Bio/Biochem Kaplan FL 7
Q: A gene consisting of 45 nucleotide bases undergoes a mutation that results in the deletion of two codons. How many nucleotides does the resultant mutant have? A: 39
Explanation: A codon is a set of 3 nucleotides. Deletion of two codons corresponds to a deletion of 2*3=6 nucleotides, so the resulting mutant will be 45-6=39 nucleotides long.
Q: A man with type AB blood marries a woman with type O blood. Which of the following are blood types that their children might inherit? A: Type A and type B
Explanation: A man with type AB blood has the genotype AB and can therefore produce games of either the A allele or the B allele. A woman with the type O blood has two O alleles and therefore only produce gametes of the O allele. If this couple has children, the only two genotypes their children can possibly inherit with respect to the blood groups are AO and BO, which corresponds to the phenotypes type A blood and type B blood.
Q: Based on figure 3, IRF1 mRNA levels are: A: Increased in the presence of IFN-Y
Explanation: According to Figure 3, IRF1 mRNA levels are significantly higher in the presence of IFN-Y compared to control groups for all incubation times.
Q: According to the passage and Figure 1, what will be a possible outcome of constant high levels of IFN-Y in a patient with CD? A: Gluten peptides will more easily pass through the epithelium
Explanation: According to paragraph 2, high levels of IFN-Y affect the tight junctions, leading to a leaky epithelium. As a result, gluten peptides can more easily cross the epithelium.
Q: The diagram below illustrates the relationship between interstitial (extracellular) fluid pressure and lymph flow. Would an increase in interstitial fluid protein cause an increase in lymph flow? A: Yes, because fluid movement out of the capillaries would increase interstitial fluid pressure.
Explanation: According to the diagram, lymph flow increases when interstitial fluid pressure increases. Thus, to determine whether an increase in interstitial fluid protein would increase lymph flow, you have to determine whether an increase in interstitial fluid protein would increase interstitial fluid pressure. Remember how osmosis works; water diffuses from a region of lower solute concentration to a region of higher solute concentration. When proteins leak out of the capillaries into the interstitial spaces, this makes the interstitial spaces a region of higher solute concentration. Fluid will then flow out of the capillaries and into the interstitium, increasing both the interstitial fluid volume and the interstitial fluid pressure.
Q: What does CTx toxin cause the ribosylation of? A: G proteins
Explanation: According to the passage, ADP-ribosylation results in a 100-fold increase in the production of cAMP. cAMP is a second messenger that is regulated by the activation of G protein-coupled receptors. Thus, CTx likely causes the ribosylation of G proteins.
Q: Assuming that all STAT molecules are activated through receptor-associated phosphorylation, which class of enzymes most likely activates STAT1? A: Kinase
Explanation: According to the question stem, all STAT molecules are activated through receptor-associated phosphorylation. A kinase is an enzyme that catalyzes the transfer of phosphate groups from an ATP molecule to a specific substrate.
Q: Which of the following is most similar to the type of feedback portrayed in Figure 1 with respect to the IRF1 gene assuming that IRF1 is expressed to a small degree under normal conditions? A: A positive inducible operon
Explanation: According to the question stem, the IRF1 gene is expressed to a small degree under normal conditions. This is aligned with a positive system. Paragraph 1 states that gluten peptides induce the expression of the IRF1 gene, leading to the production of IL-18 and IFN-Y. IFN-Y acts back on the IRF1 gene through a transcription activator, STAT1, further increasing IRF1 production. This is analogous to an inducible system. Thus, a combination of these two mechanisms results in a positive inducible system.
Q: Branched-chain amino acids (BCAAs) are aliphatic branched amino acids that are used as exercise supplements for their anabolic properties. Which of the following is NOT a proteinogenic BCAA? A: Alanine
Explanation: Aliphatic amino acids have nonpolar side chains. Valine, leucine, and isoleucine all have side chains that have a beta carbon attached to three other carbonds, thereby forming a branch. Alanine, while aliphatic, has a side chain containing a beta carbon attached to three hydrogens and is therefore not branched.
Q: Which of the following methods could be used to determine the influence of cell position on differentiation during embryological development? A: Reverse the position of two cells from opposite poles at the blastula stage, then allow development.
Explanation: An experiment should be conducted which observes the interaction of cells before the fetal period is reached. It should also manipulate the position of those cells. The correct answer meets both criteria.
Q: Which of the following is the best explanation for the lack of a control group that received only placebos? A: The risks of giving patients placebos outweighed the benefits of adding a placebo group.
Explanation: As the question stem notes, the ADVANCE trial did not use a placebo group (a control group that received no experimental treatment at all). While placebos are common in clinical trials, they are not used in cases where there is known beneficial treatment, as it is unethical to deny patients treatment. In the case of diabetes, the risks of leaving it untreated (eg, possible blindness, kidney failure, amputations) are so high that using a placebo cannot be justified.
Q: Which stage of embryonic development consists of a hollow ball of cells surrounding a fluid-filled center? A: Blastula
Explanation: Blastulation begins when the morula develops a fluid-filled cavity, which by the fourth day of development becomes a hollow sphere of cells called the blastula.
Q: Study participants were also categorized into two major haplotypes, PAV and AVI. If a woman homozygous for the PAV haplotype bears a child fathered by a man heterozygous for the PAV haplotype, what is the probability that the child will be homozygous for the AVI haplotype? A: 0.00
Explanation: Both parents would need to carry an allele for the AVI haplotype in order for their child to be homozygous for the haplotype. Because the question stem states that the mother is homozygous for the PAV haplotype, 0.00 is the best choice.
Q: What effect did the conjugation of CTB and RTB to VLPs have on the host immune response against FM1-6, respectively? A: Both improved the immune response.
Explanation: Compare the viral titers of CTB-VLPs and RTB-VLPs with other groups (Mock and VLP) to understand their impact on the immune response. Lower viral titers indicate a better immune response. Figure 2 shows that in the case of FM1-6, both CTB-VLPs and RTP-VLPs had lower viral titers compared to the Mock and VLP groups, indicating that both CTB-VLPs and RTB-VLPs improved the immune response against FM1-6.
Q: DNA with which of the following characteristics would be most easily denatured? A: Complementary strands which have a higher A-T content than G-C content
Explanation: DNA denaturation occurs when hydrogen bonds between the two complementary DNA strands are broken, usually by heat, which separates the two individual DNA strands from one another. The melting temperature of DNA, or the temperature at which 50% of the DNA has been denatured into single strands, depends on a variety of factors, including the length of the DNA strand, the salt concentration of the surrounding solution, and the nucleotide composition. Longer DNA strands have more base pairs between the complementary strands, so more energy is needed to break the increased number of hydrogen bonds. An increase in the concentration of sodium, or other cations, in the solution also stabilizes DNA and increases the melting temperature. DNA strands are negatively charged due to their phosphate backbones and thus complementary strands repel one another. However, the positive charge of cations shields the negative phosphate groups and decreases the repulsion between the two strands, leading to increased DNA stability. Finally in base pairing, adenine (A), and thymine (T) form two hydrogen bonds with one another, while guanine (G) and cytosine (C) for three hydrogen bonds when paired, meaning the energy needed to break G-C- base pairings is higher than the energy needed to break A-T base pairings.
Q: Which of the following substances would be included in the buffer used for experiment 2? A: GTP
Explanation: Experiment 2, which is described in paragraph 4, uses "a buffer supplemented with the necessary biological components needed for translation." Therefore, the answer choice should include a substance that is used in the process of translation, or the process of creating proteins from mRNA. GTP is used as a source of energy for the elongation and translocation steps of translation.
Q: Based on the data presented in the passage, which conclusion can be drawn from the junior researcher's experiment? A: The sample contains both aspartame and protein X.
Explanation: Figure 2 confirms the presence of protein X in the sample, but not aspartame. Since aspartame is a dipeptide and protein X is described as a polypeptide, aspartame is likely much smaller than protein X. The western blot cannot be used to conclude that aspartame is not in the sample. The ELISA in table 1 is specific against aspartame, confirms the presence of aspartame. Therefore, the sample contains both protein X and aspartame.
Q: Based on the passage, aspartame is most likely to be digested by: A: Protease
Explanation: Given aspartame is a dipeptide, it is likely to be digested/cleaved by a protease, which cleaves peptide bonds in proteins by hydrolysis.
Q: Which of the following is true regarding the genotype of females exhibiting X-linked recessive disorders? A: It is similar to the genotype of females exhibiting autosomal recessive disorders.
Explanation: In order for a female to have an X-linked recessive disorder, that individual must have two X chromosomes that carry the mutation. Autosomal recessive disorders are similar in that they too require both inherited alleles to be mutated for the disorder to be expressed.
Q: If the two cysteines investigated in figure 1 were substituted for alanines, which of the following results would be expected? A: Loss of hemolytic activity
Explanation: In paragraph 3, it is hypothesized that two cysteine residues in TlyA are crucial for proper hemolytic activity. According to figure 1, in reducing conditions, hemolytic activity is significantly decreased compared to non-reducing conditions. Reducing environments prevent cysteine residues from being able to make disulfide bonds, meaning the data from figure 1 suggests TlyA hemolytic activity requires disulfide bond formation at these cysteines. Mutation of these cysteines to alanine, which does not form disulfide bonds, would abolish the ability of TlyA to form disulfide bonds, leading to an expected loss of hemolytic activity.
Q: What is a possible molecular weight of the form of TlyA seen bound on target cell membranes? A: 84kDa
Explanation: In paragraph two, the passage states TlyA is approximately a 28kDa protein that oligomerizes upon binding to a red cell membrane. Therefore, a possible molecular weight of the oligomerized form of TlyA on the red cell membrane would be a multiple of 28. 84 is a multiple of 28 and would be the molecular weight of three TlyA molecules.
Q: Given the data in the experiment described in the passage, increasing the temperature of a solution of unwound (ss) DNA a few degrees would most likely: A: Slow the renaturing process so that dsDNA takes longer to form.
Explanation: Increasing temperature leads to the breaking of the hydrogen bonds between DNA strands, which also makes it more difficult for the hydrogen bonds to form. Thus, it can be assumed that high temperatures would prevent renaturation of the unwound single-stranded DNA molecules from occurring as rapidly as it might otherwise occur.
Q: The membrane receptor for MSG opens an ion channel. The movement of ions through this channel is best described as: A: Facilitated diffusion
Explanation: Ion channels are proteins that create specific pathways for charged molecules. The movement of ions through a channel down their concentration gradient is an example of a passive process. Passive transport includes diffusion, facilitated diffusion, and osmosis. Diffusion and osmosis are movements of solute and of solvent through the membrane, respectively. Facilitated diffusion requires a carrier molecule or channel.
Q: The structure of palmitic acid is shown below: What is the net reaction from the complete oxidation of palmitic acid in an individual with MCAD deficiency? A: (See Kaplan for Answer)
Explanation: It is not necessary to know the full net reaction of beta-oxidation to answer this question. Palmitic acid is a 16 carbon fatty acid. Each round of beta-oxidation removes 2 carbons from the chain, creating 1 acetyl-CoA. Patients with MCAD deficiency cannot oxidize medium-chain fatty acids 6 to 12 carbons long. That means only 4 carbons can be removed from palmitic acid before it can no longer be broken down. The answer will be the net reaction of only two rounds of beta-oxidation, which would result in only 2 acetyl-CoA molecules being produced from palmitic acid.
Q: Genetic analysis of a heritable form of MCAD shows that point mutations at A100, F252, and E376 result in disease. However, mutations in R164 and M165 have no apparent effect. Which conclusion is best supported? A: A100, F252, and E376 dictate the folding of the enzyme's active site.
Explanation: MCAD deficiency might arise if the enzyme's active site were disrupted. If A100, F252, and E376 dictate the folding of MCAD's active site, mutations at these residues would cause the enzyme to lose its ability to function correctly.
Q: Based on the biochemical pathway of fat breakdown, a physician screening a patient for MCAD deficiency using a mass spectrometry would see which metabolite level elevated when compared to that of a healthy individual? A: (See Kaplan for Answer)
Explanation: MCAD is an enzyme that catalyzes the breakdown of medium-chain fatty acids, between 6 and 12 carbons in length. If a patient is deficient in this enzyme, the metabolites of fatty acid oxidation that would accumulate would be between 6 and 12 carbons.
Q: Excessive blood glucose levels lead to polyuria, an increase in urine output. A possible explanation for this is that the presence of glucose in the renal tubule directly causes a decrease in: A: Movement of water through cells in the descending limb of the loop of Henle.
Explanation: One of the classic symptoms of type 2 diabetes is polyuria, excess water excretion. Higher than normal amounts of glucose in the urine means that the urine osmolarity will be higher than usual. This will result in less reabsorption of water from the filtrate in the nephrons. The ascending limb is almost completely impermeable to water.
Q: Below is an incomplete table of partial pressures of oxygen (pO2) and carbon dioxide (PCO2) in alveoli and alveolar capillaries. A: 100
Explanation: Oxygen flows from the air in the alveolus into the alveolar capillaries. In order for this to happen, the partial pressures of oxygen in the air must be higher than that in the capillary.
Q: Eukaryotic proteins may be toxic to bacteria when produced in large amounts. As such, scientists find it useful to insert eukaryotic genes into bacterial DNA at a site where the gene may be induced.Scientists use the lac operon to control eukaryotic gene expression. Provision of allolactose induces expression of the eukaryotic gene. What occurs in the absence of allolactose? A: The repressor binds to the operator, preventing RNA polymerase from transcribing the eukaryotic genes.
Explanation: Regulation of gene transcription enables prokaryotes and eukaryotes to control the synthesis of proteins. Operons, which are found in nature specifically in prokaryotes, contain a promoter site, a regulator gene, an operator site, and structural genes. The promoter is the site of binding for the RNA polymerase, the regulator gene code for a repressor molecule, the operator is the binding site for the repressor, and the structural genes encode for the proteins of interest. The lac operon is a type of inducible system, where the repressor is bound to the operator until an inducer, which in the case of the lac operon in lactose or a lactose analogue, binds to the repressor and stops it from binding to the operator. This releases the blockage that stops RNA polymerase, and transcription can continue. In the question stem, we are told allolactose induces transcription, suggesting it is an inducer of the lac operon. Therefore, in the absence of allolactose, the repressor would be expected to bind the operator site.
Q: Based on figure 1, at temperatures below 85 degrees celsius, the renaturation of the organism's DNA is: A: Spontaneous and exothermic
Explanation: Renaturation is another word for the reannealing of two complementary DNA strands. To break hydrogen bonds between DNA strands, energy must be supplied, making denaturation an endothermic process. This means energy is given off when hydrogen bonds form during DNA renaturation/annealing and is therefore an exothermic process. Furthermore, since single-stranded complementary DNA strands will reanneal to reform double-stranded DNA without additional input of energy, the process is spontaneous.
Q: In a person with MCAD deficiency, if the body has excess glycogen, which of the following correctly represents the levels of electron transfer intermediates before and after glycogen depletion? A: (See Kaplan for Answer)
Explanation: Since MCAD normally oxidizes fats to produce NADH and FADH2, a person that is MCAD deficient would not be able to produce appropriate levels of NADH and FADH2, so they would rely on glycogen (stored glucose) for the production of these electron carriers. With excess glycogen, the presence of a large proton gradient prevents the transfer of electrons in the ETC, thus causing a buildup of NADH and FAHD2. However, when the body is in an energy-deficient state such as hypoglycemia or reduced levels of glycogen, the need for the ATP causes a decrease in these carriers and a corresponding increase in NAD+ and FAD.
Q: An otherwise healthy patient was diagnosed with MCAD deficiency. Genetic screening of the patient's parents revealed that the father is a carrier for MCAD, while the mother is not. Given the results of the genetic screening, which of the following most likely explains the patient's diagnosis? A: Maternal germline mutation in the gene coding for MCAD.
Explanation: Since the father is a carrier of the condition but does not display symptoms of MCAD deficiency, it must be a recessive allele. The mother is homozygous for the unaffected allele, meaning the only way the child might obtain two copies of the affected gene is for a mutation in the maternal DNA to be passed on to the child. To distinguish between a complete loss of the chromosome and a germline mutation in the gene, recognize that the patient in question has no other health conditions aside from MCAD deficiency. Monosomy of any chromosome, as a result of nondisjunction, is a serious medical condition that would manifest with multiple side effects. Thus, the most likely cause of this patient having MCAD deficiency is a maternal germline mutation in the MCAD gene.
Q: A patient who has gained 40 pounds in the past 3 months complains of fatigue to her physician. She is found to have a goiter and a decreased metabolic rate. Based on this information, the patient most likely has a deficiency of: A: Thyroxine
Explanation: Since thyroxine is a thyroid hormone that plays an important role in regulating metabolism, and since the patient's metabolism is affected, thyroxine is correct. -Furthermore, she has a goiter, which is an increase in the mass of the thyroid. Goiters occur in response to low thyroxine; the pituitary gland secretes thyroid-stimulating hormones in an attempt to trigger an increase in thyroid hormone production. When the thyroid gland is unable to increase thyroxine synthesis, hypertrophy results.
Q: At the end of the experiment described in the passage, how many patients in the intensive treatment group exhibited albuminuria? A: 2,181
Explanation: The answer to this question is in table 1. The total number of people with albuminuria is the number of people who has albuminuria BEFORE the trial, plus the number who DEVELOPED albuminuria, minus the number of people whose albuminuria resolved during the trial. -According to the table, that number is 1623+1480-922= 2181
Q: A hydropathy plot indicates the hydrophilicity (negative values) and hydrophobicity (positive values) of different regions of a protein. According to the hydropathy plot below, how many cellular membrane-spanning regions are likely in this protein? A: 2
Explanation: The cell membrane is a bilayed composed of phospholipids, which are composed of a hydrophilic head and two hydrophobic fatty acid tails. The fatty acid tails create the innermost part of the cell membrane, making the cell membrane hydrophobic. Therefore, the regions of membrane-associated proteins that span the bilayer are expected to also be hydrophobic. On the flip side, the parts of the protein that interface with the cytoplasm or the outside of the cell, and are not associated with the membrane, tend to be hydrophilic. The hydropathy plot indicates two regions of increased hydrophobicity, peaking around amino acid 40 and amino acid 80. This suggests that there are two regions that span the cell membrane.
Q: A researcher studying transposition discovered a new strain of cells that have a 1000-fold greater rate of transposition than other strains. Which of the following explanations would best account for this observation? A: The new strain lacks the dam methylation system.
Explanation: The dam methylation system, mentioned in the fourth paragraph, regulates the rate of transposition and the activation of Tn10 with DNA replication. The lack of methylation on the daughter strand of newly synthesized DNA activates the transposon by increasing the transcription of the transposase gene and enhancing the binding of the transposase enzyme to the transposon. Therefore, if a strain of cells lacked the dam methylation system, there wouldn't be any methylation of the strain's DNA, and the incidence of transposition would be expected to be much higher than in strains with the methylation system.
Q: Suppose the junior researcher contaminated the sample with aspartame. How would this change the results of the experiment? A: This would lead to higher absorbance values in the fourth column of Table 1.
Explanation: The elisa data in Table 1 is the only data that will be affected by contamination with aspartame since aspartame, a dipeptide, is smaller than 10kDA and would not appear on the western blot. Adding aspartame to the sample would cause the absorbance values of the ELISA to increase for the sample in column 4 of Table 1.
Q: Why is the viral titer of VLP included in Figure 2? A: VLP serves as a negative control for the immune response.
Explanation: The experiment aimed to investigate the effect of CTB- and RTB- conjugation to VLPs on the viral titer against viruses UI182 and FM1-6. The VLP group is included as a negative control to assess the normal immune response against the respective miruses without any conjugation of CTB or RTB. Thus, the effectiveness of CTB-VLP and RTB-VLP in improving the immune response against the viruses can be compared by examining the difference in viral titer between the control VLP group CTB-VLP and RTB-VLP groups.
Q: The binding of oxygen to hemoglobin does not follow the rules of michaelis-Menten kinetics, primarily because: A: Hemoglobin exhibits allosteric effects.
Explanation: The key assumption on which Michaelis-Menten kinetics is based is that there is one active site capable of binding a substrate. Hemoglobin, however, is an allosteric protein with four subunits, and exhibits cooperativity: when oxygen binds, the other sites bind oxygen more easily. While all of the answer choices are true statements, hemoglobin exhibiting allosteric effects is the only statement that explains why Michaelis-Menten kinetics doesn't apply to the minding of oxygen to Hb. **Allosteric enzymes are an exception to the michaelis-Menten model. **Michaelis Menten kinetics assumes that there is a single binding site for the substrate. Hemoglobin has four subunits and four binding sites that impact the binding of the substrate, so it has allosteric effects on binding.
Q: When sildenafil inhibits PDE-5, it: A: Increases the Km of the enzyme
Explanation: The last sentence of the passage states that sildenafil "is a reversible competitive inhibitor" of the enzyme PDE-5. By definition, competitive inhibitors compete with the substrate to bind to an enzyme. Because of this competition, they increase Km, which is the substrate concentration at which the reaction has half-maximal velocity.
Q: What is the net charge of aspartame at a pH of 7.4? A: -1
Explanation: The net charge of aspartame is based on the individual amino acid. According to figure 1, aspartame consists of aspartic acid and phenylalanine. Aspartic acid carries a. -1 charge and phenylalanine is neutral at pH 7.4. Therefore, the net charge of aspartame is -1.
Q: Multiple sclerosis is a central nervous system disease that results from autoimmune damage to oligodendrocytes. How do neurons that are impacted by this disease differ from healthy neurons? A: The electrical signals generated by healthy neurons tend to dissipate less rapidly than the electrical signals generated by multiple sclerosis neurons.
Explanation: The oligodendrocytes produce myelin sheaths for neurons in the central nervous system. Myelin sheaths are important for providing insulation to the axons, preventing dissipation of electrical signals. Damage to the myelin sheath thus results in more rapid dissipation of electrical signals generated by action potentials leading to slower transmission of electrical signals.
Q: It can be inferred from the passage that, aside from being involved in the regulation of transposon activity, the presence of hemimethylated DNA also allows cells to: A: Distinguish between old and new strands of DNA.
Explanation: The passage discusses the issue of hemi-methylation in the context of the dam methylation system in the fourth paragraph. It's stated that following replication, the new strand, which is complementary based-paired to the parent strand, lacks methyl groups until the dam system has a chance to act. The reason that the parent strand has methyl groups attached to the DNA, while the new strand lacks them, is rooted in the semi-conservative mechanism by which DNA is replicated. During DNA replication, the parent strands unwind and act as a template for complementary base-pairing with free nucleotides. The end result is two daughter DNA helices- each helix consists of one of the original parent strands and one newly synthesized strand. Thus, following replication of methylated DNA, each daughter helix will consist of one newly synthesized strand and one methylated parent strand- thus, hemi-methylated DNA.
Q: Based on the passage, CTx most likely has: A: Hydrophobic and hydrophilic sequences in both CTA and CTB
Explanation: The passage indicates that noncovalent interactions are critical for the association of CTA and CTB. Noncovalent interactions include hydrophobic and hydrophilic interactions. CTx, a protein, is likely to have both hydrophobic and hydrophilic sequences that helps proteins fold into their proper conformations and mediate interactions with other molecules.
Q: Researchers tag the TlyA protein with a fluroscent marker and transfect HeLa cells, a human cell line, with the construct. All of the following cellular compartments would be expected to show the fluorescent labeling EXCEPT: A: The nucleus
Explanation: The passage says that TlyA is an rRNA methyltransferase as well as a hemolysin. This suggests TlyA is expected to both act on ribosomes in the cytosol and be excreted into the environment where it can form pores in the membranes of red blood cells. This indicates TlyA would be both in the cytoplasm of a cell and in the parts of the cell that are involved in the secretory pathway. This includes the rough endoplasmic reticulum (RER), where proteins slated for excretion are created, and the Golgi apparatus, where proteins are modified and packaged for secretion. TlyA, however, is not expected to be localized to the nucleus based on the information in the passage.
Q: Based on the information in the passage, what effect does rRNA methylation most likely have on translation in the presence of capreomycin? A: Loss of rRNA methylation will increase translation efficiency.
Explanation: The passage states that TlyA is an rRNA methyltransferase. Thus, higher levels of TlyA are expected to increase rRNA methylation. Figure 2 shows that while translation efficiency is almost at 100% in the absence of capreomycin with mock vector or Rv1694 (TlyA) transfection, transfection with Rv1694 significantly lowers translation efficiency compared to the mock-transfected controls in the presence of capreomycin. Together, this suggests increased rRNA methylation decreases translation efficiency in the presence of capreomycin, while lack of rRNA methylation would be expected to increase translation efficiency in the presence of capreomycin.
Q: Multicopy inhibition is a regulatory mechanism that limits the expression of the transposase gene by blocking: A: Translation
Explanation: The passage states that the net result of multicopy inhibition is that IN RNA is prevented from attaching to a ribosome. Ribosomes are the site of protein synthesis, or translation, and consist of a small subunit and a large subunit. Translation is initiated by the binding of an mRNA transcript to the small ribosomal subunit. Therefore, if mRNA is not able to attach to a ribosome, translation cannot occur. Hence, multicopy inhibition must repress translation.
Q: Asymmetric dimethylarginine (ADMA) is a potent endogenous non-competitive NO synthase inhibitor. Based on information from the passage, increased plasma levels of ADMA would most likely be associated with: A: Decreased cGMP levels
Explanation: The question says that ADMA is a strong endogenous non-competitive inhibitor of nO synthase; in other words, ADMA blocks NO synthesis. The second paragraph, as well as Figure 1, explains that NO activates GC which transforms GTP into cGMP. Therefore, a reduction in NO levels would most likely also reduce cGMP levels.
Q: The graph illustrates plasma concentrations, over time, of sildenafil (474.6 g/mol) in healthy males after a 100mg oral dose. Clinical effectiveness of sildenafil begins to decrease significantly at plasma concentrations below 4*10^-7 M. This implies that the therapeutic benefits of this oral dose would most likely: A: Begin to fade significantly about four hours after oral administration.
Explanation: The question says that the effects of sildenafil decrease significantly once its concentration falls below 4*10^-7 mol/L. To see where this value falls on the graph, it must be converted to ug/L. According to the question, sildenafil weighs about 475 g/mol: (4*10^-7 mol/L)(475g/mol)(10^6)ug/g)= 200ug/L -Applying this result to the graph makes it apparent that the effects of sildenafil will begin to wear off about four hours, after oral administration.
Q: Unlike the movement of NO between cells, the release of serotonin from neurons involves the simultaneous exit of between 1,000 and 10,000 molecules. Serotonin release illustrates which of the following transport mechanisms? A: Exocytosis
Explanation: The question states that neurons release a large number of serotonin molecules simultaneously. Among the mechanisms listed, only exocytosis-- in which a vesicle fuses with the cell membrane and expels its contents- transfers large numbers of molecules at once.
Q: Why is the administration of taste samples without L-Arg included in figure 1? A: The plain taste samples provide a baseline against which the response to samples containing L-arm can be measured.
Explanation: The study aimed to delineate how the taste perception of the five primary flavors varied with L-arg administration, as a function of the subject's PROP-tasting phenotype (paragraph 2). Figure 1 addresses the question of whether and how much taste perception of each of the five major flavors was affected when co-administered with L-arg. The administration of the flavors on their own served as a baseline against which that difference could be measured.
Q: According to the passage, the binding of TlyA to red cell membranes most likely: A: Increases the number of intermolecular disulfide bonds between TylA molecules.
Explanation: This question is asking about what likely occurs when TlyA binds to red cell membranes. In paragraph two, the passage states, "when free in solution, the TlyA protein exists as a monomer, but when bound to target cell membranes, quickly oligomerizes." The results from figure 1 also indicate that in reducing conditions, less hemoglobin is released from red blood cells. Together, this suggests that upon binding to red cell membranes, TlyA oligomerizes through disulfide bonds using cysteine residues. This would increase the number of intermolecular disulfide bonds between TlyA molecules.
Q: Based on the information in the passage, the co-administration of which of the following amino acids would most likely alter PROP sensitivity? A: (See Kaplan for Answer)
Explanation: This question requires identification of which of four amino acids is most likely to alter PROP sensitivity. Since the passage explains that L-Arg alters PROP sensitivity, look for the amino acid that is most similar to L-arg. Lysine, like arginine, is a positively charged basic amino acid under physiological conditions.
Q: A biologist notices that E. coli that were cultured in space lost the ability for surface adherence and finds that there are large sequences of DNA interrupting the genes coding for adhesion proteins. If surface adherence returns after the bacteria are cultured on Earth for a few generations, which of the following would most likely account for that finding? A: The large disrupting sequences removed themselves from the adhesin genes once the environmental conditions became normal.
Explanation: Transposons are generic elements that can insert and remove themselves from the genome. Insertion of a transposon can disrupt gene function, and removal of the transposon can restore gene function. In this case, when surface adherence is not required due to low gravity conditions, adhesin genes are disrupted by transposons. Once the bacteria return to Earth, the transposons remove themselves from the adhesin genes, restoring adherence function.
Q: Which of the following is NOT supported by the experimental data? A: Co-administration of L-Arg with sucrose enhances the sweet flavor perception across all participants.
Explanation: When L-Arg increases the perceived intensity of another flavor, individuals report an enhanced perception of that flavor. While co-administration of L-Arg with sucrose does enhance the sweet flavor perception of super-tasters and medium tasters with respect to their baseline perception scores, it decreases the relative flavor perception of non-tasters. Thus, the correct answer is not supported by the data in figure 1.
Q: Suppose that instead of studying the denaturation of DNA, the experimenter wanted to study the temperature at which the organism's DNA polymerase denatured. To do this the experimenter would most likely want to measure: A: The change in the mass of DNA in the sample over time.
Explanation: When an enzyme is denatured, it loses its three-dimensional shape. Because of this, the enzyme also loses its enzymatic activity. Therefore, because DNA polymerase assembles DNA, measuring the amount of DNA being synthesized can be used as a measure of the enzymatic activity of DNA polymerase. If the rate of DNA being synthesized stays the same, then the enzyme is working as it usually does and has not been denatured. If the rate of DNA production decreases, then the enzyme is no longer working and has been denatured.
Q: When transcribing the transposon, DNA polymerase requires a magnesium ion to help keep the DNA in place and align the bases. This magnesium ion is classified as a : A: Cofactor
Explanation: When an enzyme needs another molecule to help its enzymatic activity, and that molecule isn't an enzyme itself, it is called a cofactor. Here, magnesium (a divalent metal ion) acts as a cofactor for the DNA polymerase.
Q: It was discovered that the source of a unique human illness was a new pathogen. Scientists identified the pathogen to be a bacterium rather than a virus because: A: It grew in medium containing only nutrient broth and agar.
Explanation: in differentiating between bacteria and viruses, it is necessary to find a characteristic that is exclusive to one category. Viruses are obligate parasites, meaning viruses are unable to self-replicate and require a host cell to reproduce. If the pathogen in question is able to reproduce in the absence of host tissue to grow. it would indicate the pathogen is viral.
Q: Which technique employs the use of a stationary pH gradient inside the gel to separate proteins? A: Isoelectric focusing
Explanation: Isoelectric focusing separates proteins based on their isoelectric points by using a gel with a pH gradient in addition to an electric field. The positively charged proteins will migrate towards the cathode, while the negatively charged proteins will migrate towards the anode. When the protein reaches the portion of the gel where the pH is equal to the protein's pI, the protein will become neutral and stop moving.
