Bio/Biochem MQL

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 B) type O and type AB C) type B and type O D) type A and type AB

A; A man with type AB blood has the genotype AB and can therefore produce with either the A allele or with the B allele. A woman with type O blood has two O alleles and therefore only produces gametes of the O allele. So if this couple has children, the only two genotypes their children can possible inherit with respect to the blood groups are AO and BO, which corresponds to the phenotypes type A blood and type B blood, respectively. Thus, (A) is the correct answer.

Which of these techniques is most useful in determining the primary structure of a protein? A) Edman degradation B) Bradford (Coomassie Brilliant Blue) protein assay C) X-ray crystallography D) Gabriel reaction

A; Determining the primary sequence of a protein means determining the exact arrangement of amino acids in the protein (that is, which comes first, which comes second, and so on). This is most easily accomplished using a technique that removes amino acids one at a time; that method is the Edman degradation, choice (A) . Choice (B) . The Bradford protein assay uses Coomassie Brilliant Blue dye to determine protein concentration , not its primary sequence. Choice (C) . X-ray crystallography is generally used to determine three-dimensional structure, not primary structure. Choice (D) . The Gabriel reaction is used to synthesize amino acids, not sequence peptides.

Despite the importance of mitochondria, scientists have demonstrated that human mitochondrial DNA mutates at a fairly high rate. These mutations are most likely to be: A) point mutations B) frameshift mutations C) lethal mutations D) nondisjunctions

A; Mitochondrial DNA directs the synthesis of mitochondrial proteins, which ultimately play a major role in cell survival. Since mitochondria are so essential to eukaryotic cell life, one would therefore expect replication of its DNA to be highly accurate. Mutations that would cause a dramatic change in its DNA and its ability to produce proteins needed for ATP formation would be lethal to the cell. Since the question states that mutations do occur, the most likely type to occur would be one that causes the least damage. Point mutations are defined as those in which only one nitrogenous base is affected; for example, a cytosine is substituted for an adenine during replication. Point mutations are not usually lethal because of the redundancy of the genetic code; that is, each amino acid is typically coded for by more than one codon. So a point mutation is the type of mutation least likely to affect their productivity, and choice (A) is correct. Choice (B). A frameshift mutation is a mutation in which bases are inserted or deleted during DNA replication or transcription. This produces a shift in the reading frame of the mRNA strand being translated, usually leading to the formation of nonsense polypeptides. Changes in protein synthesis would most likely be dangerous for the mitochondria and the cell itself. Choice (C). Lethal mutations are, by definition, those that would cause the mitochondria to become nonfunctional. So if a lethal mutation occurs, the entire cell will likely die because ti could not produce enough ATP without the mitochondria. Choice (D). Nondisjunction is the failure of homologous chromosomes to separate during meiosis. Mitochondria have one circular chromosome, there aren't any homologous chromosomes, so crossing over can't occur at all. Also, mitochondria do not undergo meiosis. Only specialized eukaryotic cells in sexually reproducing organisms undergo meiosis.

A patient with renal failure has nephrons which lack the ability to actively secrete or reabsorb any substances. Which of the following actions will the patient's kidney most likely still be able to perform? A) excretion of salt in urine B) production of hypertonic urine C) regulation of blood osmolality D) conservation of glucose

A; Recall that the fluid in Bowman's capsule is isotonic to plasma. Without reabsorption or secretion, isotonic urine can still be produced. In other words, this patient's nephrons filter the blood at the glomerulus, but that initial filtrate then becomes urine without any further modification. This process will still allow excess salt to be removed from the body, because that salt diffuses through the glomerulus, which is still functioning in this patient. therefore, choice (A) is correct. Choice (B) is wrong because the production of hypertonic urine requires the concentration gradient between the medulla and cortex of the kidney. This gradient is generated by active secretion and reabsorption. Choice (C) is wrong because regulation of blood osmolality requires reabsorption and secretion. Choice (D) is wrong because glucose is normally filtered and then actively reabsorbed in the proximal convoluted tubule, so this patient could not conserve glucose.

When UDP-glucose reacts with the nonreducing end of a lengthening glycogen polymer, which type of bond is most commonly formed? A) alpha-1,4-Glycosidic bond B) alpha-1,6-Glycosidic bond C) beta-1,4-Glycosidic bond D) beta-1,6-Glycosidic bond

A; The bond that links most glucosyl moieties in glycogen is the alpha-1,4-glycosidic bond, (A).

Which of the following correctly orders the anatomical features encountered by oxygen during inhalation? A) pharynx -> larynx -> trachea -> bronchioles B) larynx -> pharynx -> trachea -> alveoli C) larynx -> pharynx -> bronchi -> alveoli D) nares -> larynx -> bronchioles -> trachea

A; The correct order is given in (A) . Choice (B) and (C) . The pharynx is encountered before the larynx. Choice (D) . The trachea is encountered before the bronchioles.

*See passage for #54 on Kaplan Test 1* 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 B) the new strain transcribes 500 copies of OUT RNA per copy of IN RNA C) only one copy of the target sequence is present after insertion of the transposon into the new strain's DNA D) the overlap between the transcripts from P in and P out in the new strain is twice as many base pairs as the other strains

A; The dam methylation system, mentioned in the fourth paragraph, regulates the rate of transposition and the activation of Tn10 with DNA replication. It's stated that 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 this 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. Therefore, (A) is the correct answer. Choice (B). Copies of OUT RNA and IN RNA are concerned with multicopy inhibition. OUT RNA is the mRNA transcript of the strong promoter, P out, and IN RNA is the mRNA transcript of the weak promoter, P in. In a typical cell containing Tn10, there are 100 copies of OUT RNA per copy of IN RNA. Thus, multicopy inhibition inhibits Tn10 transposition by ensuring that all IN RNA will be bound to OUT RNA and is therefore unavailable for translation. If IN RNA was translated then transposition would occur. Therefore, if the new strain of cells has an even higher ratio of OUT RNA to IN RNA, the chances of IN RNA being translated would be even lower. Therefore, this would DECREASE the rate of transposition, not increase it. Choice (C). This choice has to do with the insertion of the transposon into host DNA and the absence of a repeated sequence on both sides of the transposon. From the passage: following transposon insertion, the target sequence is duplicated and flanks the transposon. However, the passage never relates the presence of this repeated sequence in the host DNA following transposon insertion to the rate of transposition or the functional ability of the transposon. In fact, the passage never discusses ANY function for this repeated sequence. Therefore, one cannot conclude that the lack of a repeated sequence in the host DNA would affect the rate of transposition. Choice (D). This choice refers to the overlap between the IN RNA and OUTRNA of Tn10. This overlap is what allows OUT RNA to complementary base pair to IN RNA, which in turn prevents the translation of the IN RNA. If the overlap between the two transcripts were increase, then the interaction between the two molecules would be even stronger. This would mean that it would take more energy to separate the IN RNA from the OUT RNA. And this means that if anything, the IN RNA would remain paired to the OUT RNA and would be translated LESS frequently.

*See passage for #35 on Kaplan Test 1* At temperatures below 85 degrees C, the renaturation of the organism's DNA is: A) spontaneous and exothermic B) spontaneous and endothermic C) nonspontaneous and exothermic D) nonspontaneous and endothermic

A; The denaturation process is endothermic (heat must be supplied) and the renaturation process is exothermic. This rules out (B) and (D). No additional input of energy is necessary for the reannealing process, so this is best described as spontaneous.

The semi-permeable membrane of the dialysis machine functions in a manner most analogous to which part of the kidney? A) glomerulus B) ureter C) descending loop of Henle D) vasa recta

A; The glomerulus, (A), functions like a sieve, allowing the filtration (movement from the circulation into the nephron tubule) of small molecules while blocking the filtration of the plasma proteins. The semi-permeable membrane serves an analogous function in the dialysis machine. Choice (B) is wrong because the ureter is merely a tube connecting the kidney to the bladder. Choice (C) is wrong because the descending loop of Henle does not serve a filtration function. The primary action of the descending loop is reabsorption of water. Choice (D) is wrong because The vasa recta are the capillaries that supply nutrients to the nephron.

Cells can enter a quiescent state, also called G0, in which they do not actively divide. Typically, a cell would be considered to be in which phase of the cell cycle immediately before entering G0? A) G1 B) S C) G2 D) M

A; The key checkpoint where the cell "decides" if it is going to actively divide or enter G 0 is at the end of G 1 , just before the start of S phase. Once the cell begins synthesizing DNA, it generally becomes committed to dividing. Thus choice (A) is correct.

Which of the following conformations most accurately represents the three-dimensional structure of the compound in aqueous solution? A) the envelope form of ribose B) the chair form of ribose C) the boat form of ribose D) the planar form of ribose

A; The lowest energy form of ribose (in solution) is the envelope form, thus (A) is the correct answer. (B) While the chair form is glucose's predominant conformation, glucose is a six-membered ring, whereas ribose generally forms 5-membered rings. (C) The boat form is not a likely conformation for ribose. (D) Most ribose molecules will adopt a non-planar conformation to alleviate ring strain.

Which of the following hormones elevates calcium blood concentration? A) parathyroid hormone B) calcitonin C) glucagon D) aldosterone

A; The parathyoid hormone (PTH) causes calcium blood concentration to increase in response to low blood calcium levels. PTH causes the osteoclasts in bone to break down bone, referred to as bone resorption, so that calcium is released into the blood, therefore (A) is the correct response. Choice (B) . Calcitonin antagonizes PTH and causes calcium levels to decrease in the blood. Choice (C) . Glucagon has no effect on blood calcium levels. Choice (D) . Aldosterone has no effect on blood calcium levels.

*See passage for #6 on Kaplan Test 1* Which of the following best describes the response of the skeletal system to increased levels of PTH? A) Osteoclast activity increased because bone breakdown will raise blood calcium levels B) Osteoclast activity decreases because high levels of blood calcium trigger bone formation C) Osteoblast activity increases because bond remodeling will raise blood calcium levels D) Osteoblast activity decreases because high levels of blood calcium trigger bone breakdown

A; The passage states that the systems that respond to PTH do so with the goal of raising blood calcium. Recall that osteoclasts are cells that break down bones, and osteoblasts are cells that lay down bone. Choice (A) is therefore correct because stimulation of osteoclasts breaks down bone, releasing calcium from bone into blood. Choice (C) is wrong because increasing osteoblast activity will lower blood calcium by placing it into bone, contrary to the purpose of PTH. Choices (B) and (D) are incorrect because they pair the cells with incorrect roles.

Obesity linked to leptin gene mutations is inherited in an autosomal recessive manner. Which of the following represents the percentage of offspring expected to develop obesity when a mouse homozygous for the dominant allele of the leptin gene is mated with a mouse that exhibits obesity linked to leptin mutation? A) 0% B) 25% C) 50% D) 100%

A; The question stem states that a homozygous dominant mouse is mated with a homozygous recessive mouse (a mouse exhibiting the recessive phenotype of obesity linked to leptin gene will have two recessive alleles). In such a case, 100% of their offspring will be heterozygotes. The question stem stated that obesity linked to leptin gene mutations is autosomal recessive, so 0% of heterozygotes will exhibit the obese phenotype, therefore (A) is the correct answer.

Hydrophobic interactions have the LEAST significant role in which of the following? A) the secondary structure of globuar proteins B) the stability of intracellular vesicles C) the spontaneous formation of micelles D) the immiscibility of oil and water

A; The secondary structure of proteins (globular or not) is determined primarily by local hydrogen bonds; hydrophobic interactions play a more significant role in the tertiary structure of proteins, thus (A) is the answer. (B), (C), and (D) rely on hydrophobic interactions.

Which of the following is a true statement concerning muscle? A) cardiac muscle combines attributes of both smooth and skeletal muscle B) skeletal muscle lacks striations C) muscle tissue only carries out aerobic respiration D) the presence of Ca 2+ will inhibit muscular contractions

A; The three types of muscle tissues are skeletal, cardiac and smooth. Skeletal muscle is striated and is under voluntary control, while smooth muscles lack these striations and are involuntary. Cardiac muscle contains attributes of both skeletal and smooth since they are striated, but are under involuntary control. (A) is the correct answer. Choice (B) . A defining histological feature of skeletal muscles is their striations. Choice (C) . Muscle tissue can undergo fermentation in anaerobic conditions and produce lactate from pyruvate. This lactate or lactic acid is responsible for muscle fatigue. Choice (D) . The release of Ca 2+ from the sarcoplasmic reticulum is responsible for muscle contraction.

*See passage for #26 on AAMC Test 3* If HDAC inhibition by βOHB is a physiological response in living animals, the information in the passage indicates it is likely to occur when: A) there is sustained fatty acid oxidation. B) the pentose phosphate shunt is activated. C) there is increased gluconeogenesis. D) the Cori cycle is occurring.

A; This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is A because fasting leads to glycogen breakdown and gluconeogenesis, then continued fasting leads to the production of ketone bodies by sustained fatty acid oxidation. βOHB is a component of ketone bodies, so serum levels of βOHB are elevated when there is sustained fatty acid oxidation. It is a Scientific Reasoning and Problem Solving question because it requires determining which biological process would result in a particular physiological process in the body.

*See passage for #8 on AAMC Test 1* Based on the information in the passage, a mature AT1 mRNA is most likely to contain a sequence coding for which genetic factor(s)? A) signal sequence B) introns C) promoter D) nuclear localization signal

A; This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is A. The passage indicates that AT1 is a transmembrane protein. Transmembrane proteins enter the endomembrane system by docking at the rough ER. This is facilitated by using a signal sequence. Mature mRNAs are not likely to contain introns (B) or promoter sequences (C). A nuclear localization signal permits proteins to enter the nucleus. This is not likely for a transmembrane protein (D). This is a Scientific Reasoning and Problem Solving question because you are asked to predict which genetic factor is coded for by a sequence within AT1 mRNA based on the information provided in the passage.

*See passage for # on AAMC Test 1* Based on the passage, which of the following is LEAST likely to be a symptom of diabetes mellitus? A) loss of appetite B) sweet-tasting urine C) unexplained weight loss D) feelings of fatigue

A; This is a Biology question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is A, because based on the passage, the symptoms of diabetes mellitus are frequent urination and catabolism of fatty acid as well as proteins as alternative energy sources. In diabetic patients, high blood glucose results in excretion of excess sugar into the urine, hence sweet-tasting urine. Similarly, the catabolism of fatty acids and proteins results in weight loss and feelings of fatigue respectively. This leaves loss of appetite as the only correct answer. This is a Scientific Reasoning and Problem Solving question because you are asked to evaluate the information from the passage and identify possible symptoms of diabetes mellitus.

*See passage for #42 on AAMC Test 1* Pericytes were growth-arrested in Experiment 1 so that the: A) pericyte growth would not interfere with the measurement of EC growth B) pericytes would not inhibit EC growth C) metabolic wastes of the pericytes would not interfere with the measurement of EC growth D) ECs would not cause the pericytes to grow

A; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The question asks the examinee to identify the reason for using pericytes that were growth-arrested in Experiment 1. The results of the experiment were expressed in terms of the number of endothelial cells. Had the pericytes been allowed to divide, the observed increase in the number of cells in the culture would have been due to increases in the number of pericytes and in the number of endothelial cells. By inhibiting the division of pericytes, the observed change in cell number is attributable solely to the growth of endothelial cells. The reason for using growth-arrested cells is therefore to simplify the measurement of the number of endothelial cells in the culture. Thus, A is the best answer. This is a Reasoning about the Design and Execution of Research question as it asks why the researchers performed a specific step in the experiment.

Certain viruses contain RNA as their genetic material. One of the ways these RNA viruses replicate themselves is to: A) code for or carry a transcriptase that copies viral RNA. B) infect microorganisms possessing RNA as their genetic material. C) alter the host cell's polymerase in order to synthesize progeny viral RNA from the viral RNA template. D) stimulate the transcription of specific sequences of the host's DNA, which, in turn, direct the assembly of viral particles.

A; This is a Biology question that falls under the content category "The structure, growth, physiology, and genetics of prokaryotes and viruses." The answer to this question is A because RNA viruses require a type of transcriptase (reverse transcriptase) to replicate themselves. It is a Scientific Reasoning and Problem Solving question because it requires determining which mechanism is most reasonable for the replication of RNA viruses.

An RNA molecule has 1500 bases. What is the maximum number of amino acids it can encode? A) 500 B) 1000 C) 1500 D) 4500

A; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is A because each amino acid is encoded by three bases, so 1500/3 is 500, which is the maximum of amino acids the RNA molecule can encode. This is a Knowledge of Scientific Concepts and Principles because it addresses basic concepts of gene to protein knowledge.

*See passage for #40 on AAMC Test 3* Which statement best accounts for the hereditary transmission of SDH-linked paraganglioma in a parent specific manner? SDH is: A) an imprinted gene. B) a Y-linked gene. C) an X-linked gene. D) a tumor suppressor gene.

A; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is A because imprinted genes are expressed in a parent-specific manner. B is incorrect because based on the passage both sexes may carry the mutated allele. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of regulation of gene expression.

*See passage for #6 on AAMC Test 3* Based on the information in the passage, what is the most likely mechanism of inheritance for HPRCC? A) autosomal dominant B) autosomal recessive C) X-linked recessive D) Y linked

A; This is a Biology question that falls under the content category "Transmission of heritable information from generation to generation and the processes that increase genetic diversity." The answer to this question is A because the allele must be inherited in a autosomal dominant pattern in order for individuals II-1 and II-2 to produce both affected and unaffected offspring. The inheritance pattern for the offspring of these individuals is consistent with autosomal dominant if the parents are heterozygous. The rest of the pedigree is consistent with this mode of inheritance as well. In addition, the offspring of individuals II-1 and II-2 rule out autosomal recessive, X-linked recessive, and Y-linked. It is a Scientific Reasoning and Problem Solving question because in addition to interpreting the pedigree shown in the figure, it requires forming a hypothesis regarding the genetic transmission pattern of HPRCC.

The average osmotic pressure of ocean water is 28 atm corresponding to a concentration of 0.50 M solutes (approximated as NaCl). What is the approximate concentration of solutes (also approximated as NaCl) present in blood with an osmotic pressure of 7 atm? A) 0.12 M B) 0.25 M C) 2.0 M D) 3.5 M

A; This is a General Chemistry question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is A because osmotic pressure is directly proportional to solute concentration. Since the osmotic pressure of blood is one-fourth that of ocean water, the solute concentration is also one-fourth that of ocean water, or 0.25 × 0.50 M = 0.12 M. It is a Scientific Reasoning and Problem Solving question because it requires working with scientific models to solve a problem.

What is the net number of ATP molecules synthesized by an obligate anaerobe per molecule of glucose? A) 2 ATP B) 6 ATP C) 8 ATP D) 36 ATP

A; This question cannot be answered from the passage so it is a "pseudo-discrete" question. An obligate anaerobe is an organism that must live WITHOUT oxygen in order to survive. Obligate anaerobes produce ATP via fermentation, which includes both glycolysis and the reactions necessary to regenerate the NAD+ necessary for glycolysis to continue. Fermentation leads to a net production of 2 ATP; this ATP is generated during glycolysis. Therefore, an obligate anaerobe will produce 2 ATP per molecule of glucose, which matches with choice (A). The other answer choices are wrong because they state the wrong number of ATP, but choice (D) is a particular trap. Aerobic organisms produce a net of 36 ATP per molecule of glucose, as shown in the equation provided in the passage. However, the question is asking about obligate anaerobes.

In a person with MCAD deficiency, if the body has excess glucose stores, which electron transfer intermediates are likely to accumulate? How does this change if the body is energy depleted? A) NADH and FADH2 when high glucose, NAD+ and FAD+2 when depleted B) NADH and FADH2 when high glucose, NADH and FADH2 when depleted C) NAD+ and FAD+2 when high glucose, NAD+ and FAD+2 when depleted D) NAD+ and FAD+2 when high glucose, NADH and FADH2 when depleted

A; While the question stem specifically asks about MCAD deficiency, this is a bit of a distractor because the reactions responsible for controlling the ratio of electron transport substrates are in the ETC downstream of the MCAD reaction. With excess glucose stores, the presence of a large proton gradient prevents the transfer of electrons in the ETC, thus causing a buildup of NADH and FADH2 while in a low energy state the need for ATP causes a decrease in these carriers and a corresponding increase in NAD+ and FAD+2. This corresponds with choice (A).

Which of the following descriptions correctly characterizes a micelle? A) a single-layered structure with hydrophobic groups pointing outward B) a single-layered structure with hydrophilic groups pointing outward C) a double-layered structure with hydrophobic groups on the exterior surfaces D) a double-layered structure with hydrophilic groups on the exterior surfaces

B; A micelle forms when lipids reach a critical concentration in water. Unlike a membrane, a micelle consists of a single layer of lipid molecules, which eliminates (C) and (D) . Since the micelle's surface is exposed to water, it should have hydrophilic groups on the outer surface, which matches choice (B) . Choice (A) . This would not be expected, as hydrophobic groups are relatively unstable in water. Choice (C) . This is not observed in nature. Choice (D) . This is the structure of a vesicle , not a micelle.

A patient diagnosed with MCAD deficiency and no other health conditions is found to have a father that carries the same mutation, but a mother without evidence of mutation when a buccal swab is taken. Which of the following is the most likely reason? A) paternal nondisjunction conferring two copies of the gene for MCAD B) maternal germline mutation in the gene coding for MCAD C) maternal nondisjunction deleting one copy of the gene coding for MCAD D) multiple somatic mutations in the fetus of the gene coding for MCAD

B; A single event is almost always more likely than multiple events. Therefore the consideration should be isolated to events that would occur before division of the zygote. In this case, any loss of the fully functioning gene is likely to cause the symptoms. However, this still leaves maternal MCAD loss and maternal MCAD mutation as possibilities. To distinguish between them, recognize that the child does not have any other medical conditions. Monosomy in any chromosome, as would result from maternal nondisjunction resulting in gene loss, is a serious medical condition. Therefore, choice (B) is correct.

*See passage for #22 on Kaplan Test 1* In the experiments described in the passage, how does water most likely react to the polar protein groups? A) by not forming a solvation layer at all around any of the polar protein groups B) by forming a solvation layer to a lesser extent than around non-polar groups C) by forming a solvation layer to a greater extent than around non-polar groups D) by forming a solvation layer to an equal extent as that of the non-polar groups

B; Because the number of hydrogen bonds per unit of mass in water is greater than that of water with any polar group, as well as the fact that even the most polar groups have solubility limits, a solvation layer will form around even polar groups, though not to the extent that it does around non-polar groups because hydrogen bonding can still occur to a limited degree.

Oxaloacetate, the precursor of citrate in the citric acid cycle, has: A) 3 carbons B) 4 carbons C) 5 carbons D) 6 carbons

B; Citrate (citric acid) has six carbons. It is generated in the first step of the Krebs cycle from oxaloacetate and acetyl CoA, which has two carbons. Therefore, oxaloacetate has 6 - 2 = 4 carbons, which matches choice (B)

Which of the following conditions is most likely to inhibit the Krebs cycle? A) low [NADH]/[NAD+] ratio and low [ATP]/[ADP] ratio B) high [NADH]/[NAD+] ratio and high [ATP]/[ADP] ratio C) high [NADH]/[NAD+] ratio and low [ATP]/[ADP] ratio D) low [NADH]/[NAD+] ratio and high [ATP]/[ADP] ratio

B; Each answer choice asks about two ratios, [NADH]/[NAD + ] and [ATP]/[ADP]. If the Krebs cycle is inhibited , then the cell does not need energy at that point. Starting with the latter, a high ratio of ATP to ADP indicates that the cell has a lot of energy stored, and is not likely to need more energy immediately. Thus, we should expect a high [ATP]/[ADP] ratio, which eliminates (A) and (C) . Similarly, the Krebs cycle generates NADH, so an excess of it—that is, a high [NADH]/[NAD + ] ratio—should inhibit the Krebs cycle by Le Châtelier's principle. The correct answer is choice (B)

*See #1 on Kaplan Test 1 to see images* Antibodies to leptin were generated and injected into the wild type mice. Within hours of the injection, leptin was completely cleared from blood and remained undetectable for three weeks. Which of the following figures best represents daily food intake before the injection and one week after the injection of antibodies? A) graph showing 15g food intake before injection, 13g after injection B) graph showing 13g food intake before injection, 15g after injection C) graph showing 14g food intake before injection, 12g after injection D) graph showing 13g food intake before and after injection

B; From the information presented in the figures, it can be seen that daily injections of leptin will reduce the amount of food consumed by the wild type mice. The last sentence of the passage indicates that leptin, among other things, decreases appetite. thus it can be expected that the elimination of leptin from the blood by antibodies would increase appetite and food intake in wild type mice. Of all the answer choices, only choice (B) represents increased food intake after administration of antibodies to leptin and is thus the correct answer. Choices (A) and (C) show that food intake has decreased. Food intake after the clearance of leptin is expected to increase. Choice (D) shows that food intake has stayed the same. Food intake after the clearance of leptin is expected to increase.

Which of these changes would be most likely to stimulate hormone-sensitive triacylglycerol lipase? A) an increase in aldosterone levels B) a decrease in insulin levels C) a decrease in cortisol levels D) a decrease in epinephrine levels

B; Hormone-sensitive triacylglycerol lipase breaks down triglycerides in fat tissue when the body has relatively low levels of glucose. Both (A) and (B) are observed in that case. Fat cells, however, do not respond to glucagon levels, so the most likely trigger, among the answer choices listed, would be a decrease in insulin levels, choice (B) . Choice (A) . An increase in aldosterone would not be predicted to have any effect on lipase activity. Choice (C) . An increase in cortisol levels activates the lipase. Choice (D) . An increase in epinephrine levels activates the lipase.

In order to prevent the net movement of water between the blood and the dialyzing fluid, the dialyzing fluid: A) is hypoosmotic to blood B) is isoosmotic to blood C) contains a high concentration of solutes D) contains hydrophilic proteins

B; If the dialyzing fluid is isoosmotic, it has the same concentration of particles and thus, the same osmotic pressure exists on either side of the membrane. Therefore, with an isoosmotic dialyzing fluid, there will be no net flow of water by osmosis between the blood and the dialyzing fluid. This means choice (B) is correct. Choice (A). A hypoosmotic dialyzing fluid would lead to flow of water into the circulation from the dialyzing fluid. Choice (C). This choice is too vague to be correct. Since "high" is an imprecise term, it's impossible to know whether the exact concentration would be hypoosmotic, isoosmotic, or hyperosmotic to blood. Choice (D). Hydrophilicity has nothing to do with the net flow of water. A hydrophilic protein is "water-loving" because ti contains polar amino acids.

After an infection, such as by C. difficile, the normal gut flora "re-colonize" the digestive tract. Those re-colonizing bacteria are most likely harbored by the: A) terminal ileum B) cecum B) jejunum D) rectum

B; Normal gut flora "hang out" in the cecum, the blind outpouching of the large intestine. So choice (B) is correct. Choice (A). The terminal ileum is part of the small intestine. Choice (C). The jejenum is the middle portion of the small intestine, which is primarily involved in absorption. Choice (D). The rectum is the site of solid waste storage, not a harbor for bacteria.

*See passage for #37 on Kaplan Test 1* Which of the following statements regarding recombination frequency is supported by information in the passage? A) double-crossovers occur more frequently than single-crossovers B) recombination can underestimate distance between loci C) loci with recombination frequencies of 1.5% are more common than those with a recombination frequency of 1.0% D) recombination frequency increases when introns are spliced out between two loci

B; Recombination frequency is used to determine genetic distance, but cannot determine physical distance. If this is unclear, think about two alleles that are very far apart. The passage states that these alleles are likely to crossover multiple times; if a crossover event occurs an even number of times, the recombination frequency will appear to be zero. This thinking shows why choice (B) is correct: the recombination can make the alleles look closer together. Choice (A) is incorrect because the frequency of a double crossover is the product of each single crossover. Choice (C) is not correct since the passage never states which recombination frequencies are more common, only that they are related to distance. Loci with a recombination frequency of 1.5% recombine more often than those with a frequency of 1.0%, but who knows which one is more common. Finally, recombination occurs with DNA, while introns are spliced out during RNA processing: so (D) is incorrect as well.

*See passage for #42 on Kaplan Test 2 (short)* At the end of the experiment described in the passage, how many patients in the intensive treatment group exhibited albuminuria? A) 1,480 B) 2,181 C) 3,103 D) 4,649

B; The answer to this question is in Table 1. The total number of people with albuminuria is the number of people who had 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 1,623 + 1,480 - 922 = 2,181. (Note that the answers are far enough apart to allow estimation such as 1,600 + 1,500 - 900 = 3,100 - 900 = 2,200.) This matches with choice (B). Choice (A). This is just the number of patients who developed albuminuria. Choice (C). This is the number of patients who had albuminuria before the trial plus those who developed it during the trial; it also incorrectly includes the patients whose albuminuria disappeared. Choice (D). This is all the patients minus the ones whose albuminuria disappeared.

In the nephron, increasing the permeability of the collecting duct to water has the effect of: A) diluting the urine B) concentrating the urine C) increasing the rate of glomerular filtration D) buffering urine pH

B; The collecting duct of the nephrons is where the majority of water is reabsorbed and is under the control of the hormone antidiuretic hormone or ADH. ADH causes the filtrate to become more concentrated by increasing the permeability of the collecting duct so that more water is reabsorbed. Choice (A) . The collecting duct would have to be impermeable to water to form dilute urine. Decreased levels of ADH would cause this. Choice (C) . Increasing the permeability of the collecting duct would have no effect on the rate of glomerular filtration. Glomerular filtration occurs at the Bowman's capsule, which is the furthest structure from the collecting duct in the nephron. Choice (D) . Changing the tonicity of the filtrate by concentration and dilution has no effect on the urine pH. The active secretion of H + ions into the filtrate only affects urine pH.

Which of these immune defenses has both antibody-dependent and antibody-independent mechanisms of action? A) lysozyme B) complement system C) interferon D) prions

B; The complement system has a classical pathway that requires antigen binding, and an alternative pathway that is antibody-independent. Choice (B) is correct. Choice (A) . Lysozyme is an enzyme that does not depend on antibodies at all. Choice (C) . Interferon is a cytokine that can recruit immune responses. Choice (D) . Prions are not part of the immune system.

Which of the following statements regarding histones is FALSE? A) the exterior of histones has a high proportion of positively charged side chains B) histones form a tetramer C) a nucleosome consists of approximately 200 base pairs wrapped around a histone core D) histones are highly conserved across species

B; The easiest way to attack this question is to cycle through the answer choices looking for an incorrect statement about histones. That statement is choice (B) : while four different histones form the core around which DNA winds in a nucleosome, there are two copies of each protein, so the histone core is an octamer .

A particular genus of plant inhabits 20 islands in an island chain. Which of the following, if true, would most plausibly have been caused by the founder effect? A) on one island, a species of that genus went extinct for 500 years, and then returned to the island B) on one island, all plants of that genus lack a compound produced by all other members of that genus on the other islands C) on one island, a new species arises incorporating genes from two other species within that genus D) on one island, researchers find the common ancestor of all the plants of that genus

B; The founder effect is a form of genetic drift—a change in gene frequencies that occurs as a result of chance. Specifically, the founder effect occurs when a small population is isolated from a larger population, and therefore has a substantially different allele distribution. This might result in, for example, a particular recessive allele being found almost everywhere, as in choice (B) . Choice (A) . This can happen, but is not an example of the founder effect. Choice (C) . This can happen, but is not an example of the founder effect. Choice (D) . This suggests radiation of the species from one island to the others, but this is not evidence for the founder effect.

*See passage for #9 on Kaplan Test 1* Which of the following statements regarding surgical removal of a parathyroid gland is supported by information in the passage? A) Surgical therapy is necessary in people with high PTH levels B) Blood PTH levels fall rapidly following the removal of an overactive parathyroid gland C) Kidney, bone, and intestinal disease result if PTH levels fall following surgery D) Circulating calcium levels will be symptomatically low following the removal of a parathyroid gland

B; The key to answering this question is the information that the half-life of PTH is only a few minutes. Choice (B) is correct because once the source of the excess PTH is removed, the hormone circulating in the body will be exhausted quickly. Choice (A) is incorrect because the passage states that surgical therapy is not necessary unless the patient is symptomatic -- simply having high levels of PTH is not enough. Choice (C) is incorrect because it's not mentioned in or supported by the passage. And choice (D) is incorrect because the passage tells us that the other three parathyroid glands compensate adequately.

*See passage for #21 on Kaplan Test 1* As non-polar protein groups cluster in water, the water experiences a(n): A) decrease in entropy B) increase in entropy C) decrease in hydrogen-bonding D) increase in hydrogen-bonding

B; The organization of the solvation layer causes a decrease in entropy, so the clustering of non-polar groups, by virtue of diminishing the layer, causes a favorable increase in entropy. In fact, this increase in entropy is the predominant thermodynamic influence resulting in the clustering of nonpolar groups in polar solvents like water.

*See passage for #55 on Kaplan Test 1* It can be inferred from the passage that, aside from being involved in the regulation of transposon activity, the presence of hemi-methylated DNA also allows cells to: A) divide more rapidly B) distinguish between old and new strands of DNA C) incorporate free methyl groups into other biological molecules such as amino acids and cofactors D) transport proteins into and out of the nucleus

B; The passage discusses the issue of hemi-methylation in the context of the dam methylation system in the fourth paragraph, where 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 kick into action. The reason why the parent strand has methyl groups attached to the DNA, while the new strand lacks them, is due to the semi-conservative mechanism by which DNA is replicated. During DNA replication, the parent strands unwind and act as templates 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 - and there's how you get your hemi-methylated DNA. Choice (B) is the correct answer. The cell is able to distinguish between the old strand - the strand with methyl groups, and the new strand - the one without methyl groups. Choice (A). Although the hemi-methylated DNA arises after replication, it does NOT affect the rate at which the cell replicates, or divides. The replication of DNA, and hence the presence of hemi-methylated DNA, is only an indication of cell division. Choice (C). Choice (C). The use of free methyl groups in the synthesis of amino acids and cofactors is unrelated to the hemi-methylation of DNA. The passage does not mention the source of the methyl groups used in the dam system, nor does it address other possible uses for methyl groups. Choice (D). Besides the fact that there is nothing in the passage that leads to this conclusion, transportation across the nuclear membrane depends on carrier proteins, transport proteins, and the pores found in the nuclear membrane. And although DNA is believed to lend a helping hand towards directing the transport process, this "help" is independent of its methylation status.

*See passage for #39 on Kaplan Test 1* Suppose that classical Mendelian genetics predicts that the probability of inheriting two alleles, which are 5 centiMorgans apart, is 25%. The actual frequency with which the two alleles are inherited together is expected to be: A) lower than 25% B) greater than 25% C) equal to 25% D) fluctuating around 25%

B; The passage states that Mendelian genetics does not take into account linkage of alleles. In this question, the alleles that are 5 centiMorgans are certainly linked, since 5 centiMorgans means that the alleles are rarely recombined (i.e. they are so close together, one rarely moves from one chromosome to another without the other allele). Therefore, linked chromosomes are more likely to be inherited together than originally predicted by Mendelian genetics: the probability should be greater than 25%, which matches choice (B). As a side notes, for those who argue that 5 centiMorgans could mean that the two alleles are so far apart that they often crossover and then crossover again, the answer is still choice (B), since on average the alleles end up sharing a chromosome.

The vital capacity of the lungs is: A) the amount of air exchanged in a normal breath B) the maximum amount of air that can be moved in a single respiratory cycle C) the minimum amount of air that remains in the lungs at all times C) the total amount of air the lungs can hold

B; The vital capacity (VC) of the lungs equals the tidal volume plus the inspiratory and expiratory reserve volumes. It thus represents the difference between the maximum and minimum amounts of air that the lung can hold, which is closest to choice (B) . Choice (A) . This is the definition of tidal volume (TV). Choice (C) . This is the definition of the residual volume (RV). Choice (D) . This is the definition of total lung capacity (TLC).

*See passage for #24 on AAMC Test 1* Which mechanism best describes how P-gp facilitates drug resistance? A) P-gp binds to antitumor drugs in the presence of ATP and degrades the drugs B) P-gp serves as a pump and uses active transport to move antitumor drugs outside the cell C) P-gp prevents the entry of anti-tumor drugs into the cell D) P-gp causes increased membrane permeability, which causes antitumor drugs to exit the cell

B; This is a Biochemistry question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer is B because P-gp is an ABC transporter protein, which uses ATP to actively transport antitumor drugs out of the cell. This is a Scientific Reasoning and Problem Solving question because you must apply knowledge of a transport protein to a situation of drug resistance.

Pellagra also results from a deficiency of nicotinamide, which is synthesized from tryptophan. Nicotinamide nucleotides are neither oxidized nor reduced during which step of cellular respiration? A) glycolysis B) chemiosmosis C) citric acid cycle D) electron transport chain

B; This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer is B because NAD+ is the oxidized form, and NADH is the reduced form of nicotinamide adenine dinucleotide. NAD+ is neither reduced to form NADH, nor is NADH oxidized to form NAD+ specifically during chemiosmosis. This is a Knowledge of Scientific Concepts and Principles question because you must use knowledge of chemiosmosis to determine the role of NAD+ and NADH in that process.

*See passage for #34 on AAMC Test 3* Based on Figure 1, hypoxia induces an enzyme of: A) the respiratory chain. B) glycolysis. C) glycogenesis. D) the citric acid cycle.

B; This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is B because the data presented in Figure 1 indicate an increase in cellular levels of phosphoglucose isomerase, an enzyme involved in glycolysis. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of enzymes involved in metabolic pathways.

*See passage for #33 on AAMC Test 3* What are the effects of hypoxia on energy metabolism? In hypoxia: A) the citric acid cycle is induced to produce more NADH. B) lactic fermentation is induced to regenerate NAD+. C) the electron transport chain is induced to consume more NADH. D) glycogenesis is induced to regenerate NAD+.

B; This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is B because the passage states that in hypoxia, glycolysis is activated while mitochondrial functions (oxidative phosphorylation) are attenuated. This metabolic reprogramming leads to increased concentration of NADH inside the cell. For glycolysis to proceed, NADH must be converted back to NAD+ through lactic fermentation. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of the net molecular and energetic results of respiration processes.

*See passage for #39 on AAMC Test 3* Overexpression of which enzyme is likely to result in increased levels of HIF? A) succinyl decarboxylase B) succinyl-CoA synthetase C) succinate dehydrogenase D) succinate carboxylase

B; This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is B because the passage states that succinate modulates the level of HIF by inhibiting HIF hydroxylase, an enzyme that induces HIF degradation. Thus, overexpression of succinyl-CoA synthetase, which results in the increased production of succinate, will enhance HIF levels. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of enzymes and their products in metabolic pathways.

*See passage for #28 on AAMC Test 3* Which statement is best supported by the data using the anti-AcTubK40 antibody? A) HDACs acetylate many types of target protein besides histones. B) βOHB is not a general deacetylase inhibitor. C) βOHB is present in the nucleus but not the cytoplasm. D) βOHB inhibits different deacetylases compared to butyrate.

B; This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is B because according to Figure 1, βOHB has no significant effect upon tubulin deacetylase. A is incorrect because the data does not address this issue, since only one potential non-histone target is shown (tubulin). C is incorrect because βOHB was administered outside the cells, which means it must diffuse through both the plasma membrane and the cytoplasm to get into the nucleus and affect HDACs; therefore it is unlikely that βOHB is only in the nucleus. D is incorrect because there are no data presented indicating which HDACs butyrate acts on. Even though the activity of HDAC6 is not affected by βOHB, there are no data to confirm that butyrate regulates the activity of HDAC6. It is a Data-based and Statistical Reasoning question because it involves using the data presented in the figure to make a scientific conclusion.

Mucous secretions in the respiratory tract inhibit microbial infections. These secretions are produced by which of the following tissue types found in the lungs? A) smooth muscle B) epithelial C) nervous D) connective

B; This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is B because epithelial cells are specialized cells for secretion and are present in the respiratory tract. It is a Knowledge of Scientific Concepts and Principles question because it requires matching tissue structure and function.

*See passage for #49 on AAMC Test 1* Normally, a hypothalamic factor stimulates the release of adrenocorticotropic hormone (ACTH) from the pituitary gland. In a patient with Addison's disease, the secretion of the hypothalamic factor will: A) be lower than normal B) be higher than normla C) be unchanged D) increase before disease onset and decrease thereafter

B; This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is B because ACTH stimulates the adrenal cortex to secrete glucocorticoids. According to the passage, Addison's disease occurs when cells of the adrenal cortex are destroyed, rendering them unable to secrete glucocorticoids. As part of a hormone cascade pathway, ACTH secretion is expected to be higher than normal in a patient with Addison's disease to attempt to stimulate the adrenal cortex. This is a Scientific Reasoning and Problem Solving question because you are asked to determine how a disease affects a hormone cascade pathway.

*See passage for #41 on AAMC Test 1* A student hypothesized that EC growth might be affected by the DNA from circulating erythrocytes. Is this student's hypothesis reasonable? A) no; the DNA in circulating erythrocytes in needed to help transport O2 through the capillaries B) no; circulating erythrocytes do not contain DNA C) yes; DNA is responsible for cell division in most cells D) yes; circulating erythrocytes carry DNA nutrients through the capillaries

B; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer to this question is B because mammalian erythrocytes (red blood cells) lose their nuclei during maturation. Therefore, circulating erythrocytes do not contain DNA that could affect endothelial cell growth. This is a Scientific Reasoning and Problem Solving question because you are asked to evaluate a scientific hypothesis.

A certain bacterium was cultured for several generations in medium containing 15N, transferred to medium containing 14N, and allowed to complete two rounds of cell division. Given that the bacterium's genome mass is 5.4 fg when grown in 14N media and 5.5 fg when grown in 15N medium, individual bacteria with which of the following genome masses would most likely be isolated from this culture? A) 5.4 fg only B) 5.4 fg and 5.45 fg C) 5.4 fg and 5.5 fg D) 5.45 fg only

B; This is a Biology question that falls under the content category "The structure, growth, physiology, and genetics of prokaryotes and viruses." The answer to this question is B because DNA replication is semi-conservative. Therefore, after the first round of cell division the genome mass in each bacterium will be 5.45 fg (one DNA strand will contain 15N and the other strand 14N). Following the second round of cell division, half of the bacteria will have a genome mass of 5.4 fg (14N exclusively) and the other half a mass genome of 5.45 fg (14N in one DNA strand and 15N in the other). This is a Scientific Reasoning and Problem Solving question because you need to work with the scientific model of DNA replication.

*See passage for #20 on AAMC Test 1* Upon activation, p65 and cRel control the level of IL-6 mRNA by: A) binding RNA B) binding DNA C) replicating RNA D) replicating DNA

B; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer is B because both p65 and cRel function as transcription factors, which bind DNA and subsequently recruit RNA polymerase. This is a Knowledge of Scientific Concepts and Principles question because you must understand the role of transcription factors in genetics.

*See passage for #23 on AAMC Test 1* According to the passage, what is most likely correct about the genes that are expressed as a result of the constant activation of the NF-kB pathway? A) they cause disruption of the mitochondria B) they contain a p65/cRel binding site in their promoter region C) they have accumulated mutations that alter function D) they bind to STN in the cytoplasm

B; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer is B because the constantly active signaling pathway will result in the overexpression of genes that are under control of the p65/cRel transcription factors. Mitochondrial disruption would occur if the apoptotic machinery was stimulated, but these genes are antiapoptotic. Mutations occur to the NF-kB pathway, which cause the constant activation. The gene products of this signaling pathway do not necessarily have mutations, as their overexpression is sufficient for pathogenesis. Binding to STN is not discussed. This is a Scientific Reasoning and Problem Solving question because you must understand the causes of a biological event and reason about the other factors involved.

*See passage for #18 on AAMC Test 3* The Cgr1 and Cgr2 proteins are produced from: A) separate mRNAs transcribed from different promoter sequences upstream of each gene. B) a single mRNA transcribed from a single promoter sequence upstream of the operon. C) a single RNA transcribed from a single promoter sequence and alternately spliced to produce separate mRNAs. D) separate mRNAs translated from a single promoter upstream of the operon.

B; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is B because an operon containing two genes in prokaryotic cells is transcribed from a single promoter upstream of the first gene in the operon. It is a Scientific Reasoning and Problem Solving question because it requires scientific evaluation of the process that produces the Cgr1 and Cgr2 proteins.

*See passage for #25 on AAMC Test 3* HDACs change chromatin by: A) decreasing its coiling and promoting DNA replication. B) increasing its condensation and inhibiting transcription. C) decreasing charge repulsion between acetyl groups, which increases transcription. D) loosening the attachment of DNA to nucleosome core particles.

B; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is B because the passage states that HDACs counter the effects of histone acetyltransferases (HATs). As histone acetylation typically promotes transcription by modifying chromatin structure, HDACs would inhibit transcription by condensing chromatin structure. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of how transcription can be altered by histone and chromatin modification.

*See passage for #38 on AAMC Test 1* The lung cells of heavy smokers would be expected to have greatly increased concentrations of cP-450 and: A) DNA sequences that code for cP-450 B) mRNA sequences that code for cP-450 C) rRNA that process cP-450 D) tRNA that are specific for cysteine

B; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is B, because protein levels relate most directly to mRNA levels. This is Scientific Reasoning and Problem Solving question because it requires you to work with the scientific models of gene expression.

*See passage for #5 on AAMC Test 1* The information in the passage best supports the conclusion that Intron 8 of HSP110 most likely contains which of the following? A) stop codon B) splice acceptor site C) HSP110 gene promoter D) partial coding sequence of HSP110

B; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is B. Deletions in the T17 microsatellite located in Intron 8 of HSP110 cause Exon 9 to be omitted from the final mRNA sequence during pre-mRNA processing. This would occur during alternative splicing, which strongly suggests that the microsatellite contains sequences that influence splicing. This is a Scientific Reasoning and Problem Solving question because you need to evaluate which genetic feature correctly relates to the information presented on Intron 8 in the passage.

An 18-carbon fatty acid undergoes β-oxidation as shown below: [C18 acid]-S-CoA + FAD + NAD+ + H2O + CoA-SH → [C16 acid]-S-CoA + CH3CO-S-CoA An experimenter adds a radioactive label to C-1 of the 18-carbon acid and to the sulfur atom in the coenzyme A used as a reactant. Where will those labels likely appear in the products? A) the labeled carbon will appear in the C16 acid while the labeled sulfur will appear in acetyl coenzyme A B) the labeled carbon will appear in acetyl coenzyme A while the labeled sulfur will appear in the C16 acid C) both labels will appear in the C16 acid D) both labels will appear in the acetyl coenzyme A

B; This question is testing your knowledge of fatty acid oxidation, which is known as β-oxidation because it cleaves the carbon-carbon bond before the β carbon of the acid. (Remember that because of the peculiarities of α,β notation, the β carbon is C-3, not C-2!) The question wants us to trace the fate of two labels, a carbon atom and a sulfur atom. The carbon atom is C-1 of the acid; as noted above, this will end up in the acetyl coenzyme A. The sulfur in the labeled coenzyme A will end up attached to the C16 product. Choice (B) matches this. Choice (A) This reverses the fate of both labels. Choice (C) C-1 is the carboxylic acid carbon, not the carbon at the end of the hydrocarbon tail. Choice (D) The acetyl coenzyme A formed in β-oxidation is formed from C-1 and C-2 of a fatty acid; the sulfur label should thus end up attached to the remainder of the carbon chain.

Nitric oxide can enter cells by diffusing across the cell membrane. This is most likely because nitric oxide is: A) relatively polar B) relatively small C) a free radical D) concentrated inside the cell

B; This question wants to know why NO can diffuse across cell membranes. Diffusion generally requires that a molecule be either small or nonpolar, or preferably both. Nitric oxide, as a diatomic substance, is relatively small, and that's why it can diffuse across the cell membrane. Thus, (B) is the correct answer. Choice (A) . Polar molecules generally have difficulty diffusing across cell membranes. Choice (C) . Free radicals are not inherently more capable of diffusing across cell membranes than molecules with no unpaired electrons. Choice (D) . Diffusion moves a molecule down its concentration gradient; if this were true, NO would be moving against its gradient.

*See passage for #6 on Kaplan Test 2 (short)* Viagra™ will not produce an erection in the absence of sexual stimulation. Which of the following best explains this fact? A) the PDE-5 enzyme is not expressed until the cavernous nerves are stimulated B) Viagra does not produce cGMP C) PDE-5 cannot be competitively inhibited under low levels of NO D) Viagra increases production of NO

B; Viagra™ does not signal any portion of the mechanism from neural impulse to erection. Rather, it boosts cGMP concentration by inhibiting the enzyme that ordinarily breaks it down. In this way it does not create the signal to create an erection, but it boosts the second messenger to its maximum effectiveness. Hence (B) is correct. Choice (A) . Nothing in the passage indicates that the PDE-5 enzyme is synthesized after cavernous nerve stimulation. In any case, this would not explain sildenafil's effect. Choice (C) . This answer does not make logical sense. No nitric oxide is produced in the absence of the neural signal. Choice (D) . This answer is wrong because the passage explains this is not the mechanism of the drug and further if it were, the information in the prompt could not be correct.

When an action potential is generated, which of the following events occurs first? A) increased permeability to K+ B) increased permeability to Na+ C) increased permeability to Cl- D) decreased permeability to Ca2+

B; When an action potential is generated, the neuronal membrane is depolarized caused by the influx of Na + into the neuron, (B) is the correct answer. Choice (A) . The increased permeability to K + occurs when the neuronal membrane is repolarized. Choice (C) . Negatively charged chloride ions are used to stabilize the membrane potential because the negatively charged chloride ions will cancel out the positively charged sodium ions. Increased permeability to Cl - will cancel the depolarization and no action potential will be propagated. Choice (D) . The calcium ion permeability has no effect on the action potential. The influx of calcium causes the release of neurotransmitter vesicles into the synaptic cleft

*See passage for #5 on Kaplan Test 1* One conclusion of the study in the passage was that lower leptin levels are part of the body's response to food deprivation. Based on this information, administering leptin to mice in the Ob1 group would most likely cause which of the following physiological changes to occur? A) Increased resistance to insulin B) Decreased plasma cortisol levels C) Increased appetite D) Decreased locomotor activity

B; Without leptin, mouse behavior resembles that of extreme starvation. Physiologically, this means elevated plasma cortisol, inability to effectively undergo thermogenesis, unrestrained appetite, and resistance to insulin. Therefore, administering leptin should reverse any of the above symptoms.

The term holoenzymes refers to: A) the inactive precursor form of enzymes B) the different structural forms of enzymes that have the same function C) the complexes formed when enzymes bind their cofactors D) enzymes that have lost their tertiary and quaternary structures

C; A holoenzyme is a complex of an enzyme with its associated cofactor(s) (the enzyme that results when the cofactors are absent is an apoenzyme ). This matches choice (C) . Choice (A) . This is the definition of a proenzyme or zymogen . Choice (B) . This is the definition of an isozyme . Choice (D) . This is the definition of a denatured enzyme.

Many people experience a "food coma" - a feeling of tiredness and lethargy - after eating a large meal. This is best explained as a result of: A) increased blood flow to the digestive tract as a result of increased sympathetic nervous activity B) decreased blood flow to the digestive tract as a result of increased sympathetic nervous activity C) increased blood flow to the digestive tract as a result of increased parasympathetic nervous activity D) decreased blood flow to the digestive tract as a result of increased parasympathetic nervous activity

C; After eating a large meal, blood flow needs to increase to the digestive tract to aid in digestion. This eliminates (B) and (D) . Since the sympathetic nervous system is fight or flight , while the parasympathetic nervous system is rest and digest , it would be para sympathetic activity that increases blood flow to the digestive tract. Choice (C) is correct.

*See #23 on Kaplan Test 2 (short)* Which of the following molecules represents arginine? A) *see image B) *see image C) *see image D) *see image

C; Arginine (C) is an amino acid, not to be confused with the nucleotide adenine (D) . Adenine and Guanine are purines - and consist of a double ring structure. Cytosine and Thymine ((A) and (B) respectively) are pyrimidines. All four nucleotides are key structural components of DNA and RNA, while amino acids are the basic building blocks of proteins. (C) is correct.

Cells capable of becoming any cell type in the human body or even the placenta, are described as: A) pluripotent B) determined C) totipotent D) differentiated

C; Cells capable of becoming any cell type are totipotent stem cells, which is choice (C) . Choice (A) . Pluripotent cells are capable of becoming many cell types, but not all cell types. Choice (B) . Determined cells (also called committed cells) are tied to one lineage. Choice (D) . Differentiated cells have undergone differentiation, which means that they are no longer capable of becoming all cell types.

Fructose, although it is a ketose, can act as a reducing sugar in water. This is possible because fructose can undergo: A) mutarotation B) esterification C) tautomerization D) oxidation

C; For a sugar to be reducing, it must have a free aldehyde group. As a ketose, fructose does not normally have a free aldehyde. However, it can, in its open chain form, undergo tautomerization , which is an isomerization reaction that causes it to form an aldose. Choice (C) is correct. Choice (A) . Mutarotation is the conversion between α and β anomers. Choice (B) . Esterification is replacement of an OH side group in a carboxylic acid with an OR group. Choice (D) . Although fructose must undergo tautomerization to become a reducing sugar, this occurs through the relocation of a proton. So while the reducing sugar is oxidized in the process, the ring structure is still maintained. The fact that fructose can be oxidized through mechanisms other than tautomerization does not directly make it a reducing sugar though; this is because oxidation of fructose results in the ring structure breaking and decomposition of the sugar.

Genetic testing for MCAD deficiency can be challenging because multiple, distant sites within the gene can be mutated to produce disease, while mutations at other sites within the gene are benign. What does this suggest about the necessary enzyme structure? A) primary structure is most important for enzyme function B) secondary structure is most important for enzyme function C) tertiary structure is most important for enzyme function D) quaternary structure is most important for enzyme function

C; From outside knowledge, most inactive enzymes have modifications of the active site, which is part of the tertiary structure. This information is consistent with choice (C), the correct answer. Choices (A) and (B) are incorrect because there are mutations that are benign. These mutations would change primary structure (and possibly secondary structure), however, since the enzyme is still functional, the tertiary structure must be intact. In the second paragraph the passage states that MCAD is monomeric, thus quaternary structure is not a concern, ruling out (D).

*See passage for #4 on Kaplan Test 1* If leptin serves the same function in humans as it does in mice, which of the following groups of people is LEAST likely to lose weight due to daily administration of leptin? A) Individuals with normal body weight B) Overweight individuals with mutations in the leptin gene C) Overweight individuals with mutations in the leptin receptor gene C) Overweight individuals with no mutations in the leptin pathway

C; From the figures in the passage one can see that leptin promoted weight loss in mice without functional leptin in the serum and in wild type mice. This suggests that individuals with normal body weight, with mutations in the leptin gene and without defects in the leptin pathway, will respond to leptin administration. The leptin receptor, on the other hand, represents a component of the leptin pathway downstream of leptin. Thus if leptin is administered to individuals without functional leptin receptors, it would not be able to exert its effects and those individuals would be least likely to lose weight. This means that choice (C) is correct. There is no reason to believe that the individuals described in choices (A), (B), or (D) would fail to respond to leptin administration. Even those with mutated leptin genes, choice (B), would respond, because the administered leptin means that those individuals would not need to depend on their own bodies producing leptin.

*See passage for #14 on Kaplan Test 2 (short)* Which of the following results from the experiment mentioned in the fourth paragraph of the passage would WEAKEN the argument that umami is a unique receptor which constitutes a fifth taste? A) the receptor responds to glutamate at the same concentrations at which glutamate can be tasted B) chemicals which mimic the taste of glutamate activate the receptor C) the binding of glutamate hyperpolarizes all taste receptors D) the amount of glutamate in foods is much greater than the concentrations of glutamate found in the brain

C; If the binding of glutamate hyperpolarized taste receptors, they would be less likely to reach threshold and produce an action potential. Without an action potential, no taste would register within the brain. Hence (C) is correct. Choice (A). If the receptor responds at the same concentrations of glutamate which produce a distinct taste, this provides evidence that the glutamate is activating a unique receptor. Choice (B). If chemicals which mimic glutamate can bind the receptor and produce the taste of umami, this supports the idea that umami is a unique taste. Choice (D). This would explain why a truncated receptor was necessary. If the receptor were not truncated it would produce an overwhelming taste of umami but this is prevented by the truncation of the receptor. This supports the idea of a truncated glutamate taste receptor.

During which phase of the menstrual cycle is progesterone at the highest concentration? A) follicular phase B) ovulation C) luteal phase D) menstruation

C; In a woman who hasn't become pregnant, the level of progesterone peaks about a week after ovulation and then begins to drop along with the estrogen level, so (C) is the best answer.

Which of the following processes would NOT occur in response to reduced blood insulin levels? A) Decreased uptake of blood glucose by muscle cells B) Increased conversion of glycogen into muscle glucose C) Increased utilization of glucose as fuel D) Increased utilization of fatty acid as fuel

C; The answer to this question must be a process that does not occur when insulin levels are low. Insulin levels are reduced at times of low blood glucose concentration in order to conserve glucose for the use by the brain. Thus when insulin levels fall, tissues such as muscle and liver decrease glucose uptake and utilization, choice (A), and start using fatty acids instead, choice (D). When insulin levels are low, glucagon levels rise by default and promote conversion of glycogen into glucose, choice (B), to maintain blood glucose levels. The only process that doesn't occur in response low insulin levels, is choice (C), increase utilization of glucose as fuel.

The primary purpose of the ductus arteriosus in the fetus is to: A) reduce blood flow to the liver B) move blood from the right atrium to the left atrium C) reduce blood flow through the lungs D) connect the maternal and fetal circulations

C; The ductus arteriosus, one of the adaptations in the fetal circulation, connects the pulmonary artery to the aorta. As a result, each time the ventricles contract, blood is shunted from the pulmonary artery to the aorta, bypassing the lungs. This is important because the lungs are not fully developed early in gestation, and therefore are not capable of supporting high blood pressures. Choice (C) is correct. Choice (A) . This is the purpose of the ductus venosus , not the ductus arteriosus. Choice (B) . This is the purpose of the foramen ovale, not the ductus arteriosus. Choice (D) . This is also the purpose of the ductus venosus.

Which of the following is not a tenet of cell theory? A) all living things are composed of cells B) the cell is the basic functional unit of life C) cells arise from both preexisting cells and nonliving sources D) cells use nucleic acids to store information

C; The four tenets of cell theory are: (1) all living things are composed of cells (2) cells are the basic functional units of life (3) cells arise only from other cells (4) cells store information using nucleic acids, in particular DNA. Choice (C) contradicts statement (3), and is therefore correct.

The binding of oxygen to hemoglobin does not follow the rules of Michaelis-Menten kinetics, primarily because: A) hemoglobin does not modify oxygen B) hemoglobin can bind more tightly to carbon monoxide than to oxygen C) hemoglobin exhibits allosteric effects D) oxygen is a gas at standard temperature and pressure

C; The key assumption on which Michaelis-Menten kinetics are based is that there is one active site capable of binding substrate. Hemoglobin, however, is an allosteric protein with four subunits, and exhibits cooperativity: when one oxygen binds, the other sites bind oxygen more easily. While all of the answer choices are true statement, choice (C) is the only statement that explains why Michaelis-Menten kinetics doesn't apply to the binding of oxygen to Hb. Choice (A). Michaelis-Menten kinetics does not require that the product of the enzyme-substrate complex differ from the original substrate. Choice (B). Michaelis-Menten kinetics do not require that an enzyme be able to bind only one substrate. Choice (D). Michaelis-Menten kinetics applies to dissolved gases as well as to any other aqueous solute.

*See passage for #53 on Kaplan Test 1* Multicopy inhibition is a regulatory mechanism that limits the expression of the transposase gene by blocking: A) replication B) transcription C) translation D) post-transcriptional modification

C; The passage says that the net result of multicopy inhibition is that IN RNA is not allowed to attach 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 allowed to attach to a ribosome, translation cannot occur. Hence, multicopy inhibition must repress translation, and (C) is the correct answer. Choice (A). Replication is the process by which new DNA is synthesized using the cell's original DNA as a template. Since the question tells you that multicopy inhibition acts on RNA, not DNA, it must NOT block replication. Choice (B). Transcription is the production of RNA from a DNA template. Since IN RNA is produced, this implies that transcription must have occurred. Choice (D). Post-transcriptional modification refers to modifications made to an mRNA transcript AFTER transcription has occurred, as its name implies. Post-transcriptional modifications include removing introns and adding a poly(A) tail of a 5' cap. Since these occur PRIOR to the binding of the mRNA to a ribosomal subunit, multicopy inhibition does not block it.

*See passage for #44 on Kaplan Test 1* Paclitaxel is most likely to hamper which phase of mitosis? A) Prophase B) Metaphase C) Anaphase D) Telophase

C; The passage tells us that paclitaxel stabilizes tubulin polymers once they form. Thus, it becomes harder for them to shorten. And when in mitosis do tubulin polymers shorten? Anaphase -- when the chromosomes are pulled towards the two poles of the cell! Thus, the correct answer is choice (C). Choice (A). Prophase doesn't involve shortening of tubulin. Choice (B). Metaphase doesn't involve shortening of tubulin. Choice (D). Telophase doesn't involve shortening of tubulin.

Which of these proteins, found in microfilaments, is the most common protein in most living human cells? A) collagen B) keratin C) actin D) elastin

C; The protein that matches the description in the question stem is actin, choice (C) . Choice (A) . Collagen is a structural protein; it is not part of microfilaments, and is the primary component of the extracellular matrix. Choice (B) . Keratin is found primarily in skin and hair, and generally in dead tissue. Choice (D) . Elastin is a component of the extracellular matrix, and is not part of microfilaments.

*See passage for #27 on Kaplan Test 1* Which of the following experiments would most support the hypothesis that acetylcholine (ACh) receptors on muscle cells are necessary for susceptibility to rabies infection? A) determining the average number of ACh receptors in muscle cells in foxes, opossums, and dogs B) demonstrating that a viral G protein with proline at amino acid 333 cannot bind to ACh receptors C) showing that a muscle cell line derived from mice that lack ACh receptors cannot be infected by rabies virus D) showing that cats, which have as many muscle ACh receptors as foxes, are equally susceptible to infection as foxes

C; The question requires one to show that ACh receptors are necessary for rabies infection. The easiest way to show this is to demonstrate that a lack of ACh receptors make infection impossible, and that's exactly what choice (C) says. Choice (A). This might help with the hypothesis described in the passage, but not with the one in the question stem. Choice (B). This would not demonstrate that ACh receptors are necessary fro infection, merely that they are capable of binding rabies particles. Choice (D). This might help with the hypothesis described in the passage, but not with the one in the question stem.

Four different human cell cultures -- erythrocytes, epidermal cells, skeletal muscle cells ,and intestinal cells -- were grown in a medium containing radioactive adenine. After 10 days, the mitochondria were isolated via centrifugation, and their level of radioactivity was measured using a liquid scintillation counter. Which of the following cells would be expected to have the greatest number of counts per minute of radioactive decay? A) erythrocytes B) epidermal cells C) skeletal muscle cells D) intestinal cells

C; The question stem states that radioactive adenine is placed in a culture with these cells. During this time, the DNA in the cells will replicate incorporating the radioactive adenine. This includes chromosomal DNA, as well as mitochondrial DNA. The third paragraph states that mitochondria replicate independently of their cells. Since all autosomal human cells have the same amount of DNA in their nuclei, the only difference in radioactivity will be the amount that was incorporated into the mitochondrial DNA. This is why the cells' mitochondria were isolated via centrifugation, and the radiation from each sample was measured using a scintillation counter. The cells with the greatest number of mitochondria will have the highest radioactive count when their mitochondria are separated. The number of mitochondria in a cell is dependent on the energy needs of the tissue. Muscle cells, choice (C), need a lot of energy in order to contract. In general, muscle cells have a higher content of mitochondria than do any other type of autosomal cell, so choice (C) is correct.

*See passage for #21 on AAMC Test 1* Considering the structure of STN, what is the most likely mechanism for its entry into the cell? A) active transport B) receptor mediated endocytosis C) diffusion directly through the membrane D) passage through an ion channel

C; This is a Biochemistry question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer is C because the structure of STN shows sufficient planarity and hydrophobicity to pass through the membrane by simple diffusion. It is a Scientific Reasoning and Problem Solving question because you must predict an outcome based on scientific knowledge and observational reasoning.

*See passage for #51 on AAMC Test 1* The most rapid rate of gluconeogenesis will most likely occur in the body when: A) blood glucose levels are high B) cortisol release is inhibited C) the body's stores of carbohydrates are low D) the body's stores of proteins are low

C; This is a Biochemistry question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is C, because gluconeogenesis is the pathway for the synthesis of glucose from other metabolic compounds and thus it is activated when the body's stores of carbohydrates are low. This is a Knowledge of Scientific Concepts and Principles question because it relates to assumed knowledge regarding gluconeogenesis.

*See passage for # on AAMC Test 1* What aspect of Experiment 1 does NOT address whether membrane composition has an effect on Na+K+ATPase activity? The activity of the Na+K+ATPase: A) showed less temperature dependence in the 14:1 liposome than the 14:0 liposome B) was highest in the 14:0 liposomes at all temperatures C) increased with temperature in both the 14:1 liposome and the 14:0 liposome D) was greater at all temperatures when cholesterol was present

C; This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer is C because the researchers were not trying to show that protein activity increases with temperature, which is an expected result of all kinetic studies assuming the protein does not thermally denature. They raised the temperature to see how fluidity plays a role in activity. This is a Reasoning about the Design and Execution of Research question because you are asked to understand the purpose of an experiment based on the experimental parameters and execution.

During the production of insulin, the translated polypeptide is cleaved into the mature form and secreted from the cell. The cleavage most likely takes place in which of the following locations? A) nucleus B) ribosomes C) endomembrane system D) cytoplasm

C; This is a Biology question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer to this question is C because secreted proteins such as insulin are cleaved into mature form within endomembrane system. This is a Knowledge of Scientific Concepts and Principles question because it relates to assumed knowledge regarding the role of endomembrane system in processing of secreted proteins.

Which of the following best describes the chemical energy that is derived from the Krebs cycle? Energy is produced in the forms of: A) ATP, which directly supplies energy for many cellular processes, and NAD+, which supplies energy for the electron transport chain. B) NAD+, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain. C) ATP, which directly supplies energy for many cellular processes, and NADH, which supplies energy for the electron transport chain. D) NADH, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain.

C; This is a Biology question that falls under the content category "Principles of bioenergetics and fuel molecule metabolism." The answer to this question is C because the Krebs cycle produces both ATP and NADH but not NAD+. ATP directly supplies energy for many cellular processes, such as muscle contraction, and NADH, which is used in the electron transport chain. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of the net molecular and energetic results of respiration processes.

A certain drug blocks the activity of enzyme A by reversibly binding the enzyme's active site. Given this, the drug most likely inhibits enzyme A's activity by: A) lowering the enzyme's activation energy. B) feedback inhibition. C) competitive inhibition. D) noncompetitive inhibition.

C; This is a Biology question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is C because in competitive inhibition, the inhibitor reversibly binds the enzyme's active site. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of enzyme regulation.

*See passage for #49 on AAMC Test 3* Based on the passage, in which of the following ways is the action of serotonin on postsynaptic receptors most likely terminated? A) alteration of the receptor's affinity for serotonin B) production of mutant enzymes involved in serotonin synthesis C) transportation of released serotonin back into the presynaptic terminal D) absorption of the serotonin-receptor complex into the postsynaptic compartment

C; This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer to this question is C because the passage states that antidepressant drugs act to increase the amount of serotonin in the synaptic space. Therefore, the action of serotonin on postsynaptic receptors would be terminated by the removal of serotonin from the synaptic space. This would occur by transport of released serotonin back into the presynaptic terminal, an action that SSRIs block. It is a Knowledge of Scientific Concepts and Principles question because it requires recognizing the correct scientific principle based on the information presented in the passage.

Large amounts of protein are found in the urine of a patient. Based on this information, which portion of the nephron is most likely malfunctioning? A) collecting duct B) distal tubule C) glomerulus D) Loop of Henle

C; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ systems." The answer to this question is C because in healthy individuals, the structure of the glomerular capillaries prevents the entry of large molecules, such as proteins, into the filtrate. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of the structure and function of the different parts of the mammalian excretory system.

*See passage for #1 on AAMC Test 3* Individual M has longer limbs and muscles than Individual N. Both individuals lift the same amount of weight from the ground to their shoulders an equal number of times. If all repetitions were completed in synchrony, which statement about the amount of work completed by the individuals is true? At the end of the workout: A) both individuals will have done the same amount of work because the force was applied for the same duration of time. B) Individual M will have done less work because the object was moved a greater distance. C) Individual N will have done less work because the object was moved a shorter distance. D) Individual M will have done more work because the muscle contracted faster to achieve synchrony with Individual N.

C; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ systems." The answer to this question is C because work is equal to the product of the force applied to an object and the distance over which the distance is moved. Individual N has shorter limbs (and likely a shorter ground to shoulder distance), thus this individual will perform less work because the muscles contracted a shorter distance, and the object was moved a shorter distance. A is incorrect because the stem indicates that Individual M has longer limbs than does Individual N. Given this, Individual M will have done more work than Individual N because the distance the muscles contracted when moving the load from ground to shoulder (also a greater distance for Individual M) is greater. B is incorrect because work is equal to the product of the force applied to an object and the distance over which the distance is moved. Thus, moving an object a longer distance would result in more, not less, work being done. D is incorrect because while it is true that Individual M did more work, work is not dependent on speed, and thus, the rate of muscle contraction is not relevant to the amount of work done by the individuals. It is a Scientific Reasoning and Problem Solving question because it requires comparing the physiological form and function of two individuals.

*See passage for #1 on AAMC Test 1* Based on the data presented in the passage, which statement best describes the HSP110ΔE9 allele? A) cancer-promoting and dominant to HSP110WT B) cancer-promoting and recessive to HSP110WT C) cancer-suppressing and dominant to HSP110WT D) cancer-suppressing and recessive to HSP110WT

C; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C. According to Figure 1, as compared to HSP110WT, HSP110ΔE9 slows tumor growth, and thus, it is cancer-suppressing. According to the data in Table 2, HSP110ΔE9 counteracts the ability of HSP110WT to prevent protein aggregation and prevent apoptosis; thus, the HSP110ΔE9 allele is dominant to the HSP110WT allele. This is a Data-based and Statistical Reasoning question as it required interpretation of data in Figure 1 and Table 2.

*See passage for #4 on AAMC Test 1* Which nucleotide pairing(s) would be recognized by the MMR system during DNA replication? I. dTMP and dCMP II. dGMP and dAMP III. dAMP and dTMP A) I only B) III only C) I and II only D) I, II, and III

C; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is C. The MMR system recognizes and repairs nucleotide mismatches during DNA replication. The nucleotide pairs dTMP and dCMP (I) and dGMP and dAMP (II) are mismatches. dAMP and dTMP (III) is a correct pairing. This is a Knowledge of Scientific Concepts and Principles because it requires assumed knowledge of nucleotide pairing during DNA replication.

The GTP used by tubulin could come directly from the products of: A) glycolysis B) the pentose phosphate pathway C) the citric acid cycle D) oxidative phosphorylation

C; This is a fairly straightforward question: which of the four processes listed directly produces GTP? That's the citric acid cycle, choice (C). Choice (A). Glycolysis does not directly produce GTP. Choice (B). The pentose phosphate pathway does not directly produce GTP. Choice (D). Oxidative phosphorylation does not directly produce GTP.

DNA with which of the following characteristics would be most easily denatured? A) complementary strands with an equal composition of A, C, T, and G base pairs B) complementary strands which are identical to one another C) complementary strands which have a higher A-T content than G-C content D) any DNA found in a sodium-rich solution

C; This question calls upon the knowledge that adenines (A) and thymines (T) only form two hydrogen bonds with each other, while guanines (G) and cytosines (C) form three. This means that A-T rich areas of DNA are held together by fewer bonds and will separate more easily than G-C rich areas, so choice (C) is correct.

*See passage for #43 on Kaplan Test 1* Based on the data in Table 1, what is the net charge on the R-groups of the five amino acids listed for S. cerevisiae tubulin at pH 7.4? A) -2 B) -1 C) 0 D) 1

C; To answer this question, it's necessary to know the charges on the acids listed under S. cerevisiae in Table 1. The basic amino acids -- which are positively charged at pH 7.4 -- are arginine and lysine. The acidic amino acids are aspartic acid and glutamic acid. Since none of these amino acids appear, the net charge on those five amino acids should be zero, which is choice (C). Choice (A). -OH is not negatively charged. Choice (B). This answer may be chosen if an error is made concerning the structures. Choice (D). None of the amino acids is positively charged.

Which of these forms of transferring information between bacterial species requires a virus as a vector? A) conjugation B) transformation C) transduction D) transposon

C; Transduction is the transfer of genes from one bacterium to another using a viral vector, so choice (C) is correct. Choice (A) . Conjugation is analogous to sexual reproduction in eukaryotes; it requires a sex pilus , physical bridge between cells, rather than a viral vector. Choice (B) . Transformation involves the uptake of "naked" DNA through the cell membrane. It does not require a virus. Choice (D) . Transposons are "jumping genes" that can insert and remove themselves from the genome; they do not require a vector.

Maternal lineage can be traced by sequencing the mitochondrial DNA because the mitochondrial genome is derived primarily from the mother. The best explanation for this phenomenon is that: A) sperm have no mitochondria and thus cannot contribute to the mitochondrial genome of the offspring B) the zygote divides rapidly, diffusing the paternal mitochondria amongst many cells C) following penetration, the sperm derived mitochondria disintegrate within the egg D) all the genes coding for mitochondria are located on the X chromosome

C; Upon penetration of the egg, the sperm mitochondria fall apart and are not replicated when divisions begin. Choice (C) is correct. Choice (A) . Sperm have large numbers of mitochondria that are used to drive locomotion. Choice (B) . Mitochondria are replicated prior to cell division. Also, this choice would not explain the presence of the maternal genome which should also be diffused by division. Choice (D) . Many of the genes coding for mitochondrial proteins are located in the mitochondrial DNA.

*See passage for #28 on Kaplan Test 1* Which of the following is most likely to be true regarding cells in the muscles that spasm in a patient suffering from hydrophobia? A) they need an efflux of Na+ ions from the sarcoplasmic reticulum to contract B) they are involved in the absorption of nutrients from the bolus C) they receive innervation via the autonomic division of the nervous system D) they contain an inner circular layer which contracts and causes the esophagus to shorten

C; While swallowing may initially be under voluntary control, the autonomic nervous system plays a key role in the coordinated movement through the esophagus. Thus, Choice (C) is correct. Choice (A). Muscle contraction is dependent on release of calcium, not sodium. Choice (B). The esophagus is responsible for transport and does not play a significant role in absorption. Choice (D). The esophagus does contain an inner circular layer of muscle, but upon contraction it constricts the esophagus. The longitudinal layer is responsible for shortening.

Cerebrosides and gangliosides are best characterized as: A) triacylglycerides B) steroids C) waxes D) sphingolipids

D; Cerebrosides and gangliosides are subclasses of sphingolipids, which are derivatives of the fatty acid derivative sphingosine , which contains two fatty acid groups, one of which is bound to the amino group in sphingosine. Choice (D) is correct. Choice (A) . Sphingolipids are not triacylglycerides, since, by definition, the latter have three fatty acid groups. Sphingolipids also do not contain glycerol. Choice (B) . Sphingolipids are not steroids, as they do not have the ring structure normally observed in steroids. Choice (C) . Sphingolipids are not waxes, as these are highly-insoluble esters of fatty acids and alcohols.

Why does the human body store energy for long-term consumption as lipids rather than as carbohydrates? A) carbohydrates are more dense than lipids B) lipids are more hydrophilic than carbohydrates C) the carbon in carbohydrates has a lower average oxidation state than lipids D) the carbon in lipids has a lower average oxidation state than carbohydrates

D; Fats are a better source of energy (~9 kilocalories per gram) than carbohydrates (~4 kilocalories per gram). This is not because fats are less dense per se , but rather because the carbons in lipids are in a lower oxidation state (almost all carbons are bonded to hydrogen only in lipids, while most carbons are bonded to at least one oxygen in carbohydrates). Since aerobic respiration is just a highly regulated and drawn out form of combustion, the larger the change in oxidation state, the more high-energy electrons that will be available for generating energy. This matches choice (D) . Choice (A) . This is true, but irrelevant. It's the energy density, not the mass density, that matters. Choice (B) . This is untrue. Choice (C) . This is incorrect; the oxidation state of carbon in carbohydrates (typically 0) is higher than in lipids (typically -2).

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) absorbance of DNA at a different wavelength B) the total amino acid content over time C) the molecular weight of the protein before and after denaturation D) the change in the mass of DNA in the sample over time

D; If we're going to measure denaturation of an enzyme, what change would we want to make? The question wants to determine the temperature at which the denaturation is complete. When that happens, enzyme activity should be essentially zero. Therefore, because DNA polymerase assembles DNA, the easiest way to figure this out is to measure when the amount of DNA in the sample becomes constant, which would be when no additional DNA is being assembled by DNA polymerase. This matches choice (D). Choice (A). The absorbance of DNA would not be useful, since it would not be clear when the enzyme has stopped, and one would expect the wavelength to change as the composition of nucleotides in the system changes. Choice (B). The total amino acid content should not change, despite denaturation. Choice (C). There is no reason to believe that the polymerase will change its molecular mass when it denatures; moreover, it would be difficult to measure this while heating the enzyme.

Which of these classes of enzymes involved in DNA synthesis is NOT involved in RNA synthesis? A) helicases B) topoisomerases C) polymerases D) ligases

D; In DNA synthesis, DNA ligase is required to join the Okazaki fragments on the lagging strand. RNA synthesis does not create a lagging strand, so there is no need for a ligase in RNA synthesis. Choice (D) is correct. Choice (A) . Helicases are required in RNA synthesis to separate the template and non-template strands. Choice (B) . Topoisomerases are required in RNA synthesis to remove the supercoiling from DNA. Choice (C) . RNA polymerase is needed to add the nucleotides to the growing RNA chain.

*See passage for #3 on Kaplan Test 1* Leptin acts to prevent hypothalamic secretions of neuropeptide Y, a potent feeding stimulant. As a result, the levels of neuropeptide Y are most probably: A) reduced in the Ob2 but not in the Ob1 strain of mice B) reduced in both the Ob2 and the Ob1 strain of mice C) elevated in the Ob2 but not in the Ob1 strain of mice D) elevated in both the Ob2 and the Ob1 strains of mice

D; In order for levels of neuropeptide Y to be reduced, mice must produce functional leptin and be capable of responding to it. Since the Ob1 strain is defective in the former and the Ob2 strain is defective in the latter, both strains will have elevated levels of neuropeptide Y. This matches with choice (D).

Which of the following statements regarding beta- oxidation of the 16-carbon fatty acid hexadecanoic acid is correct? A) the oxidation takes place in the cytosol, and produces an 18-carbon product B) the oxidation takes place in the cytosol, and produces an 14-carbon product and acetyl CoA C) the oxidation takes place in the mitochondria, and produces an 18-carbon product D) the oxidation takes place in the mitochondria, and produces an 14-carbon product and acetyl CoA

D; In β- oxidation, a fatty acid loses a two-carbon group. Thus hexadecanoic acid, which has 16 carbons, should end up with 14 carbons. This eliminates (A) and (C) . Since β- oxidation takes place in the mitochondria, not the cytosol, the correct answer is choice (D) . Choice (A) . β- oxidation does not take place in the cytosol, and reduces the number of carbon atoms. Choice (B) . β- oxidation does not take place in the cytosol. Choice (C) . β- oxidation reduces the number of carbon atoms.

Which of the following molecules can be utilized as a source of acetyl CoA in humans? I. Fatty acids II. Amino acids III. Ethanol A) I only B) II only C) I and II only D) I, II, and III

D; Item I, fatty acids, can be used to generate acetyl CoA via β-oxidation; that eliminates (B) . Item II, amino acids, can also be used, if they are ketogenic . That eliminates (A) . Item III, ethanol, can be used, too; if that were not the case, alcoholic beverages could not be used as a substitute for glucose, as some alcoholics do. Choice (D) is correct.

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) the number of aquaporins closing in the collecting duct B) reabsorption of Na+ in the distal convoluted tubule C) movement of water through cells in the ascending limb of the loop of Henle D) movement of water through cells in the descending limb of the loop of Henle

D; One of the classic symptoms of type 2 diabetes is polyuria, excess water excretion. But how does this happen? 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. Choices (C) and (D) are the only statements that are consistent with this reasoning. However, the ascending limb is almost completely impermeable to water, which means choice (D) is the correct answer. Choice (A). An increase in open aquaporins would increase reabsorption of water from urine, leading to a decrease in urine output, which is the opposite of polyuria. Choice (B). Glucose does not directly affect Na+ reabsorption, aldosterone does. Choice (C). The ascending limb of the loop of Henle is impermeable to water.

All of the following are characteristics of the human respiratory system EXCEPT: A) ventilation is controlled by the medulla oblongata B) gas exchange occurs across the alveolar-capillary membrane C) the rate of breathing is increased by sympathetic stimulation D) concentration of the diaphragm contracts the thoracic cavity and permits the lungs to fill

D; The diaphragm is a muscle extending across the bottom of the ribcage. It separates the thoracic cavity (lung and heart) from the abdominal cavity (liver, stomach, intestines). In the relaxed state, the diaphragm is shaped like a dome. In order to draw air into the lungs, the diaphragm contracts and flattens causing the thoracic cavity to enlarge rather than contract. This reduces intra-thoracic pressure causing air to be sucked into the thorax. When the diaphragm relaxes and returns to the dome shape, the shortening of the thoracic cavity exhales air. Choice (A) . Respiratory functions are under the control of the medulla oblongata. Choice (B) . Deoxygneated blood in the pulmonary artery is oxygenated at the alveolar-capillary boundary. Choice (C) . Sympathetic stimulation is the "fight or flight" response, which is accompanied by increased oxygenation of muscle tissue by increased breathing rate.

Genes A and B on the same chromosome recombine 10 percent of the time. Genes B and C recombine 3 percent of the time. Is this information sufficient to determine the order of the three genes on the chromosome? A) yes, the order is A-B-C B) yes, the order is A-C-B C) yes, the order is B-A-C D) no, the information is insufficient

D; The information in the question tells us that A and B are 10 map units apart, while B and C are 3 map units apart. This is consistent with two possible orders: A-B-C (with A and C 13 map units apart), or A-C-B (with A and C 7 map units apart). Choice (D) is correct. Choice (A) . This would be true if we knew that A and C are 13 map units apart. Choice (B) . This would be true if we knew that A and C are 7 map units apart. Choice (C) . This is impossible from the information given.

*See passage for #21 on Kaplan Test 2 (short)* Why is MCAD deficiency more likely to appear earlier (as soon as a few days after birth) in breastfed infants? Initially, breastfed infants receive a: A) larger supply of medium chain fatty acids than bottle fed infants B) smaller supply of medium chain fatty acids than bottle fed infants C) larger number of carbohydrate calories than bottle fed infants D) smaller number of carbohydrate calories than bottle fed infants

D; The initial period of breastfeeding occurs before milk let-down in a woman. Therefore the infant is receiving a limited amount of calories from colostrum, mostly consisting of protein. In bottle-fed infants, there is immediate consumption of a higher calorie, higher carbohydrate diet. This diet is consistent with the MCAD deficiency treatment, therefore preventing symptoms in these infants. Additionally, a switch to fat metabolism would trigger the use of the MCAD enzyme, and a low carbohydrate load would facilitate that switch by activating glucagon and other hormones. Therefore (D) is correct. The amount of medium chain fatty acids should not contribute either way to MCAD deficiency symptoms because they will not be broken down. The only symptom that is expected to increase in the case of medium chain fatty acid excess is hepatomegaly as they accumulate. Therefore (A) and (B) are incorrect.

The osteoclasts in bone are best classified as: A) B cells B) T cells C) symbiotic bacteria D) macrophages

D; The osteoclasts in bone dissolve bone to release calcium; to do this, they must have significant digestive activity, which is most suggestive of macrophages, choice (D) . Choice (A) . Osteoclasts do not respond to infections. Choice (B) . Osteoclasts do not respond to infections. Choice (C) . Osteoclasts are not symbiotic bacteria.

*See passage for #25 on Kaplan Test 1* The L protein codes for a nucleic acid polymerase. Based on information in the passage, it most likely synthesizes: A) DNA from an RNA template B) RNA from a DNA template C) DNA from a DNA template D) RNA from an RNA template

D; The passage states that the virus does not integrate itself into the host genome, and it is not a retrovirus. Therefore, it does not need to synthesize DNA. Furthermore, it's stated that the rabies genome is RNA. Therefore, the only plausible answer is choice (D): it must synthesize RNA from RNA. Choice (A). There is no need for the rabies virus to create DNA. Choice (B). The passage states that the rabies virus genome is RNA. Choice (C). There is no need for the rabies virus to create DNA.

Which of the following mitochondrial genome characteristics differs most from the characteristics of the nuclear genome? A) mitochondrial DNA is a double-helix B) some mitochondrial genes code for tRNA C) specific mutations to mitochondrial DNA can be lethal to the organism D) almost every base in mitochondrial DNA codes for a product

D; The question asks for a major difference between nuclear and mitochondrial DNA. Before moving on to the answer choices, you should rub through major characteristics of nuclear DNA in order to eliminate answer choices. Nuclear DNA contains introns and exons, or non-coding and coding regions. Therefore, choice (D) is not a characteristic of nuclear DNA, and is a good candidate for the correct answer -- it also happens to be correct, as this is a trait of mitochondrial DNA. But even if you weren't sure of this fact, you could eliminate the other answer choices to choose (D) as correct. Choice (A). The nuclear genome is comprised of double-helical DNA that codes for mRNA, tRNA, and rRNA. Even if mitochondrial DNA is a double helix, it is not different from nuclear DNA so it can't be the right answer. Choice (B). Mitochondrial DNA is dependent on nuclear DNA to produce the proteins and nucleic acids it needs for replication. Nuclear DNA codes for tRNA, so therefore cannot be a difference between the two types of DNA. Choice (C). If a mutation occurred in the nuclear genome that rendered an essential gene non-functional such as an enzyme involved in glycolysis, the organism would die. Thus, choice (C) is also a characteristic of the nuclear genome and the mitochondrial DNA.

Which of the following is NOT a stop codon? A) UAA B) UAG C) UGA D) UGG

D; The three stop codons that end protein synthesis are UAA, UAG, and UGA. UGG, choice (D) , is the only codon for tryptophan. (Note that on Test Day, exact knowledge of the correspondence table for the genetic code is not required, other than the start and stop codons. Understanding how the code works, however, is essential.)

A scientist experimentally determines that an unknown monosaccharide rotates plane-polarized light in a clockwise direction. Based on this observation: A) the monosaccharide is of R configuration B) the monosaccharide is of S configuration C) the monosaccharide is of D configuration D) it cannot be determined if the monosaccharide is D, L, R, or S

D; The three systems described in the question and answers all describe different properties. (R)/(S) is based on absolute structure, D/L on the relationship to glyceraldehyde, and (+)/(-) on rotation of light. Therefore the scientist knows that the monosaccharide is of (+) designation, and that is all.Choice (D) is correct.

Besides ATP, products of reactions in the glycolytic pathway include: I. ADP II. NAD+ III. NADH A) II only B) III only C) I and II only D) I and III only

D; This Roman numeral question is asking about products of individual reactions in the glycolytic pathway. Each item comes up twice, so we can start anywhere. Item I, ADP, is a product of the reactions in the "investment" phase of glycolysis: in the first step of glycolysis, glucose is phosphorylated to trap it in the cell. This eliminates (A) and (B) . In the "payoff" phase of glycolysis, NAD + is reduced to NADH. The reverse reaction does not happen in glycolysis. Therefore, item III, NADH, is true, and the correct answer is choice (D).

*See passage for #26 on AAMC Test 1* What is the most likely location of P-gp within the plasma membrane? A) associated with lipids on the cytoplasmic side only B) associated with lipids on the extracellular side only C) peripheral to the plasma membrane D) within a lipid raft

D; This is a Biochemistry question that falls under the content category "Assemblies of molecules, cells, and groups of cells within single cellular and multicellular organisms." The answer D because the passage states the P-gp is found in cholesterol rich domains, which would indicate lipid rafts. This is a Knowledge of Scientific Concepts and Principles question because you must apply knowledge about membrane structures to solve the problem.

*See passage for #22 on AAMC Test 1* Assuming no mutations to the signaling pathway described in the passage, what event directly activates CARD11? A) proteolytic cleavage B) GTP binding C) membrane association D) phosphorylation

D; This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer is D because the protein that activates CARD11 is PKC, which is a kinase. This implies that phosphorylation is the activating event. This is Knowledge of Scientific Concepts and Principles question because you need to apply knowledge of kinase function to the signaling system in the passage.

Allosteric inhibition of an enzyme involves which of the following events? A) binding of an inhibitor uncompetitively in place of the substrate B) binding of an inhibitor competitively in place of the substrate C) binding of an inhibitor noncompetitively in place of the substrate D) binding of an inhibitor to a site other than the substrate binding site

D; This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is D because allosteric inhibition involves binding of an inhibitor to a site other than the substrate binding site. It is a Knowledge of Scientific Concepts and Principles question because it requires knowledge of the control of enzyme activity.

*See passage for #15 on AAMC Test 3* Which type of catalytic activity is most likely missing from cFLIP? A) oxidoreductase activity B) lyase activity C) isomerase activity D) hydrolase activity

D; This is a Biochemistry question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is D because the passage notes that cFLIP is a homolog of a protease, caspase-8. Therefore, it is most likely that the catalytic activity that is missing from cFLIP is hydrolase activity. It is a Scientific Reasoning and Problem Solving question because it requires determining which enzymatic activity corresponds to an enzyme described in the passage.

*See image for #46 on AAMC Test 1* Which of the following best described the bond that would form between the following two nucleotides if they were located adjacent to each other as shown in a single strand of DNA? A) a bond between the phosphate of the thymine and the phosphate of the adenine B) a bond between an oxygen in the thymine base and a nitrogen in the adenine base C) a bond between the phosphate of the thymine and the sugar of the adenine D) a bond between the phosphate of the adenine and the sugar of the thymine

D; This is a Biochemistry question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is D because nucleotides are linked to one another by phosphodiester bonds between the sugar base of one nucleotide and the phosphate group of the adjacent nucleotide in a way that the 5' end bears a phosphate, and the 3' end a hydroxyl group. This is a Knowledge of Scientific Concepts and Principles because it asks about the structure of nucleic acids.

*See passage for #16 on AAMC Test 3* Cells that contain a large amount of phosphorylated p38 are most likely: A) increasing in size. B) replicating their DNA. C) dividing. D) undergoing growth arrest.

D; This is a Biology question that falls under the content category "Processes of cell division, differentiation, and specialization." The answer to this question is D because the passage indicates that p38 phosphorylation levels correlates with apoptotic signaling, which most likely leads to growth arrest. It is a Scientific Reasoning and Problem Solving question because it requires determining a scientific outcome based on the information presented in the passage.

*See passage for #40 on AAMC Test 1* The pericytes used in these experiments were probably in which phase of the cell cycle? A) telophase B) metaphase C) anaphase D) interphase

D; This is a Biology question that falls under the content category "Processes of cell division, differentiation, and specialization." The question asks the examinee to identify the phase of the cell cycle in which the growth of the pericytes was arrested. According to the passage, the pericytes were treated so that they would not divide but would, from a metabolic perspective, continue to function normally. Three of the listed options (A, B, and C) are phases of mitosis and thus, occur during cell division. Interphase (D) is the phase of the cell cycle between cell divisions and is the phase in which the cell obtains nutrients, grows, reads its DNA, and conducts other "normal" cell functions. Thus, D is the best answer. It is a Scientific Reasoning and Problem Solving question because you are asked to make a prediction about a biological process based on the information provided in the passage.

*See passage for #39 on AAMC Test 1* No drowsiness was initially felt by the alcoholic because the previous abuse of alcohol had: A) denatured the cP-450 B) inhibited the cP-450 C) reduced the cP-450 D) induced the cP-450

D; This is a Biology question that falls under the content category "Structure and function of proteins and their constituent amino acids." The answer to this question is D, because the passage states that cP-450 is inducible and thus previous abuse of alcohol, a cellular toxin, results in an increased concentration of cP-450. This leads to rapid barbiturate metabolism by cP-450 and the absence of initial drowsiness. This is a Scientific Reasoning and Problem Solving question because it requires you to propose an explanation for the absence of the initial drowsiness.

What is the function of the Na+K+ATPase during a neuronal action potential? A) stimulation of the action potential B) depolarization of the membrane C) hyperpolarization of the membrane D) restoration of the resting potential

D; This is a Biology question that falls under the content category "Structure and functions of the nervous and endocrine systems and ways in which these systems coordinate the organ systems." The answer is D because the Na+K+ ATPase functions to restore the resting membrane potential by moving the ions against their concentration gradients. This is a Knowledge of Scientific Concepts and Principles question because you must use your knowledge of the role of pumps in an action potential to solve the problem.

When concentrated urine is being produced, in which of the following regions of the kidney will the glomerular filtrate reach its highest concentration? A) proximal convoluted tubule B) distal convoluted tubule C) cortical portion of the collecting duct D) medullary position of the collecting duct

D; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ system." The answer is D because glomerular filtrate is most concentrated in the medullary portion of the collecting duct, compared to the other kidney structures listed. This is a Knowledge of Scientific Concepts and Principles because it requires knowing the process of urine production in the kidney.

*See passage for #19 on AAMC Test 3* Which action(s) could contribute to the positive inotropic effect of digoxin on cardiac myocytes? I. Decrease transport of Ca^2+ to the extracellular environment. II. Increase availability of intracellular Ca^2+ to bind to troponin. III. Increase overall Ca^2+ stores in the sarcoplasmic reticulum. A) II only B) III only C) I and II only D) I, II, and III

D; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ systems." The answer to this question is D because all of these actions will lead to an increase in intracellular calcium levels, resulting in enhancement of the contractile force of the heart. It is a Scientific Reasoning and Problem Solving question because it requires scientific evaluation of various scenarios to determine their outcomes.

One potential complication of celiac disease is osteoporosis, which can occur as the body utilizes bone tissue to maintain adequate levels of calcium in the blood. The most effective way for the body to utilize bone tissue to increase blood calcium levels would be to simultaneously: A) increase osteoblast activity and decrease osteoclast activity. B) increase osteoblast activity and increase osteoclast activity. C) decrease osteoblast activity and decrease osteoclast activity. D) decrease osteoblast activity and increase osteoclast activity.

D; This is a Biology question that falls under the content category "Structure and integrative functions of the main organ systems." The answer to this question is D because while osteoblasts function to build and repair bone, osteoclasts break down bone. Therefore, to increase calcium levels in the blood, osteoblast activity should be decreased and osteoclast activity should be increased to release stored calcium from the bone to the bloodstream. It is a Reasoning about the Design and Execution of Research question because it requires the identification of the correct design strategy to alleviate abnormalities that result from a disease state.

*See passage for #37 on AAMC Test 3* Which type(s) of restriction enzyme(s) can recognize the HIF binding sequence? A restriction enzyme that has: I. a four-base recognition sequence II. a six-base recognition sequence III. an eight-base recognition sequence A) I only B) II only C) III only D) I and II only

D; This is a Biology question that falls under the content category "Transmission of genetic information from the gene to the protein." The answer to this question is D because only the CCCGGG within the HIF binding sequence is palindromic. Therefore, only a restriction enzyme that recognizes a four-base sequence or a six-base sequence can recognize this sequence within the HIF binding sequence. It is a Reasoning about the Design and Execution of Research question because it requires reasoning about use of restriction enzymes in research.

Assume that K and M are two unlinked genes that affect hearing. The dominant K allele is necessary for hearing, and the dominant M allele causes deafness regardless of the other genes present. Given this, what fraction of the offspring of parents with the genotypes KkMm and Kkmm will most likely be deaf? A) 1/4 B) 3/8 C) 1/2 D) 5/8

D; This is a Biology question that falls under the content category "Transmission of heritable information from generation to generation and the processes that increase genetic diversity." The answer to this question is D, because among the offspring of KkMm and Kkmm parents, the ones who lack a dominant K allele (necessary for hearing), or carry a dominant M allele (causes deafness) are deaf. Based on the Punnett square analysis, 10 out of 16 or 5/8 of all offspring are likely to be deaf. This is Scientific Reasoning and Problem Solving question because it requires you to calculate the number of offspring with a particular phenotype from a parental cross.

*See passage for #10 on Kaplan Test 2 (short)* When Viagra™ inhibits phosphodiesterase-5, it: A) binds irreversibly to the active site B) reduces the Vmax of the enzyme C) binds to an allosteric site D) increases the Km of the enzyme

D; This is a pseudo-discrete question asking for the role of a competitive inhibitor. By definition, competitive inhibitors bind reversibly to the active site, which eliminates (A) and (C) . Since reversible inhibition can be overcome by adding more substrate, the Vmax remains the same. The inhibitor actually increases the Michaelis-Menten constant, Km , the concentration at which the enzyme has half-maximal velocity.

Which of the following has the lowest ratio of lipid to protein? A) chylomicrons B) VLDL C) LDL D) HDL

D; This question is asking for the lowest ratio of lipid to protein. Since lipids are less dense than proteins (fat floats in water, while muscle doesn't), the lowest ratio of lipid to protein would correspond to the highest density. Among the choices listed, the one with the highest density is HDL, choice (D) . Choice (A) . Chylomicrons actually have the highest lipid to protein ratio, and the lowest density . Choice (B) . VLDL has a lower lipid to protein ratio than chylomicrons, but not as low as HDL. Choice (C) . LDL has a lower lipid to protein ratio than chylomicrons, but not as low as HDL.

An enzyme has two substrates. The enzyme has a higher Km for Substrate A than for Substrate B. This most clearly suggests that the enzyme: A) reacts with substrate A faster than it does with substrate B B) reacts with substrate A more slowly than it does with substrate B C) has a higher affinity for substrate A than for substrate B D) has a lower affinity for substrate A than for substrate B

D; This question tests understanding of enzyme kinetics—specifically, the Michaelis constant, denoted Km. Km represents the substrate concentration at which half the enzyme's active sites are bound. Therefore, the largerKm is, the less affinity the enzyme has for that particular substrate. Choice (D) is correct. Choice (A) . The speed of an enzyme's reaction is directly proportional to vmax, not Km. Choice (B) . The speed of an enzyme's reaction is directly proportional to vmax; it is not related to Km. Choice (C) . Higher Km values correspond to lower enzyme affinities.

Microtubules attach to chromosomes at: A) telomeres B) histones C) short arms D) centromeres

D; This question wants to know where the microtubules attach to chromosomes. This is a discrete question that does not require information from the passage; the correct answer is choice (D), centromeres. Choice (A). Telomeres are the ends of chromosomes. Choice (B). The DNA in chromosomes is wrapped around histones. Choice (C). Microtubules do not attach directly to the short arms of chromosomes.