Official Guide - All sections

An epitope is a region on the surface of an antigen molecule to which a specific antibody binds. The table shows the physical and biological characteristics of several different molecules. According to the information, which characteristics are most likely to be associated with a molecule's potential to elicit a strong immune response? A. High molecular weight and increased number of epitopes B. High molecular weight and reduced number of epitopes C. Low molecular weight and increased number of epitopes D. Low molecular weight and reduced number of epitopes

A A. An epitope binds one specific antibody. Therefore, a molecule with several epitopes will bind several distinct antibody molecules. The presence of several antibodies on the surface of an antigen is expected to elicit a stronger immune response than if fewer antibodies were present. The table shows that molecules with higher molecular weights are associated with higher numbers of epitopes, which will bind a greater number of antibodies. These molecules are therefore expected to elicit a more robust immune response. B. This is incorrect. Molecules with a reduced number of epitopes will bind fewer antibodies, therefore generating less potential for a robust immune response. C. This is incorrect. The table shows that molecules with a low molecular weight do not have an increased number of epitopes. D. This is incorrect. Although from the table a low molecular weight is correlated with a reduced number of epitopes, fewer antibody recognition sites will not generate a robust immune response.

Q4. The terminal electron acceptor in the metabolic pathway responsible for the chemical changes observed when Culture A was electrically stimulated is: A. pyruvate. B. oxygen. C. NAD+. D. water.

ANS A. According to the passage, Culture A uses lactic acid fermentation to provide the energy for the contractions that result from electrical stimulation. In this process, NADH reduces pyruvate to produce lactate. Therefore pyruvate serves as the electron acceptor in production of lactate. Oxygen would be the final electron acceptor if Culture A were to go through oxidative (aerobic) respiration

Q3 The addition of acetylcholine to the medium most likely induced: A. depolarization of the cell membrane that resulted in contraction. B. repolarization of the cell membrane that resulted in relaxation. C. hyperpolarization of the cell membrane that resulted in contraction. D. depolarization of the cell membrane that resulted in relaxation.

ANS A. Ach is released at the neuromuscular junction where it binds to receptors on the muscle cells and depending on the type of muscle cell, causes depolarization or repolarization of the cell membrane. In skeletal muscles, Ach binds to its receptors, which leads to depolarization of the muscle cell membrane and muscle contraction. AcH binding to ligand channels cause the influx of Na2+ across the motor end plate to cause depolarization and subsequent muscle contraction

Q2 Which steps involved in the contraction of a skeletal muscle require binding and/or hydrolysis of ATP? I. Dissociation of myosin head from actin filament II. Attachment of myosin head to actin filament III. Conformational change that moves actin and myosin filaments relative to one another IV. Binding of troponin to actin filament V. Release of calcium from the sarcoplasmic reticulum VI. Re-uptake of calcium into the sarcoplasm A. I, II, and III only B. II, III, and IV only C. I, III, and VI only D. III, IV, and VI only

ANS. C C is correct because binding of ATP dissociates the myosin head from actin filament, ATP hydrolysis then puts it in a high energy conformation, Pi leaving attaches it back to the actin filament at a closer point. The re-uptake of calcium into the SR occurs via an ATP hydrolyzing pump that moves calcium against its concentration gradient.

Q11 What are the configurations (R or S) of the chiral centers bonded to substituents X and Y, respectively, in Compound 1? A. R, R B. R, S C. S, R D. S, S

And A; guessed C For the carbon atoms bonded to the Y groups, the lowest priority group is already in the back and so the ranking of the remaining groups, which occurs in a clockwise manner, gives the Rconfiguration. For the carbon atoms bonded to X, the X group is in the back, and the relative rank of the remaining groups occurs in a counter-clockwise manner. When H is placed in the back, however, this would also be clockwise, or R

Q14 Suppose that, under the experimental conditions employed, Compound 1 is "saturated" with cations. What graph depicts the expected change in rate of K+ transport from the "IN" to the "OUT" phases at pH 2 (Trial 1 conditions) as a function of increasing K+ ion concentration in the "IN" phase? The graphs are as a picture on the opposite side

And D Compound 1 acts like a catalyst for cation transport across the organic phase. If Compound 1 is saturated with substrate, increasing the substrate concentration will not increase the rate. In a plot of velocity V versus [S] (classic Michaelis-Menten kinetic treatment) the rate of cation transport is at the limiting value, called Vmax

Q5. What control experiment was necessary to ensure that the apparent subcellular locations of the cat6xbs and lacY6xbstranscripts were NOT skewed by the location preference of the bound MS2-GFP? A. Determination of expression of MS2-GFP in cells that lacked the 6xbs insertion upstream of the cat and lacYgenes B. Use of E. coli cells that expressed only MS2 instead of the MS2-GFP fusion protein C. Insertion of the 6xbs region upstream of both the catand lacY genes in the same cells D. Determination of expression of both MS2 and GFP as separate proteins rather than as a fusion protein

Ans A A is right bc lack of the 6xbs sequence would leave the MS2-GFP protein without a binding partner. It would then be located in either the cytoplasm or the membrane (in this case, the cytoplasm) based on the properties of the MS2-GFP protein itself rather than being restricted to the location of the cat6xbs and lacY6xbs transcripts. Basically, if MS2-GFP doesn't have anything to bind to, you will know that wherever its expressing the fluorescence it where its location is. And then you can put in the binding proteins and see where the fluorescence is expressed

A researcher measures the initial rate (V0 = Δ[P]/Δt) of an enzymatically catalyzed reaction at a variety of substrate concentrations [S]0. S --(enzyme)--> P Which graph best represents the observed relationship between [S]0 versus V0?

Ans A A. For enzymatically catalyzed processes, the reaction rate will level off with increasing concentration of substrate as the enzyme becomes the bottleneck for the reaction. This is called saturation kinetics. B. Unlike first-order processes, increasing the concentration of substrate will not linearly increase the rate. C. Unlike second- or third-order processes, where increasing the substrate concentration can cause a squaring or tripling of the initial rate, enzymatically catalyzed reactions will not show such increases. D. In general, increasing the concentration of a reactant will increase the rate of its reaction.

Q16 According to the passage, one of the reasons Black men have medical mistrust is because seeking help violates their: A. gender schema. B. gender script. C. gender conditioning. D. gender adaptation.

Ans A A. Gender Schema: What is associated with each identity/what constitutes an identity. Male identity schema includes not seeking social services/help hence they dont healthcare professionals. The cognitions regarding what constitutes the male identity are an example of schema B. Script is organized information regarding the order of actions that are appropriate to a familiar situation. Gender Script: organized information regarding the order of actions that are appropriate to a familiar situation. In this situation, they are not following a gender script. or "how to act" C. doesn't exist D. doesn't exist

Q9 A researcher wants to determine whether the findings from the first study are also influenced by psychological responses to stress. To test this idea, a random sample of mothers from the first study is later given a stress assessment. What is the flaw in this research design? A. The dependent variable is temporally prior to measurement of the independent variable. B. The independent variable is temporally prior to measurement of the dependent variable. C. The updated sample contains too little variation to draw reliable conclusions. D. The updated sample is not representative of the population of mothers.

Ans A A. Hypothesized factors must be temporally prior to their assumed effects. Option A is correct because the hypothesized factor (mediating factor of psychological response to stress) was assessed after the outcome of interest (birth weight). B. This is incorrect because it identifies the correct time order of variables, so would not present a flaw. C. This is incorrect because the sample was randomly constructed so is not likely to present a flaw. In addition, the implied references to variability and reliability in the option are used imprecisely. D. This is incorrect because, as with option C, the sample is not the primary flaw in the design. Instead, the use of a random sample is likely to help ensure the representativeness of the sample.

Q1 What stereochemical designation can be assigned to naturally occurring aspartic acid? A. S B. Z C. D D. E

Ans A A. The correct stereochemical designation can be made by using either the Fischer projection of Compound 8 (with R = CH2CO2H) or the perspective drawing of Compound 1. For the Fischer projection, the priorities of the remaining three groups are NH2 > CO2H > R. This occurs in a clockwise fashion. Since the H is pointing towards the viewer in this drawing, however, the priorities would reverse to counterclockwise, or S when viewed with H (the lowest priority group) in the back. B. Z - For same side cis double bonds. Not relavent here. Z is used when the largest groups are on the same side of the double bond but are not necessarily the same C. All amino acids are in L configuration, not D D. E - for opposite side large groups on double bonds (trans). Not relavent here. Difference between E/Trans is that E is used when the largest groups are on opposite sides of the double bond and aren't necessarily the same General concept about amino acids*** Glycine - no chirality, Cystine - R, rest are S. All naturally occuring amino acids are L configuration. (all sugars are D) in eukaryotes

Q2 How could the researchers use the foot-in-the-door technique to increase the participants' likelihood of adopting a diet diary? A. Encourage the participants to sign a petition in support of diet diaries. B. Have the participants personally interact with the health buddy who adopted the diet diary. C. Tell the participants that the health buddy who adopts the diet diary is a trustworthy health expert. D. Tell the participants that by agreeing to be in the study, they have agreed to cooperate with the researchers.

Ans A A. The foot-in-the-door technique refers to convincing individuals to make a small commitment toward a cause, because this small commitment increases the likelihood of a larger commitment toward the same cause in the future. Option A is an example of this. If the participants make a small commitment toward diet diaries (signing a petition), according to the foot-in-the-door technique, they will be more likely to adopt a diet diary in the future. B. This is incorrect. It describes a situation that can plausibly increase the likelihood of adopting a diet diary but is not an example for the foot-in-the-door technique. This is not a request C. This is incorrect. It gives an example of how the peripheral route to persuasion can be used to increase the participants' likelihood of adopting a diet diary. D. This is incorrect. It suggests that compliance can be used to increase the likelihood of adopting a diet diary. The foot-in-the-door technique is used by asking a person for a small request to lead up to a subsequent larger request.

Q15 The scientists who developed the experimental protocol described in the passage chose TNBS over many potential candidates to label PE molecules. What characteristic about the rate of reaction between TNBS and outer envelope PE molecules allowed the experiment to provide useful data? The rate of TNBS reaction with outer envelope PE molecules is: A. faster than the rate of exchange between inner and outer envelope PE molecules. B. slower than the rate of phosphate transport across the membrane. C. facilitated by the additional phosphate present in solution. D. easily measured by the method of initial rates.

Ans A Explanation question (i.e., you have to answer "why"). As a first step, we can eliminate D quickly because it's really out of scope -- we have no reason to think the researchers dealt with this at all. Now we have to think more about what the question is asking for. There are two ways to approach explaining this question: the "right" way (i.e., if you really try to understand what they're getting at) and the "good enough" way, which uses shortcuts in situations where you might not completely understand the passage or the question. For the "good enough way", note that the question specifically asks about a characteristic relating to the rate of reaction between outer envelope PE molecules. A is the only answer choice that says something specific about outer envelope PE molecules, so that alone makes it likely to be a better explanation of what the question is asking about than answer choices that just say something super general. (We can definitely eliminate C for this reason - it doesn't mention the membrane at all!). In terms of the "right" way, the idea here is that there is a constant "churn" of PE molecules across the lipid membrane (this is generally the case for lipids in the membrane). TNBS allows us to take a "snapshot" of the conditions at the membrane in a given moment, and to do so, its reaction must take place really fast -- that is, before the "churn" of the membrane components could muddy the picture. Think about taking a picture with a short vs. a long exposure -- if you have a long exposure, you'll get a blurry picture because the position of your subjects has been changing

Q12 The information in the passage supports the prediction that P. falciparum creates unique protein trafficking structures outside the parasite itself for the trafficking of which parasite protein? A. PfEMP1 B. PfSET10 C. Histone H3 D. Hemoglobin

Ans A Paragraph 1 states PfEMP is a parasite protein that is present on the surface (plasma membrane) of the RBC that the parasite inhabits. This means that there must be a mechanism for transporting pfEMP1 from the parasite to the RBC plasma membrane

Q27 The pressure and volume changes that occur during a cycle of breathing are illustrated graphically in the figure shown. What does the area within the curve represent? A. Work done B. Oxygen removed C. Lung volume change D. Air pressure change

Ans A This question requires you to interpret graphically a thermodynamic process and the area it encloses when represented in pressure-volume coordinates. The area under a curve that represents graphically the relationship of two physical variables (one independent and the other dependent) has the physical dimension of the product of the measurement units of the variables. In this case, a thermodynamic process is shown in PV coordinates, hence the area under the curve has the dimensions of [N/m2] · [m3] = [N·m] = [J]. The unit of joule (J) is used for energy and work.

Q10 Which independent variable is most relevant for a study that investigates the possible impact of primary groups on the outcome measure discussed in the passage? A. Family structure B. Religious affiliation C. Occupational status D. Population density

Ans A family is primary group B. religion can encompass both primary and secondary C. more relevant to social states and social role D. not related. This is more demographics

Q10 What factor can account for the relative rate of Ca2+ transport across the membrane as a function of pH? A. Deprotonation of the X groups B. Deprotonation of the Y groups C. Protonation of the central O atoms D. Change in ring size of the macrocyle

Ans A The key is to understand that talking about changes in pH is equivalent to talking about changes in terms of which functional groups are protonated or deprotonated. This is a key concept for the MCAT. The idea is that functional groups that are weak acids or bases are more likely to be protonated at a low pH, and less likely to be protonated at a high pH. This can affect charge -- for instance, a COOH group is neutral, but its deprotonated form, COO-, is negatively charged. Charge shapes all sorts of interactions between biomolecules, so changing pH can change functionality. With this in mind, we need to look for a structure in the crown ether ring (b/c the passage tells us that the crown ether ring controls cation transport) that can meaningfully be protonated or deprotonated. Answer choice D is out b/c we know that we need to focus on protonation/deprotonation in a Q about pH. C is out because ether oxygens are super unreactive and don't engage in acid-base chemistry. This leaves the X and Y groups, which are defined in the legend of Figure 1. Y is some super-large chain with lots of hydrocarbons, an amide group, a highly substituted amine...all very unreactive in terms of acid-base chemistry. So, B can be eliminated. Then let's consider A. A is a carboxylic acid group (-COOH), which is a prototypical example of a weak acid that shows pH sensitivity in terms of shifting between its protonated and deprotonated states, so that's exactly what the question is looking for

Q30 Which aspect of equality is considered a requirement in a meritocracy? A. Equality of opportunity B. Equality of talent C. Equality of skill D. Equality of outcome

Ans A; I picked B A meritocracy is when societal rewards, status, and positions are awarded to individuals based on their own ability and work (that is, merit). In order for a meritocracy to operate, everyone within the society would need the same opportunity to succeed, so that rewards are actually based (primarily) on merit. B. This is incorrect because a meritocracy is not when everyone has the same level of talent. C. This is incorrect because a meritocracy is not when everyone has the same level of skill. d. This is incorrect because a meritocracy is not when everyone experiences the same outcome.

Q22 Of the five candidate genes, which produces a factor that most markedly increases the efficiency with which fibroblasts commit to a neuronal lineage in vitro? A. Ascl1 B. Brn2 C. Zic1 D. Olig2

Ans A; I picked D - didn't realize that the "-" in front of the genes actually means omission. A. Tuj1-positive cells are cells derived from fibroblasts that have been reprogrammed into neurons. Figure 2 shows that when all 5 genes are present (indicated by 5 factor in the figure), the number of Tuj1-positive cells is set at 1 (100%). Omission of Ascl1 reduces the number of these cells from 1.0 to 0.3; omission of Brn2, from 1.0 to 0.6; omission of Myt1l, from 1.0 to 0.66; omission of Zic1, from 1.0 to 0.66, and omission of Olig2, from 1.0 to 0.7. Because omission of Ascl1 causes the biggest reduction in Tuj1-positive cells, this gene must produce a factor that most markedly increases the efficiency with which fibroblasts commit to a neuronal lineage. B. This is incorrect. Omission of Brn2 reduces the number of reprogrammed neuronal cells by 40% relative to the 5-gene pool, whereas omission of Ascl1 reduces it by 70%. C. This is incorrect. Omission of either Myt1l or Zic1reduces the number of reprogrammed neuronal cells by 34% relative to the 5-gene pool, whereas omission of Ascl1 reduces it by 70%. D. This is incorrect. Omission of Olig2 reduces the number of reprogrammed neuronal cells by 30% relative to the 5-gene pool, whereas omission of Ascl1 reduces it by 70%.

Q6 In helping to explain the results of the study, which other concept would be most similar to a homophilous social network? A. Reference group B. Secondary group C. Out-group D. Social group

Ans A; I picked D bc in the passage they say that homophily is the tendency of social contacts to be similar to one another. So, I thought that meant the you're similar to the social group you are a part of. A. A is correct. A reference group is defined as "any group that individuals use as a standard for evaluating themselves and their own behavior," and is similar to the mechanism behind the effect of the "homophilous" group in the study. B. This is incorrect because secondary group refers to a formal, impersonal group. C. This is incorrect because an out-group is defined as a group that people do not feel connected to. D. This is incorrect because a social group, which is a general term that refers to a collection of people with common identity and regular interaction, is not as specific as a reference group.

Q15 What type of learning is taking place in Study 1? A. Operant conditioning B. Classical conditioning C. Social learning D. Observational learning

Ans A; you picked B - you knew this A is correct. In Study 1, the frequency of pressing the lever increases because it results in the delivery of heroin, which is an example of operant conditioning through positive reinforcement. This is incorrect because classical conditioning does not involve a change in behavior due to the behavior's outcome. This is incorrect because social learning refers to learning through observing a model. This is incorrect because observational learning, like social learning, refers to learning through a model. Operant conditioning is associated with BF. Skinner. Reduce/Increase response based on stimuli

Q22 At the time of the car accidents, which component of the nervous system of both Person A and Person B was NOT likely to be activated? A. sympathetic nervous system. B. parasympathetic nervous system. C. peripheral nervous system. D. central nervous system.

Ans B A. This is incorrect because the sympathetic system activation is involved in the stress response and is likely to be activated at the time of a stressful event. B is correct. The parasympathetic nervous system is an energy conserving (rest-and-digest) system and is not likely to be activated at the time of a stressful event. C. This is incorrect because the sympathetic nervous system is a part of the peripheral nervous system, and therefore, the peripheral nervous system will be activated at the time of a stressful event. D. This is incorrect because the reticular activating system (part of the central nervous system) will be activated at the time of a stressful event.

Q19 A scientist proposed that the 32P label was entering PE molecules by direct exchange (swapping phosphate groups with those found in solution) and NOT through synthesis of new PE by bacterial cells. What experimental modification can show this is NOT the case? A. Introduce TNBS prior to pulsing with 32PO43-. B. Measure the rate of incorporation of 32PO43- into acellular PE. C. Use mouse cell cultures instead of bacterial cells. D. Decrease the concentration of 32PO43- and observe the effect on incorporation rate.

Ans B Q19 is similar to Q15 in that the way you want to think about it may vary depending on how clearly you get what they're trying to do. The basic issue that the Q stem introduces is that there may be more than one mechanism through which radiolabeled P shows up on the PE groups -- the logic of the experiment assumes that P is incorporated through biosynthesis, but the question is asking, what if the P just is swapped out in solution through a non-biosynthetic process? The Q is then asking you to figure out a control experiment that would distinguish between those mechanisms. C and D can be eliminated because they're not mechanism-specific. A might be tempting, but hopefully we can realize that this would just wind up measuring nothing -- it'd be like trying to take a picture of the sunrise before the sun actually rises. B involves saying "hey, let's look at what happen if no cells are involved, and that will give us a measure of whether these P groups just get swapped out with those in solution through an equilibrium process." That's the kind of control they're looking for

Q5 Which biological molecule can react with Compound 11 so that pyridoxal phosphate (Compound 5) is regenerated? A. ATP B. Oxaloacetate C. NADH D. FAD+

Ans B A. ATP would transfer a phosphate group. Nothing in the passage suggests this B. The last sentence in the passage states that reaction in figure 1 is reversible. Before going further, notice the difference between compounds 11 and 5 (picture), which is the conversion of an amine -> carbonyl. Now refer back to the product side of figure 1 and notce that oxaloacetic acid can be used to accomplish this conversion C/D. NADH/FADH2 and their oxidated forms are high energy electron carriers than play a role in REDOX reactions in cellular respiration and metabolism. This is a transamination reaction

Q29 If sounds produced by the human vocal cords are approximated as waves on a string fixed at both ends, and the average length of a vocal cord is 15 mm, what is the fundamental frequency of the sound? (Note: Use 3 m/s for the speed of sound through the vocal cord.) A. 10 Hz B. 100 Hz C. 1000 Hz D. 10,000 Hz

Ans B A. This incorrect answer results from using the length of the vocal cords as 15 cm. B. This question requires you to apply knowledge of sound production by vibrating cords and strings, and to identify the physical properties of such cords and their relationship with sound characteristics: frequency, propagation speed, and amplitude. In particular, the relationship between the sound speed and frequency for a string fixed at both ends must be used. The fundamental frequency of a vibrating string is given by f = v/(2L) = (3 m/s)/(30 × 10-3 m) = 1/10-2 Hz = 100 Hz. C. This incorrect answer results from using the sound speed as 30 m/s. D. This incorrect answer results from using the length of the vocal cords as 1.5 mm and the sound speed as 30 m/s

Q5 If the study were modified to investigate the effect of homophily on the changes in participants' exercise patterns as well as their likelihood of adopting an Internet-based diary, how would this change the design of the study? A. A new independent variable would be added. B. A new dependent variable would be added. C. Levels of an existing independent variable would increase. D. The study would become an experimental study.

Ans B A. This is incorrect because "changes in exercise patterns" is not an example of adding another independent variable because the researchers are not interested in the effect of changes in exercise patterns on another variable. B. B is correct. "Changes in exercise patterns" corresponds to a new dependent variable because the researchers are interested in the effect of homophily (independent variable) on exercise, along with adopting an Internet-based diary. C. This is incorrect because measuring changes in exercise patterns introduces a new variable to the study rather than changing the levels of an existing variable. D. This is incorrect because adding a new variable does not render a study experimental.

Q13 The information in the passage suggests that PfSET10 has which function in var gene localization or expression? PfSET10: A. allows active and silent var genes to co-localize in the nucleus. B. marks the chromatin of the active var promoter for re-expression after mitosis. C. marks the chromatin of a silent var promoter to be expressed after mitosis. D. marks the chromatin of multiple var promoters for simultaneous expression.

Ans B A. passage says that active and silent var genes don't colocalize together B. According to the passage, P. falciparum cells contain the most PfSET10 when the intraerythrocyte parasites are in an actively dividing life cycle phase. To maintain cellular identity, there has to be a mechanism for marking genes that were transcriptionally active before mitosis for reactivation after mitosis. Because PfSET10 colocalizes with the active var gene, it is reasonable to hypothesize that PfSET10 methlytransferase activity is involved in this bookmarking c. So PfSET10 is a histone lysine methyltransferase, hence it will make an inactive gene active. This is why it is present near active var genes, not silent ones - you wouldn't want to start making active non silent var. This can also be seen in the graph to the left, where PSET10 is closer in distance for the "on" var gene" and farther for the off var gene. D. pfSET10 is involved in only var gene expression as discussed in the passage (end of paragraph 1). "Only one var gene is transcribed at a time over multiple mitotic generations"

A hot water tail-flick test measures the time it takes rats to remove their tail when it is dipped in hot water. Rats housed with a running wheel exhibit a delayed response in the test. Based on this response, which type of sensory receptors are most likely negatively regulated by exercise? A. Baroreceptors B. Nociceptors C. Mechanoreceptors D. Chemoreceptors

Ans B A. pressure C. touch D. chemical change

Q24 As one step in the estimation of the efficiency of neuronal induction, scientists calculated the average number of induced cells present in 30 randomly selected 20× visual fields. Which change to this particular aspect of the experimental protocol would increase the accuracy of the estimates of efficiency? A. Increase the magnification of the oculars used to define the field of view. B. Increase the number of visual fields counted per petri dish. C. Select visual fields from the central portion of the petri dish where cell density is highest. D. Use the presence of green fluorescence to identify cells appropriate for quantification.

Ans B Increasing the number of visual fields counted would increase the sample of cells observed from the total "population," and result in results that better approximate the true values

Q8 What is the mechanical power exerted on the protein when the retraction speed is 1000 nm/s? A. 1.5 × 10-18 W B. 7.5 × 10-17 W C. 3.5 × 10-6 W D. 5.5 × 106 W

Ans B Power = W/t or Fv. Therefore, from figure 3, at 1000 nms the answer is P = 75pN * 1000 nm/s =75*10^-12 N * 1000 *10^-9 m/s = 75000 * 10^-21 W (J/s or N*m/s)= 7.5 * 10^-17 W Concept: The equations for Power are Work/Time, Force x Velocity, or Energy/Time. Work and energy can be interchangeable because of the work-energy theorem. The unit for power is a watt (J/S)

Q4 What thermodynamic and chemical changes (if any) occur during aspartate transamination (Figure 1)? A. Since ΔG < 0, ATP is produced. B. Since ΔG = 0, no ATP is required or produced. C. Since ΔG > 0, ATP is required. D. Since ΔG > 0, ATP is produced.

Ans B Since it says that Keq is 1.0 that means deltaG = 0 according to the equation ΔG = -RTln(Keq). This means that the rxn is at equilibrium and no ATP is produced or required

Q22 Soaps are chemically modified natural products that can be derived from all of the following EXCEPT: A. fatty acids. B. cholesterols. C. triacylglycerols. D. phospholipids.

Ans B A. Fatty acids are RCO2H which can easily be converted to soaps RCO2Na by the addition of NaOH. B. While cholesterols are hydrocarbons found in cell membranes, they are not structurally similar to fatty acids. C. Triacylglycerols contain RCO2- groups that can be released by hydrolysis and precipitated as soap. D. Phospholipids also contain RCO2- groups that can be released by hydrolysis and precipitated as soap

Q25 A team of researchers wanted to test whether the James-Lange theory of emotional arousal could explain subjects' physical and emotional experiences while they viewed clips from a horror film. Which of the following scenarios is most consistent with the James-Lange theory? A. The participants felt general excitement and simultaneously experienced physical symptoms of autonomic arousal, such as a racing heart. B. The participants experienced physical symptoms of autonomic arousal, such as a racing heart, and then they reported that they felt afraid. C. The participants felt fear, and then began to experience physical symptoms of autonomic arousal, such as a racing heart. D. The participants showed physical symptoms of autonomic arousal, such as a racing heart, and then they reported that they felt general excitement.

Ans B I picked D bc I didn't read carefully. Why would they be excited after watching horror movie clips stupid fck?

Q11 The data in the passage suggest that the substrate binding domain in PfSET10 is: A. the SET domain. B. the PHD domain. C. the N-terminal domain. D. the C-terminal domain.

Ans B, I picked C A. this is wrong bc the observation that a fragment lacking the PHD domain but containing the SET domain does not bind unmodified histone H3 argues that the PHD domain, not the SET domain, is the substrate binding domain. B. According to the data in Figure 2, a PfSET10 fragment that contains the PHD domain and the SET domain binds unmodified histone H3. A fragment lacking the PHD domain but containing the SET domain does not bind unmodified histone H3. Taken together, these data best support the conclusion that the PHD domain is the substrate binding domain. C and D. This is incorrect. Because a PfSET10 fragment that contains the PHD domain and the SET domain binds unmodified histone H3, the substrate binding domain is near the center of the primary structure of the protein and is not near the N-terminus or C-terminus

Q6 Which wavelength of laser light can be used with the photodiode detector in the atomic force microscope? A. 226 nm B. 633 nm C. 1.26 μm D. 3.17 μm

Ans B; picked A bc I had no idea how to do it. I thought this was a math question but it was really a conceptual question. In the passage it says "the deflection of the cantilever was measured using visible laser light reflected off its surface onto a photodiode detector". Visible light is between: 400 - 750 nm. Red light is highest wavelength, blue light is lowest. (ROYGBIV)The only answer within this range is B. Concept: Look at the EM spectrum picture. KNOW THIS SHIT BY HEART

Q25 Enzymes catalyze chemical reactions by stabilizing: A. the substrate. B. the product. C. the transition state. D. the equilibrium.

Ans C

Q14 Which statement provides the most likely explanation for the results in Figure 1? A. ABA-induced rats show a decrease in withdrawal symptoms because an opioid agonist mimicked the effects of wheel-running. B. Rats with 24-hour access to food show a decrease in withdrawal symptoms because an opioid agonist increased the reinforcement value of food. C. ABA-induced rats show an increase in withdrawal symptoms because an opioid antagonist interfered with the effects of wheel-running. D. Non-ABA-induced rats show an increase in withdrawal symptoms because an opioid antagonist enhanced the effects of wheel-running.

Ans C A is incorrect. ABA-induced rats (the rats with restricted access to food and access to a running wheel) do not show a decrease in withdrawal symptoms. B is incorrect. The drug used in the study, naloxone, is an opioid antagonist. C is correct. ABA-induced rats (the rats with restricted access to food and access to a running wheel) show increased withdrawal symptoms because naloxone, an opioid antagonist, likely interferes with the physiological effects of wheel-running. D is incorrect. Non-ABA-induced rats do not show an increase in withdrawal symptoms.

Q10 Of the three general cell types or cell-derived structures described in the passage as binding P. falciparum-infectedRBCs, at least two of the three have which characteristic? A. They have nuclei. B. They are cell fragments. C. They are bone marrow-derived. D. They are connected by tight junctions.

Ans C A. EC have nuclei but platelets and RBCs dont. They lack most membrane bound organelles. Main reason why RBCs lack mitochondria and other membrane bound organelles is to maximize O2 transport B. Only platelets are cell fragments C. The passage states that infected RBCs adhere to platelets, EC, and other mature RBCs. These are all derived from bone-marrow D. Only ECs are connected by tight junctions. Remember: tight junction is one of three kinds of cell-cell junction - the others are gap junctions (communicating junctions) and desmosomes (anchoring junctions). Tight junctions are present in the bladder, intestine, kidney (to prevent water), desmosomes are present in skin/intestine (let water through), and gap junctions are present in cardiac muscle and neurons.

Q17 In Reaction 1, what is a possible structure for either R1 or R2 of the reactant? A. CH3 B. NH2 C. (CH2)15CH3 D. (CH2O)10CH3

Ans C A. Methyl is not sufficiently long to be part of a phosphatadylethanolamine. B. If R were NH2 the group would not be a fatty acid acyl chain. C. Because PE is a lipid, the R groups must represent the acyl chains. This is nonpolar and long, what is expected for a phosophlipid. D. The acyl chains of PE molecules do not contain oxygen atoms.

Q21 What happens to the pH of a soapy solution as a result of the introduction of hardness ions? A. The pH increases as [H+] increases. B. The pH is not changed since no acid-base reaction occurs. C. The pH decreases as [OH - ] decreases. D. The effect on pH depends on the identity M2+.

Ans C A. This can be eliminated, pH decreases as H+ increases B. An acid-base reaction does occur. Moreover this reaction will affect the pH of the solution since weak bases are eliminated from solution C. Removal of weak bases such as RCO2- results in a decline in pH and a decrease in OH-. Remember that RCO2- is a WEAK base, and the conjugate of this createas a strong acid. The way I see it is that reax 1 creates Na+ which will react with OH- thus decreasing [OH] and pH

Q13 Based on the passage, ABA-induced rats are most likely to demonstrate: A. increased sensitivity to the effects after running in their wheels. B. increased sensitivity to pain over time. C. withdrawal symptoms if they are prevented from running in their wheels. D. withdrawal symptoms if they are injected with opiates

Ans C A. This is incorrect because after running in their wheels, the ABA-induced rats likely have high levels of endorphins and will not be especially sensitive to the effects of endorphin agonists. B. This is incorrect. ABA rats are not likely to become more sensitive to pain over time, especially because the running behavior results in an increase in endorphins, which are natural painkillers. C. C is correct. Rats that have developed exercise addiction should experience withdrawal symptoms if they are hindered from running in their wheels. The running behavior leads to an increase in endorphins which, according to the passage, leads to exercise addiction. Preventing the running behavior is likely to lead to withdrawal, as it would interfere with the increase in endorphins that the ABA-induced rats are addicted to. D. This is incorrect because opiates are endorphin agonists and will not lead to withdrawal symptoms for the ABA-induced rats.

Q17 Which operationalization is most appropriate for the independent variable of the proposed follow-up experiment? A. Level of mistrust, established by an inventory that measures participants' medical mistrust B. Level of mistrust, established by independent judges who rate participants' medical mistrust C. Type of communication, established by training a doctor who is also a confederate to use patient-centered communication or a communication style that is not patient-centered D. Type of communication, established by giving doctors in the study an inventory that assesses whether their communication style is patient-centered or not

Ans C A. This is incorrect because level of mistrust is the dependent variable. B. This is incorrect because level of mistrust is the dependent variable. C. C is correct. The researchers are interested in the effect of type of communication on level of mistrust; therefore, type of communication is the independent variable. To establish a causal relationship between the two variables, the independent variable needs to be manipulated, as described in C, by training the doctor to adopt different communication styles. D. This is incorrect because the operational definition provided does not involve manipulating the independent variable.

Q18 Which finding, when combined with the data in the passage, is most likely to lead researchers to conclude that the 5-HT2A and 5-HT2B receptor subtypes mediate serotonin-dependent liver regeneration? A. Administration of 5-HT2A receptor agonist resulted in reduced Ki67 staining. B. RNA for seven different receptor subtypes was detectible in naïve liver tissue. C. Up-regulation of 5-HT2A and 5-HT2B was observed during periods of peak hepatocyte proliferation. D. Administration of 5-HT2C and 5-HT3 receptor antagonists reduced the number of Ki67-positive cells

Ans C A. This is incorrect. A reduction in Ki67 staining after administration of a 5-HT2A agonist is not consistent with the data presented in the passage. B. This is incorrect. The presence of RNA encoding all the receptor types in the liver tissues would not lead researchers to conclude that the 5-HT2A and 5-HT2Breceptor subtypes directly mediate serotonin-dependent liver regeneration. C. An observed up-regulation of 5-HT2A and 5-HT2Bduring periods of peak hepatocyte proliferation corroborates the data presented in the passage (that administration of antagonists of 5-HT2A and 5-HT2B lead to decreased hepatocyte proliferation) and, of the options presented, is most likely to lead researchers to conclude that the 5-HT2A and 5-HT2B receptor subtypes mediate serotonin-dependent liver regeneration. D. This is incorrect. That 5-HT2C and 5-HT3 receptor antagonists failed to reduce the number of positive Ki67 cells is consistent with the conclusion that the 5-HT2A and 5-HT2B receptor subtypes mediate serotonin-dependent liver regeneration, but this observation is not evidence of a role of 5-HT2A and 5-HT2B in this process. Thus, this observation provides less support for the role of the 5-HT2A and 5-HT2Breceptor subtypes in serotonin-dependent liver regeneration than does option C

Q20 Which type of enzyme catalyzes the conversion of glutamate to GABA? A. Kinase B. Transferase C. Decarboxylase D. Dehydrogenase

Ans C Comparison of the chemical formulas provided in the passage show that the conversion of glutamate to GABA involves the removal of a carbon from the carbon chain. These reactions are often catalyzed by decarboxylases A general strategy for this type of question is to draw the structures out. Their formulas are given in the passage. Assess the difference between the two and determine the answer based on your knowledge of the individual enzyme functions that are required for the MCAT

Q15 The liver synthesizes factors that act cooperatively with platelets to facilitate which physiological process? A. Cholesterol synthesis B. Glucose metabolism C. Blood clotting D. Fat digestion

Ans C Platelets are involved to form a plug at the site of blood vessel damage. Blood clotting factors that are synthesized in the liver in an inactive form (fibrinogen) which floats around in the blood. When this fibrinogen interacts with tissue factors during damage, it initiates a cascade which leads to prothrombin being activated to thrombin, the molecule incharge of converting fibrinogen to fibrin (which forms the clot). The extrinsic cascade is out of scope for the MCAT, but this is how it proceeds: III (TF) --> VIII --> X --> II (thrmbin), which activates V, VIII, XI, XIII (intrinsic pathway). which all then amplify II (thrombin). Now there is so much thrombin

Q26 Anxious about a nagging illness, a patient feels ignored by a doctor who is struggling to catch up with patient examinations on a very busy day. The doctor misinterprets the discomfort and agitation of the patient as hostility. Which sociological paradigm can best explain this scenario? A. Functionalism B. Conflict Theory C. Symbolic Interactionism D. Social Constructionism

Ans C This is incorrect because functionalism is considered a macro-level theory that understands social phenomena in terms of their function for society. This is incorrect because conflict theory focuses on the differences in material resources among groups in society. C is the correct answer since the scenario concerns the inter-subjective negotiation of symbols or meanings, which is indicative of symbolic interactionist theory. This is incorrect because social constructionism bridges the micro and macro levels but places more emphasis on how concepts emerge (such as illness), as opposed to understanding social interactions.

Q28 A researcher interested in memory of novel words shows participants unrelated words printed on a card, one after the other. Participants see 20 words in total, wait for 2 minutes, and then are asked to write down all of the words they can remember. The researcher finds that 95% of the subjects remember the first three words. This finding is an example of: A. the recency effect. B. proactive interference. C. the primacy effect. D. retroactive interference.

Ans C This is incorrect because the recency effect refers to improved memory for the later information that is still in working memory. This is incorrect because proactive interference refers to earlier information interfering with memory for later information. Improved memory for earlier information is an example of the primacy effect. This is incorrect because retroactive interference refers to later information interfering with memory for earlier information.

Q27 A teacher rewards his students by distributing plastic chips. Students receive a chip for each instance of desirable behavior. At the end of each month, they can exchange their chips for prizes. The teacher sees major decreases in undesirable behaviors as a result of this system, which is known as: A. aversive conditioning. B. operant extinction. C. a token economy. D. an unconditioned stimulus.

Ans C This is incorrect because the students are being trained by positive reinforcement, not punishment. This is incorrect because although operant learning is taking place, the stem specifies the acquisition of desirable behaviors, not extinction. C is correct. Rewarding individuals with secondary reinforcers that can be exchanged for appetitive stimuli is typical of a token economy. This is incorrect because unconditioned stimuli are involved in classical conditioning, not operant conditioning.

Q7 Using knowledge of Michaelis-Menten kinetics, what effect would the addition of chloramphenicol have on the kinetics of its target enzyme? A. Vmax decreases, and KM increases. B. Vmax decreases, and KM remains unchanged. C. Vmax remains unchanged, and KM increases. D. Vmax remains unchanged, and KM decreases.

Ans C easy memorization about competitive inhibition

Q28 Human speech is generated in the vocal cords as the lungs push air past them. What property of the vocal cords is changed so that the frequency of sound can be altered? A. Volume B. Density C. Tension D. Number

Ans C A. A change in the volume of the vocal cords cannot affect the sound frequency because the frequency only depends on the propagation speed (for a fixed-length cord), and the speed is independent of the volume. B. A change in the density of the vocal cords cannot affect the sound frequency because the frequency only depends on the propagation speed (for a fixed-length cord), and the speed is independent of the density, as it propagates as a transverse wave through the cords. C. The frequency of the sound produced in vibrating cords and strings (such as vocal cords) of fixed length is proportional to the propagation speed of the sound through the cord. In turn, the propagation speed of a transverse wave (such as the sound wave in the vocal cords) is directly proportional to the tension applied along the cord. D. The number of vocal cords used in generating the speech is constant.

Q19 Based on the passage, unpleasant healthcare experiences act as: A. positive reinforcement. B. negative reinforcement. C. positive punishment. D. negative punishment.

Ans C C is correct. According to the passage, negative healthcare experiences reduce the frequency of seeking health care. This suggests that negative healthcare experiences punish seeking health care. It is positive punishment as opposed to negative, because it involves introducing an unpleasant experience rather than removing a pleasant stimulus

Q25 The overall reaction for glycolysis: Glucose + 2NAD+ + 2ADP + 2Pi → 2Pyruvate + 2NADH+ 2H+ + 2ATP + 2H2O can be broken down into two separate processes (reactions 1 and 2). Glucose + 2NAD+ → 2Pyruvate + 2NADH + 2H+ Reaction 1: ΔG° = -146 kJ/mol ADP + Pi → ATP + H2O Reaction 2: ΔG° = +30.5 kJ/mol What is ΔG° for glycolysis? A. -207.0 kJ/mol B. -176.5 kJ/mol C. -85.0 kJ/mol D. -54.5 kJ/mol

Ans C Glycolysis is the net sum of Reaction 1 and two rounds of Reaction 2. The overall free energy change can therefore be calculated as (-146 kJ/mol) + 2(30.5 kJ/mol) = -85 kJ/mol.

Q23 While Person A was in a coma, researchers considered stimulating her brain to bring her out of the comatose state. The researchers would most likely have stimulated the: A. Wernicke's area. B. parietal lobes. C. reticular activating system. D. somatosensory cortex.

Ans C This is incorrect, as Wernicke's area is involved mostly in speech processing. This is incorrect, as the parietal lobes do not directly control alertness. C is correct. The reticular activating system is involved in controlling alertness and is most likely to be stimulated to bring someone out of a coma. This is incorrect, as the somatosensory cortex is involved in receiving the sensory signals from the skin.

Q21 What is the most likely reason why Tuj1 was used to assess the phenotype of cells that have incorporated the five candidate genes? A. Tuj1 induces expression of the TauEGFP protein. B. Tuj1 is expressed in fibroblasts and neurons. C. Tuj1 is an early marker of neural differentiation. D. Tuj1 is present in embryonic and adult cells in culture.

Ans C; I picked A A. Tuj1 is a tubule protein according to the passage. The TauEGFP is a gene that engineered in mice to express a green fluorescent protein I their neuronal tissues only not in the tubules B. The goal of the experiment was to turn fibroblasts into neurons. If both cells expressed the gene then you wouldn't be able to tell the difference btw the two C. Again, goal is to go from fibroblast to neurons so it would make sense to make the Tuj1 protein into a reliable marker for selecting cells that after a short time in culture show signs of having been covenanted into neurons. Therefore, it must serve as an early marker for neurons D. This is incorrect. Tuj1 is used to select for cells that have differentiated into neurons. Since fibroblasts were isolated from postnatal mice, no embryonic cells were used in the experiment, which involves the conversion of one adult cell type into another adult cell type

Q29 Researchers conducted an experiment to study Weber's Law. Going from 10- to 12-pound weights created the just noticeable difference for one participant. For this participant, how many pounds need to be added to a 20-pound weight to create the just noticeable difference? A. 1 pound B. 2 pounds C. 4 pounds D. 8 pounds

Ans C; I picked B A. This is incorrect because it suggests that a smaller amount of change will lead to the just noticeable difference in a more intense stimulus, whereas the amount of change needed to produce the just noticeable difference increases as the original stimulus intensity increases. B. This is incorrect because it suggests that the intensity needed to create the just noticeable difference is constant. C is correct. Weber's Law states that just noticeable difference is a ratio of the existing stimulus intensity. Two pounds are 1/5 of 10 pounds and they create the just noticeable difference for 10 pounds. To create the just noticeable difference for 20 pounds, 4 pounds (1/5 of 20 pounds) are needed. Simple math/ratio. 2/10 = x/20 . X = 2/10*20 = 4 D. This is incorrect because it suggests an intensity that is greater than the minimum needed to create the just noticeable difference.

Q7 What is the mechanical work done by the cantilever when the extension increases from 10 nm to 15 nm? A. 2.00 × 10-22 J B. 4.70 × 10-20 J C. 2.50 × 10-19 J D. 5.00 × 10-18 J

Ans C; got it wrong bc I didn't know how to do it To solve this, you had to know triangular area formula which is (.5 x base x height). Nano = 10^-9 and pico = 10^-12 Look at figure 2. Extention from 10-15, means the area under that triangle since W = Fd. Therefore, the answer is 0.5*100pN * 5nm = 0.5*100*10^-12 N * 5 *10^-9 m = 250 *10^-21 = 2.5*10^-19 J Imp unit: J = Nm Work on notebook

Q18 The tendency of doctors to use a physician-centered communication style more often with Black patients is an example of: A. prejudice. B. stereotyping. C. discrimination. D. ethnocentrism.

Ans C; picked B - was debating btw these 2 A. Prejudice is just an attitude, no action/behavioral component. Prejudice as an emotional component. B. Sterotyping is just a cognition and there is no behavioral component. Leads to prejudice. C. Discrimination is the taking acttion based on prejudice. The doctors are actally treating the black male patients differently based on their racial backgrounds D. Ethnocentricm: ones culture is greater than others. This is not the case here.

Q16 Which statement correctly describes both PO43- and TNBS? A. Both TNBS and PO43- are hydrophobic. B. TNBS is hydrophobic and PO43- is hydrophilic. C. PO43- is hydrophobic and TNBS is hydrophilic. D. Both TNBS and PO43- are hydrophilic

Ans D The structure of TNBS shows that it contains three polar substituents on a benzene ring, which makes the molecule hydrophilic. Also, it does not cross the hydrophobic membrane. Phosphate is a negatively charged hydrophilic ion. Phosphate is a charged species and TNBS is also hydrophilic because it does not cross the lipid membrane (according to the passage) and it also has polar substitutents such as the nitrate groups.

Q17 Where are the serotonin receptors 5-HT2A and 5-HT2B most likely to be located in hepatocytes? A. In the nucleus B. In the cytosol C. Embedded in the mitochondrial membrane D. Embedded in the cell membrane

Ans D A. This is incorrect. The hydroxyl group and amine group of serotonin makes the molecule polar and thus, it does not readily cross the phospholipid bilayer of the cell membrane. Therefore, it is not likely to be located in the nucleus. B. This is incorrect. The hydroxyl group and amine group of serotonin makes the molecule polar and thus, it does not readily cross the phospholipid bilayer of the cell membrane. Therefore, it is not likely to be located in cytosol. C. This is incorrect. The hydroxyl group and amine group of serotonin makes the molecule polar and thus, it does not readily cross the phospholipid bilayer of the cell membrane. Therefore, it is not likely to be located in the mitochondrial membrane. D. The hydroxyl group and amine group of serotonin makes the molecule polar and thus, it does not readily cross the phospholipid bilayer that is the cell membrane. It is most likely that the serotonin receptors on hepatocytes are embedded within the cell membrane to facilitate serotonin transport.

Q20 The hardness ions described in the passage are: A. alkaline earth metals. B. strongly acidic cations. C. formed in nature by reduction of other cations. D. derived from atoms upon loss of two electrons.

Ans D A. Fe2+ is a transition metal, not a alkaline earth metal B. Only small molecules that are highly charged cations are strongly acidic. Lewis acids are electron pair acceptors. To increase the likelihood (i.e. acidity) of a base accepting an electron pair then increase the positive charge. After a +3 charge then the smaller the cation gets the more acidic it becomes C. Reducition is the gaining of electrons. Anions are formed by reduction not cations D. All ions described in the passage have a +2 charge, meaning they lost two electrons

Q1: Which conclusion is best supported by the findings in Figure 1? A. Non-obese participants experience more cognitive dissonance than obese participants. B. Participants experience more cognitive dissonance in homophilous groups. C. Non-obese participants conform more than obese participants. D. Participants conform more in homophilous groups.

Ans D A. Misidentifies the concept that is being assessed as cognitive dissonance. B. Misidentifies the concept that is being assessed as cognitive dissonance. C.. Suggests that non-obese participants conform more than obese participants, but this is not correct across all conditions. Figure 1 shows that in the homophilous condition, obese participants conform more than non-obese participants. D. The question requires identifying the concept that is assessed in the study as conformity and combining this information with the numeric information in Figure 1. The "Fraction of diary adopters" refers to the fraction of participants who adopted a diary after they were told that their buddy had started using one, which is an example of conformity. The figure shows that all participants conform more in homophilous groups (indicated by the gray bars). Concept: Cognitive dissonance is when an individual holds two beliefs that are contradictory in nature. In order to relieve cognitive dissonance -- the individual has to make some sort of change in belief structure or behavior. Conformity is usually due to peer pressure. If a buddy is using a diary -> then you are more likely to as well

Q3 Which statement is NOT compatible with the hypothesis that the self-serving bias can account for participants' explanations of their body weights? A. Obese participants view their unhealthy weight as a result of having too many fast food restaurants near home. B. Non-obese participants view their healthy weight as a result of having strong willpower. C. Obese participants view their unhealthy weight as a result of not having time to exercise regularly. D. Non-obese participants view their healthy weight as a result of not having any fast food restaurants near their home.

Ans D A. This is compatible with the self-serving bias. An obese individual is likely to attribute his or her obesity to an external source (availability of fast food). B. This is compatible with the self-serving bias. A non-obese individual is likely to attribute his or her obesity to an internal source (strong willpower). C. This is compatible with the self-serving bias. An obese individual is likely to attribute his or her obesity to an external source (not having time to exercise). D. Self-serving bias suggests that when explaining their own behavior, individuals attribute positive behaviors to internal, stable sources, but attribute negative behaviors to external sources. A non-obese individual would attribute his or her healthy weight to an internal, stable source, such as strong willpower. However, a non-obese individual would not attribute his or her healthy weight to an external source, such as not having any fast food restaurants near home. Therefore, D is incompatible with the self-serving bias.

Q24 Which conclusion is supported by Figure 1? A. Person A showed improvement in both types of memory at post-test. B. Person B showed improvement in memory for childhood events at post-test. C. Both Person A and Person B showed improvement in both types of memory at post-test. D. Person A showed improvement in memory for events right before the brain injury at post-test.

Ans D A. This is incorrect because Person A's memory for childhood events did not improve at post-test. B. This is incorrect because Person B's memory for childhood events is the same at pre-test and post-test. C. This is incorrect because both Person A and Person B did not show improvement in both types of memory. The only improvement was seen in Person A's memory for events right before brain injury. D is correct. Figure 1 shows that Person A showed an improvement in memory for events right before the injury at post-test.

Q13 What modification of Compound 1 will favor transport of Na+ relative to K+? A. Replace X with CH3. B. Replace Y with CH3. C. Increase the ring size by adding one -CH2CH2O-. D. Decrease the ring size by removing one -CH2CH2O-.

Ans D A/B. The X or Y group would not favor one molecule over another because the "CENTRAL portion of the molecule is what selectively binds to cations based on their ionic radius" as discussed in the passage. C. Increasing the ring size would favor larger cations, not smaller ones D. Reducing the ring size could favor one molecule transported over another that are chemically similar. Looking at the periodic table, Na+ is smaller than K+ (K+ is below Na+ on the periodic table). Reducing the ring size would favor this

Q24 Which experimental approach can be used to analyze the metal content of soapy precipitate produced by Reaction 1? Dissolve the solid in a known volume of: A. 0.1 M NaHCO3(aq), then titrate with standardized 0.1 M HCl(aq) using an indicator. B. 0.1 M NaOH(aq), then titrate with standardized 0.1 M HCl(aq) using an indicator. C. 0.1 M NaCl(aq), then titrate with standardized 0.1 M NaOH(aq) using an indicator. D. 0.1 M HCl(aq), then titrate with standardized 0.1 M NaOH(aq) using an indicator.

Ans D Acidic or basic precipitate -> Break it up Mg(FA)COO = Mg2+ and (FA)COO- Mg2+ isn't acidic because it does not contain an acidic hydrogen However, FA-COO- is basic because it can remove hydrogens from solution Therefore you need an acidic solution to pronate the base and make it an acid. Also, adding NaOH would result in MgOH2, which is very soluble, and just result in the dissociation of more Mg2+ ions back into solution, which would continue to react with the FA and precipitate it. You don't want to make something that will immediately dissolve back into its former components. If you add acid, the small amount of FA-COO- (from the dissolution of the precipitate) will react with the H and form FA-COOH. This will not immediately dissolve back into solution. This will also remove FA-COO- ions from solution, pushing the formation of more ions, which will continue to react with H+ if you keep adding HCl.

Q20 Another researcher reviews the study described in the passage and suggests that the medical mistrust experienced by Black men can be explained, in part, by the concept of institutional discrimination. Which statement best describes that concept? A. Discrimination is not systematic, except when observed within institutions. B. If they have a history of unfair treatment, institutions are labeled as discriminatory. C. When several individuals exhibit prejudiced attitudes within an institution, then that institution will also be discriminatory. D. As opposed to discriminatory acts committed by individuals, there are institutional policies that disadvantage certain groups and favor others.

Ans D D is correct. Medical mistrust could be considered part of a general mistrust in institutions (for example, medicine, government, economy) that some social groups experience. Of the options, D is the best description of institutional discrimination, since it identifies the contrast with individual discrimination.

Q23 A pump is used to force an aqueous solution through a pipe at high temperature according to Poiseuille's Law: (refer to notebook for the equation. Can't copy paste here) where ΔP is the pressure difference applied by the pump, ris the radius of the pipe, L is the length of the pipe, and ηis the viscosity of the solution. Which graph depicts the rate of energy consumed over time in order to maintain constant flow through a pipe subject to boiler scale? (graphs on the opposite side

Ans D The energy consumed by the pump is used to create the pressure difference that maintains the viscous flow through the pipe. For a constant flow to occur while the pipe's diameter decreases, the pressure difference should increase in time. According to Poisseuille's Law, at constant flow rate, the pressure difference is inversely proportional to the radius of the pipe to the fourth power. The energy used is equal to the work done in moving the fluid, hence is proportional to the pressure difference. In other words, energy is inversely proportional to the radius of the pipe to the fourth power. As the radius decreases in time, the energy used increases more due to the fourth power dependence of the radius. In other words, more power (energy per unit time) is required to maintain the same flow through a pipe with decreased radius. Hence, the only choice that depicts such a power increase that is getting larger in time is shown in graph D you're basically looking for a graph that will give you a positive, nonlinear relationship based on the form of Poiseuille's law they show you -- i.e., you'll need more energy to push through a constricted pipe, and the radius term is to the fourth power.

According to the hypothesis presented in the passage, which drug is most likely to cause a decline in the wheel-running behavior of ABA-induced rats? A. Alcohol B. Cocaine C. Marijuana D. Morphine

Ans D The hypothesis is the passage is basically the fact that exercise induce endorphins. D is correct. The hypothesis suggests that exercise addiction results from the fact that exercise increases endorphin levels. If this neurotransmitter is stimulated by a drug, then ABA-induced rats would be less likely to run in their wheels (because their endorphin levels would be elevated by an alternative method). The only drug in the list that is an endorphin agonist is D, morphine.

Q3 Which change to an equilibrium mixture of compounds 1-4 will increase the ratio of Compound 4 to Compound 1? A. Adding Compound 3 B. Increasing the temperature C. Adding more catalyst and pyridoxal phosphate D. Adding Compound 2

Ans D This was an easy question bc its referring to le chateliers principle. A. C1 +C2= C3+ C4. Adding C3 will shift reaction to the left. Increasing C1. This is the opposite answer B. Increasing the temperature will not do anything because the reaction is thermoneutral. Heat is neither a reactant nor a product C. Adding a Catalyst (enzyme) will not increase or decrease concentration. It will only change the rate of the reaction D. Correct Answer. C1 +C2= C3+ C4. Adding C2 will shift reaction to the right, increaseing the C4 concentration

Q7 A study finds that low birth weight is associated with delayed cognitive development in childhood. Based on this result and the studies in the passage, children born to women living in which neighborhoods are likely to be at greater risk of falling behind in elementary school? A. Asian American enclaves B. Neighborhoods next to Asian American enclaves C. Neighborhoods with extensive reciprocal exchange D. Neighborhoods with high rates of violent crime

Ans D in the passage it literally said it

Q14 Lysine and amino acids with similar chemical characteristics in histones most likely promote the interaction of histones with which DNA components? A. Purines B. Pyrimidines C. Deoxyribose D. Phosphate groups

Ans D picked it bc lys is basic and + so it will interact with neg phosphate groups

Q26 Positron Emission Spectroscopy (PET) imaging involves injecting a patient with a positron (anti-electron) emitting isotope. The emission of positrons occurs during: A. alpha decay. B. alpha absorption. C. gamma absorption. D. beta decay.

Ans D Beta decay is the decay in which a beta particle (electron or positron) is emitted from a heavy-atom nucleus

Q12 Compound 1 remains mainly confined to the central phase during the experiments because it is: A. hydrophilic with only polar groups. B. hydrophilic with both polar and nonpolar groups. C. lipophilic with only nonpolar groups. D. lipophilic with both polar and nonpolar groups

Ans D Y group of the molecule has an ether (CH2OCH2) and amide (CO-N---). Ethers is nonpolar. Also, the entire main backbone of the molecule itself is a giant multi-ether complex (18 crown-6 O ether). Hence, this is nonpolar molecule (liphophilic loving) molecule. This is further asserted when the passage states that Compound 1 dissolves in organic liquid layer. Secondly,, it has polar groups on the X substituent (CO2H, carboxylic acid) Concept: Crown ethers have the ability to solvate metal cations in nonpolar/organic solvents. If you look at the figure containing the picture of the crown ether, you can see that the face of the oxygens and its lone pairs are all on the inside of the ring. This is where the oxygen atoms will bind metal cations and what allows the ether to be nonpolar and dissolve in organic solvents

Q18 The 32P label was generated from naturally occurring phosphorous by: A. removing a neutron from the nucleus. B. adding a proton to the nucleus. C. adding three electrons to the atom. D. adding a neutron to the nucleus.

Ans D - got this wrong bc I looked at the atomic number not the atomic mass Check the periodic table; see that P is normally 31. Therefore to get P32, you need to add a neutron

Q9 Chloramphenicol did NOT inhibit translation in E. coli cells containing the cat6xbs expression plasmid. What experimental parameter could be changed in order to affect translation inhibition? A. Increase the chloramphenicol concentration. B. Increase the chloramphenicol incubation time. C. Alter the incubation temperature by a few degrees. D. Use an alternate antibiotic.

Ans D, I picked A (honestly was confused about it) A, B and C are wrong bc chlorophenicol is ineffective. Any type of alteration other than the change of the type of antibiotic will not be effective D. Using an alternative antibiotic other than chloramphenical because E.coli is resistant against chloramphenicol antibiotic because it has an enzyme encoded for by the cat gene - chloramphenicol acetyltransferase - renders chloramphenical ineffective and gives E.coli resistance against this antibiotic. E.coli would not be resistant against all antibiotics though.

Q16 According to the passage, platelets are LEAST likely to contain: A. transmembrane serotonin transporters. B. ribosomes. C. serotonin. D. Ki67.

Ans D. A. The passage states that platlets carry 95% of the seretonin synthesized outside of the paltelets, therefore it must have a transmembrane seretonin trnasporter. B. This is incorrect. Platelets are formed from large cells called megakaryocytes. Platelets consist of plasma membrane-encased megakaryocyte cytoplasm, which contains ribosomes. This option is very likely to be true; therefore, it is not the correct answer C. Again, passage states that platelets carry serotonin D. Platelets are cell fragments without nuclei and therefore would not be expected to contain a protein like Ki67 that is detected exclusively in the nuclei of proliferating whole cells (stated in the passage). Because this option presents a situation that is unlikely to be true, it is the correct answer to the question

Q23 Does the experimental approach described in the passage yield cells that could be used in an animal model of Parkinson disease to replace dopamine-deficient neurons in the brain? A. Yes, because the cells obtained have the functional characteristics of nerve cells. B. Yes, because the cells obtained are similar to cells in the central nervous system. C. No, because the cells obtained may contain tumorigenic pluripotent cells. D. No, because the cells obtained lack the correct neurotransmitter phenotype.

Ans D; I picked B A. This is incorrect. It is not sufficient that the cells are functional neurons, the cells have to be dopamine-synthesizing neurons. b. This is incorrect. It is not sufficient that the cells are similar to neurons that may be found in the brain, the cells have to be dopamine-synthesizing neurons. C. This is incorrect. The cells obtained are differentiated neuron-like cells that are derived from differentiated fibroblasts; no pluripotent intermediate is involved in the production of the reprogrammed cells. D. Although the cells obtained under the experimental conditions are similar to CNS neurons and may therefore be similar to neurons found in the brain, the neurons that are generated produce either glutamate or GABA. Since Parkinsonism is associated with deficiency of dopamine-producing neurons, neither of these 2 types of cells would alleviate the symptoms of this disorder.

Q2 The side chain of which amino acid is used by transaminases to bind to pyridoxal phosphate in the enzyme's resting state? A. Val B. Asp C. Phe D. Lys

Ans D; I picked B - I got this wrong bc I didn't read carefully. In the passage it shows the enzyme that pyridoxal phosphate uses to do the reaction and the structure is lys

Q9 What are the units for the rate constant ku discussed in the passage? A. M-1•s-1 B. M•s C. M•s-1 D. s-1

Ans D; picked B For this question, you had to know the units for the 0th, 1st, 2nd rate laws and also figure out which one applies here. You had to know that this is a unimolecular unfolding mechanism bc the unfolding of a protein only depends on itself not anything else. This makes this unfolding process first order. A. This is the unit for second order rate constant. Second order graphs are [A]/t (inverse parbola) or 1/[A] vs time where k = slope (linear positive slope) B. not units for any rate constant, all rate constants are 1/s C. This is order for zeroth order where graph is depicated as [A] vs time where k = -slope D. This is the unit for a first order reaction. We know that this is a first orde reaction because the passage states that the unfolding occurs in a unimolecular mechanism

Q21 After taking the experimental drug, Person A showed an improvement in: A. anterograde memory. B. retrograde memory. C. semantic memory. D. short-term memory.

Ans b; picked C was thinking B but then I thought that both childhood memories and memories right before the event counts for retrograde and there was no increase in the childhood memories so didn't pick that but B is the right answer bc retrograde memory refers to the ability to remember the information before brain injury. Figure 1 shows that Person A showed a considerable increase in memory for events right before brain injury during the post-test.

Q19 The amino acid precursor of serotonin is best described as having which type of R group? A. Nonpolar, aliphatic B. Polar, uncharged C. Aromatic D. Negatively charged

Ans c passage says serotonin comes from tryptophan which is aromatic

Q8 Which concept is most relevant for explaining the relationship between social resources and birth-weight outcomes that was found in the first study in the passage? A. Poverty B. Heredity C. Social networks D. Socialization

Ans is C; I picked A A. This is incorrect because the description of the study in the passage specifically mentions a lack of social resources but does not refer to a lack of material resources. It is possible that poverty is a contributing factor, but it was not specifically identified in the passage. B. This is incorrect because, although a genetic component is possible with low birth weight, the passage is specifically about social and environmental associations. (no mention in the passage) C. Option C is correct because reciprocal exchange, as it is described as a social resource with the first study in the passage, is best identified as a social network process. D. This is incorrect because socialization refers to the lifelong process by which norms and values are learned, and thus does not specifically address a social resource (such as reciprocal exchange).

Q4 All the participants in the study are given information regarding the benefits of a healthy diet. According to the cognitive dissonance theory, which hypothetical finding is most likely? A. Obese participants will change their unhealthy eating behaviors. B. Non-obese participants will change their unhealthy eating behaviors. C. Obese participants will question the validity of the information provided. D. Non-obese participants will overemphasize the importance of the information provided.

Ans is C; I picked A Basically, I thought that they would change their behavior to align their attitudes when there is a discrepancy btw attitudes and behaviors but thats not the case. When people experience cognitive dissonance, you'd think that they would change their behavior to match their attitude, but instead they change their attitudes to match their behavior. If you think about it, it's true; if you're eating fast food and someone tells you that it's bad for your health, you're more likely to say "screw health" or "it's not that bad" than to stop eating fast food. A. This is incorrect because it is not the most likely outcome. It suggests that obese participants will change their behavior to align with their attitudes when there is a discrepancy between their attitudes and behaviors. B. This is incorrect because non-obese participants are less likely than obese participants to experience cognitive dissonance and change their behavior due to cognitive dissonance. C. C is correct. According to the cognitive dissonance theory, when an individual's attitudes are incongruent with his or her behavior, this leads to cognitive dissonance. To eliminate cognitive dissonance, the individual can either change his or her attitudes or his or her behavior. The theory posits that individuals are more likely to adjust their attitudes to align with their behavior than the other way around. Therefore, obese participants are likely to question the importance of the information provided. D. This is incorrect because non-obese participants are less likely than obese participants to experience cognitive dissonance and change their attitudes due to cognitive dissonance.

Q8 The bglF transcript is known to have a short half-life within the cytosol. What mechanism is most likely responsible for transport of this transcript to the cytoplasmic membrane once it is synthesized? A. Diffusion across the cytoplasm B. Transport via attachment to the mitotic spindle C. Active transport along cytoskeletal filaments D. Transport from the endoplasmic reticulum in vesicles

Ans: C A. this is wrong bc this would take a long time to occur and so the protein would degrade in the cytoplasm before reaching the intended target B. This is wrong bc prokaryotes dont have a mitotic spindle bc they dont undergo mitosis. Instead they divide by binary fission C. bglF is a protein that is coded for by the bgl gene. For a protein to be made, the following process must occur: Transcription from DNA (to mRNA) then translation (to a protein). In prokaryotes, this process occurs simultaneously as there is no compartmentalization (nuclear membrane). In eukaryotes, once the mRNA transcript is synthesized from the DNA in the nuclear region, it leaves the nuclear region, it is readily exposed to cytoplasm. Both eukaryotic and prokaryotic cells have posttranscriptional modifications to mRNA. The posttranscriptional modification that allows the mRNA strand of both eu/prokaryotes to survive in the cytoplasm is a 3' polyA tail addition (repeating Adenosine nucleotide repeats). Prokaryotes (such as E.coli as discussed in the passage) typically have a shorter polyA tail, hence the transcript of prokaryotes is considered less stable/degrades quicker (few minutes). To prevent degradation and to get the protein to the appropriate location (at the cell membrane), the best way would be to involve active transport along cytoskeleton filaments straight to the cytoplasmic membrane once the transcript is synthesized. Also, remember that cytoskeleton lies right under the plasma membrane. D. This answer choice is tempting, although E.coli (a prokaryote) does not have any membrane bound organelles. Remember that: prokaryotes have no nuclear membrane, no nucleolus, no golgi apparatus, no ER, and are incapable of phagoctyosis/pinocytosis/ exocytosis (no vessicles), lack sterol in their membrane, They also have different ribosome subunits (smaller, 30+50 = 70S while eukaryotes have 40+60 = 80S). Because they lack Endoplasmic reticulum and vessicles, this is not a possibility as it would be for eukaryotic cells. Background: cytoskeletal filaments include microfilaments, microtubules and intermediate filaments. Prokaryotes have similar features but just not identical

Q6 Which other cellular components are likely to be located near the lacY6xbs transcript in the cell membrane? A. Proteins and glycolipids B. Glycolipids and sterols C. Sterols and phospholipids D. Phospholipids and proteins

Ans: D A. prokaryotes like E. coli contain proteins (75%) but not glycolipids. Only eukaryotes contain both glycolipids and glycoproteins which serve as recognition molecules B. Contain neither. Only eukaryotes contain sterols (cholesterol) in their phospholipid bilayer for fluidity purposes C. Contain only phospholipids D. The passage states lacY is located near the cell membrane, so the ans will have to be the cell components of an E. coli membrane (a prokaryote). E. coli is a gram neg cell which contain a thin layer of peptidoglycan and a outer cell wall which contains LPS. E. coli membranes, like all cells, have plasma membrane (inner most) and this plasma membrane contains 75% protein and 25% phospholipids. Typically, these are the major two components of all cell membranes Extra information: Most prokaryotes cell membranes also do contain include spingolipids (which can be a phospolipid when head group contain a phosphates - this is then called a spingomyelin. The other types of spingolipids are eramide (OH head group), or glycolipids (sugar head group) such as cereboside (one sugar), globoside (2+ sugar), and ganglisoide (oligosacharide with NANA/Salic acid)

QUESTION 1 The muscle subtype represented by Culture C is LEAST likely to be characterized by: A. a fast rate of muscle contraction. B. the ability to engage in oxidative and anaerobic respiration. C. the presence of medium-sized motor units. D. low densities of mitochondria and capillaries.

Ans: D According to the passage, "IIa" in the table is Culture C. I got this question wrong bc I didn't read the question carefully and the fact that they asked for the "LEAST" presentative characteristic. A. Wrong bc table shows the "twitch duration" is moderately fast B. This one can be tricky bc in the "activity used for section" it says anaerobic but remember that type IIA muscle fiber still exhibits "high oxidative capacity" which means it would be able to perform BOTH anaerobic and aerobic processes. C. Wrong bc it says it in the table D. RIGHT ANS: Since oxidative capacity is high, the # of mitochondria and capillaries should be high not low CONCEPT This question requires you to understand the different btw Type I and Type II muscle fibers. Type I (slow-twitch) - mitochondria rich, slow twitch meaning slow conduction velocity, slow contraction speed, long, low power, aerobic, long duration and fatigue free Type IIA and Type IIB- Type IIA intermediate btw I and IIB. IIB are white, fast contaction speed, fast twitch (fast conduction activity), anaerobic, shirt, easily fatigue, power, ATP from creatine phosphate.

Q29 Increasing the volume of air that reaches the alveoli and takes part in gas exchange will cause blood pH to: A. increase, because the neural mechanisms that remove acid from the blood will be activated. B. increase, because the partial pressure of CO2 in the blood will decrease. C. decrease, because the affinity of hemoglobin for oxygen will be increased. D. decrease, because the work associated with increased ventilation will come more O2.

B

Q30 Scientists have hypothesized that mitochondria evolved from aerobic heterotrophic bacteria that entered and established symbiotic relationships with primitive eukaryotic anaerobes. According to this hypothesis, the bacteria that entered primitive eukaryotic cells were able to carry out which function(s) that the primitive eukaryotic cells could not? A. Glycolysis B. Citric acid cycle and electron transport C. Cell division D. Transcription and translation

B

Q26 The Gibbs free energy equation can be used to predict whether a reaction will proceed spontaneously. For which relative values of ΔH and ΔS will a spontaneous reaction always occur? A. A positive ΔH and a negative ΔS B. A positive ΔH and a positive ΔS C. A negative ΔH and a negative ΔS D. A negative ΔH and a positive ΔS

D

Q27 Which type of molecule is LEAST likely to be found in a eukaryotic cell membrane? A. Phospholipid B. Cholesterol C. Glycoprotein D. Peptidoglycan

D

Difference between heterochromatin and euchromatin

Heterochromatin (METHYLATION) - INACTIVE - in the center where its TIGHTLY bound to HISTONES EuchromatiN (ACETYLATION) - ACTIVE - outskirts of town where its available to be transcribed. The passage even says "The expressed var gene is located in a different place in the periphery of the nucleus than are silent var genes."

What is lipophilic?

lipid soluble (hydrophobic)