MCAT Missed Q's on Test 10/17/2020

In which steps in Figure 3 is the fatty acid oxidized? I. Step 1 II. Step 2 III. Step 3 A. I only B. III only C. I and III only D. I, II, and III

C. I and III only This question is asking us to determine which steps in Figure 3 are oxidations. We are told that steps 1 and 3 convert FAD2+ and NAD+ to FADH2 and NADH + H+. This means that the fatty acid must have been oxidized. Step 2 is simply the addition of water across a double bond, which adds two protons and one oxygen to the fatty acid, meaning there is no net oxidation or reduction.

That participants in Brazil might have interpreted the classifications of "Black" and "White" to be a reflection of social class, rather than race, represents what type of critique of the methodology? A. A critique of the construct validity B. A critique of the external validity C. A critique of the criterion validity D. A critique of the randomization

A. A critique of the construct validity The study purports to examine how social class influences perception of race. However, if the categories of "Black" and "White" are viewed by participants as being categories of social class rather than race, then they are not reflecting what the researchers intend to measure (participants' views of race). This is a problem with construct validity, or the manner in which the terms of the study are defined.

The class of medications described in the passage are similar to endogenous hormones produced in what endocrine gland? A. Anterior pituitary B. Posterior pituitary C. Adrenal cortex D. Adrenal medulla

A. Anterior pituitary Opioid analgesics mimic the effect of a group of hormones called endorphins, which naturally inhibit pain by blocking the release of certain neurotransmitters at nociceptors. Endorphins are produced in the anterior pituitary gland, along with other hormones such as luteinizing hormone (LH), follicle stimulating hormone (FSH), and growth hormone (GH).

Which of the following statements about human-pet interactions best reflects the Schachter-Singer theory of emotion? A. As Julia pets her cat, her blood pressure decreases and her brain releases oxytocin. She then thinks of how much she loves her cat, and experiences happiness as a result. B. Timothy's dog runs out of the house without his collar or leash. Timothy's heart rate increases and he shouts for the dog to come home. His body's cues and behavior lead him to understand that he is in a scary situation, and he feels afraid. C. Annabel is holding her guinea pig on the couch and scratching its ears. As she does this, her brain releases hormones associated with reward. At the same time, she interprets the situation as happy and calm. She then smiles. D. Luke is training for a marathon and runs with his dog, Max. He feels moderately happy and excited while he runs with Max, and records his best time yet.

A. As Julia pets her cat, her blood pressure decreases and her brain releases oxytocin. She then thinks of how much she loves her cat, and experiences happiness as a result. The Schachter-Singer theory states that emotion processing has three distinct steps: physiological arousal, cognitive interpretation of the situation, and the experience of the emotion, in that order. Julia pets her cat, which leads to physiological changes in her brain. She then thinks about her love for the cat and feels an emotion — happiness.

According to the experimental results, which enzyme is most sensitive to inhibition by CTZ? A. Glucose B. Glucose 6-phosphate dehydrogenase C. PFK D. HK

C. PFK Table 1 shows that PFK levels are reduced by both types of CTZ, even the less soluble form (uCTZ). Therefore, we can infer that PFK is sensitive to CTZ even if it is poorly soluble in the cytosol.

A psychologist assessing the willingness of subjects to spend a full weekend volunteering to clean up a local park decides to measure the effect of the foot-in-the-door technique. Which of the following would be a way to use this technique? A. Ask subjects to first sign a petition in favor of cleaning up the park B. Tell the subjects that their next-door neighbor has already agreed to volunteer to clean up the park C. Ask the subjects to recall the last time they personally made use of the park D. Remind the subjects that as members of the community, one of their civic duties is helping to maintain public spaces

A. Ask subjects to first sign a petition in favor of cleaning up the park The foot-in-the-door technique says that when someone has agreed to make a small commitment towards something, they are then much more likely to follow up with a greater commitment. Here, getting them to sign a petition would be a small commitment to increase their willingness to volunteer for the whole weekend.

Which type of social influence do medical students experience when they engage in behavior that they privately feel is unacceptable and unethical, but that is considered normative in medical school? A. Conformity B. Obedience C. Self-fulfilling prophecy D. Informational social influence

A. Conformity In the example given, the student privately disagrees with the behavior but publicly goes along with the behavior of a normative social group (in this case, other medical students and/or residents and attendings). This is a classic example of conformity.

Which of the following is the Gestalt principle of perception that the child is demonstrating by perceiving the image as a smiling face rather than as a series of disconnected dots? A. Law of Closure B. Law of Symmetry C. Law of Similarity D. Law of Proximity

A. Law of Closure The child is taking an incomplete figure - a smiley face comprised of disconnected pieces-and is perceiving it as a complete whole. This is an example of the law of closure.

What variable could be introduced to the study to evaluate attitudes towards race and class from a symbolic interactionism perspective? A. Participants' level of exposure to different races and classes B. Participants' mental states during the study C. The purpose of participants' classifications D. Participants' views of biology versus social constructs

A. Participants' level of exposure to different races and classes Symbolic interactionism is the view that an individual's experiences influence his or her perceptions. Thus, an individual's experience with race and class would influence how he or she perceives the images.

Researchers later isolate another residue that tends to be phosphorylated at position 259 of the AQP5 protein. If they desire to replicate the results described in the passage with a phosphomimetic mutant construct, this construct is most likely to be: A. T259D. B. Y259A. C. F259E. D. G259W.

A. T259D. This questions asks us to pick the mutation that most resembles the mutant used in the passage (S156E). The third paragraph tells us that the phosphorylation state of the residue is important to its regulation. That means our new mutant construct alter a residue that is often phosphorylated. In eukaryotes, the residues most prone to phosphorylation are serine (S), tyrosine (Y), and threonine (T). We can now eliminate Choices C and D and are left with choices A and B. To decide between A and B, we must examine the second (mutant) residue. The passage tells us that "phosphomimetic" mutations mimic the effect of phosphorylation, so this new residue must somehow resemble a phosphate group. If you failed to notice this information or are unsure what to do with it, do not panic. Remember that we are looking for a mutation that resembles S156E used in the experiment. "E" denotes glutamic acid, which is most similar to D (aspartic acid), the only other acidic residue. Thus, we can conclude that T259D (choice A) is most similar to S156E. This also makes sense scientifically, as both D and E are acidic residues and thus likely to be negatively charged at physiological pH - just as a phosphate group would be.

A private school sets up a system by which students may advance to the next grade solely on the basis of their individual performance on an exam. Given the wide variety of talents each student has, some students are able to advance to the next grade months or even years before other students. This system is: A. a meritocracy. B. an oligarchy. C. guaranteed to generate student failure. D. an example of spatial discrimination.

A. a meritocracy. Meritocracy is a system under which individuals are rewarded on the basis of individual skill, talent, or achievement. The testing system the school has set up rewards students for success on this test, thus creating a meritocracy.

After reviewing past charts. The methodology employed in the study can best be described as: A. a retrospective chart review. B. a prospective chart review. C. an embedded field study. D. a longitudinal study.

A. a retrospective chart review. "Retrospective" indicates that past records are examined, which is indicated in the description of the study.

Boiling chips and vacuum distillation, respectively, are used in distillations to: A. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; lower the boiling points of the substances to be distilled. B. lower the boiling points of the substances to be distilled; work synergistically with the vacuum system to further lower the boiling points. C. lower the boiling points of the substances to be distilled; provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating. D. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; speed up the distillation process by vacuuming the first distillate out of the apparatus.

A. provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; lower the boiling points of the substances to be distilled. Let's begin with the first part of this question, the function of boiling chips. When transitioning from liquid to gas during boiling, the liquid needs nucleation sites, or places to start forming bubbles. This is typically achieved either by scratching the inside of the flask or by introducing boiling chips. Next, remember that boiling occurs when the Pvap of the substance in question equals the Patm. Typically, we boil substances by increasing the temperature, thereby increasing Pvap. Alternatively, however, we can lower boiling point by reducing Patm, which can be accomplished through the introduction of a vacuum. Vacuum distillation is often used when components have very high boiling points and would otherwise be difficult to distill.

Stimulation of the iris dilator muscle is a result of activation of: A. sympathetic motor neurons. B. parasympathetic motor neurons. C. the fifth cranial nerve. D. sympathetic sensory neurons.

A. sympathetic motor neurons. Dilation of the pupils is a classic fight-or-flight response. The fight-or-flight response is part of the sympathetic nervous system.

Which of the following are products of the decay of I-131? A. Ionizing radiation, Xe-131, and an electron B. Ionizing radiation, Te-131, and an electron C. Non-ionizing radiation, Xe-131, and a positron D. Non-ionizing radiation, Te-131, and a positron

A. Ionizing radiation, Xe-131, and an electron Paragraph 3 mentions that the nuclei emit gamma rays, a form of electromagnetic radiation that consists of high-energy photons. Gamma rays represent ionizing radiation. The passage also states that I-131 undergoes β-minus decay. In β-minus decay, a neutron is converted to a proton as an electron is emitted. Therefore, iodine must be converted to an element with one additional proton, which can only be Xe. β-minus decay is depicted below. Note that we can think of this process in two ways: either simply as the emission of an electron, or as the conversion of a neutron into a proton, an electron, and a neutrino (shown in the circle on the bottom right).

Laws passed to introduce specific penalties for cyberbullying and online harassment can be described as attempts to transform: A. informal norms into formal norms. B. folkways into mores. C. mores into informal norms. D. folkways into formal norms.

A. informal norms into formal norms. Even without specific laws forbidding them, cyberbullying and online harassment are behaviors that are not favored by society. The passage provides evidence for this by confirming that most internet users condemn these behaviors. In the absence of specific legal penalties, norms against such behaviors are informal. Since formal norms are those that are encoded by law, legislation introducing specific legal penalties for these behaviors can be conceptualized as converting informal norms into formal norms.

Given information provided in the passage, which of the following is the most likely standard voltage change for the ½ O2/H2O redox pair? A. +0.82 V B. +0.14 V C. -0.18 V D. -0.45 V

A.+0.82 V In the electron transport chain, electrons are passed from species with less positive reduction potential to those with more positive reduction potential. O2 serves as the final electron acceptor of the electron transport chain and must possess a standard reduction potential that is more positive than any other acceptor in the chain. Of the standard reduction potentials mentioned in the passage, the greatest is that of Fe3+/Fe2+ in cytochrome c, for which E° = +0.22 V. Only choice A exceeds this value.

Which of the following conclusions is most strongly supported by the data presented in the passage? A. ALDH expression is associated with additional factors that promote invasiveness distinct from platinum resistance. B. Platinum resistance is the mechanism through which ALDH promotes invasiveness. C. ALDH expression is exclusively responsible for the platinum-resistant phenotype. D. BAX expression is promoted by ALDH.

A.ALDH expression is associated with additional factors that promote invasiveness distinct from platinum resistance. The left-hand side of Figure 2 shows that ALDH knockdown by itself does not meaningfully reduce invasiveness. In contrast, the left-hand side of Figure 1 shows that natively ALDH-overexpressing cells in this cell line do show higher invasiveness. ALDH is also shown to be involved in platinum resistance by the right-hand side of both figures. Combining these observations, we can conclude that platinum resistance is associated with ALDH expression, but it cannot be the whole story of how ALDH is associated with greater invasiveness. That is, native ALDH overexpression is a marker of an invasive phenotype, meaning that there must be other factors associated with ALDH expression that promote invasiveness

Lathyrism is known to target and degrade collagen in the lower limbs. Which pair of amino acids would best provide disulfide links to stabilize the folded form of collagen? A. C and C B. M and M C. M and C D. S and S

A.C and C Disulfide linkages only form between the side chains of cysteine (C) residues. When not part of such a linkage, cysteine side chains contain a thiol group (S-H). Through the process of oxidative folding, two thiol groups can connect and form an S-S bond, as shown below. These linkages are an important part of protein tertiary and quaternary structure.

The transcription factor AP-1 is a heterodimer consisting of c-jun and c-fos. C-jun and c-fos are soluble proteins that can be localized to either the cytosol or nucleus of a cell. C-jun and c-fos dimerize through a leucine zipper motif. In a leucine zipper motif, every 7 amino acid residues, or 2 full turns of an alpha helix, are leucine resides. Leucine and other amino acids on one face of the helix come together to form an opposite alpha helix that has a similar arrangement of leucine and other amino acids. Which solvent would be LEAST favorable for c-fos/c-jun dimerization? A. Hexane B. Ethanol C. Water D. Phosphate-buffered saline

A.Hexane For long questions like this one, begin by summarizing exactly what the question is asking. In short, leucine residues on different parts of a molecule are coming together to form a dimer. Leucine is hydrophobic, since its side chain contains only carbon and hydrogen. If the solvent were also hydrophobic, the face of a leucine zipper could interact just as favorably with the solvent as with the opposite alpha helix. Some leucine residues would likely interact only with the solvent, preventing formation of the dimer entirely. Hexane, with its nonpolar hydrocarbon structure (shown below), is the least polar solvent listed.

Which of the following is most likely NOT a symptom of acylcarnitine translocase deficiency? A. Hyperglycemia B. Muscle weakness C. Liver damage D. High ammonia levels in blood

A.Hyperglycemia This question asks us to determine the likely effects of a malfunction of acylcarnitine translocase. This enzyme is essential for the catabolism of fatty acids to occur. This means that without it, there will be a great abundance of fatty acids in the body and less energy available. It will also mean that glucose will be relied upon much more heavily as a source of ATP. This implies that there will be little glucose in the blood, and that hyperglycemia will not be a symptom of acylcarnitine translocase deficiency.

Which of the following reagents could be used to complete the final step of retinol synthesis shown in Figure 2? A. LiAlH4 B. O3 C. H2 with Pd D. KMnO4

A.LiAlH4 The final step in Figure 2 is the reduction of an aldehyde (retinal) to a primary alcohol (retinol). This is a classic case of reduction, which can be conceptualized as the gain of bonds to hydrogen or the loss of bonds to oxygen (or other electron-withdrawing species). LiAlH4 is a reducing agent capable of performing this reaction.

According to the passage, which feature of CTZ presents the most significant obstacle to its use as a cancer drug? A. Low solubility in hydrophilic media B. Low solubility in hydrophobic media C. Aromatic structure D. Electron delocalization

A.Low solubility in hydrophilic media The passage states that there is a structural feature of CTZ that inhibits its use as a cancer drug. We must infer this feature from the passage. Figure 1 shows that CTZ is a nonpolar compound with multiple hydrocarbon groups (shown below), lots of electron delocalization, and no charge, meaning that it is poorly soluble in polar (hydrophilic) solvents like water. Because blood and the cytosol are largely made of water, the drug likely has trouble traveling in the bloodstream.

CK isoforms containing M1 and M2 subunits migrate nearer the anode in gel electrophoresis than do CK isoforms containing the unmodified M subunit. The region near the C-terminus of the unmodified M subunit is enriched in which of the following amino acids? A. Lys, Asn, Arg B. Leu, Cys, Pro C. Lys, Ala, Asp D. Gln, Asp, Glu

A.Lys, Asn, Arg The passage states that once released from tissue, the CK M subunits are converted into the modified subunits M1 and M2 by successive cleavage of C-terminal amino acid residues. If the modified subunits migrate nearer the positive electrode than do unmodified subunits, then the modified subunits possess a lower pI than the unmodified subunits and should be enriched in lower-pI amino acids via the removal of higher-pI residues. Only this choice—lysine, asparagine, and arginine—contains two higher-pI, basic residues and is the correct answer. (Note that asparagine is not a basic amino acid.) All of the other answer choices contain fewer basic residues.

OX-LDL most likely forms through the reaction of LDL with: A. free radicals. B. FADH2 and NADH. C. hydrogen gas. D. saturated fatty acid tails.

A.free radicals. From the passage, we can infer that OX-LDL is formed through the oxidation of LDL. Free radicals are molecules or atoms that contain one unpaired valence electron. As such, they typically serve as highly reactive oxidizing agents.

Which concept is most useful in explaining the results of the relationship between race and social class in the study? A. Heredity B. Socialization C. Poverty D. Conflict theory

B. Socialization Socialization refers to how individual attitudes are shaped by social factors. As was stated, race is viewed as a fluid concept in Brazil.

Based on Figure 1, adding salt to water causes the boiling point of the water to: A. increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. B. increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure that is greater than the external pressure. C. decrease, requiring a lower average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. D. decrease, requiring a lower average kinetic energy of the liquid to produce a vapor pressure that is less than the external pressure.

A.increase, requiring a greater average kinetic energy of the liquid to produce a vapor pressure equal to the external pressure. As stated in the passage, adding salt reduces the vapor pressure of the liquid. Specifically, as the solute concentration is increased, the rate at which water molecules can break through the liquid surface decreases. Remember that boiling point is defined as the temperature at which the vapor pressure of a solution is equal to the atmospheric pressure. A decrease in vapor pressure makes this point more difficult to achieve, resulting in a higher boiling temperature.

Which of the following correctly describes the orbital hybridization of XeF4 and NH3, respectively? A. sp3d2, sp3 B. sp3, sp3 C. sp3, sp2 D. sp3d2, sp2

A.sp3d2, sp3 One quick way to determine the orbital hybridization around the central atom is to simply count up the number of bonds and lone pairs. For example, ammonia has three bonds and one lone pair around its central nitrogen atom, for a total of four regions of electron density. This orbital hybrid therefore needs four orbitals to hybridize: s, p, p, and p. Thus, ammonia is an sp3 hybrid. We can narrow it down to choices A and B. Next, XeF4 has two lone pairs and four bonds around the central xenon atom. We know this because the central Xe atom has eight valence electrons. It gets four of these eight electrons from the two lone pairs and four from the bonds — one electron from each bond. Alternatively, we could try drawing the Lewis structure of XeF4, knowing that 8 electrons must come from the Xe atom and 7 from each of the F atoms, for a total of 8 + 7(4) = 36 electrons. Drawing XeF4 with Xe as the central atom and single bonds between the Xe atom and each F atom produces a structure with only 32 electrons, so we must add two lone pairs to the central Xe atom. Two lone pairs and four bonds gives us six total regions of electron density. So we need six hybrid orbitals to hold those electrons: s, p, p, p, d, and d. That's an sp3d2 hybrid.

The passage suggests that the cargo cult myth is founded largely on the: A. work of early anthropologists such as Schwartz and Lawrence. B. popularity of the film Mondo Cane C. desire of Melanesian tribesmen to acquire manufactured goods. D. need of Melanesian "Big Men" to maintain their status.

A.work of early anthropologists such as Schwartz and Lawrence. The flawed Western view of cargo cults sprang from the early work of anthropologists. The passage specifically mentions Schwartz and Lawrence as laying down the foundational ideas - paternalism and materialism - that would come to characterize the cargo cult myth.

A mass of 10 kg is dropped from a height of 20 m. Ignoring air resistance, what is the maximum speed achieved by the mass? (assume g = 10 m/s2) A. 10 m/s B. 20 m/s C. 200 m/s D. 400 m/s

B. 20 m/s The mass starts with gravitational potential energy: PE = mgh = 10 kg x 10 m/s2 x 20 m = 2000 J The maximum speed is achieved just before impact when all of that potential energy is converted to kinetic: KE = ½ mv2 2000 J = ½ x 10 kg x v2 4000 = 10 v2 400 = v2 20 = v Thus, 20 m/s is the correct answer. You also could have arrived at the correct answer using the kinematics equation vf2 = vo2 + 2a∆x, where ∆x = 20 m, vo = 0 m/s, and a = 10 m/s2.

Which of the following is NOT a core component of emotion? A. Physiological arousal B. Conditioned responses C. Expressive displays D. Subjective experiences

B. Conditioned responses There are three core components of emotion - physiological arousal (how your body reacts to emotions, emotional information or stimuli), expressive displays (how you express your emotions), and subjective experiences (how you feel and interpret your emotions, which is extremely personal and subjective). Conditioned responses are not a core component of emotion.

A rapid mechanism is thought to govern the localization of AQP5 in response to changes in extracellular osmolarity. If this mechanism is independent of both PKA activity and S156 phosphorylation, which of the following will most likely be observed? A. HEK cells exposed to the most hypotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow out of the cells. B. HEK cells exposed to the most hypotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow into the cells. C. HEK cells exposed to isotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow out of the cells. D. HEK cells exposed to isotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow into the cells.

B. HEK cells exposed to the most hypotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow into the cells. According to the passage, aquaporins "transport water in response to osmotic gradients." An osmotic gradient is simply a concentration difference; in this case, it is the difference in solute concentration between the cell interior and its external environment. A very hypotonic, or dilute, environment will create a greater osmotic gradient than an isotonic environment, which would be similar in concentration to the cell. Thus, we can infer that aquaporins, including AQP5, will localize to the plasma membrane to a greater extent in hypotonic than in isotonic conditions. (Note that no choices even included the possibility of hypertonic conditions.) Water tends to flow down its concentration gradient, from areas of low solute (and high water) concentration to areas of high solute (and low water) concentration. Here, then, water should flow from the hypotonic exterior of each cell into the more solute-rich interior, as depicted below. Note that even if you do not know exactly what is happening here, you can eliminate choices that violate basic science (which leaves only this option).

It can be inferred that a person who has lost sight in one eye, will come to predominantly use which of the following type of information to perceive depth? I. Binocular vision II. The clarity of objects III. Retinal disparities A. I only B. II only C. III only D. I and II only

B. II only If there is only one eye available for sight, the individual will likely use context clues, such as the clarity of objects, to perceive depth (II).

Based on the information from the passage, what conformity process or factor most likely impacts the behavior of Internet users who decide to engage in harassment without examining evidence for themselves? A. Normative influence B. Informational influence C. Compliance D. Ingratiation

B. Informational influence Informational influence is an influence to accept information from others as evidence about reality, and can come into play when we are uncertain about information or what might be correct. The Internet users relied on the information from the vigilantes when making a decision to conform with the group and harass the supposed subject.

In what way did the passage study depart from Asch's line experiment study of conformity? A. There were varying numbers of people conforming in the study. B. There were reasons provided for having one's view in the study. C. Participants made decisions in groups in the study. D. Participants in the study varied in status.

B. There were reasons provided for having one's view in the study. In 1951, Solomon Asch conducted an experiment investigating the extent to which social pressure from a majority group could influence a person to conform. Asch used confederates who were instructed to give clearly incorrect answers regarding the lengths of various lines. He then measured the number of times each unknowing participant conformed to the majority view. In Asch's study, the confederates never provided an explanation for their choices to the group. In the study in the passage, however, the minority juror had a chance to convince the rest of the group of their position using reasoning.

Casinos maximize the amount of money that people are willing to put into slot machines by making sure that the slot machines pay out jackpots on a reinforcement schedule that is the most resistant to behavior extinction. These machines use which reinforcement schedule? A. Fixed-ratio B. Variable-ratio C. Variable-interval D. Continuous

B. Variable-ratio Variable-ratio reinforcement schedules tend to produce the highest response rates that are the most resistant to extinction, which is exactly why casinos use them.

The tendency of bystanders to avoid intervening in conflict, even though they would want others to intervene on their behalf, can be best explained by: A. opponent-process theory. B. cognitive dissonance theory. C. drive reduction theory. D. expectancy-value theory.

B. cognitive dissonance theory. Cognitive dissonance theory holds that people desire consistency between their thoughts, values, and actions, and seeks to explain the justifications people use for actions that do not align with their values and evoke cognitive dissonance. This would apply to instances of bystanders not intervening (an action), even though they would want others to intervene for them (a value).

When professors meet with individual students during office hours, they often exhibit communication styles that vary widely based on the person they're talking to. A study found that professors speaking with minority female students were much more likely to use shorter sentences with more instructions, and to ask questions that revealed an assumption of lack of academic skill. These professors were demonstrating: A. prejudice. B. discrimination. C. stereotyping. D. egoism.

B. discrimination. The professors are demonstrating a change in their behavior - speaking and questions - based on race and gender. Such changes are discrimination.

Consider the net reaction leading to formation of fructose 6-phosphate from glucose and ATP. ΔG° for the reaction equals: A. +3.96 kcal/mol, and Keq for the reaction is greater than 1. B. -3.60 kcal/mol, and Keq for the reaction is greater than 1. C. -3.60 kcal/mol, and Keq for the reaction is less than 1. D. +3.96 kcal/mol, and Keq for the reaction is less than 1.

B.-3.60 kcal/mol, and Keq for the reaction is greater than 1. The ΔG° for the formation of fructose 6-phosphate from glucose 6-phosphate and ATP is equal to -3,600 cal/mol. This is the sum of the ΔG° for the formation of glucose 6-phosphate and ADP from glucose and ATP (-4,000 cal/mol) and the ΔG° for the reaction to which it is coupled, the formation of fructose 6-phosphate from glucose 6-phosphate (+400 cal/mol). -3,600 cal/mol can be easily converted to -3.60 kcal/mol. ΔG° < 0 for a given reaction indicates that reaction is spontaneous under standard conditions, and that at equilibrium, the products of the reaction predominate over the reactants. This is indicated by Keq > 1 and can be clearly seen in the relationship between ΔG and Keq: ΔG = -ln Keq where Keq > 1 returns a value of ΔG < 0, and Keq < 1 returns a value of ΔG > 0. Here, then, since ΔG is less than 0, Keq for the reaction is greater than 1.

How many moles of captopril were present in the original analyte solution tested? A. 7.5 × 10-5 moles B. 1.5 × 10-4 moles C. 7.5 × 10-3 moles D. 1.5 × 10-2 moles

B.1.5 × 10-4 moles In a titration, the analyte — the substance whose quantity or concentration is to be determined — is reacted with a carefully-controlled volume of standard solution, of which the concentration is known. When analyzing a titration curve, look for the equivalence point(s), which are located halfway along the steep portion(s) of the curve. For a monoprotic acid at the equivalence point, Mbase × Vbase = Macid × Vacid = moles acid. In Figure 2, we see that the equivalence point of the reaction occurs when approximately 7.5 mL of 2 x 10-2 M NaOH solution was added. However, captopril is not a monoprotic acid; it contains both a carboxylic acid group and a thiol group and is thus diprotic. So which equivalence point is this? Look back to paragraph 2, which states that a hydrogen will produce a visible inflection (equivalence) point "only if it is not very weakly acidic, and if its ionization constant differs from that of any other acidic hydrogen of the acid by at least a factor of 104." Examining Figure 1, we see that captopril has a pka1 of 3.7 (the carboxyl group) and a pka2 of 9.8 (the thiol group). Since these values differ by approximately 6, ka1 and ka2 will differ by approximately 106. This means that the second proton is much weaker than the first, and only the first proton should produce an inflection point. Let's return to our original equation: MNaOH × VNaOH = (2 x 10-2 mol NaOH/L)(7.5 x 10-3 L) = 1.5 x 10-4 mol NaOH Remember, we have confirmed that the inflection point in question is the first equivalence point, and that our acid can be treated as a monoprotic species. Since 1 OH- ion is required to neutralize each proton, the molar amount of NaOH required to reach this first point must be equal to the molar amount of captopril present. Therefore, the total number of moles of captopril present in the original analyte solution was: (1.5 x 10-4 mol NaOH)(1 mol captopril/1 mol NaOH) = 1.5 x 10-4 mol captopril

Which of the carbon(s) in 2-methylundecanal below is (are) chiral? A. 1 only B. 2 only C. 1 and 2 D. 2 and 3

B.2 only A chiral carbon must be attached to four different substituents. Carbon 1 is a carbonyl carbon and is only bonded to three other atoms. Carbon 3 is bonded to two hydrogen atoms (not pictured) and thus cannot be chiral. Only carbon 2, which is bonded to a hydrogen, a carbonyl carbon, a methyl group, and a 9-carbon group, is chiral.

What is the chemical formula of gypsum? A. CaSO3 B. CaSO4 C. Ca2SO3 D. Ca2SO4

B.CaSO4 The passage states that gypsum is calcium sulfate, an ionic compound. Calcium cation always has a +2 charge, while sulfate anion carries a -2 charge. These ions can thus be combined in a 1:1 ratio to balance charge, leading to an overall formula of CaSO4.

Which of the statements below is supported by the experimental results, as shown in Figures 1 and 2? A. The duration of Eos co-culture with NK cells directly and linearly correlates to the amount of ECP found in the supernatant after centrifugation. B. Cells cultured with a 1:1 NK-to-Eos ratio displayed statistically similar levels of activation to cells cultured with a 5:1 NK-to-Eos ratio, as measured by CD69 expression. C. NK co-culture stimulates Eos activation while inhibiting degranulation. D. Co-culture with NK cells significantly increased Eos degranulation in all groups, as compared to Eos cells cultured alone.

B.Cells cultured with a 1:1 NK-to-Eos ratio displayed statistically similar levels of activation to cells cultured with a 5:1 NK-to-Eos ratio, as measured by CD69 expression. First, let's discuss the conventions used by these graphs to denote statistical significance. (The AAMC will throw a wide variety of graphs at you, so it is important to be able to analyze them effectively.) Each asterisk (*, **, or ***) shows that the results from one group are different from the results of another to a statistically significant degree. The horizontal line beneath the asterisk stretches from the first group to the group with which it is being compared. For example, the horizontal line under the triple asterisk (***) on the left-hand graph of Figure 1 shows that there is a statistically significant difference between the 0:1 group and the 10:1 group. Even if we did not know this off the top of our heads, we can conclude that it must be true, because that line cannot possibly mean that all four groups are statistically different from each other — otherwise, there would be no need for the other horizontal lines and asterisks shown on that same plot. Regarding this option, note that Figure 1 shows eosinophil (Eos) activation as measured by expression of CD69. Thus, we should be able to use Figure 1 to evaluate the accuracy of choice B. The figure shows no statistically significant difference between samples with a 1:1 NK-to-Eos ratio and those with a 5:1 ratio, as evidenced by the fact that no horizontal line stretches from directly above the 1:1 group to directly above the 5:1 group (shown below). As such, the 1:1 and 5:1 groups are statistically similar.

Injection of insulin into the bloodstream is LEAST likely to result in which of the following? A. Increased glycogen synthesis B. Decreased lipid synthesis C. Increased esterification of fatty acids D. Decreased gluconeogenesis

B.Decreased lipid synthesis Insulin is secreted in response to high blood sugar. If the body detects that there's plenty of blood sugar, then it would want to stop making more sugar, to store that sugar as glycogen, and to build up fatty acids into fats for storing up energy. The thing the body is LEAST likely to do is to stop storing up energy. In general, think of the function of insulin as causing the body to build up large molecules to store up energy (glycogen, lipids) and to stop the body from breaking down large molecules to provide energy.

Increasing plasma concentration of aldosterone is most likely to be followed by which of the following? A. Increased water reabsorption through increased aquaporin channels in the collecting duct B. Increased sodium reabsorption in the distal tubule C. Decreased water reabsorption in the collecting duct D. Decreased plasma calcium concentration

B.Increased sodium reabsorption in the distal tubule Aldosterone is released from the adrenal cortex in response to low blood pressure. Its primary function is to increase sodium reabsorption in the distal tubule and collecting duct. Aldosterone upregulates the sodium-potassium pumps along the lining of the nephron, pumping three sodium ions OUT of the nephron lining (and toward the blood) for every two potassium ions it pumps IN (toward the nephron and away from the blood). Since we have a net solute movement out of the nephron, aldosterone also increases the gradient that favors water reabsorption. The human nephron is shown below, with the distal tubule (a primary target of aldosterone) labeled.

If necessary to design a new experiment, which of the following best explains why researchers could use measurements of intracellular lactate levels (ILL) in cancer cells to assess efficacy of cancer drugs? A. High ILL would indicate that glycolysis is significantly inhibited. B. Low ILL would indicate that glycolysis is significantly inhibited. C. High ILL would indicate that the pentose phosphate pathway is significantly inhibited. D. Low ILL would indicate that the pentose phosphate pathway is significantly inhibited.

B.Low ILL would indicate that glycolysis is significantly inhibited. Lactate is a product of fermentation, and the first paragraph states that cancerous cells rely heavily on fermentation for energy production. Inhibiting glycolysis or fermentation would reduce ILL and indicate that the metabolism of the cancer cell is being effectively inhibited by the drug.

Given the information presented in the passage and the results of the serum electrophoresis, what is the most likely subunit composition of the CK 2 isoenzyme? A. MM B. MB C. BB D. The subunit composition cannot be determined by the given information.

B.MB The passage indicates three possible isoform compositions: BB, MB, and MM, each of which corresponds to either CK1, CK2, or CK3. The passage states that the cytosolic CK B subunit contains a significantly smaller proportion of hydrophobic residues and a greater proportion of acidic residues with low pI values than does the cytosolic CK M subunit, which is enriched in asparagine and lysine. The BB dimer, then, must have the lowest pI of the three possibilities and will migrate farthest toward the positive electrode in gel electrophoresis. (Remember, acidic residues tend to be negatively-charged, so the BB dimer will remain negative at a lower pH than the other options.) Based on Figure 2, we can deduce that the BB dimer is CK1. By a similar rationale, the MB dimer should be intermediate in pI, and should migrate closer to the positive electrode than MM, but not as close as BB. Thus, the MB dimer must be CK2.

Which of the following is most likely to be a trait of cancer stem cells? A. Ability to differentiate into any cell type in the body B. Self-renewal C. Greater reliance on aerobic glucose metabolism D. Elevated levels of BAX expression

B.Self-renewal In general, stem cells have the ability to undergo self-renewal. This is further supported by the statement in paragraph 1 that cancer stem cells may be responsible for recurrence after chemotherapy, which implies that they can resist chemotherapy, replicate, and then differentiate as needed.

If we start with the molecule below instead of palmitic acid, which step will be interrupted and which enzyme will be used to correct the issue? A. Step 1, phosphatase B. Step 1, isomerase C. Step 2, reductase D. Step 2, kinase

B.Step 1, isomerase This question is asking us to consider the implications of starting with a palmitic acid with a double bond at the 7 carbon. This will mean that after the initial conversion of the acid to a CoA precursor for entrance into the beta oxidation steps shown, we will be left with: Since the presence of an alkene is something that is supposed to be achieved by Step 1, this molecule will interrupt the formation of the double bond in step 1 (eliminate choices C and D). To fix the problem, we need to form a version of the molecule formed by step 1 with its alkene bond between the alpha and beta carbons instead of where it is now. Since the two molecules have the same atomic ingredients and only differ in the configuration of these atoms, they are isomers. Thus, an isomerase will do the trick nicely.

While the blood is buffered primarily through the equilibrium between carbon dioxide and carbonic acid, coupled with hemoglobin, the blood may also be buffered through other plasma proteins. Which of the following is true? A. A shift in the pH can alter the tertiary or quaternary structure of the protein, allowing it to buffer the pH by precipitating out of plasma in response to pH shifts. B. The amino acid residues that make up the protein may act as Brønsted acids or bases, reducing shifts in pH. C. Plasma soluble proteins have enzymatic function allowing them to sequester hydronium ions from the blood inside membrane-bound organelles in the podocytes lining the capillaries. D. In the presence of altered pH, any plasma-soluble proteins will undergo either acid- or base-catalyzed cleavage, thus depleting the acid or base causing the disruption to blood pH.

B.The amino acid residues that make up the protein may act as Brønsted acids or bases, reducing shifts in pH. The amino acids that make up a protein may include many acidic or basic side chain groups. Those side chains can either release or absorb protons, allowing them to help buffer the blood through action as a Brønsted-Lowry acid or base.

According to passage information, is the energy released by the transport of an electron pair delivered by NADH through the electron transport chain sufficient to produce three ATP molecules from three ADP and three inorganic phosphate molecules if the efficiency of the conversion of the energy released via the transport of an electron delivered by NADH into useful work in the body is found to be approximately 50%? A. Yes, with the additional calories of energy released by the electron transfer and not used in the phosphorylation employed to drive the removal of excess hydrogens from the matrix. B. Yes, with the majority of calories of energy released by the electron transfer and not used in the phosphorylation lost mainly as heat. C. No, the energy released in the process is sufficient to drive the formation of only two ATP from two ADP and two inorganic phosphate molecules because Complex I is bypassed. D. No, the enthalpy change of the reaction is negative and the formation of ATP from ADP and inorganic phosphate requires the input of outside heat energy.

B.Yes, with the majority of calories of energy released by the electron transfer and not used in the phosphorylation lost mainly as heat. This is a long question, but don't get intimidated! To obtain the answer, we only need to find two quantities: the amount of energy required to produce three molecules of ATP and the amount of energy released by the transport of one electron pair by NADH (given the efficiency value shown). In fact, we can make this problem easier by considering moles rather than molecules, since one mole of any substance must always be the same number of molecules. According to the passage, ∆G° for the transfer of two electrons from NADH to oxygen is -52.5 kcal/mol, and ∆G° for the phosphorylation of ADP to ATP is +7.3 kcal/mol. First, let's account for the 50% efficiency, so we don't forget to do that later. Of the 52.5 kcal/mol released, only approximately 26 kcal/mol of useful energy will be available to perform work. The amount of energy required to phosphorylate 3 moles of ADP to ATP is (3 moles)(7.3 kcal/mol), or approximately 22 kcal/mol. Thus, sufficient free energy is made available by transport of one mole of electron pairs to drive the phosphorylation of three moles of ATP from ADP. One electron pair, then, would provide sufficient energy to phosphorylate three ADP molecules into ATP. Most of the free energy available to do free work (~26 kcal) was consumed by the phosphorylation (~22 kcal), with only a small amount (~4 kcal) available to do additional, biologically useful work. The remaining energy (~26 kcal) which was not involved in useful work was likely dissipated as heat energy.

When the researchers connected the solution-filled glass plates of the flow chamber to the AC generator, the ITO-coated plates mostly likely functioned as: A. a resistor. B. a capacitor. C. a galvanic cell. D. an electrolytic cell.

B.a capacitor. According to the experimental procedure of paragraph 2, "The plates were connected to an AC generator (10 Hz), and a 1.5 V potential difference between the plates was sustained for 3 h." The paragraph also indicates that ITO is a transparent conducting material. When a voltage is applied, charges will accumulate on the plates, exactly as what occurs on a parallel-plate capacitor.

Eosinophils play a major role in the promotion of the inflammatory response, which is marked by increased vasodilation, blood vessel permeability, and pain. The inflammatory response is classified as: A. a function of the adaptive immune system. B. a function of the innate immune system. C. a function of passive immunity. D. a function of antibody-mediated immunity.

B.a function of the innate immune system. The innate immune system includes nonspecific immune responses. In other words, its responses provide general protection, rather than protection against specific pathogens that have been previously encountered and "remembered." Inflammation is a function of the innate immune system, as it serves as a general response to infection or injury. While that is all we need to know to answer this question, for the MCAT, you should also know some details about the various types of immune cells. Eosinophils (the focus of this question) are granulocytes, or cells that contain small particles termed granules in their cytoplasm. Other granulocytes include basophils and neutrophils, both of which also are involved in innate immunity.

Decreased number of alveoli in the lungs leads to respiratory distress because: A. damage to the respiratory epithelium reduces the ability of the epithelial cells to actively transport O2 into the body and CO2 out of the body. B. reduced surface area in the lungs reduces the rate at which O2 and CO2 can diffuse through the lung epithelium. C. passive expiration depends on the inherent elasticity in the walls of the alveoli and reduced expiratory volume prevents or reduces subsequent inspiration. D. the cilia lining the lungs which move mucus and other debris out of the lungs and keep the respiratory epithelium clear cannot function as well.

B.reduced surface area in the lungs reduces the rate at which O2 and CO2 can diffuse through the lung epithelium. Many biological processes depend on a very high surface area-to-volume ratio. The large number of tiny alveolar sacs in the lungs, the presence of microvilli in the small intestine, and the folding of the inner mitochondrial membrane are all examples in which an increased surface area allows for biological processes to take place more quickly. When the individual alveoli break down (as, for example, in emphysema) the lungs lose the necessary surface area to allow for effective diffusion of respiratory gases. The alveoli and their larger environment (the human lungs) are shown below.

A book rests horizontally on a table. The book experiences a gravitational force of mg due to the earth's gravity. According to Newton's third law: A. the book experiences a normal force of mg pushing up due to the table. B. the earth experiences a gravitational force of mg from the book. C. the table exerts a gravitational force of mg on the earth. D. the earth exerts a normal force up on the table equal to mg plus the weight of the table.

B.the earth experiences a gravitational force of mg from the book. Newton's third law can be expressed as: FA on B = −FB on A Here, the force of the earth pulling on the book is equal to and opposite of the book pulling up on the earth.

A typical eosinophil would be expected to differ from an erythrocyte in that the eosinophil: A. would not contain a membrane-bound nucleus. B. would contain a membrane-bound nucleus. C. would not be expected to be found in human blood. Show Explanation D. would be expected to be found in human blood.

B.would contain a membrane-bound nucleus. Unlike almost all cells in the human body, erythrocytes (red blood cells) do not contain DNA or a membrane-bound nucleus. This allows these cells to have more space for oxygen-carrying hemoglobin molecules. Thus, eosinophils differ from erythrocytes in that eosinophils do contain nuclei and DNA.

In a study about attitudes towards higher education, four subjects are asked to rank the relative importance of higher education. All four participants rank higher education as "very important" or "extremely important" on the survey. Which of the following participants is most likely to experience cognitive dissonance as a part of this response? A. A master electrician who has significant advanced on-the-job training but no degree B. A day laborer whose own education stopped at seventh grade and whose family includes no one who has attended any college C. A high school math teacher who encouraged his own son to skip college and focus on learning a trade to "save all that wasted tuition money" D. A university professor with two doctoral degrees who has actively advocated for significant reform in the country's higher education system

C. A high school math teacher who encouraged his own son to skip college and focus on learning a trade to "save all that wasted tuition money" Cognitive dissonance is the unpleasant feeling a person experiences when holding two contradictory beliefs at the same time. Here, a person who rates college as very important, but then tells his own child that college tuition is "wasted" would likely experience some cognitive dissonance.

A drive-reduction and cognitive theorist would argue that depression is most strongly correlated with a deficiency in which component of fulfillment? A. Socialization B. Stigma C. Arousal D. Self-efficacy

C. Arousal Drive-reduction theories suggest that depression stems from a reduction in the motivating forces of arousal. A cognitive theorist would argue that arousal is essential to sustaining most behaviors.

High levels of LDL and OX-LDL increase the proportion of cholesterol in cell membranes. If the trisomy 21 data in Figure 2 can be attributed to the effect of cholesterol on these membranes, which of the following statements is most likely true? A. hTERT cells treated with OX-LDL display more rigid membranes than hTERT cells treated with LDL. B. Untreated hTERT cells display more rigid membranes than hTERT cells treated with LDL C. At moderate to high temperatures, ethanol increases membrane fluidity. D. hTERT cells simultaneously exposed to HDL and LDL display increased membrane fluidity relative to untreated cells.

C. At moderate to high temperatures, ethanol increases membrane fluidity. Figure 2 shows that LDL- and OX-LDL-treated cells have a markedly higher incidence of trisomy 21 than untreated hTERT cells. The question stem also indicates that cells treated with these lipoproteins have more cholesterol in their membranes. At moderate to high temperatures (including normal physiological temperature), cholesterol increases the rigidity of cell membranes by attracting adjacent phospholipid tails. Thus, a more rigid membrane appears to correlate with a higher incidence of trisomy 21. From Figure 3, however, we see that ethanol appears to counteract the trisomy-inducing effects of LDL and OX-LDL. It is therefore reasonable to conclude that ethanol decreases the rigidity (or increases the fluidity) of cell membranes under the conditions in this study.

In cells with elevated low-density lipoprotein levels, ethanol has been shown to act directly on these lipoprotein molecules in a way that decreases their trisomy-inducing effects. On the basis of this information, compared with untreated hTERT cells, hTERT cells incubated with ethanol alone would most likely: A. have a higher probability of displaying trisomy 21. B. have a lower probability of displaying trisomy 21. C. have a similar probability of displaying trisomy 21. D. have a higher probability of displaying trisomy 21 than hTERT cells incubated with HDL and ethanol.

C. have a similar probability of displaying trisomy 21. From Figure 3, we see that the incubation of LDL- and OX-LDL-treated cells with ethanol decreases the incidence of trisomy 21 to levels comparable with the control (hTERT alone). However, do not jump to the conclusion that EtOH must always decrease trisomy 21 incidence. The question stem tells us that ethanol has been observed to act directly on lipoproteins "in cells with elevated low-density lipoprotein levels." We have no reason to believe that hTERT cells not incubated with LDL or OX-LDL would have such elevated levels, so we have no reason to think that ethanol incubation would have any effect on them. Note also that neither the LDL + EtOH nor the OX-LDL + EtOH show trisomy 21 levels significantly below the control levels. It is therefore very possible that ethanol incubation "canceled out" the trisomy-inducing effects of the extra low-density lipoprotein, but would have no effect in cells that did not have high low-density lipoprotein concentrations to begin with.

How might someone who contends that race is biologically based respond to the results of the study? A. There is a great deal of variability in how individuals are classified by social class. B. There is a great deal of variability how individuals are classified by race. C. There is greater variation in race within social classes than between social classes. D. There is greater variation in race between social classes than within social classes.

C. There is greater variation in race within social classes than between social classes. If race were entirely a social construct based on social class, then the findings should be 100% based on social class. In other words, one race would be completely associated with one social class, and the other race would be completely associated with the other social class. Because this is not the situation, then there is at least some biological component in the perception of race.

If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground? A. 325 N B. 650 N C. 750 N D. 1075 N

C. 750 N To turn while running, the foot must push off the ground, applying a shearing force while simultaneously supporting the weight of the body. When faced with problems like this, always consider the forces involved. Here, we must account for the normal force exerted by the ground on the foot; this is a vertical force which occurs as a result of the runner's weight. We also must consider the acceleration force, which (since the person is turning) is horizontal. These two force vectors are perpendicular and will form a right triangle. We are looking for the overall force experienced, so we must find the hypotenuse. Specifically, we need to find the hypotenuse of a triangle with legs of Fnormal = mg = (65 kg)(10 m/s2) = 650 N and Fturning = (65 kg)(5 m/s2) = 325 N. The combined vector will be bigger than either component alone, so eliminate choices A and B. To solve, let's round 650 N to 700 N and round 325 N to 300 N. With this in mind, this calculation can be approximated as: √(3002 + 7002) = √(90000 + 490000) = √(580000) = √(58 x 104) = (√58) x 102 The square root of 58 falls between 7 and 8, so the overall value of our answer falls between 700 and 800, meaning that choice C must be correct (the actual value is 761).

Which of the following is closest to the bond angle between the carbons in a molecule of acetone? A. 90º B. 109.5º C. 120º D. 180º

C.120º Acetone (shown below) has a central carbon atom that is double-bonded to an oxygen and single-bonded to two carbon atoms. Since it is bound to three substituents and has no lone pairs, the hybridization around this carbon is sp2, creating a trigonal planar geometry and a bond angle of 120º.

In a population of Amish people, the frequency of the recessive autosomal allele for polydactyly is 1.2%. What percent of the population are heterozygotes for the polydactyly allele? A. 0.0144% B. 1.19% C. 2.37% D. 97.6%

C.2.37% We can use the Hardy-Weinberg equation to solve this question. Remember that the total number of alleles in the population has to add up to 1: A + a = 1 And the total number of genotypes in the population must also add up to 1: AA + 2Aa + aa = 1 We're told that a = 0.012. (Note that the question gave the frequency of the recessive autosomal allele, not the frequency of individuals in the population that are homozygous recessive!) By the first equation above, A = 0.988. The carriers are the heterozygotes with the genotype Aa. Their frequency is: 2Aa = 2 x 0.988 x 0.012 = 0.988 x 0.024 At this point we've got what looks like a tough calculation to do, so we should probably back up and start estimating. Our calculation is telling us to take 98.8% of 0.024, so our answer is going to be really close to 0.024 (since 100% of any number is just that number itself [e.g. 100% of 56.7 is 56.7]). If we look at the answer choices, the only one that's remotely close is 0.0237, or 2.37%.

According to the data in Table 1, what mass of captopril must be dissolved in 3 L of plasma at pH 7.4 to inhibit 50% of ACE enzyme activity in vivo? Assume an equal volume of distribution. A. 7.9 pg B. 7.9 ng C. 7.9 µg D. 7.9 g

C.7.9 µg From the answers alone, we can immediately tell that this is a units question, but let's work through it from the beginning. The passage defines the in vivo IC50 value of captopril to be "the minimum plasma concentration needed to inhibit 50% of [ACE] activity in vivo." From Table 1, the mean in vivo IC50 value of captopril in the pH range which includes pH 7.4 is 0.012 µM. Assuming a plasma volume of 3 L, the necessary gram weight of dissolved captopril can be calculated as follows: (0.012 x 10-6 mol captopril/L plasma) (3 L plasma) (220 g captopril/mol captopril) = 7.92 x 10-6 g captopril = 7.92 µg captopril Remember, 1 µg = 1 x 10-6 g.

Which of the following enzymes should the researchers add to the cell samples if they want to reverse the general catalytic effects of protein kinase A? A. Glycogen phosphorylase B. Phosphoglucomutase C. Protein phosphatase 1 D. Lactate dehydrogenase

C.Protein phosphatase 1 Protein kinase A has an extremely wide variety of specific effects, but this question asks about its general catalytic function. As a kinase, PKA functions to add a phosphate group to its substrate. The opposite of this activity is the removal of phosphate from a substrate, a function which is performed by phosphatase enzymes. We do not need to know the exact reactions catalyzed by each enzyme to answer this question; nomenclature alone shows us that only this option represents a phosphatase.

Nitrogen primarily exists in the atmosphere as a diatomic gas. Which of the following is true about this form of nitrogen? A. The presence of a lone pair of electrons on each nitrogen atom in the molecule allows it to act as a strong Lewis base. B. The triple bond of electrons creates a region of high electron density that allows N2 to be very reactive as a nucleophile. C. Diatomic nitrogen gas is relatively inert and can be used as the atmosphere in laboratory conditions to prevent unwanted side reactions. D. Atmospheric nitrogen reacts spontaneously with carbon dioxide, which keeps atmospheric CO2 levels at a relatively low 0.04% (on a molar basis) of the atmosphere.

C.Diatomic nitrogen gas is relatively inert and can be used as the atmosphere in laboratory conditions to prevent unwanted side reactions. The MCAT will expect you to be familiar with N2 as a very inert gas. It makes up approximately 80% of the air you breathe, yet has no significant chemical reactions with your lungs - or with anything other than nitrogen-fixing plants. This information implies that nitrogen is very inert (unreactive). As such, it would serve as a good artificial atmosphere when working with reagents that might react with oxygen or other gases. The Lewis structure of N2 is shown below.

The researchers chose to co-culture all samples in the presence of interleukin-5, a cytokine. What is the most likely reason for this decision? A. Interleukin-5 facilitates degranulation in NK cells. B. Interleukin-5 inhibits the cytotoxic effects that NK cells have against eosinophils. C. Eosinophils die rapidly when not exposed to interleukin-5. D. The researchers were directly testing the effect of interleukin-5 on eosinophil activity.

C.Eosinophils die rapidly when not exposed to interleukin-5. The passage never gives specific information about interleukin-5 (IL-5), so this question must be answered using reasoning about the experimental design. Choice C is the only sensible option; if eosinophils die very quickly when not exposed to IL-5, then most or all cells may die before the 3- or 12-hour interval is even complete, making it difficult to assess the impact of NK co-culture. In reality, IL-5 does facilitate eosinophil survival.

The author assumes which of the following about obtaining a broad view of language? A. The more that scholars compare languages to each other, the further the study of linguistics develops. B. Normative grammars are only interested in figuring out and enforcing certain rules and their enactment. C. Focusing on rules and what is or isn't proper prevents a holistic understanding of how language works. D. Ritschl's revision of the text of Plautus is a prime example of the best technique for procuring this view.

C.Focusing on rules and what is or isn't proper prevents a holistic understanding of how language works. The author states that traditional grammar, because it is "dominated by a preoccupation with laying down rules, and distinguishing between a certain allegedly 'correct' language and another, allegedly 'incorrect'...precludes any broader view of the language phenomenon as a whole." Thus focusing on rules prevents an overall understanding of a language.

Which of the following do NOT have proteins with a nuclear localization signal? I. E. coli II. Homo sapiens III. Fungi IV. Archaea A. I only B. III only C. I and IV only D. I, III, and IV only

C.I and IV only E. coli (a species of bacteria) and archaea do not have nuclei and thus do not have a need for nuclear localization signal on their proteins (I and IV).

A ray of white light moves through the air and strikes the surface of water in a beaker. The index of refraction of the water is 1.33 and the angle of incidence is 30º. All of the following are true EXCEPT: I. the angle of reflection is 30º. II. the angle of refraction is 30º. III. total internal reflection will result, depending on the critical angle. A. I only B. I and III only C. II and III only D. I, II, and III

C.II and III only The question asks "all of the following are true EXCEPT," indicating that we need to find the false statements - or, alternatively, that we need to eliminate the true ones. The question ends up being much easier than we expect. The angle of incidence always equals the angle of reflection (I). This immediately lets us narrow it down to choice C. Light entering a more dense medium will bend towards the normal. With an angle of incidence of 30º, the angle of refraction must be less than 30º (II). Total internal reflection can only result when a ray of light begins in a higher-index material and reaches a boundary with a lower-index one (e.g. starting in water and moving towards air). Here, the light ray started in air (n = 1) and moved into water (n ∼ 1.3), making total internal reflection impossible (III).

What is the correct order of the 5 para substituents on the carbocation intermediate, if arranged from most stabilizing to least stabilizing? A. F > H > CH3 > OCH3 > NO2 B. CH3 > NO2 > OCH3 > H > F C. NO2 > OCH3 > CH3 > H > F D. NO2 > CH3 > OCH3 > F > H

C.NO2 > OCH3 > CH3 > H > F The stabilizing effects of the substituents can be determined from the relative stabilization energy (RSE), which can be found in Table 1. The more negative the RSE, the greater the stabilization; thus, the substituents should be ordered from most negative to least negative (or most positive) RSE. From Table 1, this corresponds to compound 5 > compound 1 > compound 2 > compound 3 > compound 4, which matches this answer choice.

The results in Figure 1 and the information in the passage most strongly support which of the following conclusions? A. Phosphorylation of S156 by protein kinase A promotes the immediate localization of AQP5 to the plasma membrane. B. Phosphorylation of S156 promotes the internalization of AQP5 in the short term. C. Protein kinase A promotes the internalization of AQP5 in the short term. D. 30 minutes of exposure to protein kinase A stimulates the internalization of AQP5, a process that is upregulated when S156 is phosphorylated.

C.Protein kinase A promotes the internalization of AQP5 in the short term. From Figure 1, we see that in both cell types, inhibition of protein kinase A results in significantly greater AQP5 expression. This supports the idea that stimulation of PKA activity would decrease membrane AQP5 expression, at least at some point. Paragraph 3 supports this idea with the statement that increased cAMP levels have a "biphasic" effect, decreasing AQP5 expression in the short term and increasing it in the long term. We should know (by test day) that PKA is stimulated by cAMP, so it makes sense that it would also follow this pattern. Additionally, paragraph 3 outlines the difference between translocation to the membrane and internalization from the membrane to the interior of the cell. If cAMP and PKA signaling have the short-term impact of increasing AQP5 internalization, this would bring AQP5 proteins away from the plasma membrane, explaining the results (taken after 30 minutes) shown in Figure 1.

What aspects separate single-crossover events from double-crossover events? A. Single-crossover events result in one-way displacement of chromosomal content from one chromosome to another, while double-crossover events always reverse this one-way displacement, resulting in chromosomes identical to the pre-crossover chromosomes. B. Single-crossover events occur during mitosis when a cell splits into two cells, while double-crossover events can only occur during meiosis when a cell splits into four cells. C. Single-crossover events affect only the ends of chromosome arms, while double-crossover events can affect segments in the middle of chromosome arms. D. Single-crossover events only affect one arm of each chromosome, while double-crossover events affect two arms of each chromosome.

C.Single-crossover events affect only the ends of chromosome arms, while double-crossover events can affect segments in the middle of chromosome arms. A double-crossover event is one in which chromosomal arms of homologous chromosomes cross over in two different places along the arm. This results in a section in the middle of each chromosome being exchanged. A simplified schematic of a single- vs. a double-crossover event is shown below.

The reaction to produce the "X" form of compounds 1-4 from standard-state elements will be spontaneous under what conditions? A. The reactions will be spontaneous under all conditions. B. The reactions will be spontaneous at high temperatures. C. The reactions will be spontaneous at low temperatures. D. The reactions will not be spontaneous under any conditions.

C.The reactions will be spontaneous at low temperatures. When determining the conditions under which a reaction is spontaneous, always consider the change in Gibbs free energy. If the ∆G for a reaction is positive, it is nonspontaneous, and if the ∆G for a reaction is negative, it is spontaneous. The ∆G for a reaction can be determined from the change in enthalpy and entropy based on the equation ∆G = ∆H - T∆S. Based on Table 1, the ∆H values for all of the formation reactions (Hf) are negative, which favors spontaneity. However, each reaction starts with the elements in their standard states, which include C (s), O2 (g), H2 (g), and F2 (g). The formation of a more organized compound from these reactants will result in a decrease in entropy (negative ∆S), which will favor a positive ∆G and a nonspontaneous reaction. So, if the magnitude of ∆H is greater than T∆S (which occurs at low temperatures), then the reaction is spontaneous. If the magnitude of ∆H is smaller than T∆S (which occurs at high temperatures), then the reaction is nonspontaneous. For reference, the relationship between ∆H, ∆S, and reaction spontaneity is summarized in the table below.

Another possible method of separating 2-methylundecanal and 2-methylundecanoic acid could be based on: A. their differences in the rotation of plane-polarized light. B. a mass spectrometry analysis. C. an extraction based on their differing solubilities. D. the very different scent profiles of each molecule.

C.an extraction based on their differing solubilities. Note that the compounds discussed in this question are the compounds mentioned in paragraph 2, not the enantiomers depicted in Figure 1. (The compounds that are relevant to this question are pictured below.) With that said, this question asks about separating an acidic compound from a fairly neutral one. The carboxylic acid group, if deprotonated, will be much more soluble in water than the aldehyde group. Thus, a careful extraction with a dilute weak base (such as a solution of sodium bicarbonate) would separate these two molecules. Remember, the more polar a compound is, the higher its solubility in aqueous solution!

Those species that are capable of both sexual and asexual reproduction will typically prefer sexual reproduction because it: A. increases the likelihood of each individual offspring surviving. B. increases the likelihood of beneficial mutations. C. creates more variation in the next generation. D. takes less time to complete.

C.creates more variation in the next generation. Sexual reproduction involves "shuffling the genetic deck" when recombining the genes of two parents. This significantly increases variation in the next generation. This increase in variability helps improve the survival of the whole species by allowing it to adapt more quickly to changing selection pressures.

Based on the passage, studies of language prior to the late 19th century were flawed due to the fact that they: A. replaced an earlier philosophical view of language with a comparativist one. B. did not include comparative grammar as a means of determining correctness. C. focused heavily on comparison rather than developing principles. D. reflected the understanding that languages could be compared to each other.

C.focused heavily on comparison rather than developing principles. The passage focuses on the stages that came before "linguistics proper," arguing that they were flawed because they focused on rules and comparison and had neither a "broader view of the language phenomenon as a whole" nor a focus on "principles" which would "opened up ....new horizons." In addition, it is always helpful to focus on the author's tone, and with words like "flawed" and "absurd," he clearly views language comparison very negatively.

What type of conflict are the jurors in the study likely experiencing if they are unsure of the defendant's guilt? A. Approach-approach conflict B. Avoidant-avoidant conflict C. Approach-avoidant conflict D. Double approach-avoidant conflict

D. Double approach-avoidant conflict Double approach-avoidant conflicts consist of two options with both appealing and negative characteristics, which seems to represent the jury's dilemma. If they rule the defendant guilty, they would either be punishing a criminal (approach) or punishing an innocent (avoidant). If they rule the defendant innocent, they would either be letting a criminal walk away unpunished (avoidant) or freeing an innocent (approach).

Based on passage information, which of the following strategies would NOT be reasonable for an online community to implement as ways to reduce the bystander effect to instances of online vigilantism and harassment? A. Preventing anonymous interactions by mandating that users participate under their real names B. Providing public recognition to users who report online vigilantism and harassment C. Emphasizing the role of the users themselves in maintaining the community space \ D. Rewarding users who report online vigilantism and harassment with small gift cards via email

D. Rewarding users who report online vigilantism and harassment with small gift cards via email Although this may seem reasonable at first glance, according to the passage, research supports the following ways of reducing the bystander effect in online communities: (1) decreasing the degree to which users believe that moderators will solve the problem for them, (2) reducing anonymity, and (3) spotlighting users' own behavior. A reward distributed through email would not satisfy any of those criteria.

If the students perform an enzyme inhibition assay using captopril, which of the following changes in the kinetic parameters of ACE should be expected? A. Vmax decreased; Km unchanged B. Vmax decreased; Km increased C. Vmax unchanged; Km decreased D. Vmax unchanged; Km increased

D. Vmax unchanged; Km increased According to the passage, captopril is a "competitive inhibitor of ACE." Competitive inhibitors increase the Km of their associated enzymatic reactions without altering the Vmax value, as shown in the Michaelis-Menten graph below. Here, the blue curve represents the uninhibited reaction, while the red curve is associated with the competitively inhibited reaction.

In response to the point that gender inequality can contribute to the spread of HIV, a Western European policymaker suggests that Western European gender roles should be directly imposed on African and Asian countries in order to reduce HIV transmission. This proposal is most likely to be criticized for being an example of: A. groupthink. B. deindividuation. C. stigma. D. ethnocentrism.

D. ethnocentrism. Ethnocentrism is the tendency to view one's group and its cultural expectations and norms as right, proper, and superior to others. This can manifest as a desire to impose the social norms of one's group onto another group, as shown by the policymaker described in this question. In contrast, cultural relativism points out the need to understand beliefs, practices, and values in the context of a certain culture. In public health research and policy, it is important to avoid developing initiatives that reflect ethnocentrism. Instead, policymakers and researchers are urged to work together with colleagues from other cultures to develop policy initiatives that will make sense and be effective within a certain cultural framework.

Two groups of cats are placed in separate rooms. Room 1 is large with lots of exercise toys and laser-pointer devices that operate automatically on timers. Room 1 cats exhibit exercise levels 350% greater than the control cats in Room 2. Cats from both rooms are then placed in a room with an electrified floor and a safe shelf while the voltage in the floor is slowly increased. Cats from Room 1 remain on the floor for 32% more time than the control cats. This suggests that exercise downregulates: A. baroreceptors. B. stretch receptors. C. olfactory receptors. D. nociceptors.

D. nociceptors. The shock from the floor would create pain in the cats. The cats with exercise were able to withstand a longer shock and thus probably don't feel as much pain. Nociceptors are pain receptors; thus, exercise likely downregulates nociceptors.

An experiment is arranged in which participants spend 10 seconds, 20 seconds, or 30 seconds looking at a large, complex image. The image is then removed and replaced with a second image which is identical except for one small change. Researchers then measure how long the subjects look at the new image before finding the change. In this scenario, the independent variable is: A. whether the subjects ever find the change. B. the time spent looking for the change. C. the size and prominence of the change. D. the time spent looking at the original image.

D. the time spent looking at the original image. The independent variable is the one the researchers can control, and the dependent variable is the one they measure. Here, the experimenters can control the time the subject gets to study the first image (10 seconds, 20 seconds, or 30 seconds).

A researcher analyzes a nucleic acid sample. After looking only at its nucleotide composition, he decisively concludes that this sample represents single-stranded DNA and not another form of nucleic acid. Which of the following compositions most likely represents this sample? A. 17% A, 17% T, 33% G, 33% C B. 29% A, 14% U, 11% G, 46% C C. 4% A, 4% U, 46% G, 46% C D. 12% A, 12% T, 30% G, 46% C

D. 12% A, 12% T, 30% G, 46% C We are told that the researcher can "decisively conclude" that this sample represents single-stranded DNA using only its nucleotide composition. We are therefore looking for the option that could logically represent single-stranded DNA but could not represent another type of nucleic acid. Both options A and D could be the compositions of single-stranded DNA, but choice A has equal proportions of adenine and thymine as well as equal proportions of cytosine and guanine. Choice A, then, could easily represent double-stranded DNA, so the researcher would not be able to conclude whether that sample were single- or double-stranded. Choice D, then, is the only answer remaining.

Anterior forces of 500 N or higher lead to abnormal translation and result in injury. During the force test with the mechanical stirrup, at what distance from the ankle joint was the force applied? A. 57 mm B. 60 mm C. 110 mm D. 133 mm

D. 133 mm Paragraph 3 discusses this test and tells us the associated applied force and torque values. For test day, you should know the equation for torque: τ = rFsin(Ɵ). The passage shows us that the force is applied perpendicular to the rotation of the ankle joint, so we can use sin (90) = 1. Now, we solve for distance (r) using r = τ/F = (13 Nm) / (98 N) = 0.133 m, or 133 mm.

A chemistry class conducted an experiment to separate 2-methylundecanal from 2-methylundecanoic acid by carrying out a distillation of a liquid consisting solely of these two components. Due to the high boiling points of these compounds, the class was instructed to carry out a vacuum distillation. Students began by placing 200 mL of the liquid mixture in a round-bottom flask and adding several boiling chips. The solution was slowly heated with distillate collected in the receiving flask. Students noted that the round-bottom flask still held approximately 50 mL of liquid at the end of the distillation. Key terms: 2-methylundecanal, 2-methylundecanoic acid Cause and effect: high boiling point requires vacuum distillation. The liquid remaining in the round-bottom flask at the end of the procedure was most likely: A. a mixture consisting of roughly even amounts of the two components. B. 2-methylundecanal. C. water condensed from the air in the lab. D. 2-methylundecanoic acid.

D. 2-methylundecanoic acid. In distillation procedures, the component with the lower boiling point boils off first, leaving the remaining components in the original flask. Thus, this question is basically asking which component has the higher boiling point. The hydrogen bonding created by the carboxylic acid functional group in 2-methylundecanoic acid (shown below) would lead to a vastly higher boiling point than the aldehyde (over 300ºC compared to 170ºC).

Vinblastine is a microtubule-disrupting drug that inhibits tubulin polymerization. Which of the following processes would be directly inhibited upon vinblastine treatment? I. Phagosome transport to the lysosome II. Mitosis III. Meiosis IV. Electron transport A. II only B. I and IV only C. II and III only D. I, II, and III only

D. I, II, and III only Microtubules are used in the transport of vesicles and the positioning of organelles within the cell. As part of this role, they form structures that assist in the transport of phagosomes (vesicles that contain particles that have been engulfed via phagocytosis) to the lysosomes of the cell, to which the phagosomes fuse (I). Microtubules form the spindle apparatus that is an essential part of both mitosis and meiosis (II and III). The general structure of a microtubule is shown below; note that these structures are composed of dimers of the protein tubulin.

If true, which of the following findings would most directly challenge the results of this study? A. ALDH expression correlates with the expression of other determinants of tumor invasiveness. B. Targeting a different pathway, instead of ALDH, yields better improvements in platinum-sensitivity. C. No significant relationship is found between ALDH expression levels and clinical behavior in non-carboplatin-resistant ovarian cancer cell lines. D. ALDH overexpression results in no significant differences in platinum resistance compared to control cells.

D.ALDH overexpression results in no significant differences in platinum resistance compared to control cells. Figures 1 and 2 imply a dose-dependent relationship between ALDH expression and platinum resistance, which would be directly challenged by this finding.

According to the experimental procedure, which of the following describes the physical properties of indium tin oxide? I. Opaque II. Electrically conducting III. Solid at standard temperature A. II only B. III only C. I and III only D. II and III only

D.II and III only The second paragraph of the passage states, "The solution was then spread on the conductive faces of two transparent glass plates coated with indium tin oxide (ITO)." The word "conductive" refers to electrical conductivity, and since the device is hooked up to a current generator, RN II is correct (eliminate B and C). Additionally, the plates are described as solid objects. Furthermore, both indium and tin are metals, and the vast majority of metal oxides are solids at standard temperature. Unless told otherwise by the MCAT, we can assume that experiments take place at or near standard conditions, making RN III correct (eliminate A).

Suppose that an unusually high degree of aurora kinase activity is observed in the cervix. What could we reasonably conclude about C5M treatment efficacy across all cervical cancer cell lines, using this new information and the data within the passage? A. C5M will significantly inhibit mitotic progression in cancerous cervical cells. B. C5M will significantly inhibit mitotic progression in cancerous cervical cells, but only if p53 gene expression is knocked out. C. C5M will not significantly inhibit mitotic progression in cancerous cervical cells. D. Insufficient information is available to make conclusions about C5M treatment efficacy.

D.Insufficient information is available to make conclusions about C5M treatment efficacy. From Table 1, it can be seen that C5M has a low IC50 in cells from the HeLa cell line, which are cervical cancer cells. This is a promising sign that C5M could be used as a cancer therapy to reduce mitotic progression. However, note also that there is differential efficacy in Mahlavu and Focus cell lines, which also represent cancer cells (although from a different organ). Therefore, it cannot be concluded that C5M will be effective across all cervical cancer cell lines, as promising results from only one cell line are demonstrated.

Assuming that ECP was named for its electrical charge at physiological pH, which of the following must be true? A. The primary structure of ECP contains more acidic residues than uncharged residues. B. The primary structure of ECP contains more basic residues than uncharged residues. C. The primary structure of ECP contains more acidic residues than basic residues. D. The primary structure of ECP contains more basic residues than acidic residues.

D.The primary structure of ECP contains more basic residues than acidic residues. The passage tells us that ECP is the abbreviation for eosinophil cationic protein. From the question stem, we can assume that ECP must be cationic at physiological pH (7.4). At this pH, a typical protein will have a -1 charge on its deprotonated carboxylic acid terminal and a +1 charge on its protonated amino terminal, for a total of 0 net charge from its termini. If ECP is cationic, then, its positive charge must be due to its side chains. Only basic amino acids have the potential to become positively charged; in contrast, acidic amino acids have the ability to become negatively charged. In fact, virtually any acidic residue should have a -1 charge at a pH of 7.4, so for ECP to have a positive net charge, it must have more basic residues than acidic ones.

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is 0.25. When the block is halfway down the ramp, the child pushes down on the block perpendicular to the plane, halting it. What is the minimum force the child must apply to keep the block from starting to slide down the ramp? A. mg sin θ B. 0.25 mg cos θ + mg sin θ C. mg sin θ - 0.25 mg cos θ D. [(mg sin θ) / 0.25] - mg cos θ

D.[(mg sin θ) / 0.25] - mg cos θ The gravitational force pulling the block down the ramp is mg sin θ. (This is the typical value for the gravitational force that acts on an object to drag it down an incline.) To stop the block from sliding down the ramp, we must have an equal and opposite frictional force. Since these forces are equal and opposite, we can set them equal to each other as follows: Ff = mg sin θ Now, remember that frictional force is equal to the product of the appropriate coefficient of friction and the normal force: Ff = μFN Ff = 0.25 x FN = 0.25 mg cos θ The block itself has a mass m and thus generates a normal force of FN = mg cos θ. Again, this is the standard value for the normal force on an object positioned on an inclined plane. From this information alone, we may be tempted to pick C, which is the difference between the gravitational force and the frictional force. If the child were pushing the block upwards along the plane in a parallel fashion, this choice would be correct. However, the child is actually pushing down on the car, perpendicular to the plane. Thus, the force exerted by the child (Fa) will add to the force created by the mass of the car itself and alter the value for the normal force. Our total FN = mg cos θ + Fa. Substituting, we get: Ff = 0.25 x (mg cos θ + Fa) = mg sin θ mg cos θ + Fa = (mg sin θ) / 0.25 Thus, the force with which the child must push down on the car is Fa = [(mg sin θ) / 0.25] - mg cos θ. Graphically, we can set this up by first visualizing the problem:

When normal human cells are grown in culture, they will divide a limited number of times - typically 50 rounds of mitosis. After this number is reached, the cells become apoptotic. This cell death is a result of: A. decreasing number of membrane-bound organelles per cell. B. decreasing number of non-membrane-bound organelles per cell. C. decreasing levels of growth hormone. D. chromosomal telomeres shortening after each round of division.

D.chromosomal telomeres shortening after each round of division. As cells undergo mitosis, the telomeres, the portion of the DNA on the ends of the chromosomes, gets progressively shorter. Eventually the DNA loses its telomeres entirely and is unable to reproduce. The cell then dies, undergoing apoptosis. The position of telomeres on the chromosome is shown below.

Positron Emission Tomography (PET) scans follow the movement of a radioactively labeled compound throughout the body and are often used to detect metabolic activity in cancer cells relative to normal cells. The labeled compound is most likely: A. pyruvate. B. acetyl-CoA. C. ATP. D. glucose.

D.glucose. The question asks us to determine which compound would be tracked by PET and identify cancers. The passage states that cancerous cells have a very high rate of glycolysis relative to normal cells. PET scans using labeled glucose would allow for the visualization and easy comparison of glucose movement towards cancerous cells.

The enzyme listed in step 1 of the retinol synthesis listed is most likely classified as a(n): A. transferase. B. lyase. C. isomerase. D. oxidoreductase.

D.oxidoreductase. The first step shows beta-carotene being oxidized. Thus, the monooxygenase listed can best be classified as an oxidoreductase.

Based on the passage, all of the following were objections philologists had to comparativists EXCEPT: A. the lack of focus on principles that allowed new understandings of language. B. the focus on comparison as an end in and of itself. C. the belief that comparison is the primary method of obtaining information. D. the inclusion of comparative grammar in the study of language.

D.the inclusion of comparative grammar in the study of language. At the end of the passage, the author describes a "linguistics which included comparative grammar and gave it a new direction," suggesting that using comparative grammar in and of itself was not a major problem. Comparative grammar is also never mentioned as something the philologists did not want to include in their studies.