with the hydrophobic patch on another hemoglobin molecule. Once the HbS polymerizes, it forms a rigid rod within the red blood cells, which deforms the cell and gives it the "sickle" appearance. Once sickled, the red blood cells cannot easily deform and pass through narrow capillaries, leading to loss of oxygen to certain areas of the body, which is what leads to the pain experienced by the patient. The sickling is not due to increased or decreased ionic interactions between HbS molecules, or to phosphorylation of the HbS monomers.