MCAT Course Test 1 - section 3

Reduction of temperature would have which of the following effects on the second trial? A. It would impair catalase activity. Correct Answer (Blank)

A. Enzyme activity normally increases with increased temperature (up to a point) and decreases with decreased temperature (choice A is correct and choice B is wrong). It is inherent in the nature of a catalyst, however impaired its activity, that it is regenerated at the conclusion of catalysis (choice C is wrong), and there is no basis on which to suppose that reducing the temperature of a protein will convert its chemical structure (choice D is wrong).

In another experiment, female White Leghorn chicks are administered 1.0 mg of stilbestrol for one week in the presence or absence of clomiphene citrate, an estrogen antagonist. Which of the following results would NOT contradict the findings of the first experiment? A. Clomiphene decreased the average oviduct weight from maximum levels. Correct Answer (Blank)

A. If stilbestrol is an estrogen receptor agonist (i.e., has effects like those of estrogen) and clomiphene citrate is an estrogen antagonist, then clomiphene citrate should reverse the increase in oviduct weight that is induced by stilbestrol (choice A is correct and eliminate choice C). Choices B and D contradict the passage since Figure 1 shows that stilbestrol increases oviduct weight so they should be eliminated.

Which of the following statements best describes pyruvate kinase? A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. Correct Answer (Blank)

A. Pyruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP (choice A is correct). Paragraph 3 discusses how PKM2 can exist as a dimer or tetramer, each of which would contain more than one peptide chain and would therefore have quaternary protein structure. Hemoglobin also has quaternary structure (since it contains four peptide chains) but myoglobin does not (eliminate choice B). Since PKM2 is one isoform of pyruvate kinase, there are presumably others. Isoforms are different forms of the same protein and can be due to gene duplication or alternative splicing. However, the ribosome has no function in splicing (eliminate choice D).

The fundamental difference between the 34-, 17-, and 14-amino acid forms of gastrin can be attributed to: A. primary structure. Correct Answer (Blank)

A. These are very small polypeptides. Since they are single polypeptides, they cannot have quaternary structure, which requires interactions between different polypeptides chains (eliminate choice D). Tertiary structure and secondary structure are unlikely to be achieved with such short polypeptides (eliminate choices C and B, respectively). Primary structure is simply the linear amino acid sequence, which certainly will be different since the peptides have different numbers of amino acids (choice A is correct).

If the measurements taken in Step 4 of Experiments 1 and 2 had shown no change in bacterial lactose concentration, then it is likely that within the medium: A. hydrogen ion concentration had remained unchanged in Experiments 1 and 2. Correct Answer (Blank)

A. These experiments show that a gradient of either potassium or protons can drive the transport of lactose into the cell against its gradient. If there is no lactose transport, it is safe to assume that the concentration in the media of the driving ion will remain the same (choice A is correct). There is no reason to assume that hydrogen ion concentration would increase (regardless if lactose transport had occurred or not) since this is not the case in any of the experiments (eliminate choice B). Step 4 involves a proton gradient, not a potassium gradient (eliminate choices C and D).

Aldosterone's mechanism of action on its target cell is most similar to that of which of the following hormones? A. Testosterone Correct Answer (Blank)

A. Two general classes of hormones are those that are small hydrophobic molecules like steroid hormones and those that are peptides. The steroid hormones, which include aldosterone and testosterone, diffuse through the plasma membrane to bind to a receptor which enters the nucleus to regulate transcription of a specific set of genes (choice A is correct). Peptide hormones, such as glucagon, insulin, and ACTH, cannot diffuse into the cell since they are large and hydrophilic, so they bind to cell-surface receptors to transduce a signal into cells (choices B, C, and D are wrong).

Which of the following describes the molecular geometry of a carbon dioxide molecule? A. Linear Correct Answer (Blank)

A. VSEPR theory predicts (and experiments have verified) that the carbon dioxide molecule, O=C=O, is linear since the central carbon atom contains no lone-pair electrons (choice A is correct and eliminate choice D) and the two regions of high electron density (the two double bonds) are most stable on opposite sides of the central atom. In order to be trigonal planar or tetrahedral, the central carbon would need to be bonded to three or four atoms, respectively (eliminate choices B and C).

As described in the passage, which of the following best describes how PrPSc causes disease formation? A. PrPSc induces changes in the secondary structure of PrPc proteins by causing alpha helices to convert to beta sheets. Correct Answer (Blank)

A. While a defect in chaperone proteins could plausibly cause disease, the question asks for the method by which PrPSc causes disease as described in the passage, and chaperone proteins were not discussed (eliminate choice D). The passage does, however, explain how PrPSc causes changes in the structure of normal proteins. The passage states that the amino acid sequence (the primary structure) does not change (choices B and C can be eliminated, and choice A is correct). Changes in prion proteins take place at the alpha-helix and beta-sheet levels of protein structure; this is the secondary structure of a protein.

Extrapolating from Figure 1 above, if an experimental group of 10 chicks were given a daily dose of 0.75 mg of stilbestrol for seven days, and a separate control group of 10 chicks were observed over the same period but given no stilbestrol, average oviduct weights would be approximately: B. 900 mg for the experimental group and 75 mg for the control group. Correct Answer (Blank)

B. A stilbestrol dose of 0.75 mg per day would lie midway on the x axis between the 0.5 and 1.0 mg doses. By extrapolating from the x axis to the y axis along the curve, it appears that 900 mg would be the oviduct weight predicted at 0.75 mg/day. One thousand mg oviduct weight is the response observed at 1.0 mg, but it would not be expected at 0.75 mg/day since the curve increases between these points (eliminate choices C and D). The control group in the graph has less then 200 mg but is not zero (eliminate choice A and choice B is correct).

An activated aldosterone receptor most directly regulates activity of which of the following enzymes? B. RNA polymerase II Correct Answer (Blank)

B. Aldosterone binds to aldosterone receptors to regulate transcription of a specific set of genes. The enzyme that synthesizes mRNA is RNA polymerase II, so it is this enzyme that would be most directly affected by the activated aldosterone receptor (choice B is correct). DNA polymerase is used in replication, not transcription (choice A is incorrect), and while aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin, these would be indirect effects (choices C and D are incorrect).

Which of the following is a common feature of people with Down, Patau, Edwards, and Klinefelter's syndromes? B. They all have 47 chromosomes. Correct Answer (Blank)

B. All of the listed syndromes are trisomies: Down-21, Patau-13, Edwards-18, Klinefelter's-sex chromosomes (choice B is correct). There is no mention of heart defects in Klinefelter's syndrome (choice A is incorrect). There is no gender difference in the autosomal trisomies (choice C is incorrect). There is no mention of life expectancy in any of the diseases (choice D is incorrect).

Extreme high temperature would have which of the following effects on the second trial? B. It would increase the required amount of potassium permanganate to approximately 28 mL. Correct Answer (Blank)

B. Although enzyme activity normally increases with increasing temperature, extreme high temperature denatures the enzyme and renders it inactive. Twenty-eight milliliters of potassium permanganate were required to titrate the hydrogen peroxide in the absence of enzyme (Finding 1). With catalase rendered inactive by extreme high temperature, the reaction vessel would be functionally devoid of enzyme. As in the first trial, 28 mL of potassium permanganate solution would be required to titrate the hydrogen peroxide (choice B is correct and choices A, C and D are eliminated).

A woman who is 20 weeks pregnant presents for an amniocentesis. With a needle and ultrasound guidance, the physician withdraws some of the amniotic fluid for genetic analysis. A karyotype analysis performed on the amniotic fluid reveals an extra copy of chromosome 21. Which of the following is true of the child after birth? B. Digestion will be impaired since there is no communication between the small and large intestines. Correct Answer (Blank)

B. As per the passage, the child with trisomy 21 may suffer from a prematurely ending small intestine (a blind-ending jejunum), and therefore, digestion will be impaired if not surgically corrected. Kidney dysfunction is not listed as a condition in trisomy 21 (choice A can be eliminated) nor is an underdeveloped sensory cortex (choice C can be eliminated). The passage says hypotonia is involved, which is a decrease in muscle tone, not an increase (choice D can be eliminated).

Which portion of the central nervous system confers balance by coordinating the activity of various motor units? B. Cerebellum Correct Answer (Blank)

B. Coordination of motor skills is one of the primary functions of the cerebellum (choice B is correct). The cerebrum triggers skeletal muscle contraction, but the cerebellum coordinates it (eliminate choice A). The medulla regulates many homeostatic functions (eliminate choice C) and the hippocampus is a component of the limbic sysmte (eliminate choice D).

Gastrin secretion is inhibited by low pH in the stomach. This is an example of: B. negative feedback. Correct Answer (Blank)

B. Gastrin stimulates acid production, and its release is inhibited by stomach acid. This is a negative-feedback loop designed to maintain acid within a certain pH range (choice B is correct). Competitive inhibition does not apply in this case, because there is no enzyme to be inhibited (eliminate choice A). Since gastrin, a hormone, is not excreted into the stomach lumen, it will not be degraded by stomach acid (eliminate choice C). A hormone cannot be secreted in an exocrine manner (eliminate choice D).

It has been suggested that one reason tumor cells have novel metabolism is to generate the additional biomolecules required to increase biomass, which is needed to support high rate of proliferation. Which is the best explanation of how this could occur? B. Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides. Correct Answer (Blank)

B. Glucose-6-phosphate (not fructose-6-phosphate) is shuttled from glycolysis to the pentose phosphate pathway (choice A is wrong), although this does support nucleotide and fatty acid biosynthesis. Paragraph 2 describes how glutamine is converted into the Krebs cycle intermediate α-ketoglutarate, which can then be used to generate other Krebs cycle intermediates to power biomolecule synthesis (choice B is correct). The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, choice C is wrong). Paragraph 2 describes how glutamine is first converted into glutamate, not the other way around (choice D is wrong).

A highly proliferating cell would most likely: B. express high levels of the lactate transporter and the glutamine transporter. Correct Answer (Blank)

B. Highly proliferating cells would express large amounts of hexokinase (a key glycolysis enzyme) but would not also overexpress fructose-1,6-bisphosphatase because this is an enzyme involved in gluconeogenesis. Even highly proliferative cells will avoid concurrently running reciprocally regulated pathways (eliminate choice A). The passage says that rapidly dividing tumor cells produce and export large amounts of lactate, and import large amounts of the amino acid glutamine (choice B is correct). The focus of this passage is how highly proliferative tumor cells power growth via glycolysis and fermentation, making choice C an unlikely correct answer (eliminate choice C). Based on information in the last paragraph, pyruvate dehydrogenase kinase activity is high in some rapidly growing cells, but a high rate of glycolysis will lead to high PFK expression (eliminate choice D).

According to the scheme depicted in Figure 1, if the carbon dioxide concentration increases within the cell: B. more HCO3- will be secreted into the interstitium. Correct Answer (Blank)

B. In Figure 1, chloride moves inward from the interstitium (eliminate choice A). The more CO2, the more bicarbonate which is formed and transported into the interstitium in exchange for chloride (choice B is correct). If more bicarbonate is produced and excreted into the interstitium, more chloride will enter the cell and be secreted into the lumen (eliminate choice C). The more CO2 in the cell, the more H+ will be produced and driven into the lumen (eliminate choice D).

Which of the following is expected to be true of children with Down syndrome? B. Aortic arterial blood oxygen saturation is lower than normal. Correct Answer (Blank)

B. In children with Down syndrome, one of the primary cardiac defects is a truncus arteriosus, which presents as a common arterial trunk coming off both the left and right ventricles. This allows for mixing of blood from the left and right circulations. Since the right circulation is relatively high in carbon dioxide and the left circulation is relatively low in carbon dioxide, the resultant mix will be somewhere between (i.e., not normal, choice A is wrong). By similar reasoning, the right circulation that is typically low in oxygen will mix with the left circulation that is typically higher in oxygen, so the overall aortic arterial oxygen saturation will be less than normal (choice B is correct) and the pulmonary arterial oxygen concentration will be higher than it is normally (choice D is wrong). Since a common arterial trunk is receiving blood from both the right and left ventricles, the pulmonary artery will be receiving more blood than usual and therefore will have a higher than normal blood pressure (choice C is wrong).

A drug is discovered that reduces smooth muscle contraction by inhibiting the muscle's response to acetylcholine. This is likely to: B. reduce the hypertensive effect of angiotensin II. Correct Answer (Blank)

B. In the absence of acetylcholine signaling, the overall tone of smooth muscle in artery walls will be reduced, reducing blood pressure (choice B is correct). Transcription is not related to smooth muscle contraction (eliminate choice A), and the drug acts on smooth muscle, not cardiac muscle (eliminate choice C). If the smooth muscle in blood vessels fails to contract, then blood vessels would dilate. This would increase, not decrease, renal blood flow (eliminate choice D).

A patient with a history of HBV and HDV coinfection would be expected to have antibodies to the following antigens: HDAg. HBV capsid proteins. HDV capsid proteins B. I and II only Correct Answer (Blank)

B. Item I is correct: the passage states that anti-HDAg antibodies made by the host can be used to test for HDV infection (choice C can be eliminated). Item II is correct: a normal host would be expected to make antibodies to viral capsid proteins and HBV is a true virus with a capsid (choice A can be eliminated). Item III is false: HDV is a viroid, which the passage states do not have capsid proteins (choice D can be eliminated and choice B is correct).

During spermatogenesis, spermatids: B. have already undergone meiotic recombination. Correct Answer (Blank)

B. Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (eliminate choice A). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (choice B is correct). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (eliminate choice C). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (eliminate choice D).

Can young White Leghorn female chicks respond to estrogenic stimulation? B. Yes, because their oviduct tissue enlarges in response to stilbestrol injection. Correct Answer (Blank)

B. The animals used in the experiment shown in Figure 1 were young White Leghorn chicks. These animals responded to stilbestrol with a large increase in oviduct weight, indicating that these animals can respond to estrogenic stimulation such as that of stilbestrol (choice B is correct). It is certainly true that chick weight will increase during gestation, but the relation of this to estrogen is not clear (eliminate choice A). After puberty, estrogen will play a key role in reproduction, and it is likely that oviducts will be regulated by estrogen (eliminate choice C). Stilbestrol is only a synthetic compound used in this experiment to mimic the effects that endogenous estrogen would normally produce, of which reproductive functions are particularly important (eliminate choice D).

A botanist studying viroids has isolated a new viroid capable of causing disease in potato plants. The genome was isolated and studied. In order to fit the definition of a viroid as defined in the passage, all of the following must be true EXCEPT: B. a 1:1 ratio of cytosine to guanine. Correct Answer (Blank)

B. The passage states that viroids are single stranded RNA molecules. RNA lacks thymine (choice A is true and can be eliminated) and uses ribose instead of deoxyribose (choice D is true and can be eliminated). The passage states that viroids lack capsids, so would not have coding regions for these proteins; in fact, the passage states that viroid do not produce proteins at all (this is what makes HBV unusual and not a true viroid, choice C is true and can be eliminated). However, the ratio of cytosine to guanine only has to be 1:1 if the nucleic acid is double stranded. Since viroids are single stranded, this ratio could be different (choice B is false and the correct answer choice).

Which form of the amino acid GABA does not predominate at any pH? B. The uncharged form B does not predominate at any pH since the pKa of the carboxyl group is lower than the pKa of the protonated amino group, making the former group the stronger acid. The titration of GABA from acidic to basic pH would therefore result in the following changes in its protonation state:

B. The uncharged form B does not predominate at any pH since the pKa of the carboxyl group is lower than the pKa of the protonated amino group, making the former group the stronger acid. The titration of GABA from acidic to basic pH would therefore result in the following changes in its protonation state:

Based on the test results in Figure 1 and Table 1, Patient 3 is most likely suffering from: B. prion disease. Correct Answer (Blank)

B. This question requires the use of Figure 1 and Table 1 to determine whether Patient 3 is suffering from a prion disease, hepatitis, or both. According to Figure 1, an analysis of Patient 3's blood using SOFIA showed high levels of light emittance. As described in the passage, this means that Patient 3 had high levels of PrPSc, which is consistent with a prion disease (eliminate choices A and D). According to Table 1, Patient 3 tested negative for any antibodies associated with Hepatitis B or Hepatitis D (eliminate choice C). Patient 3 is suffering from prion disease, and not hepatitis (choice B is correct).

In which of the following structures is E. coli most likely to cause infection?

Bladder Correct Answer (Blank) The passage states that E. coli only causes disease outside of the intestinal tract (eliminate choices C and D), in the urinary tract, the biliary tract, and the nervous system in particular. The bladder (choice B is correct and eliminate choice A) is the best answer since this is part of the urinary tract, mentioned as a potential site of infection.

A bacteriologist initiated an E. coli culture from one E. coli cell and hypothesized that some of the progeny in the culture would be genetically different from the original parent cell. Is this hypothesis true? C. No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. Correct Answer (Blank)

C. Bacteria reproduce asexually, by one cell replicating its genome and then splitting into two cells that are genetically identical to the original cell. The only potential sources of genetic variation in bacteria are mutation and the transfer of genetic information through conjugation, transduction, or transformation, none of which are linked to reproduction. In the absence of mutation, all progeny of a cell will be identical to the original cell (choice C is correct). Bacteria only perform recombination under special circumstances such as through the presence of Hfr plasmids that replicate a portion of the bacterial genome to make it transiently diploid (eliminate choice A). There is no indication of a role for Hfr in this case and in a clonal cell line, it could not play a role. Bacteria do not perform the recombination, independent assortment and independent segregation that create genetic diversity in eukaryotes that reproduce sexually (eliminate choice B). They also do not perform meiosis (eliminate choice D).

The substrate in the experiment is: C. hydrogen peroxide. Correct Answer (Blank)

C. Catalase is an enzyme that acts on hydrogen peroxide, reducing it to water and oxygen (choice A is wrong and choice C is correct). In an enzymatic reaction, the substrate is that on which the enzyme acts. Potassium permanganate serves in the titration but is not the substrate (choice B is wrong). Choice D refers to water, a product.

Arachidonic acid is a fatty acid contained in cell membranes. The structure of arachidonic acid is: In which component of the cell membrane would arachidonic acid most likely be found? C. Phospholipids Correct Answer (Blank)

C. Cholesterol is an abundant component of animal cell membranes, but it is a steroid, not derived from fatty acids such as arachidonate (eliminate choice A). Triglycerides could contain a fatty acid like arachidonate, but they are not membrane components (eliminate choice B). Peptidoglycans are not found in eukaryotes, only bacteria, and do not consist of fatty acids (eliminate choice D). Phospholipids is the correct answer. Phospholipids are abundant components of the plasma membrane and contain esters of many different fatty acids (choice C is correct).

Which of the following best describes the role of fructose-2,6-bisphosphate? C. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase. Correct Answer (Blank)

C. Fructose-2,6-bisphosphate exerts reciprocal control on glycolysis and gluconeogenesis through phosphofructokinase and fructose-1,6-bisphosphatase, and has no impact on hexokinase activity (eliminate choices B and D). Fructose-2,6-bisphosphate stimulates phosphofructokinase, which is used in glycolysis, and inhibits fructose-1,6-bisphosphatase, which is used in gluconeogenesis (choice C is correct and eliminate choice D).

If pirenzepine is administered to a Zollinger-Ellison syndrome patient, which of the following will occur? C. Luminal H+ secretion will decrease, because low intracellular Ca2+ will lead to lowered protein kinase activity. Correct Answer (Blank)

C. If acetylcholine is blocked, calcium levels will be low, protein kinase activity low, and H+ secretion low (choice C is correct). Calcium will be low (eliminate choice A) and H+ secretion will decrease, not increase (eliminate choices B and D).

Tryptophan, an essential amino acid found in banana, turkey, and milk proteins, can induce sleep in some people. Warm milk causes greater sleepiness than cold milk because heating the milk: C. releases free tryptophan from proteins, causing more rapid intestinal absorption. Correct Answer (Blank)

C. If heating the milk reduced tryptophan solubility, this would decrease, not increase the sleep-inducing properties of milk (eliminate choice A). Lactose is a disaccharide, not a protein, and its hydrolysis cannot release an amino acid (eliminate choice B). Amino acids, like most nutrients, are absorbed in the small intestine mostly, not the stomach (eliminate choice D). Although heating the milk does not create more tryptophan, it might help to hydrolyze some of the milk proteins and release tryptophan so it can be absorbed more rapidly after ingestion and cause greater sleepiness (choice C is correct).

Which of the following are products of the pentose phosphate pathway? NADH NADPH Ribose-5-phosphate C. II and III only Correct Answer (Blank)

C. Item I is false: NADH is an electron-carrier produced in catabolic reactions, like cellular respiration (choices A and D can be eliminated). Note that since both remaining choices include Item III it must be true and we can focus on Item II, which is true: NADPH is produced by the pentose phosphate pathway (choice B can be eliminated and choice C is correct). Item III is in fact true: ribose-5-phosphate is a primary product of the pentose phosphate pathway.

Which of the following could explain the Warburg effect? Metabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors. Cells with low proliferation rates often have a high ratio of glycolysis to mitochondrial respiration. Some oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death C. I and III only Correct Answer (Blank)

C. Item I is true: tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor. Since glycolysis and fermentation don't require oxygen, they could facilitate cell growth in anaerobic conditions (choice B can be eliminated). Item II is false: the passage is about highly proliferative cells that use glycolysis and fermentation to power growth so a statement about slow growth does not explain the Warburg effect discussed in the passage (choice D can be eliminated). Item III is true: if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell (choice A can be eliminated and choice C is correct).

Fluoroquinolones are a new class of antibiotics that are extremely effective against a wide range of bacteria. It has been determined that fluoroquinolones enter cells along with water via a passive transport mechanism. After exposing both bacterial and human cells to high concentrations of fluoroquinolones, a researcher discovered that, in the bacterial cells only, the DNA was nicked and supercoiling had been disrupted. Of the following, the most likely explanation for this is that fluoroquinolones: C. interfere with DNA gyrase, which is not present in human cells. Correct Answer (Blank)

C. Since fluoroquinolones diffuse into cells with water, they are able to enter all cells, eukaryote and prokaryote alike; they diffuse through special water channel proteins called porins (eliminate choice A). Both eukaryotes and prokaryotes utilize DNA polymerase and helicase (eliminate choices B and D). However, only prokaryotes use DNA gyrase to supercoil their DNA; eukaryotes wind DNA around histones through the action of other topoisomerases (choice C is correct).

Based on the results of Experiments 1 and 2, it is reasonable to assume that all components of the mechanism necessary for concentrating lactose intracellularly within E. coli are located in the: cell membrane. Correct Answer (Blank)

C. The key experiment is Experiment 2, in which vesicles formed from the cell membrane had the same ability to concentrate lactose as whole cells (choice C is correct). The genome will encode protein components, but the active components are not themselves physically located in the genome (eliminate choice B). The ER and Golgi are not found in bacteria (eliminate choices A and D).

Eye color, skin color and hair color in humans are examples of: C. polygenism, where the trait is determined by several genes with several alleles. Correct Answer (Blank)

C. The passage discusses many proteins and genes that control pigmentation in humans (MC1R, MSH, ASIP, TYR and OCA2) and mentions that the OCA2 gene alone has 58 alleles that have so far been documented. This information best matches choice C, since several genes are involved in pigmentation phenotypes. All other answer choices indicate that only a single gene is involved (choices A, B, and D are wrong). Note that while examples in genetics often use eye color to demonstrate monogenism, based on information in the passage, this would be an oversimplification (choice A is wrong).

A FISH probe should be made of: C. a segment of single-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorochrome. Correct Answer (Blank)

C. The passage states that during FISH, fluorescently labeled probes are hybridized to chromosome clusters. This is similar to how probes are used in Southern and Northern blots. The probe must be single-stranded DNA or RNA if it is going to hybridize or bind to the chromosomes (which are denatured first, also similar to the process in Southern or Northern blotting), and should be covalently linked to the fluorescent molecule (choice C is correct). An antibody is not typically used to bind DNA, and in any case, there is no easy way to generate a "fluorescent segment of DNA." It is much easier to use fluorescent molecules or dyes (choices A and D are wrong). A double-stranded probe will not bind the chromosomes because it will be complementary to itself (choice B is wrong).

In a population of 18,000 Caucasians, how many are expected to be carriers of cystic fibrosis? 590 Correct Answer (Blank

C. The passage states that the frequency of the autosomal recessive condition cystic fibrosis, q2, is 1 in 3600. The frequency of the recessive allele, q, then is 1 in 60. The frequency of the dominant non-disease producing allele, p, is 59 in 60. The carriers of a population are determined by the expression 2pq. In the given population, the number of carriers would be (2)(1/60)(59/60)(18000) or 590. Thus, choice C is correct and choices A, B, and D are eliminated.

Based on the data collected from the medical students' study, what percentage of the isolates were methicillin-resistant Staphylococcus aureus? C. 115/345 Correct Answer (Blank)

C. The table lists the number of cultures of each type of organism. The number of positive isolates for methicillin-resistant Staphylococcus aureus is 115. The total number of cultures reported is the number of Gram-(+) plus the number of Gram-(-) plus the fungal cultures: 160 + 180 + 5 = 345. Therefore, choice C is correct (and choices A, B, and D are eliminated).

If radiolabeled stilbestrol were administered to the experimental chicks, given its effect on the oviduct, the stilbestrol would be found most heavily concentrated: C. in the nuclei of oviduct tissue. Correct Answer (Blank)

C. When stimulated by the addition of a ligand such as stilbestrol, estrogen receptor will localize within the nucleus, where it regulates genes by binding to enhancers and promoters. Radiolabeled stilbestrol would localize with estrogen receptor in the nucleus (choice C is correct). There is no estrogen receptor in the plasma membrane or mitochondria (eliminate choices A and D). Some estrogen receptor may be located in the cytoplasm, particularly in the absence of ligand, but it will localize mostly in the nucleus when it has ligand bound (eliminate choice B).

In addition to behavioral changes consistent with encephalopathy, Patient 2 begins exhibiting symptoms of liver failure, caused by infection. Is this consistent with the information provided in the passage, including the patient's test results? D. Yes, because the patient is positive for HBV and HDV, which infect liver cells. Correct Answer (Blank)

D. According to Figure 1, Patient 2 is negative for PrPSc, and does not have prion disease (choices A and B can be eliminated). Additionally, Table 1 shows that Patient 2 was positive for both anti-HBV and HDV antibodies, meaning they have been infected with both HBV and HDV. According to the passage, HBV and HDV infect hepatocytes, or liver cells. The passage states that HBV-HDV coinfection is characterized by encephalopathy (choice D is correct and choice C is wrong).

Some vitamins are essential to humans because they act as precursors of: coenzymes. Correct Answer (Blank)

D. Auxins are plant hormones not found in humans (eliminate choice A). Glucose is not made from vitamins (eliminate choice B). Enzymes are proteins made from amino acids (eliminate choice C). Many coenzymes (which are not themselves amino acids and so cannot be made into enzymes) are required for enzyme activity and are derived from water-soluble vitamins, such as thiamine, biotin, folate, and niacin (choice D is correct).

Which of the following is a true statement? D. Tumor cells can use cap-independent translation to make proteins when growth conditions are not ideal. Correct Answer (Blank)

D. Cap-independent translation uses internal ribosomal entry sites (IRESs) and allows a cell to translate proteins during sub-optimal growth conditions (because less regulation is necessary). Since tumors typically grow quickly, they are often short of oxygen and nutrients and must deal with acidic and CO2-rich growth conditions. Activating cap-independent translation would allow the tumor cells to continue proliferating even under these less-than-optimal conditions (choice D is correct). Based on information in the last paragraph, tumors uptake large amounts of FDG, which can be imaged via PET. This means that if an anticancer treatment is working, less FDG should be detected (choice A is wrong). Normal somatic cells do not express telomerase. This enzyme is only expressed in the germ line, by some white blood cells, and in some tumor cells (choice B is wrong). Glutamine is a hydrophilic polar amino acid (choice C is wrong). Note that explicit knowledge of cap-independent translation was not necessary, only the ability to recognize the other three statements as false.

Which of the following intermolecular attractions will exhibit the greatest strength? D. Hydrogen bonds Correct Answer (Blank)

D. Choices A, B, and C are identical, so they can all be eliminated. The answer must be D. The various intermolecular forces, in order of decreasing strength, are the following: Hydrogen bonding > Dipole-Dipole interactions > Dipole- Induced dipole interactions > Induced dipole-Induced dipole interactions (London forces).

If a patient with cystic fibrosis receives a double-lung transplant from a non-cystic fibrosis donor, would the new lungs be expected to develop cystic fibrosis? D. No, since cystic fibrosis is due to a gene defect, the cells of the new lungs will have the normal CFTR gene. Correct Answer (Blank)

D. Cystic fibrosis is a genetic disease based on the abnormal protein CFTR. In a set of lungs from a person without cystic fibrosis, the lung cells presumably have normal-functioning CFTRs. Therefore, the new lungs should not be subject to the development of cystic fibrosis (eliminate choices A and B). Cystic fibrosis is a multi-organ disease since the CFTR is used in secretions from several glands, but this does not cause normal CFTRs to become abnormal. The primary defect is the protein CFTR, not the pulmonary secretions or an infectious cause (choice C is incorrect and choice D is correct).

The HCO3-/Cl- exchange is an example of: D. facilitated diffusion. Correct Answer (Blank)

D. Facilitated diffusion involves movement of molecules down a gradient with the involvement of a protein. The passage states that both bicarbonate and chloride ions are moving down a gradient, making this a case of facilitated transport (choice D is correct). Exocytosis does not generally involve ion transport and does not utilize membrane channels (eliminate choice A). Active transport involves moving ions or other molecules against a gradient (eliminate choice B). Simple diffusion is the movement of molecules down a gradient without a protein involved (eliminate choice C).

With reference to the experiment, which of the following did NOT occur during the three-minute period following the introduction of catalase? D. A new enzyme was synthesized. Correct Answer (Blank)

D. Finding 2 indicates that much hydrogen peroxide disappeared from the reaction vessel without participation of potassium permanganate. The increased disappearance was attributable to the presence of catalase and represents an enzyme-catalyzed reaction (choice C occurred and can be eliminated). An enzyme is not changed during the course of the reaction it catalyzes, and can catalyze its reaction over and over again (choices A and B would occur and can be eliminated). However, nothing in the experimental procedure or findings indicates that a new enzyme was synthesized (choice D is not true and is the correct answer choice).

Based on Figure 1, which of the following is least supported? D. Humans and chimpanzees are more closely related than fission yeast and the wild boar. Correct Answer (Blank)

D. Homologous proteins or genes are those that have evolved from a common ancestor. This matches the information in Figure 1, since the OCA2 protein in all organisms on the figure originated with the ancestor protein represented by the point at the bottom (choice A is supported and can be eliminated). The eight organisms on the figure are all eukaryotes and from diverse families (fungi, mammals, fish; choice B is supported and can be eliminated). Remember that yeast are fungi and therefore eukaryotic. OCA2 in zebrafish and Japanese killifish share a common ancestor which is not far away in evolutionary terms (i.e., is not very far down the diagram). In contrast, the common ancestor protein between human and horse OCA2 protein is the point at the bottom of the diagram; this is farther away in evolutionary terms (choice C is supported and can be eliminated). While choice D may be true based on logic and background information on evolution, it is not supported by Figure 1; this phylogenetic tree contains information on how the OCA2 proteins are evolutionarily related, not how organisms are related (choice D is the least supported and the correct answer choice).

The diaphragm plays an important role in respiration. During inspiration, the diaphragm: D. contracts, causing alveolar pressure to drop below atmospheric pressure. Correct Answer (Blank)

D. Recognize theh opportunity to treat this question as a 2 X 2 elimination. Inspiration is the drawing of air into the lungs. The diaphragm contracts and flattens during inspiration (eliminate choices A and C), expanding the chest cavity; the lungs (which are stuck to the inside wall of the chest cavity) expand as well. The expansion of the lungs decreases the pressure in the alveoli, causing air to move into the lungs from the exterior (eliminate choice B and choice D is correct).

In a given population, what is the frequency of carriers of Patau syndrome? D. None of the above Correct Answer (Blank)

D. The Hardy-Weinberg equation can only be used to describe the frequencies of autosomal recessive or dominant traits or conditions, not changes in chromosome number. Since trisomies are not autosomal recessive conditions, the Hardy-Weinberg equation cannot be used, and one cannot predict the occurrence of the carriers in a population (choice D is correct and choices A, B, and C are eliminated). Also, because trisomies are not based on dominant or recessive expressions, there is no "carrier" state.

The removal of the adrenal glands will result in a reduction of all of the following EXCEPT: D. renin secretion by juxtaglomerular cells. Correct Answer (Blank)

D. The hormones secreted by the adrenal glands include epinephrine from the medulla, and cortisol, aldosterone, and low levels of sex steroids from the adrenal cortex. As stated in the passage, aldosterone increases potassium secretion and also increases sodium reabsorption (choices A and B are true and thus are eliminated). The increased Na+ reabsorption leads to increased water reabsorption, thereby increasing the blood volume (choice C is true and thus is eliminated). The loss of aldosterone would cause water and sodium loss, decreased blood volume and decreased blood pressure. In response to decreased blood pressure, renin secretion would increase, not decrease (choice D is incorrect and thus the correct answer choice).

After an adult with cystic fibrosis ingests a carbohydrate-rich meal, which of the following would you expect to occur? D. High extracellular concentration of glucose Correct Answer (Blank)

D. The passage states that people with cystic fibrosis tend to have autodestruction of the pancreas. Autodestruction would eliminate the islets of Langerhans which secrete both glucagon and insulin (choices B and C are incorrect). After a carbohydrate-rich meal, the serum glucose concentration should rise (choice D is correct). In the absence of insulin, the serum glucose concentration will remain high until it can be cleared from the blood by the kidney (choice A is incorrect).

A researcher has discovered a novel frameshift mutation in pyruvate carboxylase and hypothesizes that this could affect tumor metabolism. Which of the following would be the best experiment to perform to either confirm or disprove her hypothesis? D. Generate a cell line that expresses the mutant form of pyruvate carboxylase and measure lactate secretion, and glucose and glutamine uptake compared to control cells that express the normal form of pyruvate carboxylase. Correct Answer (Blank)

D. The question stem states that a frameshift mutation in pyruvate carboxylase has already been found. There would therefore be little point in sequencing the gene again. In addition, "confirm[ing] which amino acid is altered in the mutant form" is a better match to confirming a point mutation rather than a frameshift mutation (choice A is wrong). Pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate in the first step of gluconeogenesis. Determining how the proton gradient is used by ATP synthase isn't relevant to this enzyme's function (choice B is wrong). Option C is a very tempting answer choice because it would be ideal to determine if the mutation is affecting enzymatic activity. Unfortunately, a western blot will not give information on this, since it can only measure protein levels (choice C is wrong). By process of elimination, the correct answer is choice D. Measuring lactate secretion, and glucose and glutamine uptake will give an indirect readout of whether basic metabolic processes are different in cells with the pyruvate carboxylase mutation versus in normal cells (choice D is correct).