NS 3

statistical power can be inferred from?

Raw data or simply observing how many patients/trials were completed.

The individual thermodynamic contribution of W (RM)chain A was found to increase the interaction energy of the MKR681H dimer. If so, what must be true for chain A? A. ΔGsolv < 0 B. ΔGsolv = 0 C. W (RM)chain A > 0 D. Hintra < 0

A is correct. Consider the following expression: W (RM)int = W (RM)dimer - W (RM)chain A - W (RM)chain B If W (RM)chain A increases the interaction energy of the MKR681H dimer, W (RM)int, W (RM)chain A must be a negative quantity, such that the -W (RM)chain A term in the equation becomes a positive value. Equation 2 also tells us that W (RM) = Hintra + ΔGsolv. If W (RM)chain A < 0, then one or both of the terms Hintra and ΔGsolv must be negative. According to the passage, the intramacromolecular enthalpy term Hintra accounts for positive terms. For this reason, it must be true that ΔGsolv < 0. B: If ΔGsolv = 0, then W (RM)chain A > 0, and the term -W (RM)chain A decreases the interaction energy of the MKR681H dimer. This contradicts the question stem. C: If W (RM)chain A > 0, the term -W (RM)chain A would decrease the interaction energy of the MKR681H dimer. D: According to the passage, the intramacromolecular enthalpy term, Hintra, accounts only for positive terms.

Which of the following strategies would likely reduce the occurrence of criminal behavior from a differential association theoretical perspective? A. A child in a high-crime area being adopted by an affluent family B. Repeated messages in school that students are expected to uphold prosocial values C. Deemphasizing the importance of values, such as material wealth, that would lead individuals to commit crimes D. Increasing penalties for crime to a degree that individuals are dissuaded from committing it

A is correct. According to differential association theory, individuals engage in criminal choices because they are exposed to it, while individuals who don't commit crimes have not been exposed to this type of behavior. Thus, a child going from an area in which he is likely to be exposed to crime to an area where he is not likely to be exposed will reduce the likelihood of this individual committing a crime. B: This is an example of reducing crime through labeling theory. C: This is an example of reducing crime through strain theory. D: This is a behavioral approach to reducing crime.

At physiological pH, what will be the principal interactions between Arg681 and Cys683? A. Ion-dipole interactions between two chain B residues B. Dipole-dipole interactions between chain A and chain B residues C. Hydrogen bonds between a chain A and a chain B residue D. Disulfide bonds between two chain B residues

A is correct. According to paragraph 2, "Arg681 engages in two direct interactions: a salt bridge with Asp145 of chain A and an intra-subunit interaction with Cys683." Since the passage describes the interaction between Arg681 and Cys683 as "intra-subunit," the two residues must be located on the same chain (eliminate choices B and C). In order for arginine to participate in a salt bridge, it must be ionized, indicating that a nitrogen of the side chain guanidinium group must be protonated. Cysteine is not ionized at physiological pH (~7.4), but it is polar. Thus, its interactions with the other residue are most likely to involve ion-dipole interactions. This is sufficient information to eliminate all answer choices other than A. B, C: Arg681 and Cys683 are both located on chain B. D: Two sulfhydryl groups present in individual cysteine side chains are required for the formation of a disulfide bond.

A woman visits her doctor to receive medical test results. While the patient anxiously waits, the doctor stops in to shake hands before going to analyze the results. The patient mistakes the handshake as a cold dismissal rather than a warm greeting. This is an example of which sociological theory? A. Symbolic interactionism B. Social constructionism C. Exchange-rational D. Feminist theory

A is correct. According to symbolic interactionsim, there are 3 core principles to communication: meaning, language and thought. These core principles lead to conclusions about the creation of a person's self and socialization into a larger community. Meaning states that humans act toward people and things according to the meanings that give to those people or things. Symbolic interactionism holds the principal of meaning to be the central aspect of human behavior. Language gives humans a means by which to negotiate meaning through symbols. Humans identify meaning in speech acts with others. Thought modifies each individual's interpretation of symbols. Thought is a mental conversation that requires different points of view. This scenario is an example of symbolic interactionism. B: Social constructionism examines the development of jointly constructed understandings of the world. It assumes that understanding, significance, and meaning are developed not separately within the individual, but in coordination with other human beings. C: Exchange-rational theory posits that patterns of behavior in societies reflect the choices made by individuals as they try to maximize their benefits and minimize their costs. D: Feminist theory is a branch of feminism that seeks to explain the nature of gender inequality. It examines women's social roles, experience, interests, and politics in a variety of fields such as education, medicine, and business.

Atrial fibrillation is the most common heart arrhythmia, causing palpitations, fainting and chest pain. According to the information in the passage, what is most likely to be observed in the ECG during atrial fibrillation? A. Missing P waves and an irregular rate B. Complete absence of P waves and a regular rate C. Complete absence of R waves and an irregular rate D. Missing R waves and a regular rate

A is correct. An arrhythmia is an irregular time period between heartbeats, not necessarily a rate that is too slow or too fast. Since the question asks about atrial fibrillation, and the P wave is attributed to contraction of the atria, the correct answer should show the P wave being effected. This does not mean that the atria are not contracting at all (eliminate choice B), but it indicates that blood is not efficiently being pumped from the atria to the ventricles. B, D: These choices incorrectly associate arrhythmia with a regular heart rate. C: Absence of an R wave would be indicative of ventricular problems, and is more dangerous problem than AFib. The R wave is a strong, sharp electrical signal coming from the ventricles. Complete absence of an R wave suggests ventricular contraction is likely not occurring or fibrillating at best, which is life-threatening and is associated with myocardial infarction, commonly known as a "heart attack."

According to the information presented in the passage, how would the absorbance at 570 nm change for enzyme A if a linear starch was used? A. It would remain relatively constant. B. It would decrease. C. It would increase. D. It would initially increase and then decrease as the starch is consumed.

A is correct. Based on Figure 3, enzyme A increased absorbance with no change in current. Thus, enzyme A likely breaks down 1,6 linkages, producing more linear polysaccharides, thereby increasing absorbance. However, if the initial starch is an entirely linear polysaccharide, then a debranching enzyme like enzyme A would have no effect on absorbance. B: Since enzyme A functions to debranch starch, and since the starch in this scenario is already linear (not branched), enzyme A will not do anything to the starch in solution. As such, we would not expect the amount of starch bound to iodine to decrease over time, so choice B is incorrect. C: This would be correct if the starch were debranched over time, allowing more iodine to bind to the linear starch product. However, in this case, the starch is linear to begin with. D: Again, we would not expect to see an increase in absorbance. Also, since enzyme A does not break 1,4 linkages, the starch in this solution would not be consumed.

If the Gibbs free energy of the liquid-crystalline phase for an aqueous dispersion of DPCC and raffinose is 17.9 kcal/mol, which of the following values is nearest the Gibbs free energy of the associated gel phase? A. 3.5 kcal/mol B. 32.3 kcal/mol C. 566.1 kcal/mol D. 601.9 kcal/mol

A is correct. By rearranging Equation 1, ΔG = Gl - Gg, an expression for the Gibbs free energy of the gel phase may be found: Gg = Gl - ΔG. The Gibbs free energy change for the phase transition of raffinose is given by ΔG = ΔH - TΔS. Substituting thermodynamic values from Table 1: ΔG = (16.0 kcal/mol) - [(3.15 x 102 K)(5.05 x 10-3 kcal/K•mol)] ΔG = 16.0 kcal/mol - 1.60 kcal/mol = 14.4 kcal/mol Substituting into the rearranged form of Equation 1: Gg = Gl - ΔG = 17.9 kcal/mol - 14.4 kcal/mol = 3.5 kcal/mol, choice A. B: This value is Gl + ΔG, not Gl - ΔG. C, D: These answers are the result of miscalculation.

If a modern portable defibrillator uses as 12 V battery and a 20 μF capacitor, what is the total charge is stored on the plates of the capacitor? A. 0.24 mC B. 24 mC C. 24 C D. 60 C

A is correct. Capacitance (C, measured in farads) is the amount of charge stored per volt, expressed with the equation C = q/V. This means that charge (q) = VC. (Do you remember the home shopping network, QVC?) The capacitance is given as 20 x 10-6 F. Substituting the voltage (12 V) and capacitance into the equation gives us the charge (in coulombs). (20 x 10-6 F)(1.2 x 101 V) = 24 x 10-5 C = 2.4 x 10-4 C = 0.24 x 10-3 C = 0.24 mC Remember when making the exponent larger, we must make the coefficient smaller by the same factor. B, C: These answers constitute order-of-magnitude errors. They may result from failing to realize that 20 μF = 20 x 10-6 F, or from confusing the relationship between coulombs (C) and millicoulombs (mC). D: This answer results from miscalculation

What fMRI-measured brain activity would be expected in a subject witnessing a high-content argument? A. The frontal lobes in high-motivation, high-knowledge people are more active than the temporal lobes in low-motivation, low-content people. B. The temporal lobes in high-motivation, high-knowledge people are more active than the frontal lobes in low-motivation, low-content people. C. The frontal lobes in low-motivation, low-knowledge people are equally as active as the temporal lobes in high-motivation, high-content people. D. The temporal lobes in low-motivation, low-knowledge people are more active than the frontal lobes in high-motivation, high-content people.

A is correct. First, you must recognize that according to the information discussed in the passage, a high-content argument will be persuading to high-motivation and high-knowledge individuals, while an argument with high sensory content will be appealing to low-motivation and low-knowledge individuals. Then, you must reason that a high-content argument will be processed logically, which is the domain of the frontal lobes, and that a high-sensory argument will be processed emotionally, which is a temporal lobe function. The image below depicts the lobes of the human brain. B: The pairings of the brain areas are backwards in this answer. C: It is unlikely that during a high-content argument the logic-processing region of the low group will be exactly as active as the emotion-processing areas in the high group. If anything, we would expect at least some activity in the frontal lobes of the low group, just not as much as the high group. D: Both the ordering and brain area are incorrect here.

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

A is correct. Informal social control happens when individuals and groups try to improve conformity to norms and laws by shaming or pressuring others. B: Formal social control is establishing or improving conformity to norms by enforcing laws or imposing sanctions. C: Mind control is a type of formal control typically attained through torture, murder, imprisonment, and exile. D: In criminal psychology, the rational choice theory adopts a utilitarian belief that man is a reasoning actor who weighs means and ends, costs and benefits, and makes a rational choice. Rational choice theory is not relevant to the details discussed by the passage.

What kind of social control is enforced by the internet vigilantes in the passage? A. Informal social control B. Formal social control C. Mind control D. Rational choice

A is correct. Informal social control happens when individuals and groups try to improve conformity to norms and laws by shaming or pressuring others. B: Formal social control is establishing or improving conformity to norms by enforcing laws or imposing sanctions. C: Mind control is a type of formal control typically attained through torture, murder, imprisonment, and exile. D: In criminal psychology, the rational choice theory adopts a utilitarian belief that man is a reasoning actor who weighs means and ends, costs and benefits, and makes a rational choice. Rational choice theory is not relevant to the details discussed by the passage.

In Figure 1, Korsakoff's syndrome patients showed a marked increase in: A. confabulation. B. declarative memory. C. procedural memory. D. aggression.

A is correct. One of the symptoms of Korsakoff's syndrome is confabulation - making up memories to fill in gaps and then believing that those memories are true. When asked to recognize sentences, Korsakoff's patients (at a rate vastly higher than normal) picked incorrect sentences, suggesting they were confabulating the memory of having heard those sentences. B: Declarative (or explicit) memory is the memory of facts and events. The sentence recall/recognition task in Figure 1 is a test of declarative memory. Korsakoff's patients perform much worse on this test than do normal patients, so we would not say that Korsakoff's patients showed an increase in declarative memory. C: Procedural memory relates to the origami task in Figure 2, not Figure 1 (which this question asks about). D: No evidence is given in the passage that the Korsakoff's patients behaved aggressively.

A group of researchers is interested in learning more about the public's attitudes toward Internet vigilantes. They organize focus groups for adults who frequently use the Internet and ask them to talk about their experiences online and with Internet vigilantism. What kind of research are they performing? A. Qualitative research B. Quantitative research C. Case studies D. Naturalistic observational research

A is correct. Qualitative research is used to gather in-depth information about a specific research question, and often involves the use of focus groups or other small samples to better understand phenomena. B: Quantitative research involves the use of tests or questionnaires to produce numerical scores that can be evaluated statistically. C: Case studies are typically extensive interviews or reports about the experience of one specific person, like someone who was accused by Internet vigilantes but later proven innocent. D: Naturalistic observational research involve observing participants in their everyday lives. For example, researchers could monitor a chat room after a violent event and see how people respond (there is no manipulation of variables in this type of research).

Which of the following best explains the tumor reduction capacities of the isotopes observed in the study? A. GKS-X emits more radiation per unit time, causing more cell death. B. GKS-X emits narrower radiation streams that target the tumor. C. GKS-Co anti-tumor emissions decompose much more slowly than GKS-X emissions. D. GKS-Co does not emit sufficient radiation to kill tumor cells.

A is correct. Table 1 shows that GKS-X is more effective at reducing tumor size (as shown by its lower RTV values over time). We need to choose a statement that can explain this improved ability of GKS-X to kill cancer cells. We must think about the differences between X and 60Co, chief among them the fivefold difference in half-life. Because X has a shorter half-life, it is logical that it would release more radiation in the same period of time because it undergoes decay more quickly. This higher dose of radiation would kill more cells, cancerous and non-cancerous, leading to a larger reduction in tumor size and more undesirable side effects. This corresponds to answer choice A. B: Nothing in the passage suggests that the radiation beams emitted by 60Co and X are different. A narrower beam would reduce the collateral damage done to tissue, but it would not explain the much greater efficacy of X with regard to killing tumor cells. C: If GKS-Co involved radioactive material that decayed much more slowly than that in GKS-X, we would expect the GKS-Co tumor-reducing effects to catch up to or surpass GKS-X as time passed. Instead, we can see in Table 1 that the disparity gets larger as time post-surgery increases. D: This answer choice is contradicted by Table 1; GKS-Co was effective at killing tumor cells, as there was a large reduction in tumor size after surgery. It just was not as effective as GKS-X.

Which subtype of schizophrenia often involves normal cognitive functioning? A. The paranoid type B. The catatonic type C. The undifferentiated type D. The disorganized type

A is correct. The defining feature of the paranoid subtype is the presence of auditory hallucinations or prominent delusional thoughts about persecution or conspiracy. However, people with this subtype are often more functional in their ability to work and engage in relationships than people with other subtypes of schizophrenia. B: The catatonic subtype involves disturbances in movement. Affected people exhibit a dramatic reduction or increase in activity. Other symptoms associated with the catatonic subtype include parrot-like repeating of what another person is saying or mimicking the movements of another person. Their cognitive function can often be impaired. C: The undifferentiated subtype is diagnosed when people have symptoms of schizophrenia that are not sufficiently formed or specific enough to permit classification of the illness into one of the other subtypes. Their cognitive abilities can fall within a wide range of ability. D: This subtype's predominant feature is disorganization of the thought processes. As a rule, hallucinations and delusions are less pronounced. These people tend to have significant impairments in their ability to maintain the activities of daily living. Even the more routine tasks, such as dressing, bathing or brushing teeth, can be impaired or lost.

Which of the following is least likely to be observed in a patient experiencing hyperventilation? A. Hypoxia B. Net exhalation of CO2 C. Increased blood pH D. Increased hemoglobin O2 affinity

A is correct. This question asks us to determine the effects of hyperventilation. During hyperventilation, there is a loss of CO2 and an increase in O2 in the blood. Hypoxia is another term for oxygen deprivation. B: During hyperventilation, CO2 is lost due to excess exhalation. C: Loss of CO2 corresponds with increased blood pH. D: Loss of CO2 corresponds with increased hemoglobin affinity for O2.

Infants are not vaccinated until they reach a certain age, as their immune systems are not strong enough to safely encounter even an attenuated virus. If an individual has a newborn, should her family, friends, and neighbors get vaccinated for the seasonal flu? A. Yes, and they should avoid direct contact with the newborn for a few days after getting the vaccine. B. No, because the vaccine will not provide any protection for the newborn. C. Yes, this will allow them to immediately have direct contact with the newborn after getting the vaccine. D. Yes, but only if they will be in contact with the family and never in direct contact with the newborn.

A is correct. Vaccination of the majority of the population helps to prevent the people around those at-risk (the young or old who are unvaccinated or with otherwise compromised immune systems) from being in contact with a contagious person. Anyone who is vaccinated is likely to find any infection by the flu virus quickly dealt with by their own immune system, before significant numbers of flu virus have a chance to propagate and cause the contagious stage of the disease. However, immediately after receiving the vaccine, the small amount of virus might have a chance of spreading to the child (or, in some cases, the vaccine will lead to a full-blown infection), so the vaccinated person should avoid contact with the child until the virus has been beaten and cleared from the system. The concept of vaccination of a large proportion of individuals to improve the resistance of the population as a whole is termed herd immunity. Herd immunity is depicted below. B, C, D: Vaccination reduces the chance of the immune-compromised person from being infected in the first place. This is true for those who will be in direct contact and for those who will be in indirect contact, as it's all about breaking the chain of potential transmission.

The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation? A. 2.5 × 101 A B. 2.5 × 10-2 A C. 5.0 × 10-2 A D. 5 A

A is correct. We are given R, V, and time, and the question asked us to determine current (charge/time, or C/s). We do not need to use the time since we should know Ohm's law, V = IR. If I = V/R and the question asks for the maximum current, we should use the smallest available resistance. I = V/R = (500 V) / (20 Ω) = (50) / (2) = 25 amps = 2.5 × 101 A B, C: These answers result from miscalculation. D: This is the answer obtained if we used the maximum resistance (100 Ohms) of the skin.

Which of the following represents a limitation to the design of this study? A. The age of the participants B. Lack of knowledge about OCD C. Ethical implications of delivering shocks to participants D. Using a non-clinical population

Any findings during this study would be more validated if the researchers could have gathered patients who had been diagnosed with OCD instead of having participants that simply had obsessive-compulsive traits.

A principal of a high school seeks to establish rules and systems in the school that reflect a meritocracy. Which of the following goals must these systems achieve if the principal is to successfully establish a meritocracy? I. Outcome equality II. Skill equality III. Opportunity equality A. I only B. III only C. II and III only D. I and III only

B is correct. A meritocracy is a society of people whose progress within the society is based on ability and talent rather than on class privilege or wealth. This requires that everyone be afforded the same opportunities to advance yet only be rewarded based upon individual outcomes due to their individual talents and/or abilities, which can vary between persons. I: Outcome equality, or "equality of outcome," refers to a state in which the lives of individuals in the society are of a similar quality. Outcome equality is often discussed with regard to goods and their distribution in an equal fashion. A meritocracy does not require outcome equality. II: Members of a meritocracy can vary with regard to their skills, so skill equality is not required for such a society. A: RN I is incorrect, while RN III is correct. C: RN II is incorrect. D: RN I is incorrect.

According to French and Raven's bases of power model, when compared to a high motivation high knowledge individual, someone with low motivation and low knowledge is more likely to be influenced by a person with: A. expert power. B. referent power. C. legitimate power. D. coercive power.

B is correct. According to the bases of power model, an individual with referent power exerts control by appealing to others' desire to belong to a group. This type of control is most likely to appeal to individuals through external factors, such as appearing desirable or feeling included and not knowledge or logic or evidence. Thus, a low-motivation, low-knowledge individual would most likely be motivated by this type of persuasion. A: An expert tends to motivate through using his knowledge of subject matter, which would likely appeal to high-motivation and high-knowledge people. C: Those with legitimate power, such as a president, exert power through the legitimacy of their role. There is no indication that this would lead either group to be more persuaded. D: Those with coercive power exert control through force or its threat. It is likely that both high and low groups would be persuaded by this.

What is the pH of a 0.10 M aqueous solution of acetylsalicylic acid? A. 1.0 B. 2.3 C. 3.5 D. 4.1

B is correct. Acetylsalicylic acid is a weak acid, with a pKa of 3.5. Therefore, the pH of this solution must be less than the pKa, because the compound is primarily in its acid form and a pH of 3.5 would mean that the concentration of weak acid and conjugate base were equal (a buffer). Choices C and D can be eliminated. A pH of 1.0 for an acid whose concentration is 0.10 M would require complete dissociation (as in a strong acid), eliminating choice A. Alternatively, the pH can be determined from the equilibrium expression: Ka = [H+][A-]/[HA] 10-3.5 = x2/(0.1-x) Since x will be small, we can approximate 0.1-x ~ 0.1, giving: 10-3.5 = x2/(0.1) 10-3.5(0.1) = x 2 10-4.5 = x2 [10-2.25][10-2.25] = x2 10-2.25 = x Since x equals the hydrogen ion concentration, taking the -log(10-2.25) = 2.25. A: This would be correct if acetylsalicyclic acid were a strong acid, but it is weak. C, D: These answers are the result of miscalculation.

Which of the following is NOT considered an organic acid? A. Folic acid B. Carbonic acid C. Ascorbic acid D. Citric acid *****with these types of questions try to think about what each has in common. if there is something in com between between choices then those are likely not the answer***

B is correct. An organic compound must contain carbon and hydrogen in its formula. Furthermore, there must be a covalent bond between a carbon and hydrogen atom in the molecular structure. Organic acids are weak acids, generally having formulas of R-CO2H, with the acidic hydrogen bonded to an oxygen atom. Even knowing this information, you may be intimidated, as you may not be familiar with the structures of all of these choices. However, you can begin with the most familiar compound, carbonic acid. The Lewis structure of this compound is shown below. Image From this structure alone, we can see that neither hydrogen atom is directly bound to carbon, so carbonic acid is not an organic acid and must be the correct answer here. A, C, D: The structures of these choices are shown below; all fit the criteria required for organic acids. Image

According to the reported responses, which of the following changed for Student E? A. Implicit attitudes towards smoking B. Explicit attitudes toward smoking C. Covert smoking behavior D. Attitude polarization

B is correct. Explicit attitudes are conscious attitudes. Student E was aware of his attitudes towards smoking and how they changed over time. He once viewed smoking as a negative habit; now he sees it as a means to an end. A: Implicit attitudes towards smoking are unconscious attitudes. C: Covert smoking behavior is unobservable behavior. Student E made changes in his overt (observable) behavior. D: Attitude polarization describes changes in attitudes among people in groups - tendency to go to the extreme.

At very low CTP concentrations, kinetic data fitted to the Michaelis-Menten equation predicts that the initial rate of the CCT-catalyzed reaction is most nearly what order with respect to CTP? A. Zero order B. First order C. Second order D. Third order

B is correct. For an enzyme-catalyzed reaction with a very low initial substrate concentration, and where Km >> [S], the Michaelis-Menten equation may be approximated as V = Vmax [S] / Km, where Vmax / Km is a constant of the reaction. Under these conditions, the reaction is approximately first order with respect to S. This can be seen graphically in Figure 1, where at low initial CTP concentrations, the enzyme activity (and the reaction velocity to which it is proportional) increases in a roughly linear fashion with CTP concentration. A generic Michaelis-Menten curve is shown below. Note that at low substrate concentrations, the reaction approximates first-order kinetics, as described above. In contrast, at very high substrate concentrations (where the enzyme is nearly or entirely saturated), the reaction approximates zero-order kinetics, since reaction rate ceases to depend on substrate concentration. A: The reaction would display zero-order kinetics only if increasing CTP concentrations did not lead to an increase in the reaction velocity. As mentioned above, this is generally true under saturating conditions, when substrate concentration is high relative to Km and when V = Vmax. C, D: A second- or third-order relationship at low substrate concentration would show a primarily non-linear relationship between CTP concentration and reaction velocity.

The results of a separate study of the thermodynamic parameters for the interactions of proteins with cyclohexanol and quartenary ammonium salts indicate that the hydrophobic solute-solute interaction is spontaneous, and that ΔH and ΔG have opposite signs. Which of the following must be true for this interaction when temperature is a positive value? I. ΔH > ΔG II. 0 < ΔS III. ΔH < ΔS A. I only B. I and II only C. II and III only D. I, II, and III

B is correct. For the interaction to be spontaneous, the free energy change of the reaction must be negative. The question states that the enthalpy and free energy changes for the interaction have opposite signs. Thus, ΔG is negative and ΔH is positive, making Roman numeral I true. For a reaction to be both endothermic (positive ΔH) and spontaneous (negative ΔG), the reaction must involve an increase in entropy (ΔS is positive). This is according to the equation: ΔG = ΔH - TΔS. Thus, II is also true because ΔS is positive. III: The quantity TΔS, not ΔS, must be greater than ΔH. For this reason, RN III does not have to be true. A: RN II is also correct. C: RN I is also correct, while RN III is incorrect. D: RN III is incorrect.

According to the data in Table 1, how many ATP molecules per carbon result from the oxidative phosphorylation of acetyl-CoA produced during β-oxidation? A. 4 ATP B. 5.5 ATP C. 8.5 ATP D. 11 ATP

B is correct. Oxidative phosphorylation, through the electron transport chain, produces ATP by using ATP synthase at the end of the electron transport chain. We're told that oligomycin blocks ATP synthase. So oligomycin also blocks energy production from oxidative phosphorylation. As an example, let's look at arachidic acid (AA). We see that AA produces 163 ATP molecules total. When we treat with oligomycin, we're blocking oxidative phosphorylation, but we still manage to produce 53 ATP. That means that when the body metabolizes one molecule of AA, it produces 53 ATP through something other than oxidative phosphorylation, leaving the remaining 110 ATP to be produced through oxphos. The question then asks us how many ATP per carbon result from oxidative phosphorylation. At this point, all we have to do is divide: 110 ATP / 20 carbons = 5.5 ATP/carbon A, C, D: These answers result from miscalculation.

Chris strongly believes that "suspects are innocent until proven guilty," but after a school shooting, he joins a group online in the pursuit of potential suspects. He identifies someone as the "shooter" and disseminates that person's contact information, which is then covered by national news outlets along with the person's picture. The next day, police announce that the person identified was in no way involved in the crime. Expressing remorse, what principle of cognitive dissonance will Chris most likely experience? A. Induced compliance B. Post-decisional conflict C. Effort justification D. Free choice paradigm

B is correct. Post-decisional conflict is the dissonance associated with behaving in a counter-attitudinal way. A: Induced compliance happens when a person is persuaded by others to behave in a way that is contrary to their attitudes. Because Chris voluntarily engaged in behavior that was against his beliefs, this is not the best answer. C: Effort justification is the state of dissonance that emerges when a person makes an effort to achieve a modest goal. This phenomenon is not relevant to the situation described above. D: Free choice reduction of conflict happens when a person has a binary choice which may conflict with their current views or beliefs. Once they make a decision and act, their attitudes can change to be more congruent with their decision.

In a third test, patients are given sentences that contain content that relate to their lives (such information having been gained through prior interviews with family members). Moreover, these sentences state facts that are either wrong (e.g. "Your niece is named Juliana" when in fact her name is "Julia") or distorted somehow. Using such sentences would likely show: A. increased recall but significantly lower recognition. B. proactive interference in the normal patients but not in the Korsakoff's patients. C. proactive interference in the Korsakoff's patients but not in the normal patients. D. proactive interference in both groups in the experiment.

B is correct. Proactive interference refers to the fact that currently existing long-term memories can interfere with the process of forming new long-term memories. This is unlikely to affect patients with Korsakoff's syndrome, as they are already unable to form new long-term memories. A: The facts in these sentences are not correct, so using them does not exemplify increased recall (retrieval of information from memory without recognition cues). C: This choice may be tempting, since we might expect the Korsakoff's patients to be more likely to display a memory problem (here, proactive interference) than the normal patients. However, normal individuals are absolutely subject to proactive interference. In contrast, Korsakoff's patients already have such impaired memory formation (as evidenced by the results in Figure 1) that they are unlikely to display additional impairment as a result of previously formed memories. After all, if memory formation is already so impaired that it is nonexistent or almost nonexistent, there is little room for more impairment due to proactive interference. D: This choice is incorrect for the reasons detailed above.

Researchers conduct an experiment to test the Cannon-Bard theory of emotional arousal. A simulated car crash is used in which the subjects were shown video of an oncoming vehicle. Which of the following results best supports the Cannon-Bard theory? A. Subject spots oncoming vehicle → Subject feels fear → Subject heart rate rises B. Subject spots oncoming vehicle → Subject feels fear and subject heart rate rises C. Subject spots oncoming vehicle → Subject heart rate rises → Subject feels fear D. Subject feels fear → Subject heart rate rises → Subject thinks "I am afraid".

B is correct. The Cannon-Bard theory states that emotion arousing stimuli simultaneously triggers 1) physiological responses and 2) the subjective experience of emotion. A: This incorrectly shows emotion and physiology occurring separately. C: This is an example of the James-Lange theory, which states that experience of emotion is in response to awareness of physiological response to arousing stimuli. D: This is unrelated to the Cannon-Bard theory. Cognitive labels such as "I am afraid" are part of two-factor theories, not Cannon-Bard

A scientist investigating the Weber-Fechner law detects a just-noticeable difference for a subject when shifting from a 5 kg to an 8 kg mass. When repeating the trial, how many kilograms must be added to a 15 kg mass to replicate the effect? A. 6 kg B. 9 kg C. 18 kg D. 24 kg

B is correct. The Weber-Fechner law states that the just-noticeable difference between two stimuli is directly proportional to the magnitude of the stimuli. If a 3 kg change creates a just-noticeable difference starting at 5 kg, tripling the initial mass will require triple the difference. Thus, the answer is 9 kg. A, C: These answers are the result of miscalculation. D: This is the final overall mass (15 kg + 9 kg = 24 kg). This question asked how many kilograms must be added, not what the final mass would be in total.

Which of the following discoveries would most support the statement that p53 is the most important protein regulator of DNA repair to have been discovered? A. When mutated, p53 continues to function to repair damaged DNA. B. More human tumors can be traced to a mutation in the p53 protein gene than in any other protein. C. Most cancerous cells lack the p53 protein. D. p53 levels are always elevated in patients with cancer.

B is correct. This question is asking us to determine the best support for a claim made in the passage. Choice B directly relates p53 to the incidence of tumors in humans. More importantly, it supports the claim that p53 is the "most important" protein that may have a role in DNA repair, since p53 mutations are implicated in more tumors than are mutations in any other gene. A: This could be true even if p53 played only a very minor role in DNA repair. C: The simple fact that cancerous cells lack a protein do not mean that protein has anything to do with the incidence of that cancer (or with DNA repair). D: This could also be true even if p53 did not function in DNA repair.

Researchers compare a DNA sequence and an identical sequence of RNA where all thymines are replaced with uracil. Which of the following describes changes that they can expect? I. The uracil base will be heavier than the thymine base because of an extra methyl group. II. The thymine base will be heavier than the uracil base because of an extra methyl group. III. The carbohydrate ring in RNA is heavier than the carbohydrate ring in DNA. IV. The carbohydrate ring in DNA is heavier than the carbohydrate ring in RNA. A. I and III only B. II and III only C. I and IV only D. II and IV only

B is correct. This question is asking us to determine the weight of a strand of genetic material that contains uracil instead of thymine. We are told in the passage that uracil can be formed from the direct deamination of cytosine, while thymine is formed from a deamination and a methylation. Therefore, thymine weighs more than uracil by one methyl group. The ribose ring of RNA will also be heavier than the deoxyribose ring of DNA because the DNA is "de-oxy," meaning it lacks the 2' hydroxyl group that is present on the ribose of RNA. A: RN I is incorrect, while RN II is correct. C: This is the opposite of the correct answer. D: RN IV is incorrect, while RN III is correct.

Conn's syndrome, also known as primary hyperaldosteronism, is most likely to cause which symptom? A. High renin concentration B. Low blood potassium C. Low blood sodium D. Hypotension

B is correct. This question is asking you to recall the effects of aldosterone and how it achieves those effects. Aldosterone increases H2O and Na+ reabsorption from the kidney while exchanging K+ ions for Na+ ions. The triggers for and results of aldosterone secretion are shown below. A: Hypersecretion of aldosterone will negatively feed back and inhibit renin production. C: Hypersecretion of aldosterone will result in high blood sodium. D: Hypersecretion of aldosterone will result in high blood pressure.

A student determined that her yield of aspirin was 3.9 g. What was her percent yield? A. 45% B. 60% C. 78% D. 92%

B is correct. We first need to determine if there is a limiting reagent and then determine the theoretical yield. The moles of acetic anhydride can be determined by using the volume and density information in step 2 of the procedure and the molecular weight given in the passage. 7 mL x 1.08 g/mL x 1 mol/102 g ~ 7 x 10-2 mol We can also calculate the moles of salicylic acid from the number of grams used in step 1 and the molecular weight given in the passage. 5.0 g x 1mol/138 g ~ (5/1.4) x 10-2 ~ 3.5 x 10-2 mol Since the stoichiometry is 1:1 from Reaction 1, the acetic anhydride is in excess and the limiting reagent is the salicylic acid. The stoichiometric ratio between salicylic acid and the acetyl salicylic acid is also 1:1; therefore, the theoretical yield can be calculated using the molecular weight from the passage. 3.5 x 10-2 mol x 180 g/mol ~ 3.5 x 2 x 10-2 x 102 ~ 7 g Since the question indicates that the actual yield was 3.9 g, which is about 4 g, the percent yield is (4/7) x 100 ~ 60%. A, C, D: These answers are the result of miscalculation.

Which of the following contradicts Weber's law? A. A weightlifter who doesn't notice a 5-pound difference in weight added in 1 pound increments, but who notices the difference in weight added in one 5-pound increment B. A non-linear relationship between the intensity of a stimulus and an individual's ability to detect it C. The existence of a threshold, above which stimuli are detectable and below which stimuli are not detectable D. Hearing a whisper in a quiet room but not hearing a shout in a noisy room

B is correct. Weber's law postulates that there is a linear relationship, not a non-linear relationship, between the intensity of a stimulus and its detection. A: This is an example of Weber's law because the change in weight at 1-pound increments may not be large enough to be noticeable. C: This is the idea postulated by Weber's law. D: Stimulus detection is relevant to background stimulus.

capacitance equatino

C = Q/V or remember QVC ---- Q = VC

The passage data regarding the thermal stability and enzyme activity of MKR681H is most consistent with what conclusion regarding the role of Arg681 in CCT? A. Arg681 replacement reduces the CTP binding affinity of MKR681H. B. Arg681 provokes increased heat sensitivity in the dimer. C. Arg681 is engaged in the catalytic function of the enzyme. D. Arg681 absence promotes dissociation of the CCT dimer.

C is correct. Be careful. This question is NOT asking about the mutant. R681H denotes that amino acid 681 (arginine, R) is changed to a histidine (H). The question is asking what we can conclude about the wild-type enzyme given the data presented in the passage. Examining Figure 1 for the enzyme activity data plotted for MKR681H and MKWT shows that substitution of arginine by histidine decreases maximum enzyme activity. Enzyme activity depends principally on an enzyme's intrinsic catalytic efficiency, its concentration, the initial substrate concentration, the presence of inhibitors or allosteric activators, temperature, and pH. In the experiment performed in the passage, all of these factors were controlled for between the enzyme variants other than their intrinsic activity. Of the choices given, this most strongly suggests that Arg681 is involved in the catalytic function of the normal enzyme. A: Enzyme activity does not specifically relate to the affinity of enzyme for substrate. B: This is a tempting answer but it is actually an opposite answer trap. The question is asking about what we can say about the normal enzyme form (the one with R). In Figure 2, the Ti for MKR681H is lower than the Ti of the wild-type enzyme. This means that the mutant loses 25% of its enzymatic activity at a lower temperature than the wild-type enzyme. We can also tell from the definition of Ti that Ti is inversely related to heat sensitivity - a lower Ti means the protein is more heat sensitive aka less heat stable. Since the mutant has a lower Ti (when the R is replaced by H) this tells us that the mutant is less heat stable (i.e. more heat sensitive) compared to the wild-type enzyme with the Arg681 in it. Thus, the arginine confers more heat stability aka decreased heat sensitivity than the histidine in the mutant. Choice B is a classic opposite wrong answer.

What psychometric property is supported if the experiences of the low-income students in the study are similar to those of other low-income students around the country? I. External validity II. Generalizability III. Internal consistency A. I only B. III only C. I and II only D. I, II, and III

C is correct. External validity describes the generalizability of the study, or the extent to which results can be applied to a wider population. III: Internal consistency refers to how well the items of a test assess the construct of interest. In this context, whatever the researchers used to measure depression, anxiety, etc. could be evaluated on its internal consistency. A: RN II is also correct. B: RN III is incorrect, while RN I and RN II are correct. D: RN III is incorrect.

In an adult, which of the following cell types is LEAST likely to enter a programmed G0 phase of the cell cycle? A. Liver cells B. Kidney cells C. Epithelial cells D. Neurons

C is correct. For this question, we need to know what the G0 phase (shown below along with the rest of the cell cycle) entails. This is the state that a cell will enter if it does not need to divide. Since epithelial cells are those that divide the most out of the options, choice C is correct. A: Liver cells are more likely to be found in G0 because they do not divide as often as epithelial cells. B: Kidney cells are more likely to be found in G0 because they do not divide as often as epithelial cells. D: Neurons in adults do not divide and are almost always found in G0.

Y32 is pH-sensitive. This can be a problem when employing it to monitor changes in the energy balance of a cell because: A. pH can differ between cells. B. the ATP-to-ADP ratio of a cell depends on pH. C. pH may change over time in the same cell. D. decreased pH favors the CPV B-state.

C is correct. If the biosensor is pH-sensitive, then changes in cellular pH over the course of a measurement could confuse changes in the energy balance of the cell, as reflected by the cell's ATP-to-ADP ratio. A: While this is true, the question specifically asks about how the fact that the biosensor is pH-sensitive could affect measurements made within a single cell over time. B: While the ATP-to-ADP ratio in most cells does to some extent depend on pH, this does not address the question of how pH changes influence the utility of the biosensor in measuring a cell's ATP-to-ADP ratio. D: Decreasing pH would favor the protonated CPV A-state. *****Decreasing pH would protonate the deprecated version and thus favor the "A" state***** come on, how many of these are you going to miss!!!

Which of the following would be LEAST useful in cellular movement? A. Flagella B. Actin polymerization C. Microtubule depolymerization D. Cilia

C is correct. In order for cells to travel to the site of injury, they need to migrate. Microtubule de-polymerization is responsible for separating chromosomes during anaphase of mitosis or meiosis I or II. It does not contribute to overall cell migration. A, D: Flagella and cilia are structures that allow simple eukaryotic and prokaryotic cells to propel themselves or nutrients in their environment. These structures are shown below; however, since both are involved in cellular movement, neither is the answer to this LEAST question. B: Rapid actin polymerization near the edge of the cellular membrane is responsible for cellular motility in complex eukaryotic cells.

Herpes simplex virus alters the DNA of a host cell, which can then be passed on to daughter cells via mitosis. In some cases, several cell generations may go by with the instructions being passed on and no virus being produced. The viral insertion to the host DNA may then be activated by an environmental trigger, resulting in a number of viruses being produced within the cell, which then escape to spread through the host body or infect new hosts. Herpes simplex is: A. lysogenic. B. lytic. C. both lytic and lysogenic. D. neither lytic nor lysogenic.

C is correct. Information about lysogenic life cycles is not in the passage. If you didn't know about lysogenic viruses, however, you can reason this question out based on the definition of lytic viruses given. Viral DNA being passed on to daughter cells through ordinary means (mitosis) occurs in a lysogenic cycle. The production of large numbers of virus particles within a cell that then escape it (with or without destroying the cell) is an example of a lytic life cycle. Herpes simplex is therefore both, switching from one to the other based on environmental conditions. Indeed, many viruses switch between both life cycles rather than sticking exclusively with one or the other. Chronic viral diseases with latency periods and outbreaks like herpes tend to fall in this group. A, B: These choices are accurate but are insufficient when taken alone. D: This is incorrect; the information in the question stem aligns with both the lytic and lysogenic life cycles

Viruses are unique from most biological organisms in that they: A. are able to reproduce inside a host. B. only use RNA as genetic material. C. are obligate parasites. D. are pathogenic entities which have been described as not being "living."

D is correct. Viruses are unique in that they occupy a gray area between living and non-living. They have been described as non-living. A: This choice is not correct, since other types of organisms (such as bacteria and fungi) are able to reproduce inside other organisms, as with a fungal infection. B: This choice is not correct, since some viruses use DNA as genetic material. C: Other types of infectious agents are obligate parasites (e.g., tapeworms).

Which regions of Figure 1 are most likely to be explained by the movement of sodium ions into a cardiac neuron? I. P II. Q III. R IV. T A. I only B. II and III only C. I, II, and III only D. I, II, III, and IV

C is correct. Let's begin with RN I. As stated in the passage, the P wave is associated with the depolarization and contraction of the atria. A rest potential is created when sodium ions are pumped to the outside surface of the membrane of a cell. When ion channels open, the sodium ions spontaneously flow back into the cell, resulting in depolarization and an action potential. Roman numeral (RN) I is thus correct. Regarding RNs II and III, the passage states that the Q and R waves are associated with the contraction of the ventricles. Like atrial contraction, ventricular contractions result from depolarization, which would involve the inward movement of cations like sodium ions. Thus, RN II and III are accurate as well. IV: The passage states that the T and U waves occur due to repolarization, which takes place when positive ions are pumped back out of the cell to return to rest potential prior to the next heartbeat. This is the opposite of what the question asks for. A: RN II and RN III are also correct. B: RN I is also correct. D: RN IV is incorrect.

A second student repeated the experiment using glucose and the equivalent enzymes of glycolysis instead of starches. How would his results compare to those shown in Figure 3? A. Purplish color with higher initial current B. Purplish color but no current because of incomplete glycolysis C. Brownish color with higher initial current D. Brownish color with lower initial current

C is correct. The first difference between the first student and the second is the use of glucose, a monosaccharide (shown below), as the fuel. Paragraph 2 states that iodine binds selectively to linear-chain polysaccharides. As a result, the iodine will NOT bind the glucose, causing the solution to remain brown. We can eliminate choices A and B. Next, we must determine the effect of using glucose and the glycolytic enzymes on the current generated by the battery. From Figure 1, we can infer that the amount of current generated should be directly related to the production of NADH. Unlike in the biostarch battery, glycolysis does not require the breakdown of starch. Thus, NADH is generated directly and more quickly via the conversion of glyceraldehyde 3-phosphate (GAP) to 1,3-bisphosphoglycerate (1,3-BPG). The enzymes should produce 1 NADH per GAP molecule, or 2 NADH per glucose molecule (since each unit of glucose breaks down into two GAP molecules). The original experiment only produced 1 NADH per glucose conversion to phosphogluconate, so we can conclude that the second student should observe a higher initial current than the first. A, B: The solution will not be purplish in color. The passage indicates that the purplish color results from iodine binding "linear chain polysaccharides." Unlike starch, glucose is not a polysaccharide (it is a monosaccharide), so the solution will remain brown. D: We would expect the initial current to be higher, not lower, when using glucose and glycolytic enzymes. As described above, glycolysis produces more NADH per glucose molecule than the starch → glucose 6-phosphate → 6-phosphogluconate conversion. More production of NADH means that more electrons will be passed along to AQDS and the electrodes of the apparatus, and current will be greater.

What is the standard cell potential for the starch battery created? A. -1.136 V B. -0.916 V C. 0.916 V D. 1.136 V PAY ATTENTION TO THE FREAKING PASSAGE AND FIGURES

C is correct. The passage states that the cell is meant to act like a galvanic cell, meaning that it proceeds in a spontaneous fashion. Since galvanic cells always have cell potentials that are greater than 0, we can eliminate any negative options (choices A and B) immediately. To decide between C and D, we need more information, namely the reduction potentials of the species involved. Paragraph 1 gives E°reduction of O2 as 0.816 V and E°reduction of AQDS as -0.10 V. Since both species cannot reduce at once, we must reverse the sign of one of the reduction potentials to find the oxidation potential of the relevant reaction. For galvanic cells, reverse the less positive reduction potential, which is -0.10 V here. The associated oxidation potential, then, is +0.10 V. Finally, since we now have a reduction and an oxidation, we can simply add the values directly to yield 0.816 V + 0.10 V = 0.916 V. A, B: Galvanic (spontaneous) cells do not have negative standard cell potentials, so these choices can be eliminated. We know the starch battery is meant to act like a galvanic cell because it is described as "generat[ing] sufficient power," meaning that it must run spontaneously. D: This choice is attained by using the reduction potentials of O2 and NAD+. However, for this question, we need to use the potentials for the two species directly involved in the battery (according to Figure 1). The figure shows AQDS at the anode of the battery and O2 at the cathode, so these are the species we must use.

The finding that Korsakoff's syndrome impairs declarative memory is most consistent with the fact that: A. it is caused by deficiency of a particular type of B vitamin. B. alcoholics are the most common patient population in the US who exhibit the syndrome. C. it is usually associated with lesions that can affect the functioning of the hippocampus. D. patients with Korsakoff's syndrome typically have a normal life expectancy if they receive treatment.

C is correct. The question asks you to find a fact consistent with damage to declarative memory. Here, the MCAT expects you to know about the central role played by the hippocampus in the functioning of declarative memory. Lesions that affect the hippocampus, as described in choice C, would be most consistent with memory impairment. A, B, D: Though these choices are all true facts about Korsakoff's, none of them directly relate to memory impairment as asked in the question.

Selenocysteine is a non-standard amino acid that is present in all domains of life. It has the structure of cysteine, but the sulfur atom is replaced with selenium. The structure of selenocysteine is shown below: What is the absolute configuration of the alpha carbon in selenocysteine? A. D B. L C. R D. S

C is correct. When assigning absolute configuration, imagine the molecule so that the lowest-priority group is facing away. In this case, it is the implicit hydrogen, which by the tetrahedral geomtry of the alpha carbon already happens to be facing away. Then, assign priority based on molecular weight of the attached atoms. Nitrogen > C-Se > C-O > H. Follow those atoms in a clockwise circle to assign the R configuration, as shown below. A, B, D: D and L are relative configurations, not absolute configurations. S is the opposite of the correct configuration.

A hospital purchases brand-new GKS-Co and GKS-X machines. Five years after installation, what is the expected ratio of the total atomic mass of radioactive material (both before and after decay) in the Co machine to that in the X machine, assuming both machines start with the same mass of radioactive material? A. 1:16 B. 1:5 C. 1:1 D. 5:1

C is correct. When reading questions, be careful not to read too quickly. In this case, fast but inefficient reading will lead us to assume that it is asking about the percentage of a certain isotope that is left after radioactive decay. However, the question is asking about atomic mass. While β-decay does cause a nuclear transmutation of protons to neutrons (β-plus) or neutrons to protons (β-minus), the atomic mass lost in these processes is negligible. This means that whether after one (Co) or five (X) half-lives, the atomic mass will be the same in both samples. A: This is the ratio of undecayed nuclei between the two samples. The GKS-Co machine has 1/2 of its nuclei undecayed and the GKS-X machine has 1/32 of its nuclei undecayed. The question, however, asks about all of the nuclei, both decayed and undecayed. B, D: These choices both involve the differences in half-live between 60Co (5 years) and X (1 year). However, since the question is asking about undecayed and decayed material, this distinction does not matter.

A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push causes the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters? A. 1 m/s B. 2 m/s C. 3 m/s D. 5 m/s

D is correct. Since the crate is initially at rest, the Vi = 0 m/s, the acceleration a = 2.0 m/s2, and the displacement = 6 m. The appropriate kinematic equation is vf2 = vi2 + 2ad. Substituting the variables gives: vf2 = 02 + 2(2)(6) vf2 = 24 vf = ~5 m/s A, B, C: These answers are the result of miscalculation.

Ecological validity

Criticism about ecological validity refers to ways that the experiment applies to the environment.

The Rorschach inkblot test is often used to identify distinct personalities in those with DPD. This test is best categorized as a: I. subjective personality assessment. II. objective personality assessment. III. projective personality assessment. A. I only B. II only C. II and III only D. I and III only

D is correct. Both RN I and III correctly describe the inkblot test. In subjective assessments, patients project their own subjective feelings, perceptions, and thoughts onto the assessment stimuli, yielding results that are open for inaccuracy. For example, physicians may reach a different conclusion despite seeing the same patient who says the same thing. Projective personality assessments require the participant to respond, and then their response is assessed for meaning. II: An objective personality assessment measures specific personality characteristics based on a set of discrete options, such as in the Meyers-Briggs personality assessment. A: RN III is also correct. B: RN II is incorrect, while RN I and RN III are correct. C: RN I is also correct, while RN II is incorrect.

Which of the following is NOT a strategy to induce compliance in a target group, as described in the passage? A. A dress code at an organization B. Referring to military recruits by their military number C. Hazing rituals in a club involving humiliation D. Pointing out differences between one group of students' performance and another group's performance

D is correct. Compliance is induced in groups that view themselves as similar. The other strategies were specified in the passage as ways to induce compliance. A, B, C: Since all of these groups view themselves as similar, these choices serve as good examples for inducing compliance.

What is the exclusion limit of the SEC column used to create the calibration curve shown in Figure 2? A. 10 kD B. 100 kD C. 600 kD D. 1000 kD

D is correct. Figure 2 indicates that the exclusion limit of the SEC column is at log (MW) = 6. If true, the molecular weight can be found by taking the inverse log of 6, 106 D = 1000 kD. A, B: These values correspond to 4 and 5, respectively, on the y-axis of Figure 2. These are well below the labeled exclusion limit. C: If log (MW) = 6, as Figure 2 indicates, then MW = 106 D = 103 kD = 1000 kD, not 600 kD.

Which of the following is NOT correct regarding the results of the FISH analysis? A. The incidence of incomplete XCI is less prevalent in patients with Klinefelter syndrome than those with Trisomy X syndrome. B. There is more statistical power for normal phenotypes than abnormal phenotypes. C. Euploid females may experience hypomethylation of their X chromosome. D. The incidence of incomplete XCI is more prevalent in patients with Klinefelter than those with Triple X syndrome.

D is correct. The FISH analysis results are found in Table 2, which shows that in the case of Triple X syndrome, about 47.6% show the XaXaXi genotype (incomplete XCI) and 52.4% show the XaXiXi genotype. For Klinefelter syndrome, 62.1% of patients show a XaXiY genotype, 37.3% show a XiXiY genotype, and 0.6% show the XaXaY genotype (incomplete XCI). Thus incomplete XCI is much more prevalent in Triple X syndrome (47.6%) than in Klinefelter (0.6%). This contradicts choice D, making it the correct answer to this "NOT" question. A: The Klinefelter group showed a 0.6% incidence of incomplete XCI, less than the 47.6% for Triple X syndrome, so this answer is true. Since this is a NOT question, this cannot be our answer. B: There are more total samples for normal phenotypes than abnormal ones, so this answer is supported. C: There are a few cases of zero signals in females, indicating that there are two active X chromosomes, meaning one has been hypomethylated.

The true weight of the p53 gene is 43.7 kDa. What effect would best explain this discrepancy from its measured weight, as mentioned in the passage? A. The protein migrated a greater distance because it contains many positively charged residues. B. The protein migrated a smaller distance because it contains many negatively charged residues. C. The protein migrated a greater distance because it contains many negatively charged residues. D. The protein migrated a smaller distance because it contains many positively charged residues.

D is correct. The final paragraph of the passage mentions that p53 is so named for its measured molecular weight using SDS-PAGE. This question is asking us to determine a plausible reason for the 9.3-kDa discrepancy. SDS-PAGE is used to grant a uniform negative charge to all proteins in an assay. However, if the protein has enough charged residues on its own, it causes the measurements to be less accurate. If there are enough positive charges, the negative charge of the protein will not be as great as anticipated and the molecule will travel a smaller distance, which, to the individual running the assay, usually means the mass of the protein is greater and would lead to the discrepancy described. A, C: The SDS-PAGE measurement is artificially large (53 kDa rather than the actual weight of 43.7 kDa). This means that the p53 protein migrated a smaller distance, not a greater distance. (Remember, the smaller the distance traveled in SDS-PAGE, the larger we would expect the protein to be.) B: In SDS-PAGE, proteins migrate toward a positively-charged pole. As such, containing many negatively-charged residues would make a protein migrate a larger distance, not a smaller one.

Which pair of enzymes, if used simultaneously, will produce the greatest amount of glucose if the experiment is repeated? A. A and B B. A and C C. B and C D. C and D

D is correct. The greatest glucose production per mass of starch will occur when there is the greatest rate of starch breakdown. For complete breakdown, this requires both enzymes that break the 1,4 linkages in linear sequences and those that break the 1,6 linkages at branch points. The enzymes that are most active at the 1,4 sites will cause the largest amount of current, as they will produce the most glucose 6-phosphate from the available linear portions of the starch. The enzymes most active at the 1,6 linkages will result in the largest absorbance of light at the peak wavelength (570 nm). Based on paragraph 3, iodine binds linear polysaccharides, so as the 1,6 linkages are broken, more linear polysaccharides are formed; this will lead to more iodine-starch interactions and an increased absorbance. Thus, the combination that maximizes total breakdown should have the maximum current and maximum absorbance at 570 nm, which matches to enzymes C and D, respectively. A, B: Enzyme A produces no current at all (0.0 mA) and does not have the highest absorbance at 570 nm of the enzymes listed. As such, we should not include it in the pair of enzymes we choose. C: This answer is tempting, as enzymes B and C are the two enzymes that produce the largest amount of current! However, current is only one aspect of this question, which specifically asks about glucose production, not current production. Paragraph 2 mentions that starch-powered batteries must break down both "the linear 1,4-D-glucose linkages and the 1,6-D-glucose linkages found at branch points." We thus need to include an enzyme that will break 1,6 linkages; otherwise, starch will not be fully broken down and less glucose will be produced. The passage tells us that iodine selectively binds linear (unbranched) polysaccharides, so the greater the absorbance of 570-nm light, the "more purple" the solution and the more linear the starch molecules must be. Thus, the enzyme that produces the highest absorbance must be the enzyme which breaks the 1,6 linkages at branch points. Initial absorbance values are very similar across the four solutions, but absorbance at 2 minutes is clearly highest in the solution that contains enzyme D. Thus, enzyme D should be included in our answer.

When Y32 is expressed within a normal cell, what is true of its nucleotide binding site? A. It is most likely to be occupied by ADP. B. It is unlikely to be occupied by Mg2+-ATP. C. It is unlikely to be occupied by Mg2+-ATP or ADP. D. It is effectively always occupied by Mg2+-ATP or ADP.

D is correct. The passage indicates that the biosensor binds Mg2+-ATP and ADP with very high affinity (with a Km ∼ 1 μM). This greatly exceeds the normal cytosolic concentrations of ADP and Mg2+-ATP (hundreds of μM and about 1 mM, respectively). Given this, it is quite likely that the nucleotide binding site of an Y32 molecule would be occupied by Mg2+-ATP or ADP under such conditions. A: The passage indicates that the normal cytosolic concentration of Mg2+-ATP somewhat exceeds that of ADP. Considering that the biosensor binds both Mg2+-ATP and ADP with great affinity relative to their physiological concentrations, it is somewhat more likely that a given binding site will be occupied by Mg2+-ATP than by ADP under physiological conditions. B, C: It is quite likely that the nucleotide binding site of the biosensor would be occupied by Mg2+-ATP or ADP under physiological conditions.

Which of the following electronic transitions for a hydrogen atom would result in the emission of a photon that would be visible to the human eye? A. n = 1 to n = 3 B. n = 2 to n = 4 C. n = 2 to n = 1 D. n = 4 to n = 2

D is correct. The visible spectrum contains electromagnetic signals with wavelengths ranging from 400 nm to 700 nm. The wavelength of light emitted during a particular electronic transition is determined by the energy difference (ΔE) between the final and initial energy levels. The energy of each level can be determined using Equation 1 and the principal quantum number in question. For light to be emitted at all, an electron must travel from a higher to a lower level, a process that releases energy. This automatically eliminates options A and B; however, we must actually perform calculations to choose between C and D. Begin with choice C, using Equation 1 and approximating the Rydberg constant as 1 ✕ 107 m-1. The energy of the n = 2 level is: E2 = -1 ✕ 107 m-1 / 22 = -0.25 ✕ 107 m-1 Since the energy of the n = 1 level is equivalent to the Rydberg constant, ΔE is: ΔE = E2 - E1 = (-1 ✕ 107 m-1) - (-0.25 ✕ 107 m-1) = -0.75 ✕ 107 m-1 Note that this value is in units of m-1, but due to the passage and the units given in the question stem, we need to convert to nanometers ( 1 nm = 10-9 m). We can ignore the negative sign, since this simply means that energy was released rather than absorbed. 1 / (0.75 ✕ 107 m-1) = 1.33 ✕ 10-7 m = 133 nm This falls in the UV range of the electromagnetic spectrum, below the visible range. This eliminates choice C, and we can choose option D and move on. However, if you would like to confirm using math, calculating the energy of the n = 4 level gives: E4 = -1 x 107 m-1 / 42 = -1/16 x 107 m-1 ΔE = E2 - E4 = (-4/16 ✕ 107 m-1) - (-1/16 ✕ 107 m-1) = -3/16 ✕ 107 m-1 To make the fraction easier to work with, round 3/16 to 3/15, or 1/5. Again, we must convert wavenumbers to nanometers: 1 / (1/5 ✕ 107 m-1) = 5 ✕ 10-7 m (5 ✕ 10-7 m)(1 ✕ 109 nm/m) = 500 nm While this is also an approximation, it falls well within the visible spectrum. The actual color of the n = 4 to n = 2 transition of hydrogen is teal (blue-green), with a wavelength of 486 nm. A, B: Moving from a lower to a higher energy level is associated with absorption of energy, not emission. As such, these choices can be eliminated right away. C: This answer is the result of miscalculation.

Underproduction of pulmonary surfactant in IRDS leads to decreased compliance of alveolar tissue. Based upon this information, which of the following must be true regarding pulmonary surfactant? A. Its adsorption to the water-alveolar interface increases surface tension, preventing alveolar collapse due to intra-thoracic pressure. B. Its absence decreases the minimum radial size of alveoli able to avoid collapse at a given pressure of inspired air. C. Its adsorption to the water-alveolar interface decreases surface tension, decreasing the pressure difference required to inflate the airway. D. Its presence increases the efficiency of gas exchange across the alveolar membrane by decreasing the surface area of the alveolus at a given pressure of inspired air.

During inspiration, contraction of the diaphragm and internal intercostal muscles leads to expansion of the thoracic cavity and a decrease in intrapleural pressure. This negative pressure, relative to atmospheric pressure at the entry of the upper airway, generates airflow through the respiratory tree and to its terminal extension—the alveoli. The elastic recoil force of the airway and the surface tension of the water lining the airway oppose expansion of the alveoli due to the influx of atmospheric pressure. Pulmonary surfactant adsorbs to the air-water-alveoli interface, reducing surface tension and the total force resisting expansion. This increases pulmonary compliance—a measure of lung volume change at a given pressure of inspired air—and decreases the work required to expand the lungs at a given atmospheric pressure. This is consistent with choice C. In general, surfactant molecules are amphipathic, meaning that they contain both hydrophobic and hydrophilic regions. The diagram below shows surfactant molecules surrounding a micelle of oil. The hydrophobic tails of the surfactant molecules mix well with the hydrophobic oil, while the hydrophilic heads point away from the oil droplet. During inspiration, contraction of the diaphragm and internal intercostal muscles leads to expansion of the thoracic cavity and a decrease in intrapleural pressure. This negative pressure, relative to atmospheric pressure at the entry of the upper airway, generates airflow through the respiratory tree and to its terminal extension—the alveoli. The elastic recoil force of the airway and the surface tension of the water lining the airway oppose expansion of the alveoli due to the influx of atmospheric pressure. Pulmonary surfactant adsorbs to the air-water-alveoli interface, reducing surface tension and the total force resisting expansion. This increases pulmonary compliance—a measure of lung volume change at a given pressure of inspired air—and decreases the work required to expand the lungs at a given atmospheric pressure. This is consistent with choice C. In general, surfactant molecules are amphipathic, meaning that they contain both hydrophobic and hydrophilic regions. The diagram below shows surfactant molecules surrounding a micelle of oil. The hydrophobic tails of the surfactant molecules mix well with the hydrophobic oil, while the hydrophilic heads point away from the oil droplet. A: Choice A indicates the surface tension would increase, which would increase the respiratory effort required to inflate the lungs. The efficiency of gas exchange across the alveolar membrane is primarily a function of the differences in partial pressures of gases in inspired air versus the partial pressure of those gases in the pulmonary capillaries. B: The absence of pulmonary surfactant increases surface tension, increasing the minimum radial size of alveoli required to overcome the collapsing force of surfactant at a given pressure of inspired air. When surface tension exceeds the average alveolar radii, collapse of the small airways occurs, eliminating choice B. D: Differences in efficiency of gas exchange wouldn't account for the increased lung-expansion difficulty IRDS patients experience, eliminating D. Content Foundations: Surfactant Surfactants are amphipathic molecules (containing both hydrophilic and hydrophobic regions) that reduce the surface tension of a liquid. For the MCAT (and in your body), the most important example is pulmonary surfactant, which reduces the surface tension in the alveoli, allowing them to remain inflated when the lung is compressed during respiration. The general category of surfactants also includes detergents, the amphipathic structure of which allows them to denature proteins (an example being sodium dodecyl sulfate, or SDS, which is used in an electrophoretic technique known as SDS-PAGE) or solubilize lipids.

Researcher bias

Researcher bias involves the researchers interjecting their own views into the experiment and biasing it.

Reflex arc

If this is still unclear, take a look at the diagram below, which depicts a standard reflex arc. Note that the motor neuron synapses on the effector muscle, which tells us that it must carry signals away from the spinal cord (efferent) and toward the muscle. In contrast, the sensory neuron in this diagram extends from the spinal cord to the peripheral receptors. Sensory neurons must transmit sensory information, and the most logical direction to transmit this information is from the periphery of the body to the central nervous system. Thus, these neurons must bring sensory information toward the spinal cord, so they are afferent. The interneuron in this diagram is found entirely within the spinal cord, so it travels neither away (efferent) nor toward (afferent) the cord.

Reliability

Reliability refers to the likelihood that results could be replicated.

nondisjunction

Note that nondisjunction refers to the failure of chromosomes or chromatids to properly separate during anaphase. Nondisjunction can occur during meiosis I or meiosis II. In the diagram below, the left image shows nondisjunction during meiosis II, while the right-hand image shows the same event occurring during meiosis I.

Restriction enzymes, some restrictions, and an example of their use.

Restriction enzymes cut near or at their recognition sequences. However, if the gene is methylated, this cutting cannot occur. The question states that they are using a common X-specific site, not the XIST site described in the passage. The mother's result shows one heavy strip at the top of the gel, indicating that it is bulky and did not travel the length of the gel, as it shows large pieces. We can infer that this is the inactivated and methylated X that was not cut in as many places. Both the mother and the father have a smear at the bottom of the gel, indicating smaller fragments, or the active X that was cut in many places. The fetus shows a band at the top, indicating an inactive X; however, it also has a much darker and broader smear, indicating more active X chromosomes. Therefore, this child has a genotype of XaXaXi, or Trisomy X with incomplete XCI.

Looking-glass self

The looking-glass self is a social psychological concept stating that a person's self grows out of society's interpersonal interactions and the perceptions of others. The term refers to people shaping their self-concepts based on their understanding of how others perceive them. If the man feels bias from society and/or police (as mentioned in paragraph 1) the looking-glass self theory states that he will internalize the bias/stigmatization directed towards him.

Cell cycle

The cell cycle can be divided into a resting phase (interphase) and cell division (mitosis or meiosis). Resting phase is also known as Gap 0 (G0). During this period, the cell just goes about its business; in fact, many fully-differentiated cells in the body remain in G0 for long periods of time. Because it can last for an essentially indefinite period of time, resting phase is often considered not to be a proper part of the cell cycle itself. Interphase is when a cell prepares for division, and it can take up approximately 90% of the time of the cell cycle. Two major things happen during interphase: growth and DNA replication. However, interphase is broken into three stages: Gap 1 (G1), synthesis (S), and Gap 2 (G2). During G1 and G2, the cell grows, and during S, DNA is replicated. The fact that S is located between G1 and G2 allows checkpoints. The G1/S checkpoint, also known as the restriction point, is when a cell commits to division. The presence of DNA damage or other external factors can cause a cell to fail this checkpoint and not divide. The G2 checkpoint that takes place before cell division similarly checks for DNA damage after DNA replication, and if damage is detected, serves to "pause" cell division until the damage is repaired. Throughout interphase, chromatin is loosely packaged (euchromatin) to allow transcription and replication. After interphase, the cell undergoes division (mitosis in non-sex/germ cells). Mitosis proceeds through prophase (where the nuclear membrane disappears, chromosomes condense, and the mitotic spindle forms), metaphase (where chromosomes line up along the metaphase plate), anaphase (where chromosome are pulled apart), and telophase/cytokinesis (where the nuclear envelope and nucleolus reappear and the cell divides). Meiosis occurs in sex/germ cells and turns a diploid (2n) parent cell into 4 haploid (n) daughter cells in a two-stage process, in which crossover between homologous chromosomes and the random allocation of maternal/paternal chromosomes to daughter cells work together to create genetic variability.

SDS-Page

The purpose of this technique is simple: it allows proteins to be separated by their mass alone. You may recall that gel electrophoresis accomplishes a similar goal for DNA and RNA: it uses an electric field to separate these molecules by length alone. So why can't gel electrophoresis also function to separate proteins? The answer is twofold. First, proteins vary widely in their structure, or folding patterns. If we simply tried to run a mixture of proteins through a gel electrophoresis apparatus, these variations would impact the proteins' migration. Additionally, simple gel electrophoresis relies on the fact that DNA and RNA molecules contain a uniform negative charge, causing them to travel toward the positive pole of the apparatus. In contrast, proteins can have positive, negative, or neutral charges, and these charges may not be uniformly distributed throughout the molecule. SDS-PAGE addresses both of these issues. First, to eliminate the effects of differences in shape, SDS-PAGE uses a strong anionic detergent: sodium dodecyl sulfate. The "sulfate" portion of this name denotes the negatively-charged head of the molecule, while the "dodecyl" refers to a long hydrocarbon chain that forms the molecule's tail. The SDS molecule denatures native proteins into their unfolded polypeptide states, which prevents protein shape from impacting the separation. To address charge differences and ensure that the proteins actually travel down the gel, SDS coats the proteins with an even distribution of charge per unit mass. Specifically, when the highly anionic SDS associates with the polypeptide backbone, the intrinsic charge of the polypeptide becomes negligible in comparison to the negative charges due to SDS. Since the protein is now highly negative, it will travel toward the positive end of the gel apparatus. Otherwise, SDS-PAGE functions similarly to standard gel electrophoresis. The larger the protein, the more hindered it is as it moves down the gel, and the shorter the distance it travels toward the positive end. In contrast, smaller proteins can travel through the pores of the gel more easily, so they migrate farther toward the positive pole.

Intersectionality example

To arrive at the correct answer, we first need to look at Figure 1. Figure 1 tells us that an older person is more likely to have experienced a year of poverty than a younger person, if for no other reason than because they've lived longer. As a result, a 60-year-old person would be more likely to experience a year of poverty than a 30-year-old person, so we can eliminate answer choices A and B. To decide between choices C and D, we need to look at Table 2. The 60-year-old Latino woman who never finished high school falls into three of the categories that are associated with higher odds of poverty (female, non-white, less than 12 years of education). The 60-year-old white man with a disability only ticks off one of the categories associated with higher odds of poverty. The 60-year-old Latino woman is a classic example of intersectionality: many facets of her identity interact and contribute to create social inequality.

when choosing a solvent for a specific reaction type

consider polity among other things C is correct. The SN2 mechanism is favored by polar aprotic solvents, such as acetone or DMSO. The structure of acetone is shown below; note that it has a dipole moment ("polar"), but does not contain O-H or N-H bonds ("aprotic"). A, B: Water and methanol are protic solvents. D: Toluene (shown below) is a nonpolar solvent. Solvents have a notable effect on the rates of these reactions. Polar protic solvents (such as water and ethanol) tend to stabilize ions in solution. Since they can stabilize the carbocation, these protic solvents are best used for SN1 reactions. However, SN2 reactions rely not on a carbocation, but on a strong nucleophile displacing the leaving group. Protic solvents tend to stabilize (weaken) this nucleophile, so they should not be used for SN2 procedures. Instead, polar aprotic solvents, such as acetone, are a better choice.

Oligodendrite

is the myelin-producing cell in the central nervous system.

Framing bias

n the social sciences, framing comprises a set of concepts and theoretical perspectives on how individuals, groups, and societies organize, perceive, and communicate about reality. Framing involves social construction of a social phenomenon - by mass media sources, political or social movements, political leaders, or other actors and organizations. It is an inevitable process of selective influence over the individual's perception of the meanings attributed to words or phrases.

R681H means?

that arginine at position 681 is substituted with histidine.