PR MCAT Course Test #1(497)

Which of the following Gestalt principles helps to explain why we tend to perceive a black circle with a striped rectangle partially covering it in Figure 2 instead of two black semi-circles with a striped rectangle in between? A) Law of continuity B) Law of proximity C) Law of solids D) Law of similarity

A. According to the Gestalt principle of continuity, we tend to see smooth, continuous patterns rather than discontinuous ones; therefore, this law would predict that we will perceive a continuous black circle partially covered by the rectangle, rather than two semi-circles (choice A is correct). The Gestalt principle of proximity states that we tend to group nearby things together; this does not apply to our perception of Figure 2 (choice B is wrong). The Gestalt principle of similarity states that we tend to group similar things together; this also does not apply to our perception of Figure 2 (choice D is wrong). There is no Gestalt law of solids (choice C is wrong).

In an experimental setting, an infant is consistently able to locate a toy underneath a red box when she observes a researcher placing the toy underneath this red box. When the researcher, in full view of the infant, places the toy underneath an adjacent blue box, the infant incorrectly looks for the toy underneath the original red box. An adherent to Piaget's theory of cognitive development might state that that the infant has made: A) an A-not-B error. B) an ambivalent attachment error. C) a trust-versus-mistrust error. D) a conservation error.

A. In an A-not-B error, the individual perseveres in looking for an object in a location where it was previously placed, even with the knowledge that it has been placed elsewhere. This is most common in infants under 12 months of age (choice A is correct). Ambivalent attachment is part of Mary Ainsworth's Stranger Paradigm. Infants with ambivalent attachment show stress when their parents leave, but do not want to be comforted when their parents return (choice B is wrong). Trust-versus-mistrust is the first stage of Erik Erikson's theory of psychosocial development, and does not explain why an infant might incorrectly look under the original red box even after observing the toy being hidden beneath the blue box (choice C is wrong). Conservation, which is attained during Piaget's preoperational stage of development (around roughly ages 4-5) refers to a child's ability to determine that a certain quantity remains the same despite adjustment of the container, shape, or apparent size; this does not explain why an infant might persist in looking under the first hiding place after a toy has been placed under a new hiding place (choice D is wrong).

In order to test this hypothesis, they create a line of mouse fibroblasts in which a gene for G418 (a biocidal agent) resistance, βgeo, is linked to the promoter of a gene expressed at high levels in ES cells, Fbx15 (Figure 1). The system shown in Figure 1 is an example of which of the following types of assay? A) Reporter gene system B) Two-hybrid screen C) Microarray D) Immunohistochemical screening

A. In attaching a gene for G418 resistance to a promoter which indicates a transformation to an ES cell, researchers created a reporter gene system in which neomycin resistance reports on iPS cell transformation (choice A is correct). Two-hybrid screening is another type of reporter gene screening usually performed in yeast and is used to study protein-protein interaction (choice B is incorrect). Microarray is a high-throughput screening method for biological molecules (usually nucleic acids; choice C is incorrect), and immunohistochemical methods involve antibody detection of a molecule (choice D is incorrect).

All of the following are true regarding learned helplessness, EXCEPT that: A) it is often linked to an internal locus of control. B) it often results in a cognitive expectation that nothing an individual does will prevent or eliminate a negative or aversive outcome. C) while learned helplessness is strongly tied to animal psychology and behavior, it can also apply to many situations involving human beings. D) it has been associated with several psychological disorders, including depression and anxiety.

A. Learned helplessness occurs when an organism is repeatedly subjected to a negative or aversive stimulus that cannot be escaped or avoided; eventually, the organism will give up trying to avoid or escape the stimulus and behave as if it is utterly helpless to change the situation—even when opportunities to escape are presented, learned helplessness will prevent any action. Individuals with an internal locus of control tend to believe that they are capable of controlling events in their lives; those with an external locus of control tend to feel as though life events are out of their control. Learned helplessness is associated with an external (not internal) locus of control (choice A is not true regarding learned helplessness, and is therefore correct). Learned helplessness does often result in a cognitive expectation that nothing an individual does will prevent or eliminate a negative or aversive outcome (choice B can be eliminated). It is also true that while learned helplessness is strongly tied to animal psychology and behavior (it was first discovered and demonstrated in extensive experiments with dogs), psychologists believe learned helplessness also applies to many situations involving human beings (choice C can be eliminated). Psychologists have also theorized that learned helplessness is associated with several psychological disorders, including depression and anxiety (choice D can be eliminated).

In which organ are sickled red blood cells most likely to be hemolyzed? A) Spleen B) Thymus C) Lymph nodes D) Kidneys

A. One of the key functions of the spleen is to remove old or deformed red blood cells from circulation. The capillaries of the spleen are much smaller than the diameter of a red blood cell, helping to induce lysis of fragile or old cells. If the spleen is removed, more damaged cells remain in circulation.

Which of the following statements about pyruvate kinase are true? A) It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. B) It has quaternary protein structure, similar to hemoglobin and myoglobin. C) It is involved in gluconeogenesis, the conversation of pyruvate to glucose during high energy states. D) It has several isoforms due to alternative splicing via the spliceosome and ribosome.

A. Pyruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP (choice A is correct). Paragraph 3 discusses how PKM2 can exist as a dimer or tetramer, each of which would contain more than one peptide chain and would therefore have quaternary protein structure. Hemoglobin also has quaternary structure (since it contains four peptide chains) but myoglobin does not (eliminate choice B). Since PKM2 is one isoform of pyruvate kinase, there are presumably others. Isoforms are different forms of the same protein and can be due to gene duplication or alternative splicing. However, the ribosome has no function in splicing (eliminate choice D).

Which one of the following best represents the phase diagram of water?

A. The characteristic that distinguishes the phase diagram for water from the phase diagram for virtually all other substances is that the solid-liquid boundary line in the phase diagram for water has a negative, rather than a positive slope. This is best illustrated by the diagram in choice A.

A cyclist collides with a tree, and the collision creates an intense compressive stress on the cyclist's left humerus (upper arm bone), causing a fracture. Which of the following changes in the initial conditions most likely could have prevented the fracture? A) A decrease in his velocity by a factor of two B) A decrease in his mass by a factor of two C) An increase in his mass by a factor of two D) A decrease in the cross-sectional area of his humerus by a factor of two

A. The cyclist's kinetic energy just before the impact will be "used" to cause damage to the tree and to the cyclist. Decreasing this kinetic energy will result in a less violent collision. Because KE = (1/2)mv2, a decrease in the cyclist's velocity would have the greatest effect (since v is squared) in lowering his KE. KE= 1/2mv2

The voltage across the membrane is -70 mV from the exterior to the interior of the cell, and in a resting axon, there is no net transfer of charge across the axon membrane. Across the membrane of an axon in its rest state: A) the potential is higher on the outside of the cell than the inside, and electric field lines run from the outside to the inside. B) the potential is higher on the outside of the cell than the inside, and electric field lines run from the inside to the outside. C) the potential is lower on the outside of the cell than the inside, and electric field lines run from the outside to the inside. D) the potential is lower on the outside of the cell than the inside, and electric field lines run from the inside to the outside.

A. The passage states that "the voltage across the membrane is -70 mV from the exterior to the interior of the cell." That is, there is a 70 mV drop in electric potential in going from the exterior to the interior; thus, the potential is higher on the exterior of the cell. This eliminates choices C and D. Since the potential is higher on the exterior than in the interior, the electric field lines must point from the outside to the inside (choice A).

The rules which govern appropriate emotional responses vary from culture to culture. These social mores are called: A) display rules. B) emotional expression. C) rituals. D) cultural emotions.

A. The rules governing emotional displays within a particular culture are called display rules (choice A is correct). Emotional expression is what is being regulated by the display rules (choice B is wrong). Rituals are a series of set activities, which can include gestures or words, that occur in a set place and in a set order; the social rules governing the expression of emotions do not (generally) address the order or place of the emotional display (choice C is wrong). Emotional responses are the experience of emotions, which is thought to be innate; there is no such thing as "cultural emotions," per se (choice D is wrong).

Step 1. A population of E. coli was incubated on a medium rich in lactose, and intracellular levels of lactose were measured. The bacteria were then transferred to a lactose-poor medium having a potassium ion concentration that was substantially greater than intracellular concentrations of this ion. As a control, some bacteria were transferred to lactose-poor media without a potassium gradient. Step 2. After a specified period of time had elapsed, intracellular concentrations of lactose were measured and an average value derived. This value was then compared to the average concentration of lactose within the bacteria at the start of the experiment. Step 3. The concentration of potassium in the lactose-poor medium was measured and compared to the potassium concentration in the medium at the start of the experiment. Step 4. Steps 1 and 2 were repeated but instead of imposing a potassium gradient, the investigators imposed an H+ gradient. The medium was acidic relative to the inside of the bacterial cells. Lactose and hydrogen ion concentration measurements were taken and compared to initial values as in Step 2. If the measurements taken in Step 4 of Experiments 1 and 2 had shown no change in bacterial lactose concentration, then it is likely that within the medium: A) hydrogen ion concentration had remained unchanged in Experiments 1 and 2. B) hydrogen ion concentration had increased in Experiments 1 and 2. C) potassium ion concentration had remained unchanged in Experiments 1 and 2. D) potassium ion concentration had increased in Experiments 1 and 2.

A. These experiments show that a gradient of either potassium or protons can drive the transport of lactose into the cell against its gradient. If there is no lactose transport, it is safe to assume that the concentration in the media of the driving ion will remain the same (A is correct). There is no reason to assume that hydrogen ion concentration would increase (regardless if lactose transport had occurred or not) since this is not the case in any of the experiments (B is wrong). Step 4 involves a proton gradient, not a potassium gradient (C and D are wrong).

Which of the following describes the molecular geometry of a carbon dioxide molecule? A) Linear B) Trigonal planar C) Tetrahedral D) Bent

A. VSEPR theory predicts (and experiments have verified) that the carbon dioxide molecule, O=C=O, is linear since the central carbon atom contains no lone-pair electrons (eliminate choice D) and the two regions of high electron density (the two double bonds) are most stable on opposite sides of the central atom. In order to be trigonal planar or tetrahedral, the central carbon would need to be bonded to three or four atoms, respectively.

BIO PASSAGE 1: The most common abnormalities of chromosome number are trisomies. These occur when there are three representatives of a particular chromosome instead of the usual two. Which of the following is a common feature of people with Down, Patau, Edwards, and Klinefelter's syndromes? A) They all have heart defects. B) They all have 47 chromosomes. C) They are all male. D) They all have a low life expectancy.

B. A normal human has 46 Chromosomes. 22 pairs of autsomes and 1 pair of sex chromosomes. All of the listed syndromes are trisomies: Down-21, Patau-13, Edwards-18, Klinefelter's-sex chromosomes (choice B is correct). There is no mention of heart defects in Klinefelter's syndrome (choice A is incorrect). There is no gender difference in the autosomal trisomies (choice C is incorrect). There is no mention of life expectancy in any of the diseases (choice D is incorrect).

How many urea groups are present in a caffeine molecule? A) 0 B) 1 C) 2 D) 3

B. A urea group is a carbonyl carbon bonded to two nitrogens, and caffeine has exactly one such group.

It appears that in their original predictions, participant wives who reported steeper drops in marital satisfaction might have used which defense mechanisms? A) Displacement B) Denial C) Repression D) Reaction formation

B. According to Freud, people use ego defense mechanisms in order to protect themselves from the effects of anxiety. The results of the current research offer the following about the women who experienced the steepest declines in marital satisfaction: (1) these women reported higher levels of external stress and (2) following this, these women had more positive predictions for their marriage. The fact that these women were more optimistic about their futures despite the presence of significant external stressors suggests the presence of a cognitive bias (as described in the final paragraph) that offers self-protection. Denial is a defense mechanism that is most related to this situation (choice B is correct). Denial is the refusal to accept external realities that are threatening to the self; for example, the refusal to consider the effect external stressors could have on marital satisfaction. Displacement is the shift of emotions, such as aggressive or sexual impulses, to safer outlets (choice A is wrong). Repression is the psychological attempt to subdue emotionally painful memories (choice C is wrong). Reaction formation is expression of the opposite of one's feelings in order to conceal unacceptable or dangerous emotions (choice D is wrong).

The attainment of object permanence most closely aligns with which of Freud's psychosexual stages? A) Oral stage B) Anal stage C) Phallic stage D) Latency stage

B. According to the first paragraph, object permanence is attained by roughly 18-24 months of age. This most closely aligns with Freud's anal stage of psychosexual development, which he postulated to span from roughly 1-3 years of age, when young children are learning how to control bowel elimination (choice B is correct). Freud's oral stage occurs from birth to roughly age 1, which means that that stage would occur before object permanence is attained at roughly 18-24 months (choice A is wrong). Freud's phallic stage (from about ages 3-6 years) and latency stage (from about 6-12 years) both occur long after the attainment of object permanence (choices C and D are wrong).

BIO PASSAGE 2: Specifically, aldosterone binds to an intracellular protein receptor, inducing an allosteric change which enables this protein to bind DNA and promote the transcription of the Na+/K+ ATPase gene. Activated aldosterone receptor most directly regulates activity of which of the following enzymes? A) DNA polymerase B) RNA polymerase II C) Na+/K+ ATPase D) Renin

B. Aldosterone binds to aldosterone receptors to regulate transcription of a specific set of genes. The enzyme that synthesizes mRNA is RNA polymerase II, so it is this enzyme that would be most directly affected by the activated aldosterone receptor. DNA polymerase is used in replication, not transcription (A is wrong), and while aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin, these would be indirect effects (eliminating C and D).

BIO PASSAGE 3: 28 mL of potassium permanganate was required to titrate 5 mL of 0.005 M hydrogen peroxide to water. The amount of potassium permanganate required for titration was one-third of the amount required in the absence of catalase Extreme high temperature would have which of the following effects on the second trial? A) It would increase the required amount of potassium permanganate to approximately 18 mL. B) It would increase the required amount of potassium permanganate to approximately 28 mL. C) It would maintain the required amount of potassium permanganate at approximately 9 mL. D) It would reduce the required amount of potassium permanganate to approximately 1 mL.

B. Although enzyme activity normally increases with increasing temperature, extreme high temperature denatures the enzyme and renders it inactive. Twenty-eight milliliters of potassium permanganate were required to titrate the hydrogen peroxide in the absence of enzyme (Finding 1). With catalase rendered inactive by extreme high temperature, the reaction vessel would be functionally devoid of enzyme. As in the first trial, 28 mL of potassium permanganate solution would be required to titrate the hydrogen peroxide (choice B is correct and choices A, C and D are eliminated).

If the absolute pressure of a gas is increased from 3 atm to 4 atm at constant volume, then the absolute temperature of the gas will increase by: A) 25%. B) 33%. C) 67%. D) 75%.

B. Assuming the validity of the Ideal-Gas law, PV = nRT, an increase in pressure by a factor of 4/3 at constant volume will result in an increase in temperature by the same factor (since P is proportional to T if V and n are constant). Multiplying the temperature by 4/3 = 133% implies an increase by 33%.

For example, in humans with blue eyes (which are typically recessive to brown eyes), the OCA2 protein tends to have arginine at amino acid 305 (305R) and glutamine at amino acid 419 (419Q); in humans with brown eyes, the OCA2 protein tends to have tryptophan at amino acid 305 (305W) and arginine at amino acid 419 (419R).The OCA2 gene is one of the major determinants of blue or brown eye color, but these same signaling and biosynthetic pathways also control other aspects of human pigmentation. The OCA2 gene overlaps with another gene called HERC2, which has two alleles: the wild type allele, and a recessive allele A1 which has been linked to a genetic predisposition to Crohn's disease. A woman with an OCA2305R : HERC2wt chromosome and an OCA2305W : HERC2A1 chromosome mates with a homozygous OCA2305R / 305R : HERC2A1 / A1 man. Their children will most likely be: A) 25% normal with blue eyes, 25% normal with brown eyes, 25% blue eyed with a risk of Crohn's disease and 25% brown eyed with a risk of Crohn's disease. B) 50% normal with blue eyes and 50% brown eyed with a risk of Crohn's disease. C) 50% blue eyed with a risk of Crohn's disease and 50% normal with brown eyes. D) 100% blue eyed with a risk of Crohn's disease.

B. Based on information in the passage, alleles of OCA2 are one of the major determinants of eye color. The woman in the question stem will have brown eyes because she is heterozygous for OCA2 and has an allele that is associated with blue eyes (305R) and an allele associated with brown eyes (305W). Since the question stem says that the OCA2 gene and the HERC2 gene overlap, they must be linked (they are 0 map units apart). This means that crossing over will not occur between these two genes and they will be inherited as a unit. The father in this question is acting like a testcross; that is, he is homozygous recessive for both genes. Therefore, the offspring produced will either get the OCA2305R : HERC2wt chromosome or the OCA2305W : HERC2A1 chromosome from the mother and an OCA2305R : HERC2A1 chromosome from the father. The children will be 50% OCA2305R / 305R : HERC2wt / A1 (blue eyed with no increased risk of Crohn's disease; note that the A1 allele of HERC2 is recessive to the wild type allele) and 50% OCA2305W / 305R : HERC2A1 / A1 (brown eyes with an increased risk of Crohn's disease). Thus, choice B is correct (eliminate choices A, C, and D). Note than choice A is what would occur if the two genes were not linked.

A converging glass lens forms a real image of a red object at a distance equal to twice its focal length. If a blue object is placed adjacent to the red object, will its image form closer to the lens or farther from the lens than the image of the red object? A) Farther, because blue light is refracted more due to its shorter wavelength B) Closer, because blue light is refracted more due to its shorter wavelength C) Farther, because blue light is refracted less due to its longer wavelength D) Closer, because blue light is refracted less due to its longer wavelength

B. Choices C and D can be eliminated immediately since blue light has a shorter wavelength than red light. The refractive index of a transparent medium increases with increasing frequency of the transmitted light. Thus, the refractive index of the lens for blue light would be slightly higher than for red light. This would imply that blue light experiences more refraction than red light. Since the lens is a converging one (because diverging lenses cannot form real images), the blue light would be bent more sharply toward the axis than the red light, so the image of the blue object would be formed closer to the lens.

Which portion of the central nervous system confers balance by coordinating the activity of various motor units? A) Cerebrum B) Cerebellum C) Medulla D) Hippocampus

B. Coordination of motor skills is one of the primary functions of the cerebellum. The cerebrum triggers skeletal muscle contraction, but the cerebellum coordinates it. MEMORY TRICK: every time you start to lose balance you hear a BELL (cereBELLum) that alerts you not to fall BELLUM=BALANCE

BIO PASSAGE 4: Gastrin stimulates the parietal cells which secrete HCl into the stomach. Gastrin secretion is inhibited by low pH in the stomach. This is an example of: A) competitive inhibition. B) negative feedback. C) acid hydrolysis of proteins. D) exocrine secretion of a hormone.

B. Gastrin stimulates acid production, and its release is inhibited by stomach acid. This is a negative-feedback loop designed to maintain acid within a certain pH range (B is correct). Competitive inhibition does not apply in this case, because there is no enzyme to be inhibited (A is wrong). Since gastrin is not excreted into the stomach lumen (it's a hormone), it will not be degraded by stomach acid (C is wrong). A hormone cannot be secreted in an exocrine manner (D is wrong).

It has been suggested that one reason tumor cells have novel metabolism is to generate biomolecules required to increase biomass, which is required when proliferation rates are high. How could this occur? A) Fructose-6-phosphate is shuttled from glycolysis to the pentose phosphate pathway to support both nucleotide and fatty acid biosynthesis. B) Glutamate can be used to produce Krebs cycle intermediates such as oxaloacetate and citrate and these molecules can be used to generate lipids, amino acids and nucleotides. C) High glycolysis rates can support the pentose phosphate pathway, which generates ribose-2-phosphate for nucleotide catabolism. D) Glutamate is converted into glutamine to provide the cell with an amino acid precursor, thus powering translation.

B. Glucose-6-phosphate (not fructose-6-phosphate) is shuttled from glycolysis to the pentose phosphate pathway (choice A is wrong), although this does support nucleotide and fatty acid biosynthesis. Paragraph 2 describes how glutamine is converted into the Krebs cycle intermediate α-ketoglutarate, which can then be used to generate other Krebs cycle intermediates to power biomolecule synthesis (choice B is correct). The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, choice C is wrong). Paragraph 2 describes how glutamine is first converted into glutamate, not the other way around (choice D is wrong).

As they embarked upon deliberations, nine jurors were leaning towards finding the defendant guilty while three jurors thought the defendant was not guilty. However, during deliberations, the nine who believed the defendant was guilty tended to confer with each other, while the three who believed he was innocent also discussed more with each other, which resulted in the jury members each becoming more entrenched in their initial positions than they had been before. This exemplifies what concept in social psychology? A) Cognitive appraisal B) Group polarization C) Informational influence D) Social categorization

B. Group polarization describes the phenomenon when groups make more extreme decisions acting together as a group than they would acting alone. Thus as individuals, these jurors believed in their points of view from the beginning, but after they formed oppositional groups they became even more entrenched in their points of view (choice B is correct). Cognitive appraisal is a process where a physiological arousal is assessed in the brain and processed for the appropriate physiological and psychological coping strategies, and thus has nothing to do with this situation (choice A is wrong). Informational influence is a group effect arising from a group's desire to be correct and to understand how to act best in a given situation. While this may play a role here, as the jury may be trying to be correct and act in the interest of justice, it has split into two opposing camps and thus one group will not "act correctly" (choice C is wrong). Social categorization is the process by which people sort themselves into categories. While this may be happening here in some small measure, as the jury breaks into two camps—those who believe the defendant is guilty and those who believe the defendant is not guilty—it is again not the primary effect (choice D is wrong).

A highly proliferating (cancer) cells would most likely: A) overexpress hexokinase and fructose-1,6-bisphosphatase. B) express high levels of the lactate transporter and the glutamine transporter. C) power cell growth by running the electron transport chain and oxidative phosphorylation. D) upregulate pyruvate dehydrogenase kinase and downregulate phosphofructokinase.

B. Highly proliferating cells would express large amounts of hexokinase (a key glycolysis enzyme) but would not also overexpress fructose-1,6-bisphosphatase because this is an enzyme involved in gluconeogenesis. Even highly proliferative cells will avoid concurrently running reciprocally regulated pathways (eliminate choice A). The passage says that rapidly dividing tumor cells produce and export large amounts of lactate, and import large amounts of the amino acid glutamine (choice B is correct). The focus of this passage is how highly proliferative tumor cells power growth via glycolysis and fermentation, making choice C an unlikely correct answer (eliminate choice C). Based on information in the last paragraph, pyruvate dehydrogenase kinase activity is high in some rapidly growing cells, but a high rate of glycolysis will lead to high PFK expression (eliminate choice D).

In Figure 1, the axon membrane can be treated as a capacitor, slow leakage channels as a 25 MΩ resistor, and the Na+/K+ pump as a voltage generator. The voltage across the membrane is -70 mV from the exterior to the interior of the cell, and in a resting axon, there is no net transfer of charge across the axon membrane. Figure 2 includes the additional Na+ influx (a 4 kΩ resistor) of an action potential. Other ion fluxes are ignored. If the active neuron in Figure 2 is clamped at a constant voltage and the total current through the Na+ channels plus the leakage channels is 2.5 µA, what is the voltage drop across the membrane? A) 1 mV B) 10 mV C) 62.5 mV D) 62.5 V

B. In Figure 2, there are two parallel resistors: one of 25 million ohms and one of only 4 thousand ohms. Since 25 million is so much greater than 4 thousand, the equivalent resistance for this parallel combination will be so close to 4 thousand ohms that the difference is negligible. Taking Req = 4000 Ω gives V = IR = (2.5 × 10-6 A)(4 × 103 Ω) = 10 × 10-3 V = 10 mV.

BIO PASSAGE 1: Some of the characteristic clinical findings include mental disability, hypotonia, blind-ending jejunum, and heart malformations, including a common arterial trunk coming off both the left and right ventricles. Which of the following is expected to be true of children with Down syndrome? A) Aortic arterial blood carbon dioxide concentration is normal. B) Aortic arterial blood oxygen saturation is lower than normal. C) Pulmonary arterial blood pressure is lower than normal. D) Pulmonary arterial blood oxygen saturation is lower than normal.

B. In children with Down syndrome, one of the primary cardiac defects is a truncus arteriosus, which presents as a common arterial trunk coming off both the left and right ventricles. This allows for mixing of blood from the left and right circulations. Since the right circulation is relatively high in carbon dioxide and the left circulation is relatively low in carbon dioxide, the resultant mix will be somewhere between (i.e., not normal, choice A is wrong). By similar reasoning, the right circulation that is typically low in oxygen will mix with the left circulation that is typically higher in oxygen, so the overall aortic arterial oxygen saturation will be less than normal (choice B is correct) and the pulmonary arterial oxygen concentration will be higher than it is normally (choice D is wrong). Since a common arterial trunk is receiving blood from both the right and left ventricles, the pulmonary artery will be receiving more blood than usual and therefore will have a higher than normal blood pressure (choice C is wrong).

Paragraph 3 states, "participants were trained to make situational attributions by rewarding participants after every third situational attribution they made"; The situational attribution training described in the study is most consistent with which of the following methods of learning? A) Classical conditioning B) Positive reinforcement C) Modeling D) Negative reinforcement

B. Paragraph 3 states, "participants were trained to make situational attributions by rewarding participants after every third situational attribution they made"; therefore, the encouragement given in the study affects learning through positive reinforcement, as it strengthens a response by rewarding it with a positive stimulus (choice B is correct). Classical conditioning refers to learning as a result of pairing an unconditioned stimulus with a conditioned stimulus; there is no information provided in the third paragraph to suggest that an unconditioned stimulus is being paired with another stimulus (choice A is wrong). Modeling refers to learning based on observing and imitating the behavior of another; again, there is no indication that modeling is occurring in this study (choice C is wrong). Negative reinforcement is the strengthening of a response because of a removal of an aversive stimulus; the third paragraph states that study participants were given a reward, not that they had an aversive stimulus removed (choice D is wrong).

A chemist mixes 4.0 g AgI(s) with 25 mL of deionized water and allows the system to come to equilibrium. Which of the following changes to the mixture will NOT affect the equilibrium [Ag+]? A) Addition of NaOH(aq) B) Dilution with 10 mL deionized water C) Addition of NaCN(aq) D) Addition of AgNO3(s)

B. Since AgI is a relatively insoluble compound, the equilibrium system described should be a saturated solution of AgI, which should contain some undissolved solid AgI at the bottom of the container. Anything added to the mixture that will react with either the Ag+ or I- ions in solution will change the ion concentrations. Addition of NaOH will precipitate silver ions to form a different insoluble compound, AgOH (eliminate choice A). Addition of a NaCN solution will form a soluble silver cyanide complex ion, dissolving more of the AgI(s) in the mixture (eliminate choice C). Addition of solid AgNO3 will increase the silver ion concentration since all nitrate compounds are highly soluble (eliminate choice D). Dilution with a small amount of water will initially reduce the ion concentrations, but the same equilibrium concentrations will be reestablished after a small amount of time by dissolving additional AgI solid.

The Q-switch momentarily interrupts the inducing light creating a build-up of energy within the crystal. This does not increase the overall energy of the laser, but concentrates it into shorter time periods or pulses. A longer interruption with the Q-switch most likely would increase the: A) total amount of work done by the laser. B) power of each laser pulse. C) wavelength of the laser light. D) frequency of the laser light.

B. Since the overall energy of the laser does not change, neither will the frequency, wavelength, nor work done by the laser. This leaves choice B: Concentrating the energy into a shorter time period increases the power of each pulse (since power equals energy delivered per unit time, by definition). Power= Work/time Energy= frequency x plancks constant velocity= frequency x wavelength

A researcher attempts to explain the phenomenon described in the passage that couples who reported greater relationship satisfaction had greater physiological co-regulation and concludes that the cognitive interpretation of the physiological state of both partners led to their reported feelings of satisfaction. This is most similar to which of the following? A) Emotional intelligence B) Schachter-Singer theory C) Cannon-Bard theory D) James-Lange theory

B. The Schachter-Singer theory posits that emotional experience is determined by physiological state and the cognitive interpretation of that state. This is most similar to the description in the question stem, in which couples experience physiological co-regulation and cognitive interpretation (choice B is correct). Emotional Intelligence is the ability to control, interpret, and understand one's own emotions and the emotions of others. This is not as specific as choice B in capturing the information in the question stem (choice A is wrong). The Cannon-Bard theory focuses on the central role of the hypothalamus in regulating emotions and is less specifically concerned with physiological activation and cognition (choice C is wrong). The James-Lange theory asserts that emotional experience is primarily based on physiological arousal, and that each different physiological state is associated with an emotion (choice D is wrong).

A tall plant with curly leaves (TT/CC) was crossed with a dwarf plant with straight leaves (tt/cc). Two of the resulting F1s were crossed with each other and produced F2s as follows: 34 tall plants with curly leaves 4 dwarf plants with straight leaves 11 tall plants with straight leaves 12 dwarf plants with curly leaves. Based on these results, are the genes for plant height and leaf shape linked? A) No, since recombination occurred. B) No, since the phenotype ratio is close to the expected ratio for this cross. C) Yes, since the recombination frequency is less than 50%. D) Yes, since there are considerably fewer recombinant plants than expected.

B. The cross between the parental plants produced an F1 generation that all have the genotype Tt/Cc (double heterozygotes). These plants were then crossed to produce the F2 generation. The expected unlinked phenotype ratio from a cross between two double heterozygotes is 9:3:3:1, with 9/16 of the offspring double-dominant, 3/16 dominant/recessive, 3/16 recessive/dominant, and 1/16 double-recessive. Based on the numbers given in the question, the actual ratio (34:11:12:4) is very close to this, so these genes are not linked (choice B is correct). Recombination can occur between linked genes; it just happens less frequently (A is eliminated). The recombination frequency (RF, the number of recombinants divided by the total number of offspring) does not have to be 50% for the genes to be unlinked; in the unlinked 9:3:3:1 ratio, the RF is only 37.5% (C is wrong). Based on the total number of F2's produced (34 + 4 + 11 + 12 = 61), the expected number of recombinants was (37.5%)(61) = (3/8)(61) ≈ 23, and 11 + 12 = 23 were produced (choice D is wrong).

The passage states that the pH range for which the eye cannot discern one of the pure-compound colors to be pKa - 1 < pH < pKa +1 Methyl orange is an indicator that is red in its undissociated form and yellow in its dissociated form. If the Ka for methyl orange is 1.6 × 10-4, which of the following gives the best pH range for the solution appears neither red nor yellow? A) 1.4 to 3.4 B) 2.8 to 4.8 C) 4.0 to 6.0 D) 6.2 to 8.2

B. The passage states that the pH range for which the eye cannot discern one of the pure-compound colors to be pKa - 1 < pH < pKa + 1. Since the Ka of methyl orange is 1.6 × 10-4, its pKa = -log Ka = -log(1.6 × 10-4) ≈ 4. Thus, the appropriate pH range is given approximately by the inequalities 3 < pH < 5, so choice B is best.

Ouabain irreversibly inhibits the Na+/K+ pump. The effect of this on the equivalent circuit model for the resting axon would be to cause: A) increased potential difference across the resistors and the capacitor. B) decreased potential difference across the resistors and the capacitor. C) decreased potential difference across the capacitor and increased potential difference across the resistors. D) decreased potential difference across the resistors and increased potential difference across the capacitor.

B. The resistors and the capacitor are in parallel, so they must have the same potential difference (eliminating choices C and D). If the source of potential difference (the Na+/K+ pump) is prevented from operating, then the common potential difference across the resistors and the capacitor would decrease.

BIO FREESTANDING Qs: GABA protein contains both an carboxyl group and an amino group on the same molecule. Which form of the amino acid GABA does not predominate at any pH?

B. The uncharged form B does not predominate at any pH since the pKa of the carboxyl group is lower than the pKa of the protonated amino group, making the former group the stronger acid. The titration of GABA from acidic to basic pH would therefore result in the following changes in its protonation state: As a result, form A predominates at pH < 2.3, form C at 2.3 < pH < 9.7, and form D at pH > 9.7.

The polypeptide +NH3-Asp-Lys-Leu-Val-Arg-Glu-Val-Lys-COO- is digested by trypsin. The products will be: A) +NH3-Asp-COO-, +NH3-Lys-Leu-Val-COO-, +NH3-Arg-Glu-Val-Lys-COO- B) +NH3-Asp-Lys-COO-, +NH3-Leu-Val-Arg-COO-, +NH3-Glu-Val-Lys-COO- C) +NH3-Asp-Lys-COO-, +NH3-Leu-Val-Arg-COO-, +NH3-Glu-Val-COO-, Lys D) +NH3-Asp-COO-, +NH3-Lys-Leu-Val-COO-, +NH3-Arg-Glu-Val-Lys

B. Trypsin and other serine proteases hydrolyze the peptide bond to the right (toward the COOH-terminus) of basic amino acid residues including lysine and arginine. The two sites of hydrolysis are between Lys and Leu and between Arg and Glu to make three fragments as shown in choice B.

See table in answer key before answering. Of the following, which is the strongest oxidizing agent, based on Table 1? A) Ag(s) B) Zn2+(aq) C) Cu2+(aq) D) Ni(s)

C. An oxidizing agent is a species that is reduced in a redox reaction, so the strongest oxidizing agent among the given choices must be the species that is the most easily reduced. Choices A and D can be eliminated because neutral metals tend to be oxidized, not reduced. Zn2+ has an Eº < 0, so its reduction is not spontaneous, making it a poor oxidizing agent. (eliminate choice B). The reduction of Cu2+, however, is spontaneous (because E > 0), so this is the best answer.

A bacteriologist initiated an E. coli culture from one E. coli cell and hypothesized that some of the progeny in the culture would be genetically different from the original parent cell. Is this hypothesis true? A) Yes; bacteria are capable of undergoing genetic recombination through a variety of mechanisms. B) Yes; bacteria reproduce sexually, and the progeny of any one cell are genetically distinct from the parent cell. C) No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. D) No; bacteria can reproduce only by meiosis, which ensures preservation of the genome.

C. Bacteria reproduce asexually, by one cell replicating its genome and then splitting into two cells that are genetically identical to the original cell. The only potential sources of genetic variation in bacteria are mutation and the transfer of genetic information through conjugation, transduction, or transformation, none of which are linked to reproduction. In the absence of mutation, all progeny of a cell will be identical to the original cell (C is true). Bacteria only perform recombination under special circumstances such as through the presence of Hfr plasmids that replicate a portion of the bacterial genome to make it transiently diploid (A is false). There is no indication of a role for Hfr in this case and in a clonal cell line, it could not play a role. Bacteria do not perform the recombination, independent assortment and independent segregation that create genetic diversity in eukaryotes that reproduce sexually (B is wrong). They also do not perform meiosis (D is wrong).

Which of the following best describes the role of fructose-2,6-bisphosphate? A) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating fructose-1,6-bisphosphatase and inhibiting phosphofructokinase. B) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting hexokinase. C) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase. D) It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating hexokinase and inhibiting fructose-1,6-bisphosphatase.

C. Fructose-2,6-bisphosphate exerts reciprocal control on glycolysis and gluconeogenesis through phosphofructokinase and fructose-1,6-bisphosphatase, and has no impact on hexokinase activity (choices B and D are wrong). Fructose-2,6-bisphosphate stimulates phosphofructokinase (glycolysis) and inhibits fructose-1,6-bisphosphatase (gluconeogenesis). Choice C is correct, and choice D can be eliminated. The most potent allosteric regulator of the glycolysis and gluconeogenesis pathways is fructose-2,6-bisphosphate (F2,6BP). As shown in Figure 16.6, F2,6BP activates phosphofructokinase (also called PFK1) - the enzyme in glycolysis that converts fructose-6-phosphate to fructose-1,6-bisphosphate. F2,6BP also inhibits fructose-1,6-bisphosphatase (F1,6BPase) - the enzyme in gluconeogenesis that accomplishes the opposite reaction. In fact, F1,6BP is ten times more sensitive to F2,6BP than AMP, another reciprocal regulator.

BIO PASSAGE 4: Zollinger-Ellison syndrome is most effectively treated by antihistamines such as cimetidine, anticholinergic drugs such as pirenzepine, and omeprazole, which blocks the H+/K+ ATPase

C. If acetylcholine is blocked, calcium levels will be low, protein kinase activity low, and H+ secretion low (C is correct). A is wrong since calcium will be low. B and D are wrong since H+ secretion will decrease, not increase.

Which of the following could explain the Warburg effect (tumor cells generate ATP mainly through glycolysis and lactic acid fermentation in the cytosol)? Metabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors. Cells with low proliferation rates often have a high ratio of glycolysis to mitochondrial respiration. Some oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death. A) I only B) III only C) I and III only D) I, II, and III

C. Item I is true: tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor. Since glycolysis and fermentation don't require oxygen, they could facilitate cell growth in anaerobic conditions (choice B can be eliminated). Item II is false: the passage is about highly proliferative cells that use glycolysis and fermentation to power growth so a statement about slow growth does not explain the Warburg effect discussed in the passage (choice D can be eliminated). Item III is true: if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell (choice A can be eliminated and choice C is correct).

Which one of the following statements concerning the SN2 reaction mechanism is true? A) It proceeds best with tertiary substrates and is a two-step mechanism. B) It proceeds with retention of stereochemistry. C) It proceeds best with primary substrates and is a one-step reaction. D) It proceeds through a carbocation intermediate.

C. SN2 cannot operate with a tertiary system because the nucleophile must attack at the same time the leaving group leaves, and tertiary substrates are too sterically hindered (eliminate choice A). SN2 goes with inversion of stereochemistry, not retention (eliminate choice B). SN1 reactions have carbocation intermediates (eliminate choice D), while reactions are one-step processes with no intermediates. Choice C is the best answer because unhindered substrates allow the nucleophile to attack while the leaving group leaves.

Fluoroquinolones are a new class of antibiotics that are extremely effective against a wide range of bacteria. It has been determined that fluoroquinolones enter cells along with water via a passive transport mechanism. After exposing both bacterial and human cells to high concentrations of fluoroquinolones, a researcher discovered that, in the bacterial cells only, the DNA was nicked and supercoiling had been disrupted. Of the following, the most likely explanation for this is that fluoroquinolones: A) are unable to diffuse into human cells. B) interfere with DNA polymerase, which is not present in human cells. C) interfere with DNA gyrase, which is not present in human cells. D) interfere with DNA helicase, which is not present in human cells.

C. Since fluoroquinolones diffuse into cells with water, they are able to enter all cells, eukaryote and prokaryote alike (they diffuse through special water channel proteins called porins), so A is wrong. Both eukaryotes and prokaryotes utilize DNA polymerase and helicase (B and D are wrong). However, only prokaryotes use DNA gyrase to supercoil their DNA; eukaryotes wind DNA around histones through the action of other topoisomerases (choice C is correct).

Which of the following statements is plausible regarding changes in the epigenetic regulation of ES (embryonic stem) cell-associated genes in MEF (induced pluripotent stem cells) cells following transfection? A) Histone acetyl transferase (HAT) is activated, leading to decreased transcription of genes for ES cell-associated proteins. B) Histone deacetylase (HDAC) is activated, leading to decreased transcription of genes for ES cell-associated proteins. C) DNA is demethylated, leading to increased transcription of genes for ES cell-associated proteins D) Nucleotide methyl transferases are activated, leading to increased transcription of genes for ES cell-associated proteins.

C. Since the goal of iPS cells is to mimic ES cells, they can be expected to express more ES cell-associated genes and thus display epigenetic changes associated with increased expression of these genes (choices A and B, discussing decreased transcription of ES genes, are incorrect). DNA methylation (caused by nucleotide methyl transferases) decreases transcription of gene products (choice D is incorrect). DNA demethylation would be expected to increase transcription of the genes for ES cell-associated proteins, since methylation decreases transcription (choice C is correct).

If researchers wanted to include a participant with severe damage to her lateral geniculate nucleus, they would need to revise the procedure by: A) using a verbal problem instead of an auditory problem. B) using a visuospatial problem instead of a verbal problem. C) reading the problems aloud rather than providing the participants with the text. D) having participants write their responses rather than verbalize them.

C. The lateral geniculate nucleus (LGN) is the area behind the retina that serves as an intermediate way station for electrochemical signals passed from the retina to the visual cortex of the brain. Accordingly, the visions of a participant with damage to the LGN would be impaired, requiring that the problems be read aloud rather than presented in text form (choice C is correct).Using a visuospatial problem or having participants write their responses would not address the basic deficit in vision caused by a damaged LGN (choices B and D are wrong). Even in its original form, the experiment entailed a verbal problem, not an auditory problem (choice A is wrong).

Cystic fibrosis is the most common inherited lethal disease of Caucasians. It is an autosomal recessive disorder, occurring with a frequency of 1 in 3600. In a population of 18,000 Caucasians, how many are expected to be carriers of cystic fibrosis? A) 50 B) 295 C) 590 D) 1180

C. The passage states that the frequency of the autosomal recessive condition cystic fibrosis, q2, is 1 in 3600. The frequency of the recessive allele, q, then is 1 in 60. The frequency of the dominant non-disease producing allele, p, is 59 in 60. The carriers of a population are determined by the expression 2pq. In the given population, the number of carriers would be (2)(1/60)(59/60)(18000) or 590. Thus, choice C is correct and choices A, B, and D are eliminated.

Which of the following is the anticodon sequence on the tRNA for the start codon? A) 5'-AUG-3' B) 5'-UAC-3' C) 5'-CAU-3' D) 5'-GUA-3'

C. The start codon on the mRNA would be 5'-AUG-3', therefore, the anticodon on the tRNA would be the complementary sequence: 3'-UAC-5' (or alternatively written, 5'-CAU-3').

C. The table lists the number of cultures of each type of organism. The number of positive isolates for methicillin-resistant Staphylococcus aureus is 115. The total number of cultures reported is the number of Gram-(+) plus the number of Gram-(-) plus the fungal cultures: 160 + 180 + 5 = 345. Therefore, choice C is correct (and choices A, B, and D are eliminated).

C. The table lists the number of cultures of each type of organism. The number of positive isolates for methicillin-resistant Staphylococcus aureus is 115. The total number of cultures reported is the number of Gram-(+) plus the number of Gram-(-) plus the fungal cultures: 160 + 180 + 5 = 345. Therefore, choice C is correct (and choices A, B, and D are eliminated).

BIO PASSAGE 7: Stilbestrol is a crystalline non-steroid with estrogenic effects often superior to those of the estrogen, estradiol. If radiolabeled stilbestrol were administered to the experimental chicks, stilbestrol would be found most heavily concentrated: A) at the cell membrane of oviduct tissue. B) in the cytoplasm of oviduct tissue. C) in the nuclei of oviduct tissue. D) in the mitochondria of oviduct tissue.

C. When stimulated by the addition of a ligand such as stilbestrol, estrogen receptor will localize within the nucleus, where it regulates genes by binding to enhancers and promoters. Radiolabeled stilbestrol would localize with estrogen receptor in the nucleus (C is correct). There is no estrogen receptor in the plasma membrane or mitochondria (A and D are wrong). Some estrogen receptor may be located in the cytoplasm, particularly in the absence of ligand, but it will localize mostly in the nucleus when it has ligand bound (B is wrong).

According to the figures, decreasing which of the following would create the greatest increase in charge stored per unit voltage on an axon membrane in its rest state? A) Leakage channel resistance B) Na+ channel resistance C) Area of the membrane surfaces D) Thickness of the membrane

D. "Charge stored per unit voltage," Q / V, is the definition of capacitance, C. The equation for the capacitance of a parallel-plate capacitor is C = κ?0A / d, where k is the dielectric constant, ?0 is a universal constant (the permittivity of free space), A is the area of each plate, and d is the distance between the plates. Of the choices given, only D, decreasing the thickness of the membrane (that is, decreasing d, distance), would increase the capacitance, C.

If a patient with cystic fibrosis receives a double-lung transplant from a non-cystic fibrosis donor, would the new lungs be expected to develop cystic fibrosis? A) Yes, once you have cystic fibrosis, it develops in every organ of the body. B) Yes, since the primary defect is with respiratory secretions. C) No, because the infectious causes of the disease will be removed when the old lungs are taken out. D) No, since cystic fibrosis is due to a gene defect, the cells of the new lungs will have the normal CFTR gene.

D. Cystic fibrosis is a genetic disease based on the abnormal protein CFTR. In a set of lungs from a person without cystic fibrosis, the lung cells presumably have normal-functioning CFTRs. Therefore, the new lungs should not be subject to the development of cystic fibrosis (eliminate choices A and B). Cystic fibrosis is a multi-organ disease since the CFTR is used in secretions from several glands, but this does not cause normal CFTRs to become abnormal. The primary defect is the protein CFTR, not the pulmonary secretions or an infectious cause (choice C is incorrect and choice D is correct).

BIO PASSAGE 4: HCl is secreted by the parietal cells through the mechanism outlined in Figure 1. H+ is extruded into the gastric lumen by an H+/K+ ATPase. Bicarbonate is transported via a protein channel into the interstitium in exchange for Cl- with both ions moving down a concentration gradient. The HCO3- / Cl - exchange is an example of: A) exocytosis. B) active transport. C) simple diffusion. D) facilitated diffusion.

D. Facilitated diffusion involves movement of molecules down a gradient with the involvement of a protein. The passage states that both bicarbonate and chloride ions are moving down a gradient, making this a case of facilitated transport (D is correct). Exocytosis does not generally involve ion transport and does not utilize membrane channels (A is wrong). Active transport involves moving ions or other molecules against a gradient (B is wrong). Simple diffusion is the movement of molecules down a gradient without a protein involved (C is wrong).

What is the pOH of an aqueous solution whose hydrogen ion concentration is measured to be 10-5 M? A) 3 B) 5 C) 7 D) 9

D. If [H+] = 10-5 M, then the pH is 5, so the pOH is 14 - 5 = 9.

Suppose that the researchers who conducted the experiment described in the final paragraph concluded, based on the results, that while experienced detectives are only moderately better at lie detection than are untrained laypeople, the polygraph is in fact a highly effective tool for this purpose. Which one of the following, if true, would cast the most doubt on this conclusion? A) Unlike most people taking a polygraph examination, the interviewees in the experiment knew that they would suffer no serious or negative consequences arising from their lies. B) Compared to the average detective, those detectives who participated in the experiment were unusually astute and adept at lie detection. C) Because the interviewees who took the polygraph were instructed to lie for purposes of the experiment, they were apt to experience less guilt than if they were lying to protect their own interests. D) The laypeople demonstrated an extremely low level of success at lie detection.

D. If the experimenters concluded that the polygraph is a highly effective instrument for detecting lies based upon the fact that it was 21% better than the detectives (who were 17% better than the laypeople), it would be necessary to know what the results were for the laypeople. If the laypeople did an extremely poor job at lie detection (let's say they scored less than 10% accuracy at lie detection), then the polygraph results would still be no better than chance. The experimenters should have offered data indicating an overall success rate, not simply a relative one (choice D would cast doubt upon the experimenters' conclusion and is thus correct). If the interviewees who took the polygraph were less likely to experience the physiological effects of stress, whether because they knew there would be no adverse consequences or because they believed they really weren't "lying" per se, this would actually strengthen the experimenters' conclusion. If the polygraph could still be that much more effective at lie detection than trained professionals, even when physiological indicators of stress (which is what the test purports to measure), are low, then it would be all the more useful and valuable (choices A and C would not cast doubt upon the experimenters' conclusion and can be eliminated). Similarly, if the detectives who participated were uncommonly skilled at lie detection and the polygraph was still substantially better, this would strengthen the claim that the polygraph is very effective (choice B would not cast doubt upon the experimenters' conclusion and can be eliminated).

The diaphragm plays an important role in respiration. During inspiration, the diaphragm: A) relaxes, causing alveolar pressure to drop below atmospheric pressure. B) contracts, causing alveolar pressure to rise above atmospheric pressure. C) relaxes, causing alveolar pressure to rise above atmospheric pressure. D) contracts, causing alveolar pressure to drop below atmospheric pressure.

D. Inspiration is the drawing of air into the lungs. The diaphragm contracts and flattens during inspiration, expanding the chest cavity; the lungs (which are stuck to the inside wall of the chest cavity) expand as well. The expansion of the lungs decreases the pressure in the alveoli, causing air to move into the lungs from the exterior.

Associations regarding African Americans were made accessible by priming them with stereotypic African American trait words. This is evident by the control group subjects' significantly faster categorization of African American photos after being primed with stereotypic trait words Which of the following concepts is most closely related to the rationale behind the study's method of measuring automatic association? A) Cognitive dissonance B) Illusory correlation C) Social cognitive theory D) Social schemas

D. Social schemas are cognitive structures that guide the information processing of ideas about categories of social events and people. When a social schema is made more accessible through priming, it can be activated and used more quickly in a particular situation. In this study, subjects' social schemas regarding African Americans were made accessible by priming them with stereotypic African American trait words. This is evident by the control group subjects' significantly faster categorization of African American photos after being primed with stereotypic trait words (choice D is correct). According to Festinger's cognitive dissonance theory, inconsistency among attitudes (cognitive dissonance) propels people in the direction of attitude change. Cognitive dissonance does not play a role in affecting the speed at which a subject categorizes photos after being primed with certain trait words (choice A is wrong). Illusory correlation refers to people's overestimation of instances that support their beliefs about the association between two things. The study does not involve having subjects estimate the number of times they have seen a person act in a way that is consistent with the stereotypes for that person (choice B is wrong). Social cognitive theory refers to Albert Bandura's theory that learning occurs in a social context with a dynamic and reciprocal interaction of the person, environment, and behavior. The study did not involve having participants learn behaviors through observing others' behavior (choice C is wrong).

The phenomenon where individuals tend to favor internal attributions in explaining others' behavior, while favoring external attributions in explaining one's own behavior is known as: A) in-group/out-group bias. B) the ultimate attribution error. C) stereotyping. D) the fundamental attribution error.

D. The fundamental attribution error is the phenomenon whereby individuals tend to favor internal attributions in explaining others' behavior, while favoring external attributions in explaining one's own behavior (choice D is correct). In-group/out-group bias refers to a pattern of favoring members of one's group over out-group members (choice A is wrong). The ultimate attribution error is the tendency for people to explain an out-group's negative behavior as flaws in their personality (as described in the second paragraph), and to explain an out-group's positive behavior as a result of chance or circumstance (choice B is wrong). Stereotyping refers to the attribution of certain characteristics to an individual on the basis of one's membership of a particular group (choice C is wrong).

Suppose a fifty-year-old man takes a polygraph as part of an insurance investigation. No deception is detected in response to any question, even when his answers are obviously and objectively false (e.g., he asserts that he was present at an event that took place before he was born). Assuming that the polygraph is in fact highly accurate at measuring the physiological indicators of stress associated with lying, which one of the following conditions, if established, would help to explain the results of this man's polygraph? Schizophrenia Antisocial personality disorder Alzheimer's disease A) I only B) I and II only C) II and III only D) I, II, and III

D. The polygraph purports to measure the physiological effects of stress (presumably due to guilt) that are associated with lying. Accordingly, the individual being tested must actually be intending to deceive, not simply stating an untruth. Item I is true: schizophrenics often suffer from delusions that they firmly believe no matter how bizarre or implausible; therefore, if this man is suffering from schizophrenia it is highly plausible that his untruths would not cause any physiological stress that could be detected by a polygraph (choice C can be eliminated). Item II is true: those with antisocial personality disorder clearly know full well that they are lying. However, since they experience little or no guilt about the lies, they may not show physiological signs of stress when they lie (choice A can be eliminated). Item III is true: individuals afflicted with brain damage or a dementia, such as Alzheimer's disease, often engage in confabulation in order to fill in the gaps left by lost memories. As with delusions, these false memories are honestly believed, no matter how strange or unlikely (choice B can be eliminated; choice D is correct).

A mixture of aspartate (polar) and phenylalanine (nonpolar) are separated into its component molecules by thin layer chromatography on a silica plate eluted by benzene. Which of the following best explains why the separation occurs? A) Aspartate will move farther with the mobile phase, because it has a polar side chain. B) Aspartate will move farther with the mobile phase, because it has a nonpolar side chain. C) Phenylalanine will move farther with the mobile phase, because it has a polar side chain. D) Phenylalanine will move farther with the mobile phase, because it has a nonpolar side chain.

D. The side chain on aspartate is -CH2COO-, which is very polar, while the side chain on phenylalanine is -CH2Ph, which is nonpolar. Since "like dissolves like," and the mobile phase is nonpolar, the molecule which is less polar (phenylalanine, in this case) will move farther with the mobile phase.

Socialized medicine is a term used to describe government regulation of health care. This public administration of health care is funded through taxation. This is also often referred to as universal health care. In the United States, there are some socialized insurance programs, such as military medicine. However, despite the implementation of the Affordable Care Act, private companies continue to provide most of the nation's health care. Public opinion has been slow to accept the notion of universal health care, noted through opposition to presidential reform efforts during the Truman, Clinton, and Obama administrations. This hesitation could be attributed to several factors, such as agreement with conservative critics, which best matches the approach to inequalities described in the theories of: A) Emile Durkheim. B) Karl Marx. C) Ludwig Gumplowicz. D) Max Weber.

D. The theoretical perspective most concerned with social inequalities is the conflict theory. Classical sociologists associated with this theory include Karl Marx, Ludwig Gumplowicz, and Max Weber. However, Emile Durkheim is more associated with structural functionalism, which is focused on contributions to social stability (choice A is wrong). The persistent opposition to forms of socialized medicine best reflect the theories of Max Weber (choice D is correct). As opposed to the other theorists, Weber argued that the presence of inequalities would not necessitate the collapse of capitalism. He suggested that responses to inequalities are moderated through additional social factors, such as agreement with authority figures (e.g., public political figures). The persistent opposition, however, challenges the theories of Karl Marx (choice B is wrong). Marx argued that social inequalities, and subsequent conflict and internal tensions as a result of power differentials, would lead to the rise of socialism. The Marxist perspective then suggests the rise of socialized medicine, as opposed to the continuation of capitalistic private systems. Finally, the theories of Ludwig Gumplowicz focused on cultural and ethnic conflicts that are not relevant to the question (choice C is wrong).

Suppose that when a student is caught lying about completing his homework, he is not allowed to participate in afternoon recess (his favorite school activity) for the rest of the week. What is this an example of? A) Positive reinforcement B) Negative reinforcement C) Positive punishment D) Negative punishment

D. Through operant conditioning, a reinforcement is used to encourage a desirable behavior to happen again; a positive reinforcement provides a rewarding stimulus (like a treat) to help encourage the behavior while a negative reinforcement removes a painful or unwanted stimulus (like an electric shock) to help encourage the behavior. Since lying about completing his homework is an undesirable behavior, this is neither an example of positive reinforcement nor of negative reinforcement (choices A and B are wrong). A punishment is used to discourage an undesirable behavior (such as lying) from happening again. A positive punishment is an unfavorable outcome or a negative stimulus (such as a spanking) following an undesirable behavior (choice C is wrong). A negative punishment would involve taking something good or desirable away (such as recess for the rest of the week) in order to reduce the reoccurrence of the undesirable behavior (choice D is correct).

If a fully saturated solution of AgI, with precipitate present, were treated with NaCl instead of NaI, which of the following observations is likely? A) As NaCl is added, all precipitates are dissolved into the aqueous solution. B) The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1. C) There is no change in the amount of undissolved AgI. D) The concentration of [I-] increases.

D. Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a decrease in the solubility for AgI with increasing concentration (elminating choice B). The following will act as a competing reaction when [Cl-] concentrations become sufficiently large. Ag+ (aq) + Cl- (aq) → AgCl (s) With this in mind, there will be no situations wherein the solution is free of precipitate (eliminatinv choice A). As NaCl as the dissolved [NaCl] concentration increasess AgCl will be precipitated from solution, which will enable additional AgI to dissolve (eliminating choice C). The increased dissolution of AgI will cause the increase in [I-], even as [Ag+] levels remain low. AgI(s) ---> Ag+ (aq) + I- (aq)

A researcher has discovered a novel frameshift mutation in pyruvate carboxylase and hypothesizes that this could affect tumor metabolism. Which of the following would be the best experiment to perform to either confirm or disprove her hypothesis? A) Isolate genomic DNA from a cancer cell line and sequence the gene for pyruvate carboxylase, to confirm which amino acid is altered in the mutant form. B) Generate an animal model that expresses the mutant form of pyruvate carboxylase, and determine how the proton gradient is used by ATP synthase isolated from these animals. C) Perform western blot analysis on lysates from a brain tumor to confirm the mutation causes increased enzymatic activity. D) Generate a cell line that expresses the mutant form of pyruvate carboxylase and measure lactate secretion, and glucose and glutamine uptake compared to control cells that express the normal form of pyruvate carboxylase.

PYRUVATE CARBOXYLASE IS THE MAIN ENZYME OF GLUCONEOGENESIS. D. The question stem states that a frameshift mutation in pyruvate carboxylase has already been found. There would therefore be little point in sequencing the gene again. In addition, "confirm[ing] which amino acid is altered in the mutant form" is a better match to confirming a point mutation rather than a frameshift mutation (choice A is wrong). Pyruvate carboxylase catalyzes the conversion of pyruvate to oxaloacetate in the first step of gluconeogenesis. Determining how the proton gradient is used by ATP synthase isn't relevant to this enzyme's function (choice B is wrong). Option C is a very tempting answer choice because it would be ideal to determine if the mutation is affecting enzymatic activity. Unfortunately, a western blot will not give information on this, since it can only measure protein levels (choice C is wrong). By process of elimination, the correct answer is choice D. Measuring lactate secretion, and glucose and glutamine uptake will give an indirect readout of whether basic metabolic processes are different in cells with the pyruvate carboxylase mutation versus in normal cells (choice D is correct).