Next Step Full Length Missed Questions

Question 7 Why did the researchers choose to study pediatric, rather than adult, thyroid cancer cases? A. Thyroid cancer is not otherwise present in children. B. Thyroid cancer is not otherwise present in adults. C. Children receive a higher relative dose of I-131 at the same contamination level. D. Children receive a lower relative dose of I-131 at the same contamination level.

C is correct. The most likely reason for the use of pediatric cancer cases is that these cases are more likely to be caused by I-131 than those in adults. Because children weigh less than adults, the same quantity of external I-131 would result in a higher concentration of the isotope in the body, making DNA damage and cancer more likely. A, B: These choices are extreme. From the passage alone, we have no reason to assume that thyroid cancers are completely absent in children or adults who have not been exposed to I-131. D: Because children weigh less than adults, they would actually receive a higher relative dose of I-131 from the surrounding environment. Content Foundations: Cancer While knowledge about specific diseases is generally not required for MCAT success, there are notable exceptions to this rule. One such exception is cancer, which can be summarized as abnormal gene expression. To make sense of this statement, let's review the steps involved in cancer development, or oncogenesis. The term "tumor" describes any abnormal proliferation of cells. Benign tumors remain localized, whereas malignant tumors (which are what the term "cancer" properly refers to) can invade other organs and tissues in the body in a process called metastasis. The first step in oncogenesis, tumor initiation, involves changes that allow a single cell to proliferate abnormally. This means that the cell must develop the ability to bypass regulatory steps of the cell cycle that normally help to limit mitotic proliferation. Tumor progression occurs as a cell develops the ability to proliferate even more aggressively, such that its descendants are selected for and come to predominate the growing tumor. In addition, malignant cells often undergo mutations that promote their own growth and the development of blood vessels to feed them (angiogenesis). Oncogenesis is most often associated with mutations that occur by random chance (and elude the normal DNA repair machinery in the cell) or as a result of mutagenic compounds known as mutagens or carcinogens. (Examples of mutagens include ultraviolet light and certain chemicals, such as reactive oxygen species.) These mutations alter the functionality of crucial genes in the cell. However, oncogenesis is also associated with dysregulation of gene expression, as the abnormally elevated expression of genes involved in growth and proliferation can help contribute to the development of a tumor. The genes involved in oncogenesis can be divided into two groups: oncogenes and tumor suppressor genes. The basic difference between them is that oncogenes function to promote abnormal growth and proliferation, leading to cancer, while tumor suppressor genes function to prevent tumorigenic properties. Oncogenes can arise from the mutation of other genes, termed proto-oncogenes. If not mutated, proto-oncogenes do not promote cancer, but certain mutations or inappropriately elevated gene expression can effectively turn them into oncogenes.

Question 20 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 is correct. 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: √(300^2 + 700^2) = √(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). A: This value is just the turning/acceleration force, not the entire magnitude of force experienced by the foot. B: This is just the normal force; it fails to take into account the turning/acceleration of the man. D: This is greater than the sum of the two force components, which means the two force vectors (normal force and turning acceleration) cannot possibly add up to this value. Content Foundations: Newton's Laws The MCAT expects you to understand Newton's three revolutionary laws of motion and to be aware of how to apply them in problem-solving contexts. Newton's first law is a statement about inertia. It states that within a reference frame, an object remains at rest or at a constant velocity unless an external force acts upon it. Essentially, this law captures the insight that forces of resistance and friction make moving objects slow down and stop. It can be summarized as Fnet = 0 at equilibrium. Newton's second law defines force. It states that the total sum of forces acting on an object is equivalent to its mass times its acceleration. This is the familiar equation Fnet = ma. Newton's third law is about how forces come in pairs. It states that when body A exerts a force on body B, body B exerts an equal and opposite force on body A: FAB = −FBA. For example, the earth exerts a gravitational force on your body, which is responsible for your weight, but your body also exerts a gravitational force on the earth. In contrast to a common misconception, weight (caused by a gravitational force) and the normal force do not form a Newton's third law pair, because they are caused by different underlying forces and do not always have to be equal.

Question 52 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.

Difficulty: 2 Medium Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 5E Kinetics and Bioenergetics C is correct. When determining the conditions under which a reaction is spontaneous, always consider the change in Gibbs free energy. If the ΔG for a reaction is positive, it is nonspontaneous, and if the ΔG for a reaction is negative, it is spontaneous. The ΔG for a reaction can be determined from the change in enthalpy and entropy based on the equation ΔG = ΔH - TΔS. Based on Table 1, the ΔH 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 ΔS), which will favor a positive ΔG and a nonspontaneous reaction. So, if the magnitude of ΔH is greater than TΔS (which occurs at low temperatures), then the reaction is spontaneous. If the magnitude of ΔH is smaller than TΔS (which occurs at high temperatures), then the reaction is nonspontaneous. For reference, the relationship between ΔH, ΔS, and reaction spontaneity is summarized in the table below. A: For this to be true, ΔS would need to be positive, which is not the case here. B: This is the exact reverse of the correct answer. D: For this to be true, ΔH would need to be positive, which is not the case here. Content Foundations: Spontaneity A spontaneous reaction is one that can proceed without the help of some additional, external force. Importantly, spontaneity does not relate to the rate of the reaction, or the speed at which it progresses—spontaneous reactions can be very slow! Rate is a kinetic parameter that is determined largely by the activation energy. In contrast, spontaneity is a thermodynamic parameter and relates to ∆G, or the change in Gibbs free energy. The most basic definition of spontaneity is ∆G < 0, meaning that the reaction has a negative change in Gibbs free energy. Spontaneity is associated with exothermic (∆H < 0) reactions and those that increase entropy (∆S > 0) through the equation ∆G = ∆H - T∆S. Note that a reaction can still be spontaneous even if it fulfills only one of these requirements. For example, a spontaneous reaction may be endothermic (∆H > 0) as long as this is balanced by a sufficiently positive ∆S. Spontaneity can also be defined in terms of other chemical concepts. By definition, a spontaneous reaction will lead to more products being present than reactants. This means that the equilibrium constant (Keq), which is broadly defined as [products]/[reactants], will be greater than 1 for a spontaneous reaction. This relationship is encoded in the equation ∆G°rxn = −RTlnKeq. Additionally, in the context of electrochemical cells, spontaneous reactions are associated with positive cell potentials (E° > 0). On Test Day, it is important to automatically recognize that spontaneity is equivalent to ∆G < 0, Keq > 1, and E° > 0.

Question 43 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.

Difficulty: 2 Medium Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 5E Kinetics and Bioenergetics B is correct. 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, ΔG° for the transfer of two electrons from NADH to oxygen is -52.5 kcal/mol, and ΔG° 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, there is 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. A: We can eliminate this option immediately because the "removal of excess hydrogens from the matrix" constitutes useful work, as it would help establish the proton gradient required for oxidative phosphorylation. If the excess energy were useful, we would not call the conversion 50% efficient, since all energy would be going toward productive cellular processes. C, D: These choices incorrectly state that the energy mentioned in the question stem is insufficient to produce three ATP molecules.

Question 19 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.

Difficulty: 3 Hard Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 1A Amino Acids and Proteins D is correct. 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, B: Uncharged residues have side chains that do not become either positive or negative. The number of these residues, then, has no impact on the overall charge of the protein. Regarding choice B in particular, it is certainly possible that ECP contains more basic residues than uncharged residues. However, the question stem asks for a statement that "must" be true. Choice B does not necessarily need to be true; a protein could contain many fewer basic residues than uncharged residues and still be cationic. C: This is the opposite of the correct answer and would be expected to lead to an anionic protein. Content Foundations: Charge of Amino Acids Free amino acids contain a minimum of two groups that can become charged. The first is the carboxylic acid terminal, which exists as neutral -COOH when protonated and negatively-charged -COO− when deprotonated. The second is the amino terminal, which becomes the positively-charged -NH3+ when protonated and the neutral -NH2 when deprotonated. Note that "protonated" is not always synonymous with "positive." A protonated position can be either positive or neutral, depending on the group in question. Similarly, a deprotonated position can be either negative or neutral. Amino acids with ionizable side chains contain a third group that can carry a charge. For the two amino acids that are commonly known as acidic (glutamic and aspartic acid), this group is a carboxylic acid. For this reason, acidic amino acids contain one group that can become positive (the amino terminal) and two groups that can become negative (the side chain and terminal carboxylic acid groups). For the three basic amino acids (histidine, lysine, and arginine), the ionizable side chain group is a basic, nitrogen-containing functional group. Like the backbone amino terminal, these groups become positively-charged when protonated and neutral when deprotonated. As such, basic amino acids contain one group that can become negative (the carboxylic acid terminal) and two groups that can become positive (the terminal amine and nitrogen-containing side chain). Importantly, not all nitrogen-containing functional groups are basic. Asparagine and glutamine both contain nitrogenous side chains (specifically, amides) but are not considered basic because amides are not readily protonated. The charge pattern of amino acids changes in response to the surrounding pH. Let's consider a simple amino acid, alanine, which has a neutral side chain. At an extremely low pH, both the carboxylic acid group and the amine group will be protonated (-COOH and -NH3+, respectively), resulting in a net charge for the amino acid of +1. At an extremely high pH, very few H+ ions will be available, and both groups will be deprotonated (-COO− and -NH2), resulting in a net charge of −1. To understand what happens at an intermediate pH, consider the relevant functional groups. Carboxylic acids are weak acids, whereas amine groups are weakly basic. This means that -COOH groups are relatively "happy" to give up their extra proton and exist as carboxylate ions (-COO−), even at relatively low pH values where there is a fair amount of H+ present. In other words, a huge amount of H+ must be present to force the carboxylate ions to accept a proton. In contrast, amine groups "like" being protonated, which means that the pH must be quite high for it to be willing to give up its proton. Thus, for intermediate pH ranges, the carboxylic acid will be deprotonated (-COO−) and the amine group will be protonated (NH3+), resulting in a net charge of zero. This is the case at physiological pH.

Question 7 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.

Difficulty: 2 Medium Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 2C Reproduction D is correct. 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. A: While this choice looks tempting, we cannot conclude that it is true for certain. Note that the question is asking us to make an extreme evaluation ("what could we conclude...about C5M treatment efficacy across all cervical cancer cell lines"). Extreme conclusions are very difficult to draw decisively; after all, there could be numerous cervical cancer cell lines that we know absolutely nothing about. B: p53-mediated effectiveness of C5M was seen only in colon cells (Hct116). It is unclear whether this also occurs in cervical cells, given the information in the passage. C: C5M could inhibit mitotic progression in cervical cells. Table 1 shows promising results with one cervical cancer cell line, but more data would need to be collected to increase confidence that it is a promising therapy in cervical cancers.

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 is correct. 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. This eliminates all options except for A and D. 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. B: The boiling chip itself does not alter the boiling point of the liquid. C: This has the functions reversed. D: The vacuum apparatus removes the atmospheric pressure, not the distillate itself. The distillate simply boils more rapidly as a result of the reduced pressure. Content Foundations: Separations Separation techniques are widely used in organic chemistry to prepare purified substances for analysis or reaction. The best technique for a given scenario often depends on the phases of the substances being separated. If all are liquids, one may be able to utilize distillation, which aims to separate liquids by utilizing the difference between their boiling points. The liquids are initially held in the same round-bottom flask, termed the distilling flask. This flask is positioned above a heat source, typically a Bunsen burner. The top of the flask is connected to a column, which leads to a downward-sloping glass condenser over a receiving flask. The condenser is held within a glass casing through which cold water is pumped. As the round-bottom flask is heated, the liquid with the lower boiling point will begin to vaporize, and its vapor will travel up the column and re-condense to fall into the receiving flask. The eventual result is a receiving flask that is rich in the lower-BP liquid, while the distilling flask will still contain the liquid(s) with the higher BP. If boiling points are very high, a vacuum may also be used to lower atmospheric pressure, which lowers the boiling points of all substances involved. Recrystallization is used to purify a solid product that contains impurities. This process involves the dissolution of the solid in a solvent and subsequent heating. The solid then dissolves and is cooled, causing it to solidify (crystallize) again. As the lattice structures of solids tend to exclude impurities, each subsequent recrystallization results in a progressively purer compound. Chromatography is a broad set of separatory techniques based on relative affinity, or tendency for a compound to attract to a certain solvent or structure. Specifically, sample molecules vary in their affinities for a mobile (moving, typically solvent-based) phase versus a stationary (static) phase. In column chromatography, the stationary phase is a vertical column packed with an adsorbent with carefully chosen properties. This adsorbent can attract sample molecules based on charge, size, or affinity for specific ligands. Finally, centrifugation utilizes a rapidly spinning apparatus to separate particles by density. More dense particles, such as cells, gravitate toward the bottom of the spun tube, while less dense substances remain at the top in a liquid termed the supernatant. This liquid can then be poured off, and further separation or analysis can be conducted.

Question 6 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 is correct. Since each answer has multiple parts, we should divide and conquer, eliminating wrong choices as we go. 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, which makes them biologically hazardous (eliminate C and D). 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). B: Te-131 would be formed if I-131 underwent positron decay or electron capture, but the passage indicates that it undergoes beta-minus decay. C, D: Again, gamma rays constitute ionizing radiation, not non-ionizing radiation. Content Foundations: Radioactive Decay and Half-life Radioactive decay is the spontaneous transformation of one atomic nucleus into another. Often, this involves a change from one element into a different element. The only way this transformation can happen is by changing the number of protons (and often neutrons) in the nucleus, since atomic identity is defined by the number of protons. There are a number of ways that this can happen, and when it does, the atom is forever changed. There is no going back - the process is irreversible. There are four primary types of decay: alpha decay, beta decay (β+ and β−), gamma decay, and electron capture. In alpha decay, an alpha particle, containing two protons, two neutrons, and a +2 charge, is emitted. In beta-minus decay, a neutron is converted into a proton in the nucleus, and a β− particle (an electron) is ejected to maintain charge balance. In beta-plus decay, a proton is converted into a neutron, and a β+ particle (a positron) is emitted to preserve charge. Gamma decay involves the emission of a gamma ray, which is a high-energy photon, from an excited nucleus. Finally, in electron capture, a nucleus "grabs" an electron, which changes a proton into a neutron. When an atom does not gain or lose any protons, and only the neutron count is changed, a new isotope is formed. An isotope is a variety of an element that is distinguished from other varieties of the same element by having a different atomic mass; however, isotopes of a given element all share the same atomic number and chemical properties. These isotopes are often used in radiolabeling techniques in the biological sciences (for example, using 2H, or deuterium (D), to track amino acid uptake in protein translation). Another high-yield fundamental concept is half-life (t1/2), which is the time required for one-half of the parent isotopes in a sample to decay into daughter (radiogenic) isotopes. If we know the half-life of a material, then we can determine how much of a sample is lost (1- ½n) or remains (½n) at any given time, as expressed in the number of half-lives that have passed (n).

Question 5 The student in charge of the experiment wishes to present his/her findings in support of the theory that motor imagery and action observation together facilitate a greater increase in motor excitability than either protocol alone. Which of the following would be the best reason to withhold this presentation? A. The procedure used did not include MEP recordings prior to each task. B. MEP amplitudes in an individual are typically highly consistent. C. The motor tasks performed in the experiment were too simple. D. The six different conditions were run in random order.

A is correct. The idea of controls is central to the testing of any medical intervention. Without a baseline value for MEPs (the control), the researchers cannot be sure that the values observed post-protocol were significantly different than what they were prior to the activity, imagined or otherwise. B: If MEPs are typically very consistent in individuals, then this would strengthen the case that the activities performed may explain the observed change in MEP amplitude. C: The motor task (opening and closing of the hand) was fairly simple, but the student wishes to discuss the effect of the observation and imagery on motor excitability. In addition, the muscle tested was performing an anatomical task, which should provide sufficient activity to bring about any change, if possible. D: Randomizing the order of the six conditions would help to reduce any confounding effects that a previously performed task might have on later results and make the group analysis results stronger. Content Foundations: Experimental Design and Variables Depending on what researchers are studying, they can either employ a cross-sectional design or an experimental design. Cross-sectional studies investigate a population at a single point in time, looking for predictive relationships among variables. A limitation of cross-sectional designs is that they can show correlations, but not causation, because looking at changes over time is necessary to assess whether a cause-and-effect relationship is present. An experimental design involves manipulating a certain variable—known as an independent variable—to see what effects it has. The measured effects are known as dependent variables. Additional types of study design exist, especially in the social sciences; of note, qualitative research focuses on analyzing experiences rather than objective metrics. In any experiment, it is important to include controls. Negative controls are treatments that are known to have no effect, and positive controls are treatments that are known to have a certain effect and can therefore be used to assess whether the experimental methodology was sound. In clinical contexts, experimental studies often use a double-blind design in which neither the subjects nor the researchers know who is receiving the treatment and who is receiving the control. In experimental design, it is important to account for confounding variables, which are external variables affecting both the independent and dependent variable. For instance, if a study analyzes the effects of stereotype bias on math performance by race, but fails to account for socioeconomic status (SES), its results may be suspect, because SES may predict math performance and be correlated with race. Moderating variables attenuate or strengthen a given relationship, and mediating variables provide an important logical link between an independent variable and a dependent variable, or outcome. Studies on humans must follow the ethical principle of informed consent, in which the subject is adequately informed about the nature of the process, and then is mentally and legally competent to give consent to it. Sample size (usually denoted with N) is simply the number of data points developed in an experiment. In the social sciences and biology, this usually means the number of people or number of organisms in the experiment. The larger the sample size, the higher the statistical power of the experiment. While more is always better, it is usually impossible to test every possible case or every possible person, so a smaller sample of the whole must be taken. Non-random sampling can introduce important sources of bias, which can limit the validity of the study.

48. Which of the following cations will form most quickly from deprotonation? See question A. CH3O substituent B. CH3 substituent C. F substituent D. H substituent

A is correct. The rate of cation formation through deprotonation is directly related to the energy required to deprotonate the compound. The more energy required, the slower the rate. This energy can be found from the column "△Hf for X → X+" in Table 1. The four cations being compared come from compounds 1-4, and of those four, compound 1 has the lowest "△Hf for X → X+," with a value of 181.8 J/mol. B: This compound has a CH3 substituent, meaning that it corresponds to compound 2 in Table 1. Compound 2 has an △Hf for X → X+ of 185.3 J/mol, which is greater than 181.8 J/mol. C: This compound has an F substituent, meaning that it corresponds to compound 4 in Table 1. Compound 4 has an △Hf for X → X+ of 191.0 J/mol, which is greater than 181.8 J/mol. D: This compound has an H substituent, meaning that it corresponds to compound 3 in Table 1. Compound 3 has an △Hf for X → X+ of 188.2 J/mol, which is greater than 181.8 J/mol.

Question 45 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

Difficulty: 2 Medium Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 4E Atomic Structure and Stoichiometry A is correct. 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. B: We can eliminate this option right away if we recognize that XeF4 and NH3 (two very different molecules) do not display the same orbital hybridization. C: XeF4 does not have an orbital hybridization of sp3. Additionally, NH3 does not have a hybridization of sp2; sp2 hybridization would better describe a molecule with three bonded regions and no lone pairs on the central atom, such as acetone. D: Again, NH3 does not have an orbital hybridization of sp2. Content Foundations: Orbital Hybridization Electrons exist in the regions around atoms in characteristic shapes depending on the angular momentum quantum number of their subshell: s, p, d, or f. The most relevant subshells for MCAT chemistry are the s subshell, which is spherical, and the p subshell, which has two lobes and a central node. When atoms combine to form a molecule, their atomic orbitals overlap to produce molecular orbitals. A single bond—consisting of two electrons—between two atoms will form a sigma (σ) bond, which has an overlapping region of electron density. Pi (π) bonds occur between two parallel p orbitals and are weaker than σ bonds. Double bonds consist of one π bond and one σ bond, while triple bonds include two π bonds and one σ bond. However, the orbitals of an atom hybridize so that they are homogenous. For example, on paper, the central carbon in methane (CH4) has four valence electrons and would appear to have one s orbital and three p orbitals. In reality, however, these orbitals combine, or hybridize, to produce four identical sp3 orbitals. Hybridization between one s orbital and two p orbitals may also occur to produce three sp2 orbitals, or between one s orbital and one p orbital to produce two sp orbitals. A quick guide to identifying the orbital hybridization of a molecule is to determine the number of regions of electron density around the atom. A region of electron density is defined as either a bond (single, double, or triple) or a lone pair of electrons. Two regions yield a hybridization of sp (common in cases of triple bonds or central atoms with two double bonds). Having three regions of electron density is associated with sp2 hybridization, and having four regions yields sp3 hybridization.

1. What is the approximate molarity of sodium chloride in ocean water, if the density of ocean water is 1.028 kg/L? A. 0.026 M B. 0.62 M C. 0.96 M D. 9.6 M

B is correct. 1 L of ocean water has a mass of 1028 g, of which 3.5% is salt. Note that the molar mass of NaCl is approximately 58 g. (1028 g)(0.035) = 36 g NaCl (36 g NaCl) / (58 g NaCl/mol) = 0.62 moles 0.62 moles / 1 L = 0.62 M A, C, D: These answer choices result from miscalculation. Content Foundations: Molarity The concentration of a solution can be expressed in multiple ways, the most frequent of which on the MCAT is molarity (M), which is the number of moles of solute per liter of solution: 1 M = 1 mol/L. A mole refers to a number of items equal to Avogadro's number (6.02 × 1023), such that a 1 M solution of glucose contains 6.02 × 1023 molecules of glucose in each liter of solution. A consequence of the definition of molarity as moles per liter is that molarity times volume (in liters) equals the number of moles of a substance that are present (mol = M × V). In biological processes, it is often the case that very small amounts of a substance are present in a solution, so you may encounter millimolar (mM), micromolar (μM), and nanomolar (nM) concentrations, which simply refer to 1 × 10−3 M, 1 × 10−6 M, and 1 × 10−9 M concentrations, respectively. A related quantity, normality (N), is the number of equivalents of reactive species per liter of solution, for which we must define the reactive species. Normality is often used to express the concentration of H+ or OH− ions produced in acid-base reactions. For example, hydrochloric acid (HCl) generates one equivalent of H+ ions and one equivalent of Cl− ions per mole, while sulfuric acid (H2SO4) generates two equivalents of H+ ions and one equivalent of SO42− ions per mole.

Question 57 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 is correct. 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. Homo sapiens and fungi are eukaryotes with nuclei. A: RN IV is also correct. B: RN I and RN IV are correct, while RN III is incorrect. D: RN III is incorrect. Content Foundations: Prokaryotes and Eukaryotes The division between prokaryotes and eukaryotes (which include both single-celled organisms such as yeast and multicellular organisms such as ourselves) is arguably the most basic fork in the tree of life. Prokaryotes (including bacteria and Archaea) are defined by the absence of a nucleus and membrane-bound organelles. Bacteria are ubiquitous and play a major role in the maintenance of human health. Most bacteria in the human body are harmless or beneficial, but some are pathogens. Although bacteria do have specific genus and species names, they are commonly described in terms of their shape. Spherical bacteria are known as cocci, rod-shaped bacteria are called bacilli, and spiral-shaped bacteria are known as spirilli. Bacteria are also classified in terms of how they use oxygen in metabolism. Bacteria that do not require oxygen for metabolism are known as anaerobes. For obligate anaerobes, oxygen is toxic. Aerotolerant anaerobes are similar to obligate anaerobes in that they cannot engage in aerobic metabolism, but oxygen is not toxic for them. Facultative anaerobes can engage in either aerobic or anaerobic metabolism, depending on the circumstances. Bacteria that require oxygen for metabolism are known as obligate aerobes. Since bacteria by definition do not have membrane-bound organelles, their structure is simpler than that of eukaryotes. However, they have some key structures, including a cell wall containing peptidoglycan that encloses the cell membrane, ribosomes that are differently sized from those of eukaryotes (30S and 50S subunits, instead of the 40S and 60S components in eukaryotes), rotating flagella, and a single circular chromosome. Structural differences between eukaryotes and prokaryotes are often specifically targeted by antibiotics.

Which steps in Figure 3 are oxidations? 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 is correct. 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. A: RN III is also correct. B: RN I is also correct. D: RN II is incorrect. Content Foundations: Reduction and Oxidation in Organic Chemistry and Biochemistry Reduction and oxidation (redox) reactions are often studied in the context of electrochemistry, generally focusing on reactions involving ionic compounds. However, reduction is defined as a decrease in the oxidation state of an atom, while oxidation is defined as an increase in the oxidation state of an atom, and these definitions can also be applied to reactions containing organic compounds. The following features are associated with reduction in organic chemistry: (1) gain of an electron, (2) decreased oxidation state, (3) formation of a C-H bond (e.g. alkene → alkane), and (4) loss of a C-O or C-N bond (or any bond between carbon and an electronegative atom). Conversely, oxidation is associated with (1) loss of an electron, (2) increased oxidation state, (3) loss of a C-H bond (e.g. alkane → alkene), and (4) gain of a C-O or C-N bond (or any bond between carbon and a highly electronegative atom). Oxygen-containing organic compounds exist on a spectrum of oxidation from alcohols (most reduced/least oxidized) to aldehydes/ketones (intermediate reduction/oxidation) to carboxylic acids (least reduced/most oxidized). A primary alcohol can be oxidized to an aldehyde by a mild oxidizing agent (such as PCC) or to a carboxylic acid by a strong oxidizing agent like NaCr2O7. A secondary alcohol will be oxidized to a ketone by either a mild or a strong oxidizing agent. A strong oxidizing agent will likewise oxidize an aldehyde to a carboxylic acid. A strong reducing agent, such as LiAlH4, can reduce a carboxylic acid directly to an alcohol, while weak reducing agents such as NaBH4 will not reduce carboxylic acids at all. A special agent, DIBAL, can reduce a carboxylic acid to an aldehyde when applied at a precise 1:1 ratio. Both mild and strong reducing agents can reduce aldehydes and ketones to primary and secondary alcohols, respectively.

Question 52 Which of the following types of motivation do the educators in the study hope to improve through the use of case-based team learning? A. Intrinsic motivation B. Extrinsic motivation C. Drive-reduction motivation D. Amotivation

Difficulty: 2 Medium Reasoning Skill: 4 Data-based and Statistical Reasoning Concept Category: 7A Biology and Psychology of Behavior A is correct. The point of case-based team learning is to boost inner motivation to succeed and learn to be a good physician (also known as intrinsic motivation). In the passage, the students have to learn concepts on their own so that they are better able to think critically in class and contribute to their learning environment. B: Extrinsic motivation is motivation to do something based on an external reward (like money or fame). C: Drive-reduction motivation is motivation based on the need to fulfill a certain drive, like hunger or thirst. D: Amotivation, as one might guess, is the lack of motivation. Content Foundations: Motivation Motivation refers to the driving force or reasoning behind our actions and behaviors. Motivation can be broadly divided into two types: extrinsic motivations are created by external forces, while intrinsic motivations are created by internal forces. These forces include attitudes, or ways of thinking or feeling about people, places, and things that are reflected in our actions and our behaviors. Our motivations and behaviors can be shaped by several factors. Instincts are innate, unlearned, and usually fixed patterns of behavior that are, in general, present in all members of a species. One example found in humans is the instinct to suckle: babies do not need to be taught how to suckle milk from their mothers or bottles. Additionally, arousal describes how even when all instincts are fulfilled, people are still motivated to do things, sometimes out of boredom and sometimes out of curiosity. That is, some behaviors are motivated by a desire to achieve an optimal level of arousal (the physiological state of being reactive to stimuli). If a person is not stimulated enough and is below their optimal level of arousal, their desire for arousal may motivate an action, decision, or behavior. A drive is an urge that results from an urge to reach a goal or satisfy a need. Basic drives stem from states of physiological need like hunger or thirst. In these basic cases, the drive would be to eat or drink. Such drives are known as primary drives and are ways that the body is alerted to being out of equilibrium. In contrast, secondary drives are not based on biological needs. Instead, they stem from learning and experiences. Secondary drives can include feelings such as love and aggression. Finally, needs are rooted in higher-level desires. For example, a young person may aspire to be a doctor not just for the financial reward, but out of as desire to help others.

Question 34 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.

Difficulty: 3 Hard Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 1C Classical Genetics C is correct. 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. A: A double-crossover event occurs in two different places along the chromosome arm, resulting in a segment exchange, not necessarily a reversal of the original crossover. B: Crossover events only occur during meiosis, not during mitosis. D: Both single- and double-crossover events will only affect one arm of each chromosome. Content Foundations: Mitosis and Meiosis In eukaryotes, the process of asexual cell division is known as mitosis. Mitosis takes place in four phases: prophase, metaphase, anaphase, and telophase. Prophase prepares the cell for mitosis: the DNA condenses such that distinct chromosomes become visible, as sister chromatids (or copies of a given chromosome) join at a region known as the centromere. The kinetochore assembles on the centromere, and is the site where microtubule fibers that extend from the centrosome and form the mitotic spindle attach to pull the sister chromatids apart in later stages of mitosis. Other microtubules known as asters extend from the centrosome to anchor it to the cell membrane. Additionally, the nuclear envelope and the nucleolus disappear, and the mitotic spindle forms. In metaphase, the chromosomes line up at the middle of the cell along an imaginary line that is known as the metaphase plate. In anaphase, the sister chromatids are separated and pulled to opposite sides of the cell by shortening of the microtubules attached to the kinetochores. Telophase can be thought of as the opposite of prophase, as a new nuclear envelope appears around each set of chromosomes and a nucleolus reappears within each of those nuclei. The process of mitosis is completed by cytokinesis. In contrast, meiosis is a form of cell division that is essential for sexual reproduction. It takes place in germ cells (also known as sex cells). Meiosis differs from mitosis in that it has two stages and results in the formation of four daughter cells, each of which has only one copy of each chromosome (haploid, n), in contrast to mitosis, which generates cells with two copies of each chromosome (diploid, 2n) that are essentially identical to their parent cell. In prophase I of meiosis, homologous chromosomes (i.e., the maternal and paternal copies of a given chromosome) pair up with each other in a process known as synapsis, forming tetrads. While paired up, homologous chromosomes may exchange genetic information in a process known as crossing over. The crossing-over points are known as chiasmata. This process results in recombinant DNA that is another source of variation in sexual reproduction, in addition to the variability inherent to the process. In metaphase I, homologous pairs, which take the form of tetrads, line up at the metaphase plate. The orientation of the homologous pairs is random in terms of which side of the metaphase plate the maternal or paternal copy of a given chromosome in a homologous pair winds up. In anaphase I, the homologous pairs are separated, and one member of each pair is pulled to each side of the cell. In meiosis II, which operates similarly to mitosis, the sister chromatids are split up into two haploid daughter cells.

A child is rolling his toy car (with mass = m) down a ramp. The coefficient of static friction between the car and the ramp is 0.25. When the car is halfway down the ramp, the child pushes down on the car perpendicular to the plane, halting it. What is the minimum force the child must apply to keep the car from starting to roll 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 is correct. The gravitational force pulling the car 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 car 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 = mg sin θ The car 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 car 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. A: This fails to account for friction in any way. Since the passage explicitly says that the coefficient of friction is 0.25, and the toy is experiencing friction with the ramp, we're going to need that 0.25 somewhere in our answer. B: This choice adds together the force of gravity (which pulls the toy down the ramp) and the force of friction (which pushes back up the ramp). Since these forces point in opposite directions, they shouldn't be added together. C: Again, this is the force the child would have to apply if he were pushing parallel to the ramp, directly up the ramp to oppose gravity. Always read the question carefully before beginning a problem, especially a calculation-based physics question.

Question 3 Which of the following processes is involved in the motor-evoked potential elicited by transcranial magnetic stimulation? I. Na+ influx II. Ca2+ release III. Cl- efflux A. I only B. II only C. II and III only D. I and II only

D is correct. With Roman numeral (RN) questions, you want to work smarter, not harder. Choose one of the RNs to evaluate first, then eliminate answers whenever possible. Paragraph 1 states that the TMS pulse causes depolarization of the motor cortex and an electrical and motor response in the muscle tissue. This indicates that the action potential travels from the brain to the muscle, where it causes a muscle response (e.g., contraction). With this information in mind, let's begin this question by examining RN II. Muscle activation does require release of calcium from the sarcoplasmic reticulum to cause a mechanical response in the muscle, so RN II is correct. Since II is true, we can eliminate choice A. Move onto RN I, which is also true; depolarization (shown below in the context of the entire action potential) is characterized by a rapid influx of Na+ into the neuron. Thus, the answer must be D, and we need not evaluate RN III at all. III: Chloride ion efflux is not involved in action potential generation or the activation of actin-myosin cross-links during muscle contraction. A: RN II is also correct. B: RN I is also correct. C: RN III is incorrect, while RN I is correct. Content Foundations: Action Potentials Action potentials are how neurons send signals to other neurons or to the neuromuscular junction. The mechanisms of action potentials involve the voltage potential difference across the neuronal cell membrane. The nerve cell maintains a resting potential of −70 mV. When the inside of the cell loses this negative potential, the cell is said to be depolarized, and when it drops to a more negative potential than −70 mV (after peaking at +40 mV in an action potential), it is said to be hyperpolarized. This resting potential is maintained by the constant action of sodium-potassium ATPase (the Na+/K+ pump), which pushes three sodium ions out of the cell for every two potassium ions it brings in. This electrochemical gradient is also maintained by the fact that the hydrophobic core of the plasma membrane does not allow ions to easily diffuse back across. If the excitatory stimulus is strong enough, and the cell's resting potential is brought up to −55 mV, the cell will undergo an action potential. In the first phase of the action potential, depolarization, the sodium voltage-gated channels open and Na+ ions rush into the cell. The sudden influx of positive charges continues until the cell membrane reaches full depolarization at +40 mV, at which point the sodium channels close and potassium voltage-gated channels open. Now that the interior of the cell is positive, the potassium is pushed by both electrical potential and its own concentration gradient to rush out, causing repolarization. Repolarization continues until the cell overshoots the −70 mV level, making the cell temporarily hyperpolarized, during what is called the refractory period. Then, the sodium-potassium pumps get back to work to re-establish the resting state of the cell. Action potentials begin at the axon hillock and move down the axon towards the synapse. This action potential can either move slowly and smoothly along the cell membrane (if the cell is unmyelinated) or can jump very rapidly down the axon from one node of Ranvier to the next (if the cell is myelinated). Nodes of Ranvier refer to gaps in the myelin sheath present in myelinated neurons.

Question 33 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.

Difficulty: 1 Easy Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 5D Biological and Organic Molecules D is correct. The first step shows beta-carotene being oxidized. Thus, the monooxygenase listed can best be classified as an oxidoreductase. A: Transferases move functional groups from one molecule to another (such as kinases that move phosphate groups onto their substrates). Here, the molecule is being oxidized in step 1. B: Lyases break molecules into two smaller molecules without using water or redox reactions. The beta-carotene is not split into smaller molecules in step 1. C: Isomerases convert a molecule from one isomer to another (including stereoisomers and constitutional isomers). Since a new atom is being added here, the reactant and product cannot possibly be isomers. Content Foundations: Key Enzyme Classifications The MCAT is not an exam of biochemistry minutiae (I know, I know, it can feel that way). However, they do expect you to know a wide variety of biochemistry fundamentals. One such high-yield area is enzyme classification. There are six primary classes of enzymes that can be used to describe all other enzymes. Oxidoreductases catalyze oxidation-reduction reactions where electrons are transferred. In metabolism, these electrons are usually in the form of hydride ions or hydrogen atoms. When a substrate is being oxidized, it is the hydrogen donor. Examples include reductases, oxidases, and dehydrogenases. Transferases catalyze transfer of a chemical group from one molecule (donor) to another (acceptor). Most of the time, the donor is a cofactor that is charged with the group about to be transferred. Examples include kinases and phosphorylases. Lyases catalyze reactions where functional groups are added to break bonds in molecules or they can be used to form new double bonds or rings by the removal of functional group(s). Decarboxylases are examples of lyases. Isomerases catalyze reactions that transfer functional groups within a molecule so that a new isomer is formed to allow for structural or geometric changes within a molecule. Hydrolases catalyze reactions that involve cleavage of a molecule using water (hydrolysis). This cases usually involves the transfer of functional groups to water. Hydrolases include amylases, proteases/peptidases, lipases, and phosphatases. Ligases are used in catalysis where two substrates are stitched together (i.e., ligated) via the formation of C-C, C-S, C-N or C-O bonds while giving off a water (condensation) molecule. Every enzyme you will ever see on the MCAT will fit under one of these labels. The test makers will not expect you to learn a bunch of random enzymes, but they will expect you to match an enzyme's name to the clues given about its function, or vice versa. Luckily, most enzymes are named for exactly what they do (e.g., pyruvate decarboxylase) and for the substrate they act upon (e.g., DNA ligase).

Question 58 Tucson, Arizona is well known for its sunny, dry weather. A 33ºC day in Tucson can feel relatively cool and pleasant compared to a similar temperature in a humid city like Tampa. The phenomenon of "dry heat" feeling subjectively cooler is a result of: A. decreased evaporation from the skin helping to keep the body cool. B. increased evaporation from the skin helping to keep the body cool. C. decreased cardiac output in response to dehydration lowering cutaneous perfusion. D. increased cardiac output in response to dehydration lowering cutaneous perfusion.

Difficulty: 1 Easy Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 5E Kinetics and Bioenergetics B is correct. One of the body's primary ways of regulating heat is through sweating. The evaporation of liquid dissipates body heat, helping to lower the body's temperature and keep it from overheating. In a dry climate, the low humidity allows water to evaporate more easily, increasing sweating. That increased sweating helps keep the body cool. A: This is the opposite of the correct answer. C, D: These are neither true nor relevant - nothing in the question suggests that people are dehydrated in one place more than the other. Content Foundations: Thermoregulation of the Body The body must maintain homeostasis of many parameters, including osmolality, pH, energy, and temperature. The maintenance of body temperature is known as thermoregulation. This process relies on multiple systems within the human body, including the endocrine, cardiovascular, integumentary, and respiratory systems. The anterior hypothalamus serves as the body's "thermostat" to maintain body core temperature. Heat exchange is determined by convection, conduction, evaporation, and radiation. Radiation, conduction, and convection are determined by the difference between the skin temperature and the ambient temperature. The rate of heat body heat loss depends primarily on the surface temperature of the skin, which is in turn a function of cutaneous blood flow. The body can lose heat by increasing cutaneous blood flow and sweating, or by decreasing the basal metabolic rate through thyroid signaling. It can gain heat by decreasing cutaneous blood flow, increasing muscular activity (through movement and shivering), increasing the basal metabolic rate through thyroid signaling, metabolizing brown adipose (in infants only), or triggering piloerection of the hair on the body (goosebumps). A forced change in body temperature results when an environmental stress is sufficient to overcome the thermoregulatory systems of the body. For example, prolonged time in cold water results in forced hypothermia, while prolonged time in hot water results in hyperthermia. A regulated change in body temperature occurs when the hypothalamic "temperature setting" is shifted, as when a fever results from infection by a pathogen.

Question 39 Which of the following is NOT an example of a demographic measure? A. Income B. Sexual orientation C. Education D. Absolute mobility

Difficulty: 1 Easy Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 9B Demographics D is correct. Demographics are statistics used to examine a population by quantifying subsets of that population. Absolute mobility means that living standards are increasing in absolute terms: you are better off than your parents and your children will be better off than you. Absolute mobility compares your income to your parent's income. If your parents make $5,000 dollars a year and you make $10,000 dollars a year, you have experienced absolute mobility. Absolute mobility is not an example of a demographic measure. A, B, C: Income, sexual orientation, and education level are examples of demographic measures. Content Foundations: Social Demographics Demography is the study of certain statistical parameters that illustrate how human populations change over time, making it essentially the mathematical branch of sociology. These statistical parameters are known as demographics and can be gathered either informally (through smaller- scale polls) or formally (through a census). Opportunities, attitudes, and outcomes can vary across several important demographic categories, including race, ethnicity, age, gender, sexual orientation, class, and socioeconomic status. The inclusion of class and socioeconomic status in this list points out the conceptual connection between demographics and social inequality. It is normal for the demographics of a society to shift over time. These shifts are most often brought about by changes in birth rates, death rates, and migration into and out of the society. Demographic transition is a specific category of demographic shifts, accounting for changes in fertility and mortality rates—and the consequences thereof—as a country develops from a preindustrial to an industrial, modern economic system.

Question 55 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.

Difficulty: 1 Easy Reasoning Skill: 4 Data-based and Statistical Reasoning Concept Category: 2A Cell Biology C is correct. 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. A: Looking at the trisomy 21 data in Figure 2, there appears to be no significant difference between OX-LDL- and LDL-treated cells. B: This is the opposite of the conclusion outlined above. LDL increases membrane cholesterol and thus increases membrane rigidity at moderate temperatures. (Note that the opposite is true at very low temperatures; cholesterol fills in the gaps between phospholipid tails, preventing the membrane from becoming too rigid.) D: The study did not include any cells that were exposed to both LDL and HDL at once, so we cannot predict what would happen in that case.

Question 58 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.

Difficulty: 2 Medium Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 1A Amino Acids and Proteins B is correct. 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. A: While changes in pH can denature proteins, this choice is wrong in reference to precipitating out of the plasma. That would create serious biological harm by potentially blocking circulatory vessels. Also, the buffering happens through controlling the pH, not through precipitation as the choice says. C, D: These choices are factually false. Content Foundations: Buffers In the body and in the lab, it is sometimes desirable to avoid the large shifts in pH that can result from the addition of acid or base to an existing solution. A buffer is a solution that resists changes in pH upon addition of acid or base. While buffers cannot protect against addition of large amounts of acid/base, they are highly effective at maintaining pH when small to moderate quantities are added. A buffer must contain either a weak acid and its conjugate base or a weak base and its conjugate acid. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log[A−]/[HA], where HA refers to a generic weak acid and A− refers to its conjugate weak base. The basic idea of a buffer is that if we add a small to moderate amount of a strong acid or base to the buffered solution, it will entirely protonate or deprotonate some of the weak acid/base. However, since the Henderson-Hasselbalch equation refers to the log of the [A−]/[HA] ratio, doing so will only minimally affect the pH. The bicarbonate buffer system (CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+) plays a major role in maintaining the pH of blood within the narrow range of roughly 7.35-7.45 that is compatible with health. It also encodes a key relationship for understanding the importance of respiration in the context of human metabolism: through Le Châtelier's principle, increasing the amount of CO2 in the blood (as occurs as a result of cellular respiration) pushes the bicarbonate reaction to the right, increasing the amount of H+ present in the blood and thereby reducing its pH. Thus, it is essential for the body to be able to breathe out CO2 and breathe in O2.

Question 59 Which of the following correctly lists a pair of analogous structures and a pair of homologous structures, respectively? A. The wing of a bee and the wing of a bird; the wing of a bird and the leg of a bird B. The wing of a bee and the wing of a bird; the arm of a human and the flipper of a walrus C. The arm of a human and the wing of a bat; the wing of a bird and the wing of a bat D. The wing of a bird and the wing of a bat; the wing of a bee and the wing of a bat

Difficulty: 2 Medium Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 1C Classical Genetics B is correct. Analogous structures are those structures that evolved independently to carry out the same function. Thus, the wing of a bee and the wing of a bird are analogous structures. Homologous structures are those that have a similar evolutionary history, arising from the same source, even if they now have different functions. The forelimbs of mammals (human arm, walrus flipper, bat wing) would all be homologous despite their different functions. Thus, this choice correctly indicates a pair of analogous structures and then a pair of homologous structures. A: The first set of structures listed here are analogous, but the second set (which come from the same type of organism) are neither analogous nor homologous. C: These are homologous and analogous structures, respectively. D: Both of these sets of structures are analogous. Content Foundations: Evolution In the 1973 words of the evolutionary biologist (and self-described creationist) Theodosius Dobzhansky, "nothing in biology makes sense except in the light of evolution." Natural selection is the mechanism through which evolution takes place. The concept of natural selection refers to the tendency of certain phenotypes to be favored in terms of reproduction. Natural selection is closely linked to the term fitness, which in the evolutionary context only refers to the chance of reproduction associated with a certain phenotype compared to baseline. Fitness must be defined in terms of specific environmental constraints. In the 20th century, the concept of group selection was proposed, arguing that natural selection could act on the level of the group, not the individual. The related concept of inclusive fitness expands the rigorous evolutionary definition of fitness (defined in terms of the differential reproduction of alleles) to account not just for individuals but also their relatives, who can be expected to share many of the same alleles. This idea helps to explain altruistic behavior. Additionally, multiple "types" of selection have been identified to describe the outcomes associated with different types of selective pressures on phenotypes that vary along a spectrum. Stabilizing selection occurs if both extremes are selected against, directional selection occurs if only one extreme phenotype is selected against and the other extreme is favored, and disruptive selection occurs when the median phenotype is selected against. The concept of Hardy-Weinberg equilibrium is often used to model stable gene pools that satisfy the following assumptions: (1) organisms must be diploid and reproduce sexually; (2) mating is random; (3) the population size is very large; (4) alleles are randomly distributed by sex; (5) no mutations occur; and (6) there is no migration into or out of the population. The Hardy-Weinberg equations allow us to use allele frequencies to predict the distribution of phenotypes in the population and vice versa. If p and q are the only two alleles of a gene present in the population, then p + q = 1. Squaring the equation yields: (p + q)2 = 12 → p2 + 2pq + q2 = 1. This second equation allows us to connect genotypes and phenotypes, because the p2 and q2 terms correspond to individuals homozygous for p and q, respectively, while the 2pq term gives the frequency of heterozygotes. Other mechanisms are associated with changes in the gene pool of a species, such as genetic drift and bottlenecks. Genetic drift refers to the role of chance, in the absence of strong selective pressures, in determining the reproductive fitness of various alleles. When no strong pressure exists for a certain allele, it may randomly happen to be reproduced more or less often. These random effects can add up over the course of evolution. A related, but more specific, concept is that of an evolutionary bottleneck. Bottlenecks occur when some external event dramatically reduces the size of a population in a way that is essentially random with regard to most, if not all, alleles. This dramatically reduces diversity in the gene pool.

Question 53 Through what form of social influence are new ideas most likely spread? A. Compliance B. Identification C. Internalization D. Minority influence

Difficulty: 2 Medium Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 7B Social Behavior D is correct. When a new idea arises, it is automatically a minority opinion. This idea can then be spread through the influence of the minority on others accepting this view. A, B, C: These may be mechanisms through which new ideas are spread, but not necessarily. These influence factors may promulgate old or new ideas. Content Foundations: Social Influences Social influence occurs when an individual's emotions, opinions, or behaviors are affected by others. Social influence takes many forms and can be seen in conformity, socialization, peer pressure, obedience, leadership, persuasion, sales, and marketing. This has led to the emergence of a field of study known as social influence theory. Social influence theory attempts to explain why the degree of change achieved by influence can vary based on how the influenced individual processes and accepts the outside "forces." There are three main ways in which people respond to external influences. Compliance occurs when individuals accept the influence—typically in response to a direct request—and adopt the behavior to gain rewards and/or avoid punishments. In the context of conformity, identification takes place when individuals accept an influence or viewpoint without deep reflection in order to produce or maintain a desired and beneficial relationship to another person or a group. Internalization occurs when individuals accept an influence and come to identify with a given behavior or mindset. Social proof refers to a phenomenon in which an individual adopts the behaviors of others—typically without a direct request being made—on the assumption that others' behavior must be correct under the circumstances at hand. The relative importance of the anticipated effect, the relative power of the influencer, and the predominance of the induced response may affect how these dynamics play out.

Question 8 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

Difficulty: 2 Medium Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 7C Learning and Behavior Change C is correct. 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. A: Socialization is not related to drive-reduction theories. B: Increased, not decreased, stigma would likely cause depression. D: While self-efficacy does affect depression, it is not related to drive-reduction theories. Content Foundations: Psychological Concept of Arousal In psychology, arousal refers to one's mental alertness, attentiveness, and general awakeness. A state of low arousal might involve drowsiness, lack of focus, boredom, and low energy overall. In contrast, a high-arousal state resembles what you may have experienced after drinking too much coffee, with an extreme feeling of being awake and even jitters and restlessness. Arousal theory posits that people behave or act in certain ways to maintain a level of optimal arousal, which varies from person to person. One particularly important element of arousal theory is the Yerkes-Dodson law, which states that performance of a behavior tends to be negatively impacted at high and low levels of arousal. The optimal level of arousal is somewhere between the two, creating an upside-down U-shaped function when arousal is graphed on the x-axis and performance on the y-axis. In other words, with too little arousal, the individual won't be sufficiently interested in the task at hand, but with too much, her efforts may be impaired by anxiety. Her performance would be best if she attained a level of moderate arousal instead. For example, imagine a student who is about to perform at his first piano recital. If he is completely unaroused - falling asleep, for example - he clearly won't perform at his best. But if he is extremely aroused to the point of being agitated, nervous, and jumpy, his performance at the recital is also likely to suffer. Note that the optimal level of arousal can vary depending on the particular task or behavior. Highly cognitive actions, like playing chess, have lower optimal levels of arousal, while less cognitive and more physical actions, like playing basketball, typically require higher levels of arousal.

Question 35 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.

Difficulty: 2 Medium Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 1A Amino Acids and Proteins B is correct. 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 (choice B). A: The MM-containing isoform is expected to migrate the shortest distance toward the anode; in fact, this isoform, CK3, actually migrates toward the negative electrode (the cathode). (Remember, in isoelectric focusing or gel electrophoresis procedures, the electrode designations match those in an electrolytic cell! In other words, the anode is the positive electrode and the cathode is the negative electrode.) Thus, choice A represents CK3, not CK2. C: Again, this is most likely the composition of CK1. D: We are provided with sufficient information to answer this question.

Question 56 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.

Difficulty: 2 Medium Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 1D Metabolism A is correct. 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. B: These species are the reduced forms of flavin adenine dinucleotide (FAD) and nicotinamide adenine dinucleotide (NAD). Since they are reduced (and are carrying electrons), they tend to serve as reducing agents and would not oxidize LDL. C: Hydrogen gas typically acts as a reductant, not an oxidant. Remember, reduction can be conceptualized as either the gain of bonds to hydrogen or the loss of bonds to oxygen. D: Saturated fatty acid tails are composed of carbon and hydrogen, with adjacent carbon atoms connected by single bonds. These structures would not serve as oxidizing agents, or electron acceptors.

Question 23 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.

Difficulty: 2 Medium Reasoning Skill: 3 Reasoning About the Design and Execution of Research Concept Category: 1A Amino Acids and Proteins A is correct. 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. B: This notation describes a mutation of tyrosine to alanine. As a nonpolar amino acid, alanine is very different from the highly polar phosphate group we are aiming to mimic here. C, D: These options begin with residues that do not tend to be phosphorylated, contradicting the question stem. Again, only S, Y, and T are typically phosphorylated in eukaryotes; you can remember that these amino acids each contain an -OH group on their side chain. Content Foundations: Amino Acid Structure and Properties Amino acids include both an amino (-NH2) and a carboxylic acid (-COOH) functional group. This can refer to a wide range of structures, but the MCAT focuses on the 20 amino acids that are coded for in human cells to build proteins. Chemically, these amino acids are known as α-amino acids because they have a single carbon at the α position, adjacent to the carbonyl carbon in the -COOH group. The central α-carbon has four substituents: -NH2, - COOH, -H, and -R. The -R group, or side chain, is the only part that differs between amino acids and determines their individual properties. Amino acid properties include polarity, acid-base behavior, chirality, and more specific chemical behaviors. Regarding polarity, all 20 amino acids can be categorized as polar (hydrophilic) or nonpolar (hydrophobic). Polar amino acids tend to be located on the exterior of globular proteins, facing the watery environment. In contrast, nonpolar amino acids are generally buried in the interior of these proteins, protected from the aqueous cytosol or extracellular fluid. Certain polar amino acids have side chains that can act as acids or bases. This behavior results from the presence of certain functional groups: carboxylic acids in the two key acidic amino acids and basic nitrogen-containing groups in the three basic AAs. Other properties are specific to a small number of amino acids. For example, glycine is the only achiral amino acid, meaning that a solution of glycine does not rotate plane-polarized light. This stems from glycine's alpha carbon, which is attached to two hydrogen atoms as well as the amino and carboxylic acid termini. Since this does not constitute four distinct substituents, glycine is achiral; all 19 other standard amino acids are chiral. Another property displayed by only one standard amino acid is the ability to form special bonds between sulfur atoms known as disulfide linkages. The amino acid that forms these bonds is cysteine, which has a thiol (-SH) group in its side chain. Disulfide bonds arise when one cysteine's sulfur atom connects to another, losing the attached hydrogen atoms in the process. These bonds are a key part of protein tertiary structure. Nonetheless, it is important to keep in mind that amino acid structure, properties, and chemical behavior encompass a very large amount of MCAT science content. To study these topics in greater depth, consult your biochemistry book or other MCAT sources.

Question 51 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.

Difficulty: 2 Medium Reasoning Skill: 3 Reasoning About the Design and Execution of Research Concept Category: 1D Metabolism B is correct. 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. A: High ILL would indicate that the cell is metabolizing effectively, so this answer is the opposite. C, D: ILL have nothing to do with the pentose phosphate pathway. Avoid these distracting answers. Content Foundations: Glycolysis Glycolysis is fundamental to all forms of cellular life. It takes place in the cytosol of cells, and its main purpose is to allow cells to obtain energy from glucose regardless of whether oxygen is present. The net reaction of glycolysis is: glucose + 2 NAD+ + 2 ADP + 2 Pi → 2 pyruvate + 2 NADH + 2 ATP + 2 H2O. Glycolysis has an initial "investment phase," consuming 2 ATP, and then a "payoff phase" generating 4 ATP. Thus, the net energy gain to the cell is 2 ATP. Pyruvate is a three-carbon alpha-keto acid that participates in multiple pathways in the body. In organisms that carry out aerobic metabolism, pyruvate is then converted to acetyl-CoA and fed into the citric acid cycle. The NADH produced by glycolysis is an electron carrier that can ultimately yield energy through oxidative phosphorylation. However, the NADH needs to be converted back to NAD+ for glycolysis to continue. This can happen in the electron transport chain or through fermentation. In lactic acid fermentation, which often occurs in human muscles, pyruvate is converted to lactate by lactate dehydrogenase in a reaction coupled to the conversion of NADH to NAD+. Of the 10 steps of glycolysis, step 3 is the most carefully regulated. In step 3, fructose 6-phosphate (formed in step 2) is transformed into fructose 1,6-bisphosphate by phosphofructokinase-1. This is the committed and rate-limiting step, and the major target for regulation. It involves the transfer of a phosphate group from ATP, and is the second of two steps where ATP is invested in the initial phase (step 1, where glucose → glucose 6-phosphate [G6P], is the first). Of note, step 3 takes place before the six-carbon carbohydrate chain is broken down into two three-carbon compounds. Glycolysis is upregulated when the cell needs more ATP, as signaled by relatively high concentrations of AMP/ADP or an abundance of inorganic phosphate (Pi). It is downregulated when the cell does not need more ATP, as indicated by a high level of ATP, abundant levels of NADH, or high levels of citrate (the product of the first step in the citric acid cycle). Glycolysis is also subject to negative regulation, in which certain products inhibit previous steps. Most notably, G6P, the first product of glycolysis, inhibits hexokinase, which catalyzes step 1, the conversion of glucose to G6P.

Question 45 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%

Difficulty: 2 Medium Reasoning Skill: 4 Data-based and Statistical Reasoning Concept Category: 1C Classical Genetics C is correct. 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%. A: This is the percent of the population with genotype aa, or 0.0122. B: This is half the right answer, and a trap for folks who forget that the heterozygote population is calculated as 2Aa. D: This is the percent of the population with genotype AA, or 0.9882. Content Foundations: Evolution In the 1973 words of the evolutionary biologist (and self-described creationist) Theodosius Dobzhansky, "nothing in biology makes sense except in the light of evolution." Natural selection is the mechanism through which evolution takes place. The concept of natural selection refers to the tendency of certain phenotypes to be favored in terms of reproduction. Natural selection is closely linked to the term fitness, which in the evolutionary context only refers to the chance of reproduction associated with a certain phenotype compared to baseline. Fitness must be defined in terms of specific environmental constraints. In the 20th century, the concept of group selection was proposed, arguing that natural selection could act on the level of the group, not the individual. The related concept of inclusive fitness expands the rigorous evolutionary definition of fitness (defined in terms of the differential reproduction of alleles) to account not just for individuals but also their relatives, who can be expected to share many of the same alleles. This idea helps to explain altruistic behavior. Additionally, multiple "types" of selection have been identified to describe the outcomes associated with different types of selective pressures on phenotypes that vary along a spectrum. Stabilizing selection occurs if both extremes are selected against, directional selection occurs if only one extreme phenotype is selected against and the other extreme is favored, and disruptive selection occurs when the median phenotype is selected against. The concept of Hardy-Weinberg equilibrium is often used to model stable gene pools that satisfy the following assumptions: (1) organisms must be diploid and reproduce sexually; (2) mating is random; (3) the population size is very large; (4) alleles are randomly distributed by sex; (5) no mutations occur; and (6) there is no migration into or out of the population. The Hardy-Weinberg equations allow us to use allele frequencies to predict the distribution of phenotypes in the population and vice versa. If p and q are the only two alleles of a gene present in the population, then p + q = 1. Squaring the equation yields: (p + q)2 = 12 → p2 + 2pq + q2 = 1. This second equation allows us to connect genotypes and phenotypes, because the p2 and q2 terms correspond to individuals homozygous for p and q, respectively, while the 2pq term gives the frequency of heterozygotes. Other mechanisms are associated with changes in the gene pool of a species, such as genetic drift and bottlenecks. Genetic drift refers to the role of chance, in the absence of strong selective pressures, in determining the reproductive fitness of various alleles. When no strong pressure exists for a certain allele, it may randomly happen to be reproduced more or less often. These random effects can add up over the course of evolution. A related, but more specific, concept is that of an evolutionary bottleneck. Bottlenecks occur when some external event dramatically reduces the size of a population in a way that is essentially random with regard to most, if not all, alleles. This dramatically reduces diversity in the gene pool.

27. 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.

Difficulty: 3 Hard Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 4A Kinematics and Force B is correct. 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. Thus, choice B is correct. A: Remember that Newton's third law applies very narrowly to the interactions between two objects. While the table does exert a normal force upwards on the book, that normal force is not what Newton's third law refers to. Notice that in the equation above, the reaction force of A on B is strictly limited to B on A. C: The question stem defines mg as the gravitational force on the book. As such, since the book is resting on top of the table, mg would not be the gravitational force exerted by the table on the earth unless the table were entirely massless (which is not stated). D: This statement, like choice A, is true. However, the question stem is asking for the Newton's third law corollary to "The book experiences a gravitational force of mg due to the earth's gravity." The two objects we should be focusing on are thus the book and the earth — not the table. Content Foundations: Newton's Laws The MCAT expects you to understand Newton's three revolutionary laws of motion and to be aware of how to apply them in problem-solving contexts. Newton's first law is a statement about inertia. It states that within a reference frame, an object remains at rest or at a constant velocity unless an external force acts upon it. Essentially, this law captures the insight that forces of resistance and friction make moving objects slow down and stop. It can be summarized as Fnet = 0 at equilibrium. Newton's second law defines force. It states that the total sum of forces acting on an object is equivalent to its mass times its acceleration. This is the familiar equation Fnet = ma. Newton's third law is about how forces come in pairs. It states that when body A exerts a force on body B, body B exerts an equal and opposite force on body A: FAB = −FBA. For example, the earth exerts a gravitational force on your body, which is responsible for your weight, but your body also exerts a gravitational force on the earth. In contrast to a common misconception, weight (caused by a gravitational force) and the normal force do not form a Newton's third law pair, because they are caused by different underlying forces and do not always have to be equal.

Question 59 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

Difficulty: 3 Hard Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 5B Bonding and Isomers D is correct. 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. As such, RN I is correct. RN II and RN III are also correct, since microtubules form the spindle apparatus that is an essential part of both mitosis and meiosis. The general structure of a microtubule is shown below; note that these structures are composed of dimers of the protein tubulin. IV: Microtubules play no direct role in electron transport. A: RN I and RN III are also correct. B: RN II and RN III are also correct, while RN IV is incorrect. C: RN I is also correct. Content Foundations: Structural Proteins The most common function of proteins in the body is not enzymatic, but structural. Structural proteins are fibrous proteins that have an elongated shape and provide structural support for cells and organ tissues. The first type of fibrous proteins are keratins, which form the skin, hair, and nails. Keratins are classified as soft or hard according to their sulfur content (i.e. the relative number of cysteine residues in their polypeptide chains). The low-sulfur keratins of the skin are much more flexible than the high-S, hard keratins. The second type of fibrous proteins you must know are the actin and myosin proteins of muscle tissue. Actin and myosin interact to form cross-linkages that allow the sliding of the filaments over each other in muscle contraction, which takes place through the contraction and relaxation of the sarcomere, the fundamental unit of all muscle fibers. When muscle contracts, the actin and myosin filaments slide over each other and the H-zone (actin-only region), Z-lines (sarcomere boundaries), and I-band (myosin-only region) all shrink, while the A-band (the entire myosin region) remains the same size. The opposite occurs upon muscle relaxation. A third type of structural protein you should know for test day is collagen, which is found in tendons, forms connective ligaments within the body, and gives extra support to the skin. Collagen is a triple helix formed by three proteins that wrap around one another. Many collagen molecules are cross-linked together in the extracellular space to form collagen fibrils to provide structural support for the cell. Elastin polypeptide chains are cross-linked together to form flexible, elastic fibers that give stretched tissues flexibility and the ability to recoil spontaneously as soon as the stretching force is relaxed.

Question 54 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

Difficulty: 3 Hard Reasoning Skill: 1 Knowledge of Scientific Concepts and Principles Concept Category: 7B Social Behavior D is correct. 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). A: In approach-approach conflicts, two options are both appealing. B: In avoidant-avoidant conflicts, both options are unappealing C: An approach-avoidance conflict is observed when one option has both positive and negative aspects, but there are two options in the question above. Content Foundations: Conflict Conflict is a very strong driver of behavior. Social conflict can occur at the level of individuals, groups, or even society as a whole. The social conflict model posits that society is unequal, and inequalities present give rise to conflicts and change. Conflict theorists, therefore, seek to explain the workings and institutions of society as a result of competition over power, resources, or other forms of capital. In contrast to conflict theorists, social functionalists argue that differentiation within society leads to order and stability. To conflict theorists, differences between social groups or inequalities within society are characterized by a struggle for power in which individuals or groups seek to redress the unequal power relationship and those with power try to protect their privilege. Not all social conflicts will lead to violence, nor is all social conflict negative in its presentation (e.g., you politely offer a seat on the bus to an older person, and they refuse, so you go back and forth). However, all social conflict involves interests. A person or group's interest is his or her attitude plus its strength, which combine to produce the desired outcomes. Social conflict can then be seen as the opposition, balancing, and resolution of two or more competing interests. An example of how conflicts can occur on an individual level is the way in which social roles can conflict when multiple different roles place conflicting expectations on a person. For example, you may need to study for the MCAT because you are a pre-med, but your family wants to you attend a party, because you are part of the family. Your decision to either stay and study or go out and please your family will reflect a choice of how to resolve this conflict.

Question 26 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 increases release of oxytocin. She then cognitively interprets 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 cognitively 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.

Difficulty: 3 Hard Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 6C Emotion and Stress A is correct. 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. B: This statement supports the James-Lange theory of emotion, which states that behavioral and physiological aspects of emotion (like increased heart rate and shouting) lead to cognitive aspects of emotion (like understanding that a situation is scary and feeling afraid). C: This statement supports the Cannon-Bard theory of emotion, which states that physiological and cognitive aspects of emotion occur simultaneously and independently. Annabel's brain released hormones at the same time that she interpreted the situation as happy and calm. The theory also states that behavioral expression is the last aspect - like when Annabel smiled. D: This statement supports the Yerkes-Dodson law, which states that people tend to perform their best when they're moderately emotionally stimulated. Luke is moderately happy when running with his dog, which leads to optimal performance. Content Foundations: Emotion Emotion refers to a person's instinctive, current state of mind, based upon mood, circumstances, and interactions with others. Emotion often influences problem-solving, decision-making, and social interaction. The three components of emotion are behavioral (action), which includes body language and facial expression, cognitive (mind), or the brain's subjective interpretation of the feeling, and physiological (body), which includes changes in heart rate, respiration, and so on, often stemming from arousal of the sympathetic nervous system. While emotions and their expression vary across cultures, certain emotions are thought to be universal, or manifested by people all over the world. These universal emotions are thought to be more biologically based than others. Paul Ekman described seven universal human emotions: anger, contempt, disgust, fear, happiness, sadness, and surprise. (Other sources may use different lists of universal emotions, since this is still very much a topic of research, but the MCAT tests Ekman's seven.) Certain emotions are associated with specific biological structures and processes; for example, fear is associated with increased activity in the amygdala, a limbic system structure found in the temporal lobes of the brain. Several theories seek to explain how the cognitive and physiological aspects of emotion are connected. An early theory of emotion, the James-Lange theory, hypothesized that a stimulus triggers a physiological response, which then leads to the subjective, conscious experience of emotion. For example, if you are watching a scary movie, your heart rate may become elevated, and that leads to the perception of fear. However, this theory is limited, since certain physiological states may be experienced as different emotions depending on context (such as either fear or excitement). A second theory, the Cannon-Bard theory, posited that cognitive and physiological responses to a stimulus occur simultaneously and independently of one another, with a behavioral response following them. However, this theory still lacks any element of cognitive appraisal, or conscious thought about how one's situation may impact the emotion(s) felt. A third theory, the Schachter-Singer theory, thus stated that a stimulus first leads to physiological arousal, then a cognitive interpretation of the circumstances, and finally a perception of emotion. This resembles the James-Lange theory, but adds the element of cognitive appraisal, which helps explain how (for example) an elevated breathing rate can be interpreted as multiple different emotions based on context.

Question 4 African-American patients who are in the pre-encounter stage of Cross's Nigrescence Model of African-American identity development might have what attitude towards Caucasian-American healthcare providers? A. They would regard them as superior practitioners and not question their medical advice. B. They would regard them with distrust and prefer to be treated by an African-American practitioner. C. They would recognize historical injustices in medical care towards racial minorities and work to empower African-American patients to self-advocate. D. They would judge the medical advice given to them in a cultural-free manner.

Difficulty: 3 Hard Reasoning Skill: 2 Scientific Reasoning and Problem Solving Concept Category: 9A Social Structure A is correct. In Cross's Nigrescence Model, African-Americans are described as progressing through several stages of cultural awareness. In the first stage, pre-encounter, African-Americans tend to view the majority Caucasian culture as being more desirable and would view a doctor of this race as being more skilled. B: This answer would be reflective of someone in the immersion-emersion stage of Cross's model. Someone in this stage would view the majority Caucasian culture with resentment and distrust and prefer to be treated by someone of his or her own race. C: This answer would be representative of someone in the internalization stage, who has integrated aspects of his own culture with that of the majority culture and is working to rectify past racial injustices. D: Cross posited that culture impacts identity and worldview.

Question 18 Which of the following biases is least likely to have affected the review of professionalism? A. Reconstructive bias B. Social desirability bias C. Attrition bias D. Selection bias

Difficulty: 3 Hard Reasoning Skill: 3 Reasoning About the Design and Execution of Research Concept Category: 8B Social Thinking C is correct. While the passage does not describe the methodology of the review, we can assume that the physicians were only surveyed once. Attrition bias occurs when participants drop out of a long-term experiment or study. A: Reconstructive bias is a type of bias related to memory. Most research on memories suggests that our memories of the past are not as accurate as we think, especially when we are remembering times of high stress. The physicians may not have accurately remembered what they witnessed during medical school. B: Social desirability bias is a type of bias related to how people respond to research questions. The physicians may have known that the researcher was examining unethical behavior and responded a certain way. Since the topic is fairly sensitive, they may have also understated unprofessional behavior or responded in a way that they felt was most socially acceptable. D: Selection bias refers to a type of bias related to how people are chosen to participate. In this case, people who witnessed unethical behavior in medical school may have been more likely to respond to the survey. Content Foundations: Cognitive Biases Cognitive biases are ways in which our perceptions and judgments systematically differ from reality. They are generally thought to be unavoidable features of our cognitive system, and may in some cases be adaptive. The closely related concept of heuristics refers to mental shortcuts or simplified iterations of principles that can help us make decisions, but can also lead to poor judgment. The representativeness heuristic is the tendency to make decisions about actions or events based upon our standard representations of those events. The closely related availability heuristic is the tendency to make decisions about how likely an action or event is based upon how readily available similar information is in our memories. Belief bias is the tendency that people have to judge things based not upon sound logic, but upon already held beliefs. Confirmation bias is a tendency that people have to focus on information that is in agreement with the beliefs they already have, rather than the information that is contrary to those beliefs. Another set of biases relate to how we interpret our and others' behavior. The fundamental attribution error refers to the tendency to place less importance on the import of a situation or context on behavior, and instead place undue emphasis on dispositional or internal qualities in order to explain behavior. In other words, we have a tendency to think that people are how they act. Moreover, we often blame our own actions on external situations but the actions of others on personality. This is called the actor-observer bias. Similarly, self-serving bias is the tendency people have to credit their successes to themselves and their failures either to the actions of others or to situations. Biases can also relate to perceptions of group belonging, as in the in-group bias where people are biased towards those viewed as being part of their in-group. Biases can also play a role in responses to questionnaires; for instance, social desirability bias refers to the tendency that people have to give socially approved responses to questions in the context of research.

Question 38 Considering the results of the laboratory testing given in the passage, which of the following diagnoses concerning the patient is most likely correct? A. The patient is unlikely to be suffering a heart attack, because the elevated CK isoenzymes are more closely associated with damage to the brain than with damage to cardiac muscle. B. Additional testing is necessary, because the CK isoenzymes that are elevated are not specific for cardiac muscle damage, and may also indicate damage to skeletal muscle. C. Additional testing is necessary, because CK isoenzymes that are markers of cardiac damage appeared at normal levels, but may have been elevated prior to the time at which the samples were drawn. D. The patient is likely suffering a heart attack, because his CK isoenzyme levels are indicative of recent cardiac muscle damage in a patient with no signs of kidney disease.

Difficulty: 3 Hard Reasoning Skill: 4 Data-based and Statistical Reasoning Concept Category: 1A Amino Acids and Proteins D is correct. Compared to reference enzyme activity levels, the patient shows dramatically elevated levels of CK3 (normal: 0-2, patient: 7241) and elevated levels of CK2 (normal: 0.5-6, patient: 219). The patient's CK1 enzyme activity levels are roughly normal (normal: 42-198, patient: 141). According to passage information, cardiac muscle contains both CK2 and CK3, while skeletal muscle contains only CK3. Thus, the fact that the patient has elevated levels of CK3 indicates that he has either cardiac or skeletal muscle damage. Add in the fact that the patient also has elevated CK2 levels then further narrows down the diagnosis to indicate that the patient does have cardiac muscle damage (such elevated CK2 levels would not have come from skeletal muscle damage). A: Neither CK2 nor CK3 contain the B subunit associated with CK1 and the brain. B: While muscle injury can account for the presence of elevated CK3 protein levels, it does not account for the presence of elevated CK2 levels. C: We have no reason to suspect that we need additional testing, as the elevated CK2 levels indicate cardiac damage.

Question 24 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.

Difficulty: 3 Hard Reasoning Skill: 4 Data-based and Statistical Reasoning Concept Category: 2A Cell Biology C is correct. 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. A: We have no evidence that protein kinase A is the enzyme that phosphorylates S156. Additionally, it is the inhibition of PKA that appears to drastically increase AQP5 localization to the membrane, as shown in Figure 1. If PKA itself "promoted the immediate localization" of AQP5 to the membrane, we would expect the opposite of these results. B: Figure 1 shows that phosphorylation of S156 (as mimicked by the mutant S156E) increases the membrane expression of AQP5. "Internalization" denotes the opposite process - the movement of membrane protein inward, away from the membrane. D: Figure 1 does not reflect the idea that S156 phosphorylation (here, denoted by the S156E mutant) enhances the activity of PKA. PKA inhibition appears to affect both the wild-type and the mutant strain similarly. Moreover, the mutant strains that mimic phosphorylation tend to be present in the membrane in a greater quantity, suggesting that, if anything, phosphorylation downregulates the internalization of AQP5.