MCAT Bio 12/30/15

Difference between (+)RNA and (-)RNA:

(+)RNA: signifies that a particular viral RNA sequence may be directly translated into the desired viral proteins. Therefore, in positive-sense RNA viruses, the viral RNA genome can be considered viral mRNA, and can be immediately translated by the host cell (-)RNA: complementary to the viral mRNA and thus must be converted to positive-sense RNA by an RNA polymerase prior to translation. Negative-sense RNA (like DNA) has a nucleotide sequence complementary to the mRNA that it encodes. Like DNA, this RNA cannot be translated into protein directly. Instead, it must first be transcribed into a positive-sense RNA which acts as an mRNA. Some viruses (Influenza, for example) have negative-sense genomes and so must carry an RNA polymerase inside the virion.

Retroviruses are a type of ________ single-stranded RNA viruses. (+ or - sense)

(+)sense

What is dictyate?

- period of arrest between the onset of oogenesis in the fetal stage and the onset of the menstrual cycle, when one cell per month will move forward with meiosis I - before meiosis 1, the cell is diploid, whereas after it enters into meiosis 1, it become haploid it's the prolonged resting phase in oogenesis. It occurs in the stage of meiotic prophase I[2] in ootidogenesis. It starts late in fetal life[2] and is terminated shortly before ovulation by the LH surge.[3] Thus, although the majority of oocytes are produced in female fetuses before birth, these pre-eggs remain arrested in the dictyate stage until puberty commences and the cells complete ootidogenesis

Types of protein modification and location of process

- phosphorylation - formation of disulfide bridges - glycosylation **can occur in both the Golgi apparatus and rough ER ***peptide bond formation occurs during translation (prior to any protein modification)

What is the role of p53?

- tumor supressor protein - it's a transcription factor that induces expression of pro-apoptotic proteins

True of False: 1) Mitochondria can undergo mitosis. 2) Prokaryotes reproduce by mitosis. 3) Eukaryotic cells divide and reproduce through mitosis.

1) FALSE 2) FALSE; reproduce by binary fission 3) TRUE

Types of protostomes:

1. Annelids 2. Arthropods 3. Molluscs *Protostome = a multicellular organism whose mouth develops from a primary embryonic opening, such as an annelid, mollusk, or arthropod; The protostomes include all other animal phyla, which means everything which is not a fungus, a plant, or unicellular. So worms, clams, insects, all of these are protostomes *Deuterostome = any members of a superphylum of animals. Deuterostomia is a subtaxon of the Bilateria branch of the subkingdom Eumetazoa, within Animalia, and are distinguished from protostomes by their embryonic development; in deuterostomes, the first opening (the blastopore) becomes the anus, while in protostomes, it becomes the mouth. Deuterostomes are also known as enterocoelomates because their coelom develops through enterocoely.

What are Mendel's Laws?

1. Law of segregation - 2 alleles of an individual are separated and passed on to the next generation singly 2. Law of independent assortment - alleles of one gene will separate into gametes independently of alleles for another gene *genes that are located in the same chromosome DON'T display independent assortment

What is the purpose of the pentose phosphate pathway?

1. produce NADPH --> used as an electron donor for reductive biosynthesis and in the process of detoxification of reactive oxygen species 2. make ribose for nucleotide biosynthesis

Role of enzymes: 1. Phosphatases 2. Nucleases 3. Proteases 4. Peptidases 5. Glucosidases 6. Lipases

1. remove phosphate groups from molecules-->part of degradation (*kinase = phosphorylase) 2. breaks down nucleic acids 3. breaks down proteins 4. breaks down proteins 5. breaks down carbohydrates 6. breaks down lipids

Alternative splicing permits somatic cells to contain the same genome while maintaining the capability to express widely differing proteins, based on the tissue in which the cell is located. Which experimental technique would be LEAST useful in detecting the differing cellular products created by alternative splicing? A. Southern blotting B. Western blotting C. Northern blotting D. ELISA (enzyme-linked immunosorbent assay)

A Alternative splicing creates different mRNA sequences leading to different proteins, thus any technique that detects changes in mRNA transcripts or final protein products could be useful. Northern blotting is used to detect RNA and both Western blotting and ELISAs can be used to detect proteins. However, Southern blotting is used to detect DNA, and a point made by the question is that the genomes are the same (therefore the least useful technique). **ELISA: purpose of an ELISA is to determine if a particular protein is present in a sample and if so, how much

Which has the most acidic environment: A) Lysosomes B) Mitochondria C) Golgi Body D) Nuclei

A At pH 4.8, the interior of the lysosomes is acidic compared to the slightly basic cytosol (pH 7.2). The lysosome maintains this pH differential by pumping in protons (H+ ions) from the cytosol across the membrane via proton pumps and chloride ion channels. - mitochondria = ~7-8 - Golgi body = ~6.5 - Nuclei = 7.3 (same with cytoplasm)

Which of the following is NOT a means by which prion infection can typically occur? A. Aerosol contact B. Spontaneous mutation C. Consumption D. Inheritance

A Directly consuming tissue contaminated with prions, especially brain tissue, can transmit the infection as can inheritance when a genetic form of the disease is passed through families. Spontaneous mutations can also arise that then give rise to the malformed proteins. Aerosol spread is highly atypical unless a person is working with cutting tools, like cranial saws, where brain matter can become aerosolized, and without appropriate personnel protective equipment. Since prions are not carried in nasal secretions, they are not spread through the common aerosol routes of coughing and sneezing.

A pregnancy test is an ELISA set up to detect human chorionic gonadotropin (hCG), a hormone specific to pregnancy. How would this ELISA be configured? A. A primary antibody specific for hCG, blood or urine sample, a secondary antibody specific for hCG B. A primary antibody specific for hCG, blood or urine sample, a secondary antibody specific for the primary antibody C. Blood or urine sample, a primary antibody specific for hCG, a secondary antibody specific for the primary antibody's antigen binding site D. Blood or urine sample, a primary antibody specific for hCG, a secondary antibody specific for the primary antibody's constant region

A ELISA tests utilize the specificity of antibodies to identify antigen or antibody in a patient's blood or urine sample. hCG in this case is acting as the antigen anti-hCG antibodies are fixed to a microtiter well plate, and the blood or urine sample is applied. If hCG is present, it will bind to the primary antibody. A wash removes any unbound substances. The secondary antibody is specific to a different region on hCG and is covalently bound to an enzyme that produces a color change; the secondary antibody will bind to hCG and application of the substrate will produce the color change.

There are many structural similarities and differences between eukaryotic and prokaryotic cells. Which of the following cellular structures are the same whether isolated from a prokaryote or eukaryote? A. Plasma membrane B. Cell wall C. Ribosomes D. Flagella

A Eukaryotic flagella are constructed out of microtubules in a "9 + 2" arrangement, whereas prokaryotic flagella are built out of a bacterial protein called flagellin (you don't need to know this name for the MCAT) and are organized in to a base, hook, and filament. Prokaryotic cell walls are constructed out of peptidoglycan, whereas eukaryotes with a cell wall utilize other carbohydrates, such as cellulose in plants or chitin in fungi. Prokaryotes have smaller 70S ribosomes (the "S" refers to the sedimentation rate during centrifugation), whereas eukaryotes have 80S ribosomes. However, the plasma membranes of eukaryotic and prokaryotic cells are organized similarly. They are phospholipid bilayers with other assorted structures (e.g., proteins, carbohydrates, etc.) embedded or attached.

Which of the following is NOT a similarity between replication and transcription? A. Both processes occur with the same fidelity B. Polymerization in both processes is based on reading a template C. a pyrophosphate is removed from every nucleoside as polymerization occurs D. both processes occur in the 5'-3' direction

A Fidelity refers to accuracy Because RNA polymerase do not proofread, transcription is less accurate (i.e. lower fidelity process) In both cases, the removal of the pyrophosphate provides the energy for polymerization to occur DNA and RNA polymerase move along the parent chain in the 3'-5' direction but new chain is made in the 5'-3' direction

The ratio of guanine-cytosine (G—C) pairs to adenine-thymine (A—T) pairs is useful in laboratory manipulation of double-stranded DNA. If a segment of DNA has a low G—C : A—T ratio, it would be reasonable to assume that this segment would: A. require less energy to separate the two DNA strands than would a comparable segment of DNA having a high G—C : A—T ratio. B. contain more guanine than cytosine. C. contain more adenine than thymine. D. require more energy to separate the two DNA strands than would a comparable segment of DNA having a high G—C : A—T ratio.

A G—C base pairs are linked by three hydrogen bonds in the double helix, while A—T base pairs are joined by only two hydrogen bonds. It takes more energy to separate G—C base pairs, and the less G—C rich a piece of double-stranded DNA is, the less energy that is required to separate the two strands of the double helix. Note that the statements "contain more guanine than cytosine" and "contain more adenine than thymine" can be eliminated since in double-stranded DNA, there must be equal quantities of G and C (and of A and T).

What cellular process can viroids exploit in order to replicate their genomes? A. Use of transcriptional machinery B. Use of DNA polymerase C. Use of ribosomes D. Use of primase

A Primase makes a short RNA sequence from a DNA template, ribosomes synthesize amino acids into proteins and DNA polymerase copies DNA from DNA so none of these would work to produce RNA in any form.

Scientists hypothesize that certain types of lung disease are the result of fragments of plasmid DNA being inserted into an exon of mRNA, leading to defective proteins. If a scientist identified a cell line that contained the mutant mRNA and they wanted to determine the actual base sequence of the mutation, which of the following techniques would be most useful in generating a large amount of genetic material to use for sequencing? A. Reverse Transcriptase Polymerase Chain Reaction (RT-PCR) B. Gas chromatography C. DNA fingerprinting D. Induction of aberrant growth with Restriction Fragment Length Polymorphism (RFLP) analysis

A RT-PCR uses reverse transcriptase with an initial mRNA sample to generate cDNA, then specific nucleotide primers to amplify specific sequences within that cDNA ["Reverse Transcriptase Polymerase Chain Reaction (RT-PCR)" is correct]. DNA fingerprinting is a method by which scientists can distinguish different DNA samples, but does NOT generate a large amount of DNA (DNA fingerprinting is wrong). Gas chromatography is used to separate very small molecular weight particles with different volatility, but is not used to increase amounts of DNA (gas chromatography is wrong). "Induction of aberrant growth with Restriction Fragment Length Polymorphism (RFLP) analysis" is a nonsense answer; RFLP analysis is a means of distinguishing different strands of DNA, but not for generating a large pool of the material. Further, it does not rely on aberrant cell growth ["induction of aberrant growth with Restriction Fragment Length Polymorphism (RFLP) analysis" is wrong].

A homozygous white bull is mated to a homozygous cow with a red coat. The offspring all have a roan coat color, which is composed of a mix of white hairs and red hairs. Which of the following is true? A) Coat color in cattle is an example of codominance B) White hair is recessive to red hair C) Coat color in cattle is an example of incomplete dominance D) White hair is dominant over red hair

A Since the offspring are neither red nor white, this trait is not inherited via simple Mendelian genetics, and neither trait is dominant or recessive. The individual hairs on the F1 cattle are either red or white, which means that both alleles are being expressed. This best matches the definition of Codominance. If the individual hairs were a blend of white and red, the best answer would have been incomplete dominance.

The amphoteric character of amino acids best explains their ability to: A) form dipolar ions. B) form peptide bonds. C) contribute directly to a protein's secondary structure. D) dissolve in nonpolar solvents.

A The amphoteric character of amino acids describes their ability to do two things: accept a proton or donate a proton. In other words, amino acids can act as either an acid or a base. When the amino portion of an amino acid deprotonates its own carboxylic acid, a dipolar ion (or zwitterion) forms. This is a direct result of the amphoteric character of the amino acid.

Which of the following lists the fungal structural hierarchy in order from smallest to largest? A. Cell Hypha Mycelium Thallus B. Cell Mycelium Hypha Thallus C. Cell Hypha Thallus Mycelium D. Cell Mycelium Thallus Hypha

A The cell is the fundamental smallest unit of all living things. Hyphae are long filaments of cells joined end to end, and can be either septate (cells are separated by cell walls called septae) or aseptate (no divisions between cells, cytoplasm and nuclei are shared). A mycelium is a network of hyphae, and the largest, most organized fungal structure (usually visible to the naked eye) is called a thallus (e.g., a mushroom is a thallus).

In the creation of an expression vector, what step is necessary to prepare both a gene of interest and the plasmid that will carry it for ligation? A. Digest both genetic sequences with the same restriction endonuclease. B. Ensure both genetic pieces are single stranded. C. Treat both with proteases to avoid contaminating interactions from proteins. D. Allow exonucleases to proofread both sequences.

A To create ends that can be connected, the same restriction endonuclease must be used to both open the plasmid and digest the ends of the gene of interest so it can be ligated into the plasmid. Both pieces need to be double-stranded, not single. Proofreading of any type would have already been done in synthesizing the sequences. At this stage, the genetic constructs are already isolated from proteins so proteases are not needed.

UV light can trigger the formation of pyrimidine dimers, which then cause malformed loops of DNA. Visible light can then trigger repair enzymes via photoreactivation. This describes a DNA repair mechanism known as: A. direct reversal. B. nonhomologous endjoining. C. excision repair. D. homologous recombination.

A Visible light triggers the enzymes involved in direct reversal. Excision repair does not utilize photoreactivation; damaged or defective bases are cut out and the strand is repaired. Homologous recombination is used with double strand breaks as is nonhomologous endjoining.

Which of the following western blotting techniques correctly pairs the material used as a probe with the material being detected? A. Labeled antibodies are used to detect certain proteins. B. Labeled nucleic acids are used to detect certain proteins. C. Labeled antibodies are used to detect certain nucleic acid sequences. D. Labeled nucleic acids are used to detect certain nucleic acid sequences.

A Western blotting uses antibodies to detect proteins. Proteins are separated via gel electrophoresis, transferred to a membrane, and then probed with a primary antibody. Northern and Southern blotting use the same general idea, but with nucleic acids; nucleic acids are first separated on a gel, then transferred to a membrane, then probed with nucleic acids. Nucleic acids are not used to detect proteins and antibodies are not used to detect nucleic acids.

Ionizing forms of energy such as X-rays and nuclear radiation cause cellular dysfunction at the location of absorbance. Extreme exposure to x-rays might result in a dramatic increase in the risk of developing: A) bone marrow cancer B) thyroid cancer C) lung cancer D) lymphogenous leukemia

A chronic radiation exposure significantly increases the risk of cancers of the tissue where the maximum amount of radiation is absorbed i.e. exposure to radioactive iodine (formed as nuclear waste) increases risk of thyroid cancer because the thyroid concentrates iodine i.e. exposure to radioactive gases (like radon) cause cancers of the throat and lungs; ingested radioactive heavy metals (i.e. strontium and plutonium) concentrate in the bone matrix leading to increases in the occurrence of bone marrow and myelogenous leukemia (cancer of white blood cells produced in the bone marrow) i.e. X-rays are absorbed by bone, increasing the risk of developing bone marrow and myelogenous leukemia --> lymphogenous leukemia = cancer of white blood cells made by lymph nodes, not bone marrow

All of the following processes occur in bacteria EXCEPT: A. mRNA processing. B. regulation of transcription. C. electron transport. D. simultaneous transcription and translation.

A mRNA processing does NOT occur in bacteria. Prokaryotic (bacterial) mRNA can be immediately translated; no processing is required (mRNA processing does not occur in bacteria and is the correct answer choice). This is one reason why transcription and translation can proceed simultaneously (simultaneous transcription and translation does occur in bacteria and can be eliminated). Although not compartmentalized as in eukaryotes, bacteria do participate in electron transport on their cell membranes (electron transport does occur and can be eliminated). Lastly, regulation of transcription does occur; in fact some of the best-understood transcriptional regulation mechanisms are found in bacteria (e.g., the lac operon, regulation of transcription occurs and can be eliminated).

Which of the following is true regarding prokaryotic flagella? A. It is the predominant form of bacterial locomotion B. it is made of microtubules connected by dyne proteins C. it allows viruses to maneuver between host cells D. it can only be located on one end of a bacterium and this defines the polarity of the cell

A prokaryotic flagella are the predominant means of bacterial locomotion only eukaryotic flagella are made of microtubules and dynein; bacterial flagella have a different structure and are made of protein flagellin viruses rely on diffusion to maneuver between host cells, not flagella bacteria can have flagella on one end (monotrichous), both ends, amphitrichous, or multiple areas (peritrichous)

(+)-Ginkgolide B is a diterpenoid trilactone with six five-membered rings, and is extracted from the root bark and leaves of the Ginkgo biloba tree. This molecule has blood-brain barrier permeability, the molecular formula C20H24O10 and a molecular weight of 424.4 g/mol. This molecule must contain: A) four isoprene units, additional functional groups and is able to perform simple diffusion through endothelial cells connected by tight junctions. B) four isoprene units, no additional functional groups and is able to perform simple diffusion through endothelial cells connected by tight junctions. C) four terpene units, additional functional groups and is able to perform simple diffusion through neurons connected by tight junctions. D) four isoprene units, additional functional groups and is able to perform simple diffusion through endothelial cells connected by gap junctions.

A. Diterpenoids contain four isoprene units (eliminate choice C) and additional functional groups (eliminate choice B). The blood-brain barrier is a highly selective permeability barrier that separates the circulating blood from the central nervous system. It is formed by endothelial cells (another reason to eliminate choice C), which are connected by tight junctions (eliminate choice D). Choice A is correct because it accurately describes the composition of the molecule.

TUNEL (terminal deoxynucleotidyl transferase dUTP nick end labeling) is a method for detecting DNA fragmentation by labeling the terminal end of nucleic acids. BrdU (5-bromo-2'-deoxyuridine) can also be used in straining protocols, as this synthetic nucleoside can be incorporated into nucleic acids during replication instead of thymidine. Tumor cells have: A) minimal TUNEL staining due to their low apoptotic rate, but strong BrdU staining due to active genome replication. B) strong TUNEL staining due to their high apoptotic rate, and strong BrdU staining due to active genome replication. C) strong TUNEL staining due to their high apoptotic rate, but weak BrdU staining due to minimal genome replication. D) minimal TUNEL staining due to their low apoptotic rate, and weak BrdU staining due to minimal genome replication.

A. Genome fragmentation is one of the hallmarks of apoptosis and tumor cells generally evade apoptosis. This means tumor cells will have minimal TUNEL staining (eliminate choices B and C). Tumor cells have high proliferation rates and so will be actively replicating the genome. This means they will likely display strong BrdU staining (choice A is correct, eliminate choice D).

Like α(1→4)-glycosidase, which of the following enzymes acts to cleave α(1→4) linkages between glucose molecules? A) Glycogen phosphorylase B) Glucose-6-phosphatase C) Phosphoglucomutase D) Pyrophosphatase

A. Glycogen phosphorylase catalyzes the rate limiting step of glycogenolysis—the release of glucose-1-phosphate from glycogen via its action on terminal α(1→4) linkages. This makes choice A true and the correct answer. Glucose-6-phosphatase hydrolyzes glucose-6-phosphate, resulting in the release of a phosphate group and free glucose for export from the cell. Choice B is false and an incorrect answer. Phosphoglucomutase is involved in both glycogenolysis and glycogenesis by interconverting glucose 1-phosphate and glucose 6-phosphate, thus choice C is false and the incorrect answer. Pyrophosphatases are members of the acid hydrolase family of enzymes that cleave diphosphate bonds. Choice D is false and an incorrect answer.

In humans, the fusion of the plasma membranes of a sperm and an ovum is followed first by which of the following? A) Release of the second polar body from the fertilized ovum B) Implantation in the uterus C) The first cell division of the zygote D) Gastrulation

A. Human ova do not complete the second meiotic cell division, including the formation of the second polar body, until after fertilization (choice A is correct). This process occurs immediately after fertilization and is a prerequisite for the first cell division of the zygote (choice C is wrong). Implantation happens a few days later, and gastrulation occurs at a stage many cell divisions after the first cell division (choices B and D are wrong).

Cells from the pigmented region of a frog gastrula are transplanted to a light gastrula region. The light region surrounding the graft is induced to form a second incomplete tadpole. Cells transplanted from a light region to the same site do not induce formation of a second tadpole. Based on the results of this experiment, the developmental fate of the region surrounding the graft most likely depends on: A) the region from which the graft cells were taken. B) the region to which the graft cells are transplanted. C) the expression of pigment genes. D) the migration of the graft cells throughout the transplant region.

A. Pigmented cells induce development of an incomplete tadpole, while cells from the light region do not when transplanted to the same area. The region from which the cells are taken appears more important for developmental fate than the area they are transplanted to (choice A is correct, and choice B is wrong). Pigmented cells appear to be the key factor, but pigmentation itself is likely to be coincidental and not the cause of the developmental fate of the region (choice A is a better answer than choice C). There is no evidence presented that migration of cells is important (choice D is wrong).

Short tandem repeat analysis for DNA fingerprinting utilizes patterns of repetitive DNA within what part of the genome to identify individuals? A. Introns B. Exons C. Sex chromosomes D. Transposons

A. Short tandem repeat analysis looks at polymorphic, repetitive sequences within introns to identify a person. Amplification via PCR is followed by electrophoresis and Southern blotting. *Transposon: a DNA sequence that can change its position within the genome, sometimes creating or reversing mutations and altering the cell's genome size. "Jump" from one position of genome to next

From Passage: Emphesyma is a form of COPD caused by immune-mediated destruction of the alveoli. Presence of irritants in the lung tissue can cause the alveolar macrophages to secrete enzymes that degrade the connective tissue components such as collagen and elastin, thus weakening the alveolar walls leading to their collapse. Therefore, smoking is a major risk factor for emphesyma. Question: The degradation of connective tissue components in emphysema is most likely caused by which of the following? A) Proteases B) Lipases C) Monoanime oxidases D) Glycoside hydrolases

A. The connective tissue components damaged in emphysema include collagen and elastin, which are extracellular proteins. Choice A is correct. Therefore, proteases released from the alveolar macrophages would be the culprits. --> Protease is the digestive enzyme needed to digest protein Lipases degrade lipids (choice B can be eliminated) monoanime oxidases degrade certain neurotransmitters such as norepinephrine and epinephrine (choice C can be eliminated) glycoside hydrolases degrade carbohydrates (choice D can be eliminated).

Which of the following salts, when dissolved in water, would produce the solution with the highest pH? A) Na2CO3 B) NH4Cl C) NaCl D) AlBr3

A. The most basic salt will produce the solution with the highest pH. The carbonate ion from Na2CO3 is the conjugate base of bicarbonate, which is itself weakly basic in aqueous solution; choice A is the most basic salt listed. NaCl is a neutral salt as neither of its ions are reactive with water (eliminate choice C). Both NH4Cl and AlBr3 are acidic salts because both cations will interact with water to increase the [H+] in solution (eliminate choices B and D).

The researcher described in the passage discovered that the HLA-A3 allele causes a rare form of hemochromatosis in anyone who is homozygous for this allele. This disorder results in dangerously high blood iron levels, causing a variety of problems, from endocrine dysfunction to liver cancer. In a certain large, randomly-mating population, the only two HLA-A alleles present are HLA-A2 and HLA-A3. If the frequency of the HLA-A2 allele is 0.9, what percentage of individuals would be expected to have hemochromatosis? A) 1% B) 9% C) 49% D) 81%

A. The phrase "large, randomly-mating" population indicates that this population meets core assumptions of Hardy-Weinberg and the equations associated with the law apply. Hardy-Weinberg states that the frequency of the homozygous genotype is simply the allele frequency squared. Since the frequency of the HLA-A2 allele is 0.9, the frequency of the HLA-A3 allele must be 0.1. Squaring this gives 0.01, or 1% (choice A is correct). Choice D is incorrect, and corresponds to the frequency of individuals who are homozygous for the HLA-A2 allele. Both choices B and C are incorrect applications of the data presented in Table 1.

Homologous vs Nonhomologous Chromosomes

A. Homologous chromosomes: - are the chromosomes which contains alleles for same type of genes {one from sperm and one from egg} or can be called similar - are a pair, one from the original female and one from the original male - The size of the chromatids and location of the centromeres are the same. The homologs carry genes for the same traits and the genes are located at the same place on both members of the pair. B. Non-homologs are not like each other - are chromosomes that contains alleles for different type of genes *recombination occurs between similar molecules of DNA (homologs) *chromosome translocation is a chromosome abnormality caused by rearrangement of parts between nonhomologous chromosomes

Contraction of cardiac muscle begins just after depolarization and lasts about as long as the action potential. During a contraction: A) Ca2+ enters the cell through voltage-gated ion channels. B) sarcomere length increases. C) polymerized actin becomes depolymerized. D) myosin binds irreversibly to actin.

A. Part of the cardiac action potential is the opening of slow voltage-gated calcium channels to create the plateau in depolarization (Phase 2) so choice A is correct. When the voltage-gated calcium channels open, calcium enters the cell and can play a role in the contraction of the cardiac muscle. Sarcomere length decreases during each contraction, and does not increase (choice B is wrong). Actin filaments in contractile tissue are not dynamic as they are in other processes and do not spontaneously depolymerize and repolymerize (choice C is wrong). Myosin binds actin reversibly, not irreversibly (choice D is wrong).

The surface of the plasma membrane facing the intestinal lumen is called the _____ surface.

Apical

What are the requirements for a complex transposon? A. Two IS elements B. An IS element and one or more genes C. Two IS elements with intervening sequence between them D. An IS element that triggers a frameshift mutation

B A complex transposon contains an IS element (the transposase and its accompanying inverted repeat sequences) and one or more genes. A composite transposon has two IS elements and intervening sequence. While transposons can trigger frameshift mutations, this is not what defines a complex transposon **There are 3 common types of transposons: 1. IS element (simplest) = transposase gene + flanked by inverted repeat sequences 2. Complex Transposon = transposase + 2 genes (like genes for antibiotic resistance) 3. Composite Transposon = 2 similar or identical IS elements (i.e. 2 transposases) with a central region in between

How does standard DNA sequencing differ from running a polymerase chain reaction to synthesize DNA? A. PCR utilizes radiolabeling which would be deleterious in DNA sequencing. B. DNA sequencing utilizes ddNTPs which lack the 3'OH group. This terminates polymerization, creating fragments. C. PCR requires electrophoresis whereas DNA sequencing does not. D. DNA sequencing is faster than PCR and yields higher copy numbers.

B DNA sequencing does use ddNTPs which randomly stop polymerization and thus create fragments of various lengths. The fragments are run on a gel so the positions of the constituent As, Ts, Gs, and Cs can be visualized. Radiolabeling is part of sequencing as is electrophoresis and its goal is not about copy numbers as is the case with PCR. *DNA Sequencing: process of determining the precise order of nucleotides within a DNA molecule. It includes any method or technology that is used to determine the order of the four bases—adenine, guanine, cytosine, and thymine—in a strand of DNA *PCR use dNTPs

During a Gram staining procedure, a bacterial culture does not appear as either Gram-positive or Gram-negative. What structural attribute could make bacteria appear Gram-indeterminate? A. Presence of an outer lipid bilayer B. Presence of a cell wall with a waxy composition C. Presence of a thick cell wall D. Presence of an inner lipid bilayer

B Gram-negative bacteria have both inner and outer lipid bilayers; it is in fact the outer lipid bilayer that stains the classic pink-to-red color that indicates Gram-negativity. Gram-positive bacteria have a thick well wall which stains purple as part of the process. A change in cell wall composition, particularly to something more waxy, would interfere with the staining process, making the results indeterminate. Mycobacterium are an example of this and alternative acid fast staining techniques are used with them. **Acid-fastness is a physical property of certain bacteria (and, less commonly, protozoa), specifically their resistance to decolorization by acids during staining procedures.[1][2] Acid-fast organisms are easy to characterize using standard microbiological techniques (e.g. Gram stain - if an acid-fast bacillus (AFB) was gram stained, the result would be an abnormal gram positive organism, which would indicate that further testing wasn't necessary)

Which of the following statements regarding the separation of proteins via gel electrophoresis is INCORRECT? A. Proteins migrate to the positive electrode of the gel apparatus because they become negatively charged during their preparation for electrophoresis. B. Decreasing the percent of polyacrylamide makes the gel less dense and decreases the ability to resolve small distances between two proteins with similar molecular weights. C. Proteins travel in denatured states through the gel. D. Proteins with smaller molecular weights travel more quickly than those with higher molecular weights when an electric current is applied.

B Imagine you are in a running race against a friend who runs at about the same speed as you do. Let's say you race your friend while running in a pool of honey. Even if you are faster than your friend, neither of you will travel very far, and the distance between you at the end of the race will be small. Now let's say you race your friend in air, a much, much less dense fluid. Now, any difference in speed is more easily appreciated, and you will be separated by a greater distance. The same is true for proteins in a gel. Decreasing the percent of the polyacrylamide (and the density) will increase your resolution capacity, and proteins with similar molecular weights will be separated by a greater distance ("Decreasing the percent of polyacrylamide makes the gel less dense and decreases the ability to resolve small distances between two proteins with similar molecular weights" is false and the correct answer choice). Proteins with smaller molecular weights travel more quickly than do proteins with larger molecular weights ("Proteins with smaller molecular weights travel more quickly than those with higher molecular weights when an electric current is applied" is true and can be eliminated). The proteins are negatively charged because during preparation they are denatured ("Proteins travel in denatured states through the gel" is true and can be eliminated) in a detergent that confers a negative charge upon them. Opposites attract ("Proteins migrate to the positive electrode of the gel apparatus because they become negatively charged during their preparation for electrophoresis" is true and can be eliminated).

All of the following are examples of enzymes which participate in hydrolytic reactions EXCEPT: A) pepsin B) insulin C) chymotrypsin D) amylase

B Insulin is a hormone! It does not itself catalyze reactions Pepsin: enzyme that breaks down proteins into smaller peptides; produced in the stomach Chymotrypsin: from pancreas; activated by trypsin Amylase: from pancreas; hydrolyzes polysaccharides to disaccharides

During a polymerase chain reaction, high heat is used to denature and separate the DNA strands while cooling allows primers to anneal. Why does the extension step require additional heating? A. Exonuclease activity requires elevated temperatures. B. The DNA polymerase used is heat sensitive and will not elongate in cooler conditions. C. Heat degrades the RNA primers so they do not become part of the permanent DNA sequence. D. Nucleotides are synthesized only in high temperature conditions.

B The DNA polymerases used in PCR are derived from thermophilic bacteria so that they will survive the heat necessary to denature the sequences in the first place and then be active only in certain temperature ranges to control their activity. Exonucleases do not require high heat, primers are removed by certain DNA polymerase activity and nucleotides are added to PCR in the form of dNTPs rather than being synthesized as part of the process.

The bacterial flora of the gut is most likely characterized as which of the following? A. Mesophilic facultative aerobes B. Mesophilic obligate anaerobes C. Thermophilic tolerant anaerobes D. Psychrophilic obligate aerobes

B The bacterial flora of the gut is most likely characterized as mesophilic obligate anaerobes. Mesophilic bacteria live in moderate temperatures and the environment of the human body qualifies as such in comparison to the thermophiles of the undersea vent (i.e. high temperatures) and the psychrophiles in the Arctic (cold temperatures). The gut's environment is oxygen poor; the flora are obligate anaerobes.

DNA from resistant bacteria in Dish 6 is extracted and placed on agar with phage-sensitive E. coli. After incubation it is determined that these E. coli are now insensitive to phage T1 infection. The most likely mechanism for their acquisition of resistance is: A) transduction. B) transformation. C) sexual reproduction. D) conjugation.

B The definition of transformation is the process in which naked DNA, not a virus, is taken into a cell and changes the genetic characteristics of that cell. That is the case in this experiment, with extracted DNA making cells resistant to the virus (choice B is correct). Transduction is mediated by a virus, conjugation involves direct transfer of DNA between bacteria, and sexual reproduction does not apply to bacteria (choices A, C, and D are wrong). *Transduction = giving new genes to prokaryotes via viral inheritance (due to faulty excision) *Transfection = giving new genes to a euk host *transformation = uptake of DNA from environment

In one strain of mouse, homozygotes for an allele of a gene develop heart defects, while in another strain of mouse, homozygotes with the same allele develop normally. Heterozygotes develop normally in both strains. What is the most likely explanation for the difference between the 2 strains? A) the allele is recessive B) the development of the heart defect is influenced by more than one locus C) the allele has pleiotropic effects on development D) the allele is codominant

B The key variable must not lie within the allele itself, since this remains the same. The genetic background of the 2 different strains of mice must affect whether or not the heart defect phenotype is expressed; Further, only one defect is observed (so it can't be pleiotropic). Therefore, the heart defect phenotype must be influenced by some other locus that is different in the 2 strains of mice

Which of the following lab techniques is described by the following steps? Step 1: Separate DNA fragments on a gel Step 2: Transfer fragments to a nitrocellulose filter Step 3: Probe the filter for the target DNA sequence with hybridized probes A. Conjugation B. Southern blotting C. Radioimmunoassay D. ELISA

B The steps describe Southern blotting. (i.e. DNA) Radioimmunoassay uses radioactively labeled antibodies to find target proteins or other target antibodies (radioimmunoassay is wrong). ELISA uses enzymatically labeled antibodies to find target proteins or other target antibodies (ELISA is wrong). Conjugation is a means by which bacteria can increase genetic diversity; it is not a lab technique (conjugation is wrong).

A researcher isolates a microbe from the human gut. She determines that it has a thick cell wall, grows only in the absence of oxygen, and requires supplemental arginine. Which of the following is true about this microbe? A. It is an arg- Gram- obligate anaerobe. B. It is an arg- Gram+ obligate anaerobe. C. It is an arg- Gram+ tolerant anaerobe. D. It is an arg+ Gram- tolerant anaerobe.

B This microbe is an arg- Gram+ obligate anaerobe. Bacteria that require supplemental amino acids are auxotrophs and the necessary amino acid is indicate with a minus sign (arg-). The thick cell wall identifies it as a Gram+ bacterium, and the fact that it grows only in the absence of oxygen indicates that it is an obligate anaerobe. Tolerant anaerobes, while they do not use oxygen, can still grow in its presence.

In addition to genotype at the HLA-A locus, the likelihood that an individual will develop iron overload is influenced by other factors, such as diet and age. Due to this, only a small portion of those who are homozygous for HLA-A3 allele will actually develop iron overload. Thus, this allele is said to have a low: A) dominance. B) penetrance. C) epistasis. D) selection.

B. A dominant allele is an allele that can mask the expression of a recessive allele. The term is unrelated to environmental factors (choice A is eliminated). Epistasis describes a situation where the expression of one gene is dependent on the presence of a particular allele at another locus. Since age and diet are not alleles, choice C can also be eliminated. The selection of a trait is related to the presence of selection pressures that increase the fitness of one allele over the other. Since the question states that the HLA-A3 genotype is already present in this individual, selection is irrelevant here (choice D is also eliminated). Choice B is correct: an allele's penetrance describes the likelihood that an individual with that allele will develop the phenotype associated with the allele. Environmental or lifestyle factors may influence penetrance through factors unrelated to the individual's genotype.

Which of the following equilibria is an example of a buffered solution? A) HCl(aq) --> H+(aq) + Cl-(aq) B) H2CO3(aq) --> H+(aq) + HCO3-(aq) C) H2O2(aq) + 2KI(aq) --> 2KOH(aq) + I2(aq) D) CaCl2(aq) --> Ca2+(aq) + 2 Cl-(aq)

B. Buffered solutions are commonly made with weak acids and their conjugate bases. The only weak acid given is carbonic acid (choice B). HCl (choice A) is a strong acid and CaCl2 (choice D) is a salt. Choice C is a redox reaction.

Diffusion capacity of carbon monoxide (DLCO) is another parameter used for diagnosis of pulmonary diseases because carbon monoxide is diffusion limited. According to the passage, which of the following would be true in emphesyma compared to a normal patient? *From Passage: Emphesyma is a form of COPD caused by immune-mediated destruction of the alveoli. Presence of irritants in the lung tissue can cause the alveolar macrophages to secrete enzymes that degrade the connective tissue components such as collagen and elastin, thus weakening the alveolar walls leading to their collapse. Therefore, smoking is a major risk factor for emphesyma A) The DLCO would higher. B) The DLCO would be lower. C) The DLCO would remain unchanged. D) The DLCO would depend on the amount of RBCs at the alveolar interface.

B. Carbon monoxide is diffusion limited; therefore, the rate of diffusion of CO is only limited by factors affecting diffusion across the membrane (choice D can be eliminated). The factors would include thickness of the diffusion barrier and the effective surface area. In emphesyma, the effective surface area is reduced due to loss of alveolar sacs, thus DLCO would be decreased (choice B is correct; choice A and C can be eliminated).

In transposition of the great arteries, a congenital birth defect, the aorta of the newborn is connected to the right ventricle and the pulmonary artery is attached to the left ventricle. This would result after birth in circulation of: A) oxygenated blood through the systemic vasculature and deoxygenated blood through the pulmonary vasculature. B) deoxygenated blood through the systemic vasculature and oxygenated blood through the pulmonary vasculature. C) oxygenated blood through both the systemic and pulmonary vasculature. D) deoxygenated blood through both the systemic

B. In the normal circulation, the right ventricle pumps blood through the pulmonary artery to the lungs. Blood returns from the lungs to the left atrium and is pumped by the left ventricle through the aorta to the rest of the body. If blood from the left ventricle passes to the pulmonary artery instead of the aorta and is then returned to the left atrium, oxygenated blood will loop around through the pulmonary system without passing to the rest of the body. At the same time, the deoxygenated blood from the systemic circulation will pass to the right atrium, then back out to the systemic circulation through the aorta rather than the pulmonary artery, causing deoxygenated blood to loop through this system (choice B is correct and choice A is wrong). Choices C and D are wrong since they have all blood either oxygenated or deoxygenated.

Muscle fibers are composed of small contractile units called sarcomeres. During contraction, which of the following occurs within a sarcomere? I. Myosin filaments shorten. II. Actin filaments shorten. III. Overlap between actin and myosin filaments increases. A) I only B) III only C) II and III only D) I, II, and III

B. Items I and II are false: During contraction, neither myosin nor actin filaments get shorter. The overlap between them increases to make the sarcomere shorter (eliminate choices A, C and D). Item III is true: The overlap between the fibers increases as part of contraction.

As electrons flow within the complexes of the electron transport chain, each intermediate carrier molecule is: A) oxidized by the preceding molecule and reduced by the following molecule. B) reduced by the preceding molecule and oxidized by the following molecule. C) reduced by both the preceding and the following molecules. D) oxidized by both the preceding and the following molecules.

B. The high-energy electrons flow downhill (energetically) from reduced carriers to more oxidized carriers. In each redox reaction one partner is reduced and another oxidized (choices C and D are wrong). The intermediate molecules are first reduced by the electrons, but then oxidized back to their original state by the next member of the chain down the line (choice B is correct and choice A is wrong). **NADH and FADH2 get oxidized while the intermediate molecules get reduced; later on, the intermediate molecules get oxidized back to their original state so process can start again

The vitelline layer in sea urchins is analogous to which of the following layers in the human ovum? *From passage: The second part of the acrosome reaction arises when the decrease in intracellular H+ concentration inactivates an inhibitory protein, the function of which is to prevent the polymerization of globular actin. The resulting actin filaments produce the acrosomal process which makes contact with the vitelline membrane. The plasma membranes of the gametes then fuse and the sperm nucleus enters the egg. A) Corona radiata B) Zona pellucida C) Plasma membrane D) Cell wall

B. The vitelline layer is analogous to the zona pellucida in human ova, since both are protective acellular layers located just outside the plasma membrane (choice B is correct). The corona radiata is a region of surrounding supportive cells (choice A is wrong), and the vitelline layer is not a membrane (choice C is wrong). The ovum, like all human cells, does not have a cell wall (choice D is wrong).

Novobiocin is an antibiotic that inhibits bacterial DNA gyrase. Its antibiotic properties are based on the fact that it: A) prevents DNA helix formation. B) inhibits DNA replication. C) prevents RNA transcription. D) inhibits protein translation.

B. DNA gyrase supercoils bacterial DNA (choice B is correct). It does not affect helix formation; the DNA is already in a double-helix before it is supercoiled (choice A is wrong). It does not affect protein synthesis, since that has to do with RNA and ribosomes, not DNA (choice D is wrong). The job of DNA gyrase is to continually introduce negative supercoils into the circular bacterial genome. Then, when the helix is opened for replication and positive supercoils are introduced (the DNA gets more tightly wound at the ends of the replication fork), the already-present negative supercoils cancel out the newly introduced positive supercoils, and DNA tension is kept to a minimum. If this does not occur, the DNA strand would become too tightly wound for replication to continue, and in the absence of replication, the bacteria cannot reproduce. Note that the same could be argued for transcription, but since that involves such a small region of the genome, the positive supercoiling is not a significant effect (choice C is wrong).

The plasma membrane on the other side of the cell facing the tissues beneath is called the _____ surface.

Basolateral

What steps are necessary to replicate a (-)RNA virus? A. Make more (-)RNA copies directly from the genome using RNA-dependent RNA polymerase. B. Use RNA-dependent DNA polymerase to synthesize a DNA version of the genome and then replicate more RNA copies using the host cell's transcriptional machinery. C. Create a (+)RNA copy of the genome using RNA-dependent RNA polymerase then replicate (-)RNA copies from that template using the same enzyme. D. Create a (+)RNA template using RNA-dependent DNA polymerase then replicate (-)RNA copies from that template using DNA-dependent RNA polymerase.

C A (-)RNA virus must synthesize a replication template (a (+)RNA copy of its genome) using RNA-dependent RNA polymerase. That same enzyme can then synthesize multiple (-)RNA copies for placement in new virions. Since nucleotide synthesis is complementary and not identical, the genome itself cannot act as template. A host cell's transcriptional machinery cannot be used to make viral genomes and an RNA-dependent DNA polymerase would produce DNA, not RNA.

Tumor supressor genes like BRCA1/2 promote anticancer processes like DNA repair. When these genes are disabled via mutation, the risk for developing cancer later in life increases dramatically. A female who inherits a single mutation in BRCA2 has a 50% and 20% chance of developing breast and ovarian cancer, respectively. Which of the following best explains why these respective chances are not 100%? A) Individuals who are homozygous for the BRCA2 mutation do not have a functional allele. B) Individuals who are homozygous for the BRCA2 mutation can still rely on other DNA repair mechanisms C) Individuals who are heterozygous for the BRCA2 mutation still have a functional allele D) Individuals who are heterozygous for the BRCA2 mutation can still rely on other DNA repair mechanisms

C A female who inherits a single mutation still has an allele which can produce functional proteins and is heterozygous, not homozygous. While an individual who is heterozygous for the mutation can rely on other DNA repair mechanisms, this is not the primary reason why these individuals will not have a 100% chance of developing breast or ovarian cancer

Which of the following transport mechanisms allow for infection of animal cells by viruses? A. Simple diffusion B. Secondary active transport C. Receptor-mediated endocytosis D. Ligand-gated channels

C Animal cells can take up viruses by endocytosis when the virus interacts with a specific cell surface receptor. Viruses, though small, are not small enough for simple diffusion. Facilitated diffusion also moves material smaller than viruses. Secondary active transport uses energy to establish electrochemical gradients, which can then be used to move material against its concentration gradient; concentration gradients are not a component of viral infections.

A strain of bacterium was infected with a phage which caused all of the bacteria it infected to become resistant to the antibiotic streptomycin. Which of the following is the most likely explanation? A. The phage is lysogenic and alters the composition of the cell wall. B. When the phage inserts into the genome, it mutates the resistance gene. C. The phage is lysogenic and carries a gene from other bacteria. D. The streptomycin resistance is conferred by a viral gene that allows the virus to replicate in the presence of antibiotic.

C As part of their life cycle, lysogenic phage incorporate their genome into their host's genome. Later, when the viral genome is excised, sometimes a portion of the host's genome is taken as well and is copied and packaged up with the viral genome. That portion of the host's genome now effectively becomes part of the viral genome and is transferred to each new host upon infection. If the portion of the host's genome happens to contain a useful gene, such as a gene for antibiotic resistance, then the new host acquires that ability. This is referred to as transduction, and is one means by which bacteria can increase their genetic diversity. Lysogenic phage do not alter the composition of the bacterial cell wall and viruses are not affected by antibiotic, so they would not carry antibiotic resistance genes as part of their own genome. Mutation of an existing resistance gene typically destroys the function of that gene; if this were the case, the bacterium would have originally been resistant to streptomycin, and then lost that resistance upon infection with the virus.

Viroids are circular pieces of single-stranded RNA, approximately 200-400 bases in length. What induces the folding of these subviral particles? A. Use of gyrase B. Double-bond linkages C. Many regions of self-complementarity D. Formation of β-helices

C Gyrase compacts bacterial genomes, β-helices are part of the folding of proteins and double-bonds create kinks and folds in fatty acids, but not nucleotide sequences.

Fluoroquinolones are a class of antibiotics that inhibit DNA gyrase. Which of the following would be inhibited? A. Releasing of ribosomal subunits after transcription B. Connecting the Okazaki fragments during DNA replication C. Condensing of DNA D. Unwinding of DNA

C Gyrase is an enzyme specific to prokaryotes that maintains bacterial DNA in its supercoiled helical state. Unwinding of DNA is accomplished primarily using helicase and topoisomerase. Okazaki fragments are connected using DNA ligase. Ribosomal separation results when a release factor enters the A site, and peptidyl transferase hydrolyzes the bond between the last tRNA and the completed peptide chain. This occurs after translation, not transcription. **DNA gyrase: - is an essential bacterial enzyme that catalyzes the ATP-dependent negative super-coiling of double-stranded closed-circular DNA - Gyrase belongs to a class of enzymes known as topoisomerases that are involved in the control of topological transitions of DNA.

A researcher is working with a certain strain of bacteria that will grow at 37°C, but only on plates that contain agar, glucose, arginine, and leucine. This organism can be classified as a(n): A. eukaryote. B. thermophile. C. auxotroph. D. psychrophile.

C If the bacteria requires arginine and leucine to grow, it is an arginine auxotroph and a leucine auxotroph. If the bacteria will grow at body temperature, it is best classified as a mesophile. Bacteria are prokaryotic, not eukaryotic. **32 degrees F = 0 degrees C so 37 degrees C is HOT

An experiment with a previously unidentified pathogen is done using guinea pigs and various means of inoculation are attempted. Within a few days, the guinea pigs begin to demonstrate a loss of coordination along with other neurological symptoms. Could the organism be a prion? A. Yes; direct inoculation is required to contract a prion infection. B. Yes; loss of coordination is specific to prion infections. C. No; the onset of symptoms was too rapid. D. No; prions do not infect guinea pigs.

C Prion incubation times are long; in humans, it can take years or even decades. The onset as described in the question is too rapid, even in the small animal model of the guinea pig. Prions can infect all sorts of animals so guinea pigs are not specifically immune and loss of coordination can be a sign of all sorts of disorders and infections. There are many means by which prion infection can be acquired, including mutation. *inoculation = treat with a vaccine

Which of the following is a difference between eukaryotic and prokaryotic translation? A. The function of the codon UAA B. The first translated codon C. The mechanism by which ribosomes recognize the 5' end of mRNA D. The number of times a single mRNA transcript can be translated

C Prokaryotic mRNA is recognized by the ribosome using the Shine-Dalgarno sequence (-10) while eukaryotic mRNA is recognized via the 5' cap that is added during post-transcriptional modification. Both eukaryotes and prokaryotes begin translation at an AUG; this specifies methionine in eukaryotes and formyl-methionine in prokaryotes. Both utilize the codon UAA to signify a stop codon rather than an amino acid. Both eukaryotes and prokaryotes can translate each mRNA transcript multiple times to generate more protein. **Kozak consensus sequence plays a major role in the initiation of the translation process; is a sequence which occurs on eukaryotic mRNA

Which of the following would NOT be required to run a polymerase chain reaction (PCR)? A. Primers B. Template DNA C. RNA polymerase D. Taq DNA polymerase

C RNA polymerase is used during transcription, not replication, so it is not needed for this technique. The requirements for the polymerase chain reaction are similar to what is required in a cell for DNA replication. Thus, DNA polymerase, template DNA, and primers would all be required.

Which of the following would lengthen Okazaki fragments? A. Increasing the rate of all aspects of replication B. Separating stop transcription sequences to a greater degree during replication C. Decreasing the number of primers generated on the lagging strand during replication D. Increasing the number of origins on the DNA strand

C This would result in DNA polymerase III traveling uninterrupted for a longer period of time and generating longer Okazaki fragments. Increasing the number of origins is likely to shorten Okazaki fragments since this would increase the number of primers, and DNA polymerase stops replicating when it runs into the next primer. Stop transcription sites do not affect replication, they stop transcription. Increasing the replication rate will not necessarily change the fragment size but will likely just generate them more quickly.

Which of the following is the most likely effect of a eukaryotic transcription factor? A. Serve to inhibit translation to allow for completion of transcription. B. Provide the energy necessary for nascent RNA polymerization. C. Increase the encounter rate of DNA with RNA polymerase. D. Bind at specific AUG sequences to start transcription.

C Transcription factors influence many processes in transcription including recruiting RNA polymerase, regulating transcription rate, and many others. In eukaryotes, transcription and translation are separated physically (transcription occurs in the nucleus and translation in the cytoplasm) so there would be no need for a transcription factor to inhibit translation. DNA does not contain AUG as uracil is only seen in RNA sequences and this sequence signifies the start sequence for translation, not transcription. The energy needed for RNA polymerization is derived from the phosphates that are cleaved (releasing pyrophosphate) from the nucleotides.

Which of the following functions is NOT typically attributed to small nuclear RNA (snRNA)? A) Processing of pre-mRNA B) regulation of transcription factors C) coordinating amino acid addition in translation D) maintaining telomeres

C transfer RNA (tRNA) is typically involved in the process of coordinating the amino acids that are added to a growing protein during translation translation does not occur in the nucleus! (which is the location of the snRNA) Processing of pre-mRNA, regulation of transcription factors, and maintenance of telomeres are all functions of snRNA and take place in nucleus

Which of the following is NOT true about bacterial conjugation? A. It changes an F- "female" bacterium into an F+ "male" bacterium. B. It increases the genetic diversity and the size of a bacterial population. C. It involves the transfer of either plasmid DNA, genomic DNA, or both, between two bacteria. D. It can be used to map the bacterial genome.

C (but I think it is B) All of the other answer choices are true. Conjugation occurs when an F+ "male" bacterium builds a conjugation bridge to an F- "female" bacterium in order to transfer DNA from male to female. Often this is just plasmid DNA (the F plasmid). However, in some cases (Hfr bacteria) the plasmid is incorporated into the bacterial genome, and conjugation results in the transfer of both genomic and plasmid DNA. Hfr bacteria can be used in genomic mapping experiments, in which conjugation is allowed to proceed for varying amounts of time, and the order in which genes are transferred to the recipient determines the order of those genes in the genome. At the end of conjugation, the F- "female", having received the F plasmid, is now an F+ "male".

Decreased secretion of aldosterone has all of the following effects EXCEPT: A) increased loss of Na+ in urine. B) decrease in extracellular fluid volume. C) increase in arterial pressure. D) increase in urine volume.

C. Aldosterone causes increased sodium reabsorption, and because of the rise in systemic Na+, there will be decreased water loss, increased extracellular fluid volume and increased blood pressure. Decreased secretion of aldosterone would cause the opposite: increased loss of sodium (choice A is true and eliminated), a decrease in the extracellular fluid volume (choice B is true and eliminated), and an increase in urine volume (choice D is true and eliminated). It will not cause an increase in blood pressure, however, since more urine is formed and extracellular fluid volume lost (choice C is false and thus the correct answer choice).

Which of the following processes do NOT occur at the placenta? I. Diffusion of amino acids from fetal blood to maternal blood II. Mixing of maternal and fetal blood and gas exchange III. Exchange of CO2 and O2 between fetal and maternal blood A) I only B) II only C) I and II only D) I and III only

C. Item I does not occur: Amino acids are charged and cannot cross the placental and fetal capillary membranes (eliminate choice B). Item II does not occur: Fetal and maternal blood do not mix, so the only way substances can cross the placental barrier is if they are lipid soluble or if they are actively transported (eliminate choice A and D). Item III is correct: CO2 and O2 are lipid soluble and easily cross the barrier (choice C is correct).

In the absence of oxygen, the oxidation of glucose produces which of the following final products? I. CO2 II. NADH III. ATP A) I only B) II and III only C) III only D) I, II, and III

C. Item I is false: CO2 is produced only by the complete oxidation of glucose through PDC and the Krebs cycle (choices A and D are wrong). Note that both of the remaining answer choices include Item III, thus Item III must be true and we can focus only on Item II. Item II is false: in the absence of oxygen, glycolysis can proceed anaerobically (fermentation) but no net NADH is produced; all NADH is oxidized back to NAD+ during the reduction of pyruvate to lactic acid (choice C is correct and choice B is wrong). Item III is true: 2 net ATP are made during glycolysis.

Which of the following would be the most effective method to determine if transcription of a gene had been silenced? A) DNA fingerprinting B) ELISA C) RT-PCR D) DNA sequencing

C. RT-PCR uses reverse transcriptase to make cDNA from mRNA. The cDNA can then amplified, and then tested for. This is the only method listed that can be used to test for the presence of mRNA and determine if transcription of the gene is occurring or not (choice C is correct). Both DNA Fingerprinting and DNA Sequencing are methods used to analyze DNA sequences; this does not provide information about transcription (choices A and D can be eliminated). While ELISA could be used to test for the presence of a protein product, it would not be the best test to use in this situation (choice B is incorrect). The absence of a protein could be due to a variety of reasons, including problems with translation, protein degradation, etc. Thus, RT-PCR would be a better method to specifically test for gene expression (i.e., transcription).

During peak cardiac activity, heart muscle utilizes the anaerobic pathway for only 10% of its energy as opposed to up to 85% for maximally contracting skeletal muscle. With respect to skeletal muscle, cardiac muscle would: A) convert more pyruvate to lactate. B) consume 1/8 as much ATP to produce the same contractile force. C) consume less glucose per ATP produced. D) require less oxygen per molecule of ATP produced.

C. Aerobic respiration includes glycolysis, the PDC, the Krebs cycle, and electron transport, to produce a total of 32 ATP per glucose (assuming the malate-aspartate shuttle is used to transfer the electrons from glycolytic NADH into the mitochondria). By contrast, anaerobic respiration involves only glycolysis and fermentation of pyruvate to lactate or alcohol, to produce 2 ATP per glucose. Anaerobic respiration is therefore much less efficient and must consume much more glucose to produce the same amount of ATP. Since cardiac muscle uses mostly aerobic respiration, it will be more efficient than skeletal muscle and uses less glucose to produce an equivalent amount of ATP (choice C is correct). Conversion of pyruvate to lactate occurs during fermentation, which cardiac muscle carries out less of, not more (choice A is wrong). The amount of contractile force generated per ATP is independent of the source of ATP used (choice B is wrong). Cardiac muscle uses more aerobic respiration, so it will use more oxygen per ATP produced, not less (choice D is wrong).

The most likely explanation for the inverse relationship between mammal size and pulse is that: *From Passage: Heart rates among different mammals are inversely related to animal size. A large animal, such as an elephant, has a pulse of approximately 25 beats/minute. The pulse of a tiny shrew, on the other hand, is twenty-four times more rapid. A) most small mammals live under anaerobic conditions. B) large animals tend to consume proportionately more food than smaller animals. C) smaller mammals have a higher metabolic rate for a given mass of tissue than larger mammals. D) smaller mammals have reduced oxygen requirements for a given mass of tissue.

C. Choice A is wrong since all mammals use primarily aerobic respiration except for short periods of unusually strenuous activity. Choice B is wrong since it does not directly involve metabolic rate or pulse and would predict that the metabolic rate and pulse rate will remain the same despite size. Choice D is wrong since reduced oxygen requirements would predict a slower pulse. Choice C is the best choice: If tissues in smaller animals have a higher metabolic rate, they will require a greater supply of blood to provide oxygen and nutrients and carry away carbon dioxide. To meet these greater needs, the pulse rate could be increased.

All of the following processes occur during the conduction of a nerve impulse across the synapse of two neurons EXCEPT: A) the release of vesicles from the presynaptic cell. B) the opening of postsynaptic ion channels. C) a retrograde depolarization of the presynaptic cell. D) the depolarization of the postsynaptic cell.

C. During the conduction of an action potential across a chemical synapse, neurotransmitter is released by the presynaptic cell, it diffuses across the synapse, binds to receptors on the postsynaptic cell, opens ion channels on the postsynaptic cell, and depolarizes (or hyperpolarizes) the postsynaptic cell. Thus, choices A, B, and D are all part of the normal conduction of an action potential across a synapse and can be eliminated. The presynaptic cell does not normally transmit a retrograde action potential, however (choice C is false and the correct response here). This is usually prevented by the refractory period after an action potential passes, causing action potentials to propagate in only one direction along an axon. **retrograde action potentials: it has also been shown that an action potential initiated in the axon can create a retrograde signal that travels in the opposite direction (Hausser 2000). This impulse travels up the axon eventually causing the cell body to become depolarized, thus triggering the dendritic voltage-gated calcium channels

Which of the following conditions leads to a reduction in water reabsorption by the kidney? A) High plasma levels of vasopressin B) Increased osmolarity of interstitial fluid in the kidney medulla C) Decreased permeability of the collecting duct to water D) An ACTH-secreting tumor

C. Factors that reduce water resorption would tend to increase urinary output, decrease blood volume, and decrease blood pressure. High levels of vasopressin (ADH) would increase water resorption and plasma volume (choice A is wrong). The greater the osmolarity of the interstitial fluid in the medulla of the kidney, the greater the water resorption as filtrate passes into the medulla in the descending loop of Henle (choice B is wrong). As filtrate passes through the collecting duct, water can be reabsorbed and the urine concentrated. If the collecting duct is impermeable to water, however, then less water will be reabsorbed, and the filtrate will remain dilute (choice C is correct). ACTH is the hormone that controls secretion of aldosterone (as well as cortisol) by the adrenal cortex. An ACTH-secreting tumor will cause elevated secretion of aldosterone, increased Na+ reabsorption, and a subsequent increase in water reabsorption (choice D is wrong).

If a population has been in Hardy-Weinberg equilibrium for ten generations, which of the following conclusions can be made? A) Evolution has occurred for the last ten generations. B) Selection factors have been changing over the last ten generations. C) The percentage of alleles in the present generation is the same as that of five generations ago. D) Allele frequencies have changed at the same rate each generation.

C. If a population is in Hardy-Weinberg equilibrium, then the allele frequencies in the gene pool will not change from one generation to the next (choice C is correct, and choice D is wrong). If the allele frequencies do not change, then the population will be genotypically and phenotypically identical after ten generations and no evolution can occur (choice A is wrong). There can be no natural selection according to Hardy-Weinberg (choice B is wrong).

What is the FRC for an individual with the following lung volumes: total lung capacity (TLC) = 5700 mL, vital capacity (VC) = 4700 mL, inspiratory reserve volume (IRV) = 3000mL and tidal volume (VT) = 500 mL? *the passage states that FRC = ERV + RV A) 3500 mL B) 1000 mL C) 2200 mL D) 1200 mL

C. The passage states that FRC = ERV + RV. TLC = VC + RV rightwards arrow RV= TLC - VC = 5700 mL - 4700 mL = 1000 mL VC = VT + IRV + ERV rightwards arrow ERV = VC - VT - IRV = 4700 mL - 500 mL - 3000 mL= 1200 mL FRC = ERV + RV = 1200 mL + 1000 ml = 2200 mL Thus choice C is correct and choices A, B and D are eliminated.

Which of the following best describes the cascade that leads to cellular apoptosis after exposure to intra- or extracellular death signals? A) Effector kinases cluster together and activate each other. The activation of effector kinases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. B) Effector caspases cluster together and activate each other. The activation of effector caspases leads to the activation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. C) Initiator caspases cluster together and activate each other. The activation of initiator caspases leads to the activation of effector caspases, which cleave proteins at aspartic acid sites, triggering apoptosis. D) Effector kinases cluster together with initiator capsases, activating the initiator capsases. The activation of effector kinases leads to the deactivation of initiator caspases, which cleave proteins at aspartic acid sites, triggering apoptosis.

C. This is a recall question pertinent to concepts discussed in the passage (apoptosis). Apoptosis is carried out by proteases called caspases, which cleave targets at aspartic acid residues. This eliminated choices A and D, which both indicate that apoptosis is started by effector kinases. Choice C is correct: Initiator caspases cluster together and initiate apoptosis (thus the name). These initiator caspases can then activate effector caspases, which cleave proteins at aspartic acid residues; this triggers apoptosis. Choice B is incorrect, and confuses the role of initiator and effector caspases.

Thirteen amino acids, including methionine, valine and proline, are glucogenic in humans. This means their α-keto acid carbon skeleton is converted to pyruvate during amino acid catabolism. After deamination, valine can therefore: I. Be converted into CO2 and H2O to generate ATP. II. Generate at least three NADH and two FADH2. III. Enter gluconeogenesis to generate glucose. A) I only B) III only C) I and II only D) I and III only

D *glucogenic amino acid = amino acid that can be converted into glucose through gluconeogenesis Item I is true: Amino acids are catabolized via deamination into α-keto acids and ammonia. Based on the information in the question stem, the α-keto acid formed from valine will be converted to pyruvate. Pyruvate can keep going through cellular respiration to generate CO2, H2O and ATP. Eliminate choice B. Item II is false: Pyruvate is converted into one acetyl-CoA (during which 1 NADH is made), and the acetyl-CoA would then generate three NADH and only one FADH2 as it cycles through the Krebs cycle. Eliminate choice C. Item III is true: Pyruvate can also enter gluconeogenesis to generate glucose. Eliminate choice A and choice D is correct.

Which of the following is specific to an animal virus and NOT to a bacteriophage? A. Destruction of the host cell during the productive cycle B. Infection of the host cell occurs via penetration of genome through host cell membrane C. Lysis of the cell membrane after the lytic cycle D. Uncoating of a genome from the capsid coat within the host cell cytoplasm

D Animal viruses can be endocytosed whole (i.e., uncoated) through a cell membrane, and then uncoating of the genome occurs in the cytoplasm. In bacteriophages, such as the T4 phage, the capsid remains outside of the cell and the central shaft injects the genome through the membrane directly. Both bacteriophages and animal viruses must get their genome into a host cell (infection of the host cell occurs via penetration of genome through host cell membrane is true of both viruses, so is wrong). There is no immediate destruction of the host cell with the productive cycle, generally, because the virus merely buds out of the cell via exocytosis (destruction of the host cell during the productive cycle is wrong). Lysis of the cell membrane does occur during the lytic cycle, but this is generally true for both animal and bacterial viruses (lysis of the cell membrane after the lytic cycle is not the best answer).

Which of the following is the most accurate, from least organized to most organized? A. Deoxyribose, nucleotide, nucleoside, nucleosome, DNA helix, chromatin B. Deoxyribose, nucleotide, nucleoside, DNA helix, nucleosome, chromatin C. Deoxyribose, nucleoside, nucleotide, DNA helix, chromatin, nucleosome D. Deoxyribose, nucleoside, nucleotide, DNA helix, nucleosome, chromatin

D Deoxyribose is the sugar component of the DNA backbone. A nucleoside is the next level of organization and is a deoxyribose with the nitrogenous base attached; a nucleotide is more organized than a nucleoside and includes phosphate groups ("Deoxyribose, nucleotide, nucleoside, DNA helix, nucleosome, chromatin" and "Deoxyribose, nucleotide, nucleoside, nucleosome, DNA helix, chromatin" are wrong). Double stranded DNA winds around histone octamers to form nucleosomes ("Deoxyribose, nucleoside, nucleotide, DNA helix, chromatin, nucleosome" is wrong), which are further packaged and condensed to form chromatin.

If an electrophoretic gel is placed in reverse, but the DNA samples are still loaded into the established wells, what will happen to the sample? A. The samples will not run at all. B. The samples will still run down the gel, but the separation will not be as good since they will be running toward the negative electrode. C. The samples will separate based on density rather than weight. D. The samples will run out of the back of the gel towards the positive electrode rather than down the length of the gel toward the negative electrode.

D Electrophoresis uses electrical current to separate based on size and weight, not density. Since DNA is negatively charged, it will still be trying to move away from the negative electrode and towards the positive electrode, thus it will end up running towards the back of the gel and off of it. DNA cannot run towards the negative electrode at all since it is also negatively charged.

Which of the following is a true statement regarding fungi? A. During budding, fungi release haploid hyphae with the intent of finding other hyphae of the same species for sexual reproduction. B. Production of endospores allows fungi to survive during times of non-ideal environmental conditions. C. Hyphae that are specialized to digest and absorb nutrients are called fruiting bodies. D. Gametes may join together and exist in a dikaryon form until fusion of the nuclei occurs.

D Gametes may join without immediate fusion of nuclei in fungi. The result is a cell with two nuclei, called a dikaryon. Fungi produce spores (not endospores, like bacteria for help with survival (production of endospores allows fungi to survive during times of non-ideal environmental conditions is wrong). Budding is an asexual process, not sexual (during budding, fungi release haploid hyphae with the intent of finding other hyphae of the same species for sexual reproduction is wrong). The hyphae that digest nutrients are called haustoria, not fruiting bodies (hyphae that are specialized to digest and absorb nutrients are called fruiting bodies is wrong).

How do chromosomal translocations end up potentially creating new gene products or enhancing the activity of existing gene products? A. Recombination occurs only between somatic and sex chromosomes. B. Recombination occurs only between the arms of the same chromosome. C. Recombination occurs between homologous chromosomes, but the exchange of gene segments is unequal. D. Recombination occurs between non-homologous chromosomes placing previously unconnected sequences in proximity.

D Incomplete recombination between homologous chromosomes would not have the same effect as the genes on homologous chromosomes are essentially the same. Recombination does not occur between the arms of a single chromosome. While recombination between a somatic and sex chromosome could be deleterious, translocations are not limited to such an exchange.

A bacterial strain auxotrophic for lysine, isoleucine, and alanine production is mixed with an Hfr bacterial strain whose genome needs to be mapped. A sample taken after three minutes is able to grow on media supplemented with isoleucine and alanine. A sample taken after six minutes is able to grow on media supplemented with only isoleucine. A sample taken after nine minutes is able to grow on minimal media. What is the order of the amino acid synthesis genes on the Hfr strain? A. Lysine - Isoleucine - Alanine B. Alanine - Isoleucine - Lysine C. Isoleucine - Lysine - Alanine D. Lysine - Alanine - Isoleucine

D It can be assumed that the bacteria are conjugating. After three minutes, the resultant new strain is able to grow in the absence of leucine, indicating that the leucine synthesis gene was transferred first. After six minutes, it was able to grow in the absence of leucine and alanine (the alanine synthesis gene was transferred second), and after nine minutes the new bacterial strain needed no supplementation, indicating that the isoleucine synthesis gene was the final gene transferred.

Dinitrophenol allows protons to freely cross the inner mitochondrial membrane. Cells treated with dinitrophenol will display which of the following activities? I. Krebs cycle II. Electron transport III. ATP synthesis A) I only B) I and II only C) II and III only D) I, II, and III

D Items I and II are true: If protons freely flow across the inner membrane, the proton gradient will be destroyed. If protons flow across the membrane without passing down a gradient through the ATP synthase, then ATP production will halt. The Krebs cycle and electron transport will actually accelerate to attempt to rebuild the proton gradient (eliminate choices A and C). Item III is also true: ATP production during glycolysis will be unhindered (eliminate choice B and choice D is correct).

Prions do not follow the principles of the Central Dogma and are referred to as self-replicating proteins. Which of the following best describes their means of replication? A. A disease-causing version of the protein recruits loaded and activated tRNAs to synthesize new prions. B. A regular version of the protein undergoes a nonsense mutation to induce a change to the disease-causing version. C. A regular version of the protein automatically becomes disease-causing when produced in neurons. D. A disease-causing version of the protein acts as the folding template to cause a regular version of the protein to become misshapen.

D Prions use the disease-causing version of the protein to induce changes in the normally present, regular version of the protein, thus being termed self-replicating. tRNAs, even when loaded with an amino acid and activated, will not synthesize new proteins with the ribosomal machinery to guide that process. The genes to encode a protein can incur a nonsense mutation, but not the protein itself. Just being in a neuron is not sufficient to alter prion folding; the disease-causing version needs to also be present. **Prions = misfolded, self-replicating proteins responsible for a class of diseases known as transmissible spongiform encephalopathies (TSEs) that cause degeneration of CNS tissues

How is a standard curve formulated for a radioimmunoassay? A. Unlabeled antigen is mixed with antibody followed by radiolabeled antigen and the increases in radioactivity are measured until they asymptotically level. B. Radiolabeled antigen is mixed with antibody followed by antigen without the radiolabel and the transfer of radioactivity between them is measured. C. Radiolabeled antigen and unlabeled antigen are simultaneously mixed with antibody and competitive binding determines the shape of the curve. D. Radiolabeled antigen is mixed with antibody followed by antigen without the radiolabel and the decrease in radioactivity is measured as the amount of unlabeled antigen increases.

D Radioimmunoassays are a type of competition assay where the radiolabeled version is offset by the unlabeled version and the decrease in radioactivity indicates the level of the unknown being measured. This process is done with known concentrations of each to establish the standard curve to which results from unknowns can then be compared.

Which of the following is a best example of an opportunistic bacterial infection? A. Ulcers caused by Helicobacter pylori B. Food poisoning cause by exotoxins C. A Streptococcal infection of the throat D. A Staphlycoccal infection in a wound

D Staphylococcus is a normal part of the skin's flora and is helpful in combating other pathogenic organisms unless it ends up on damaged or exposed tissue, such as an open wound, where it can then cause an opportunistic infection. Streptococcus is not part of the throat's flora nor H. pylori in the gut. Food poisoning is just that - the exotoxins produce the symptoms rather than an actual infection. ** opportunistic infection is an infection caused by bacterial, viral, fungal, or protozoan pathogens that take advantage of a host with a weakened immune system or an altered microbiota (such as a disrupted gut flora). Many of these pathogens do not cause disease in a healthy host that has a normal immune system. A compromised immune system, however, presents an "opportunity" for the pathogen to infect. ***Immunodeficiency or immunosuppression can be caused by: Malnutrition Fatigue Recurrent infections Immunosuppressing agents for organ transplant recipients Advanced HIV infection Chemotherapy for cancer Genetic predisposition Skin damage Antibiotic treatment leading to disruption of the physiological microbiome, thus allowing some microorganisms to outcompete others and become pathogenic (e.g. disruption of intestinal flora may lead to Clostridium difficile infection Medical procedures Pregnancy

When running an ELISA to test for the presence of anti-chickenpox antibody in a patient's blood, which of the following would be bound to the microtiter well plate? A. primary anti-chickenpox antibody B. secondary anti-chickenpox antibody C. conjugated enzyme D. chickenpox antigen

D When testing for the presence of antibody in a patient's blood, the antigen for which that antibody is specific must be bound to the microtiter well plate. Primary antibody is only bound to the well plate when testing for the presence of antigen in a patient's blood. Neither secondary antibody (the one conjugated to an enzyme) nor the enzyme itself can ever be bound to the well plate or false-positive results would occur.

In order to study the mechanism by which pyruvate is converted to acetyl CoA, the enzyme responsible for the conversion had to be located. It would most likely by found in the: A) cytosol of the cell B) plasma membrane C) endoplasmic reticulum D) matrix of the mitochondria

D describing the process of pyruvate dehydrogenase pyruvate dehydrogenase and krebs cycle occur in matrix of mitochondria

Which is true concerning viruses? A. The productive cycle is the most efficient infective cycle for phages B. Viruses that infect human cells must have an envelope C. Genetic information can be transferred between hosts via transfection D. A virus with an RNA genome must code for an RNA-dependent RNA polymerase

D In order to replicate its genome, and RNA virus must code for an RNA-dependent RNA polymerase; this enzyme will create a new strand of RNA by reading a template strand of RNA Viral host cells will not express these enzymes naturally; they have no need to make RNA by reading RNA Host cells normally produce RNA using DNA as a template Phages only infect bacteria, and can only undergo the lytic or lysogenic cycles; the productive cycle involves budding through cell membrane and can't occur in hosts with cell walls, like bacteria although viruses with an envelope (lipid bilayer coating) are restricted to infecting animal cells, the outer membrane is not required Genetic info can be transferred between hosts but that is called transduction

All of the following contribute to speciation EXCEPT: A) geographic isolation of populations. B) genetic diversity. C) natural selection. D) maintenance of Hardy-Weinberg equilibrium.

D. **Speciation is the evolutionary process by which new reproductively isolated biological kinds, species, arise Speciation requires two populations to become reproductively isolated so that they can no longer interbreed. Geographic isolation is the easiest way to allow two populations to diverge and become reproductively isolated (choice A is true and eliminated). To become reproductively isolated, the populations must diverge genetically, evolving differently, and this would require genetic diversity in the population to begin (choice B is true and eliminated). Natural selection could drive the changes in isolated populations that cause them to evolve into two species (choice C is true and eliminated). If Hardy-Weinberg is maintained, then there can be no changes in the allele frequencies in the gene pool of a population and no evolution, which would not allow speciation (choice D is false and the correct answer choice here).

The conformational change of a regulatory protein after the binding of a repressor most likely represents an alteration of the protein's: A) amino acid composition. B) primary structure. C) secondary structure. D) tertiary structure.

D. A conformational change in a protein is a change in the larger scale folding of a protein, with changes in the relative positions in space of amino acids located far from each other in the linear polypeptide chain. The tertiary structure of a protein involves large-scale structure within a polypeptide chain that is stabilized by interactions between amino acids that can be distant from each other in the linear sequence; thus, this is the most likely level of structure altered during a conformational change (choice D is correct). The amino acid composition and primary structure (the linear sequence of amino acid residues in the polypeptide chain) cannot be changed without breaking covalent peptide bonds, which does not occur in a conformational change (choices A and B are wrong). The secondary structure of a protein involves folding that is stabilized by the nearest neighbors in the polypeptide chain, including structures such as α helix and β sheet. Conformational changes can alter secondary structures somewhat, but mostly alter tertiary structure (choice D is a better answer than choice C).

In a male individual with Down's Syndrome (trisomy 21), how many chromosomes would be visible at metaphase I of spermatogenesis? A) 23 B) 24 C) 46 D) 47

D. An individual with trisomy 21 has an extra copy of chromosome 21 (three total copies). During metaphase I, the developing gametes are still diploid (separation of homologues has not yet occurred), so this individual would have the normal 46 chromosomes plus the extra copy of chromosome 21, for a total of 47 chromosomes (choice D is correct; eliminate choices A, B and C).

Which of the following represents the correct sequence for embryogenesis? A) Fertilization → gastrulation → blastulation → neural tube formation → somite formation B) Fertilization → gastrulation → blastulation → somite formation → neural tube formation C) Fertilization → blastulation → neural tube formation → gastrulation → somite formation D) Fertilization → blastulation → gastrulation → neural tube formation → somite formation

D. Fertilization is the first step, followed by a series of rapid cell cleavages to form a hollow ball of cells called the blastulam (eliminate choices A and B). Next comes the gastrula, in which cells move into the interior of embryo to form the three germ layers (eliminate choice C). Gastrulation is followed by the formation of the neural tube, which will form the nervous system, followed by the formation of other organs and tissues, such as the somites that will differentiate into bones and muscle. This makes choice D the only possible correct order of events. *Somites: blocks of mesoderm that are located on either side of the neural tube in the developing vertebrate embryo. Somites are precursor populations of cells that give rise to important structures associated with the vertebrate body plan and will eventually differentiate into dermis, skeletal muscle, cartilage, tendons, and vertebrae. Somites also determine the migratory paths of neural crest cells and of the axons of spinal nerves

2,4-dinitrophenol (2,4-DNP) is a highly toxic substance which was sold to the public as a weight loss drug in the 1930s. It acts by permeabilizing the inner mitochondrial membrane (IMM) to ions. Which of the following is true of 2,4-DNP's effects on oxidative phosphorylation? A) It causes decreased flux of electrons through ATP synthase. B) It likely leads to a decrease in body temperature. C) It leads to decreased consumption of FADH2 and NADH by the electron transport chain proteins. D) It causes a decrease in the electromotive potential built up by the electron transport chain.

D. If the inner mitochondrial membrane became permeable to ions, then hydrogen ions would not need to go through the ATP synthase in order to re-enter the matrix. This would dissipate the proton gradient established by the electron transport chain, and decrease its potential to generate ATP (choice D is correct). FADH2 and NADH would still be consumed by the transport chain proteins, since the electron transport chain is not shut down; it's just that the proton gradient would be more easily dissipated (choice C is wrong). The dissipation of the gradient would result in heat production; note that this is similar to what happens in brown fat, used by hibernating animals to stay warm (choice B is wrong). Electrons do not flow through the ATP synthase (choice A is wrong).

Most biological unsaturated fatty acids are cis and contain non-conjugated double bonds. Because of this, additional steps are required in β-oxidation. These most likely include: A) Changing a single double bond to trans via a reductase enzyme in the mitochondrial matrix. B) Combining two double bonds via a reductase enzyme (which uses NAD+ as a reducing agent), then changing the resultant trans double bond to cis via an isomerase enzyme. C) Moving a single double bond down the fatty acid chain via a translocase enzyme in the mitochondrial matrix. D) Combining two double bonds via a reductase enzyme (which uses NADPH as a reducing agent), then changing the resultant cis double bond to trans via an isomerase enzyme.

D. Monounsaturated fatty acids require an isomerase enzyme to move the double bond during β-oxidation (eliminate choice A). Polyunsaturated fatty acids require both an isomerase and a reductase enzyme to complete β-oxidation; this also requires the reducing agent NADPH (choice D is correct). Note that NAD+ is an oxidizing agent (not a reducing agent) because it oxidizes another molecule, and is thus reduced itself (eliminate choice B). β-oxidation does not require a translocase enzyme (eliminate choice C).

The blood vessel that carries the most highly oxygenated blood in the fetus is the: A) pulmonary vein. B) ductus arteriosus. C) umbilical artery. D) umbilical vein.

D. The most highly oxygenated blood would be the blood returning from the point of oxygenation to the systemic circulation. In the adult this would be the pulmonary vein, but the point of oxygenation in the fetus is the placenta, not the lungs (choice D is correct and choice A is wrong). The ductus arteriosus does carry oxygenated blood but is found after the umbilical vein, so it does not carry the most highly oxygenated blood (choice B is wrong). The umbilical artery carries deoxygenated blood from the fetus to the placenta (choice C is wrong). **pulmonary vein = carries oxygenated from lungs to heart **pulmonary artery = carries deoxygenated veins carry blood to heart artery carries blood away from heart

From Passage: The Luria-Delbruck experiment was devised to resolve a debate regarding the source of genetic variability in bacteria. It was observed that certain strains of Escherichia coli were sensitive to infection by bacteriophage T1, which leads to the lysis of the host bacteria. Q: Which of the following is the most likely sequence of events in the infection of E. coli by phage T1? A) Replication of viral genome, production of viral DNA polymerase, translation of viral lysozyme, assembly of infectious virus B) Production of viral DNA polymerase, replication of viral genome, translation of viral lysozyme, assembly of infectious virus C) Translation of viral lysozyme, production of viral DNA polymerase, replication of viral genome, assembly of infectious virus D) Production of viral DNA polymerase, replication of viral genome, assembly of infectious virus, translation of viral lysozyme

D. The production of viral DNA polymerase must occur prior to replication of the viral genome since the polymerase is needed to do this (choice A is wrong). The last step must be production of lysozyme since this will lyse the cell, bursting it open and stopping any further activities required to produce infectious virus (choices A, B, and C are wrong). The correct events are in choice D: the expression of the viral DNA polymerase must come first to then replicate the genome. Infectious virus can then be assembled, and released by cell lysis induced by viral lysozyme.

Which of the following hormones acts on target cells via a second-messenger system? A) Thyroid hormone B) Aldosterone C) Cortisol D) Epinephrine

D. Thyroid hormone, aldosterone, and cortisol are all small hydrophobic molecules that passively diffuse through the plasma membrane to bind to receptors inside the cell in the cytoplasm and nucleus. These receptors then regulate transcription, without the use of second-messenger systems (choices A, B, and C are wrong). A hydrophilic hormone like epinephrine, however, binds to cell-surface receptors and activates adenylate cyclase to make cAMP, a second messenger (choice D is correct). Note that in general, steroid hormones exert their effects by modifying transcription, while protein hormones utilize second-messenger systems. Thyroid hormone, a protein hormone, is an exception.

Vinca alkaloids are a class of anti-cancer drugs derived from the periwinkle plant. Once absorbed into a cell, they interfere with the polymerization of microtubules. These drugs can prevent cancer from spreading by disrupting: A) pseudopod formation, thereby preventing cellular locomotion. B) prophase, thereby halting mitosis. C) transcription, thereby halting production of crucial cell proteins. D) metaphase, thereby halting tumor growth.

D. Formation of microtubules is crucial for formation of the metaphase plate. Microtubules polymerize from the centrioles outward. They contact centromeres (to become kinetochore fibers) and "push" the chromosomes towards the center of the cell to form the metaphase plate. Without proper microtubule polymerization, metaphase (and the rest of mitosis) cannot occur. If mitosis cannot occur, then neither can tumor growth. Note that microtubules are also required during prophase in forming the mitotic spindle; however, some parts of prophase can still occur in the absence of microtubule formation (DNA condensation, loss of the nuclear membrane). This makes choice D a better choice than choice B, since virtually all of metaphase depends on proper microtubule polymerization. Pseudopod formation requires the growth of microfilaments (actin fibers), not microtubules (choice A is wrong), and transcription (RNA polymerization) does not require microtubules at all (choice C is wrong).

Complete reduction of acetic acid will produce which of the following? A) O2(g) B) CO2(g) C) CH3CHO D) CH3CH2OH

D. Reduction of acetic acid will initially produce the aldehyde given as choice C, where the carbonyl carbon has been reduced from an oxidation state of +3 to +2. Choice D is the result of further reduction to an oxidation state of +1, while choices A and B do not correspond to the reduction of acetic acid.

A chemist prepares a 1 M solution of sulfuric acid. Which of the following gives the relative concentrations of the species in solution? A) [H2SO4] > [HSO4-] > [SO42-] B) [SO42-] > [HSO4-] > [H2SO4] C) [HSO4-] > [H2SO4] > [SO42-] D) [HSO4-] > [SO42-] > [H2SO4]

D. Since sulfuric acid is a strong acid and dissociation of the first proton is complete, undissociated H2SO4 will have the lowest relative concentration, eliminating choices A and C. Since the dissociation constant for the loss of the second proton is well below 1 (~ 10-2), HSO4- will exist in greater concentration than SO42-, making choice D the correct answer.

HHV-8 is a Rhadinovirus and is remarkable since it has stolen numerous genes from host cells, including genes that code for complement-binding proteins, interleukin-6, BCL-2, cyclin-D, and a G protein-coupled receptor. What is the most likely mechanism by which this was accomplished? A) The lytic cycle B) Transduction C) Transformation D) Inaccurate provirus excision

D. The lytic cycle involves host cell lysis to release new viral particles and does not explain how HHV-8 could acquire host genes (choice A can be eliminated). Transformation is the uptake of DNA from the environment into a bacterium and cannot explain how HHV-8 acquired human genes (choice C can be eliminated). When the provirus leaves the host genome (excises), it can leave behind some viral DNA, or it can take some extra host DNA, altering the viral genome. This inaccurate excision could explain how HHV-8 acquired copies of some genes that originated in humans. Transduction is the term used to describe this process in bacteria (not humans), which makes choice D a better answer than choice B.

___________ is the accumulation of differences between groups which can lead to the formation of new species, usually a result of diffusion of the same species to different and isolated environments which blocks the gene flow among the distinct populations allowing differentiated fixation of characteristics through genetic drift and natural selection

Divergent evolution

T or F: DNA Polymerase can elongate DNA without a primer.

FALSE DNA polymerase needs primers in order to elongate DNA in DNA replication primer provides a free 3'OH group

T or F: A patient can be resistant to an antibiotic. T or F: An enveloped virus can help eradicate antibiotic-resistant bacteria in a patient.

FALSE the bacteria are, not the patient FALSE; can only infect animal cells

What are Hardy-Weinberg's principles?

Frequencies of alleles in the gene pool of a population will not change over time, provided that a number of assumptions are true: 1. There is no mutation 2. There is no migration 3. There is no natural selection 4. There is random mating 5. The population is sufficiently large to prevent random drift in allele frequences

Genetic penetrance vs. Genetic Pleiotropism

Genetic Penetrance: EX: Cowden syndrome is associated with loss-of-function mutations in the lipid phosphatase PTEN, and causes an increased risk of certain forms of cancer. This is an example of genetic penetrance, since a certain genotype has the potential to lead to a phenotype (cancer) but this does not occur in all instances. Genetic Pleiotropism: a single gene affects a number of phenotypic traits in the same organism. These pleiotropic effects often seem to be unrelated to each other. The usual underlying mechanism is that the same gene is activated in several different tissues, producing apparently different effects. It follows that the phenomenon must be extremely common, since most genes will have effects in more than one tissue.

Hypotonic Hypertonic Isotonic

Hypotonic = swollen EX: solution with fewer solutes than cell Hypertonic = shrunken/ shriveled EX: A solution that has more solutes than a cell. Isotonic = equal EX: solution that has the same concentration of particles as the cell

A researcher wants to clone a gene of interest into a plasmid. She plans on starting with genomic DNA, then cutting the gene of interest out. Then she plans on cutting open the plasmid and finally, ligating the gene into the plasmid. Which of the following will she need? I. Restriction endonuclease II. DNA ligase III. Taq polymerase

I and II A researcher will need I and II to clone a gene of interest into a plasmid. She plans on starting with genomic DNA, then cutting the gene of interest out. Then she plans on cutting open the plasmid and finally, ligating the gene into the plasmid. Item I is required: In order to cut the gene out of the genomic DNA sample, the researcher will require restriction endonuclease enzymes. The same enzymes will be required to cut open the plasmid (II only can be eliminated). Item II is required: In order to ligate the gene of interest into the plasmid, the researcher will require DNA ligase (I only can be eliminated). Item III is not required: Taq polymerase is used in PCR and would not be required for this cloning experiment (I, II, and III can be eliminated). *Taq Polymerase: an enzyme able to withstand the protein-denaturing conditions (high temperature) required during PCR

A researcher has an agar plate covered with a lawn of E. coli. She adds a drop of a substance, and the next day there is a clear spot on the plate where the substance was added. This substance could be: I. a virus undergoing the lytic cycle II. a virus undergoing the productive cycle III. a chemical that is toxic to prokaryotes

I and III A clear spot on a plate (known as a plaque) indicates the E. coli are dead. This could be due to the addition of a lytic virus or toxin. However, only animal viruses can go through the productive cycle because viruses cannot bud out of a cell with a cell wall, such as bacteria.

Which of the following is true about the differences in genome structure between eukaryotes and prokaryotes? I.A eukaryotic chromosome is linear and a prokaryotic chromosome is circular. II.In both eukaryotes and prokaryotes, DNA is wrapped around histone proteins to form a nucleosome. III.Prokaryotic genomes contain fewer repetitive sections than eukaryotic genomes. Choices: I, II and III II and III I only I and III

I and III Item I is true and Item II is false: Eukaryotic chromosomes are linear and wrapped around histone proteins to form nucleosomes. Prokaryotic chromosomes are circular but are not wrapped around histone proteins (II and III and I, II and III can be eliminated). Item III is true: Prokaryotic genomes are smaller with fewer sections of repetitive sequences (I only can be eliminated).

Which of the following viruses should be able to reproduce successfully if they carry RNA-dependent RNA polymerase into their host cell? I. (+) RNA virus II. (-) RNA virus III. ds DNA virus I and II only II only I, II, and III I only

I, II, and III Item I is true: (+) RNA viruses must at least code for RNA-dependent RNA polymerase in their genome. However, if they are carrying the enzyme, they should be able to use it to make their template strand of RNA (the (-) strand) for later genome replication. Item II is true: (-) RNA viruses must always carry at least one copy of RNA-dependent RNA polymerase in order to make strands of (+) mRNA that can be used for translation into protein. Item III is true: while a ds DNA virus would not require RNA-dependent RNA polymerase to make mRNA or replicate its genome, it could still reproduce if the enzyme was present. It would simply ignore the enzyme.

A strain of bacteria is isolated which consumes oxygen during metabolism and utilizes carbon dioxide as its source of carbon. Which of the following descriptions could characterize this strain of bacteria? I.Obligate aerobe II.Autotroph III.Facultative anaerobe Choices: II and III only I only I and II only I, II, and III

I, II, and III Item I is true: this strain of bacteria uses oxygen during metabolism so it could be an obligate aerobe (choice "II and III only" can be eliminated). Item II is true: since the bacteria use carbon dioxide as their source of carbon, the strain is an autotroph (choice "I only" can be eliminated). Item III is true: the bacteria could be a facultative anaerobe, since these organisms can (and will) use oxygen during metabolism when it is available (choice "I and II only" can be eliminated). Note that without knowing more about what this strain of bacteria does in the absence of oxygen, we cannot distinguish between obligate aerobe and facultative anaerobe, however the question only asks about what the bacteria could be.

Coronaviruses are enveloped, +RNA viruses that typically cause respiratory infections. A coronavirus was implicated in the SARS epidemic of 2002. Which of the following statements are true about Coronavirus? I. They can be cultured in any type of cell as long as an RNA dependent RNA polymerase is included. II. They can be cultured in animal cells. III. Their genome likely has a poly-A tail.

II and III The items "they can be cultured only in eukaryotic cells" and "their genome likely has a poly-A tail" are true about Coronavirus. Item I is false: because the viruses are enveloped, they can only grow in animal cells. Item II is true: The envelope of the virus is derived from the plasma membrane of the host cell as the virus buds out; since all other cells have a cell wall, budding through the cell membrane/wall of those cells is not possible. Note also that additional unique enzymes do not have to be introduced with the virus; since it is +RNA it can be immediately translated into whatever unique enzyme might be needed. Item III is true: since the viral host is eukaryotic, and since it has an RNA genome, it is likely that the genome has a poly-A tail to mimic eukaryotic RNA and facilitate translation.

Are oocytes diploid or haploid?

In meiosis 1, a diploid cell becomes 2 haploid (23 chromosomes) daughter cells, each chromosome has two chromatids. One cell becomes the secondary oocyte the other cell forms the first polar body. The secondary oocyte then commences meiosis 2 which arrests at metaphase and will not continue without fertilization

Aspects of Hydrogen bonding:

In order for a hydrogen bond to form, H must be bonded to F, O, or N in one molecule to give it a partial positive charge. The hydrogen will then be attracted to the lone pair of electrons on a F, O, or N atom in a second molecule.

Where does RNA splicing occur?

Nucleus

Three cycles of PCR:

PCR cycles through 3 temperatures: 1) high temperature--template denaturing 2) low temperature--primer annealing 3) medium temperature--polymerization

Vesicle Transport

Pathway in cells involves vesicles traveling from RER--> Cis Golgi--> Medial Golgi--> Trans Golgi--> cell membrane or other target sites within the cell **vesicles are not involves in transport from the nucleus

In which phase are primary oocytes arrested in?

Primary oocytes are arrested in prophase I

Reverse transcriptase (RT) is also known as _____-dependant ______ polymerase *choices: DNA or RNA

RNA DNA **About reverse transcriptase: Unlike DNA-dependant DNA polymerase, which is used in eukaryotic cellular DNA replication, RT does not have any proofreading mechanism. Thus, many errors occur during the creation of DNA from the viral RNA. While there may be a lot of non-functional viral DNA due to these errors, it also allows for slight variations in protein structure in viable viruses, rendering anti-retroviral drugs ineffective.

Will somatostatin be elevated in response to high blood glucose?

Somatostatin is secreted in response to raised glucose or other nutrients and acts to repress digestive functions, so it will not be elevated in response to decreased glucose; but will with high blood glucose so yes!

What is synapsis?

Synapsis (also called syndesis) is the pairing of two homologous chromosomes that occurs during meiosis. It allows matching-up of homologous pairs prior to their segregation, and possible chromosomal crossover between them. Synapsis takes place during prophase I of meiosis. *crossing over moves sections of DNA between homologous chromosomes and allows for independent assortment in prophase 1

T or F: Lipid soluble substances, as well as steroids, can diffuse across cell membrane and bind to an intracellular receptor.

TRUE

________ is the uptake of genetic material (plasmids or chromosomal DNA) from the extracellular environment where as _____ is the transfer of genetic info from one bacteria to another via a lysogenic phage, where as ______ is the process of deliberately introducing nucleic acids into cells. The term is often used for non-viral methods in eukaryotic cells

Transformation Transduction Transfection

Diarrhea--when does it happen?

When water is not reabsorbed from the large intestine, diarrhea is the result

Where do antibodies float around in the blood?

antibodies float around in the blood and in the extracellular fluid, the portion of the protein most accessible to host antibodies would be the part that sticks out into the extracellular space

When mRNA is transcribed, is it complementary or does it have the same sequence as the template DNA it was transcribed from?

complementary

Electron flow from NADH to O2

electron flow from NADH to O2 is facilitated by several intermediate electron carriers, for example electrons move from a reduced donor, such as malate, to an oxidized donor, such as OAA. The electrons move from carrier to carrier, down a favorable energy gradient to the final electron acceptor, O2. The favorable energy gradient is due to the fact that electrons move towards components with a more positive reduction potential and thus a higher affinity for oxygen. As the electron flow occurs, protons are flowing across the inner membrane into the intermembrane space.

The decrease in the ability to breathe air out would result in__________ (higher/lower) PCO2 in the blood

higher

Lactose, like other carbohydrates are ___________ (hydrophobic/ hydrophilic).

hydrophilic

What is RNA interference (RNAi)?

is a way to silence gene expression after a transcript has been made; mediated by miRNA and siRNA *siRNA binds to complementary sequences on mRNAs and this dsRNA is then degraded

What is an intercalating agent?

means it inserts itself between base pairs in DNA

operator operon promoter

operator = specific DNA nucleotide sequence where transcriptional regulatory proteins can bind operon = nucleotide sequence on DNA that contains 3 elements: a coding sequence for one or more enzymes, a coding sequence for a regulatory protein, and upstream regulatory sequences where the regulatory protein can bind; unit of genomic DNA that has a cluster of genes that are controlled by a single regulatory signal EX: lac operon of prokaryotes promoter = sequence of nucleotides on a chromosome that activates RNA polymerase so that transcription can take place; found upstream of start site

Where does helicase bind to?

origin of replication

hnRNA

processed in nucleus after transcription to yield a variety of mRNA transcripts (how protein diversity can be achieved)

Cloudy cultures indicate what phase of bacterial growth?

stationary phase *clear cultures indicate no growth or still in lag phase

In a hypotonic solution, does the cell have higher or lower osmotic pressure?

there will be greater osmotic pressure in the cell *more concentrated solution in medium, water will tend to flow inside cell, causing it to burst *water flows in, in order to decrease osmotic pressure

The purpose of the ductus arteriosus is:

to help shunt blood past the inactive fetal lungs

Info about transcription factors

transcription factor (sometimes called a sequence-specific DNA-binding factor) is a protein that binds to specific DNA sequences, thereby controlling the rate of transcription of genetic information from DNA to messenger RNA. Transcription factors perform this function alone or with other proteins in a complex, by promoting (as an activator), or blocking (as a repressor) the recruitment of RNA polymerase (the enzyme that performs the transcription of genetic information from DNA to RNA) to specific genes. A defining feature of transcription factors is that they contain one or more DNA-binding domains (DBDs), which attach to specific sequences of DNA adjacent to the genes that they regulate. Additional proteins such as coactivators, chromatin remodelers, histone acetylases, deacetylases, kinases, and methylases, while also playing crucial roles in gene regulation, lack DNA-binding domains, and, therefore, are not classified as transcription factors.