Princeton Review MCAT Practice Test Questions

A healthy woman has a respiratory minute volume (the amount of air inhaled and exhaled in one minute) of about 6 L. Suppose the breathing tube she uses has a cross-sectional area of 1 cm2. What is the average flow speed through the tube? A. 1 m/s B. 3.1 m/s C. 100 m/s D. 310 m/s

A. 1 m/s The volume flow rate (volume of air flowing per second) is given by the equation f = Av, where A is the cross-sectional area of the cylinder and v is the average speed of the flow. The flow rate in this case is f = 6 L/min × (1 m3 / 1000 L) × (1 min / 60 s) = 1 × 10^-4 m3/s. The cross-sectional area is 1 cm2 × (1 m / 100 cm)2 = 10^-4 m2. Thus v = f / A = (1 × 10^-4 m3/s) / (10^-4 m2) = 1 m/s.

SEE OTHER SIDE FIRST Cholesterol is an abundant component of animal cell membranes, but it is a steroid, not derived from fatty acids such as arachidonate (eliminate A). Triglycerides could contain a fatty acid like arachidonate, but they are not membrane components (eliminate B). Peptidoglycans are not found in eukaryotes, only bacteria, and do not consist of fatty acids (eliminate D). Phospholipids is the correct answer. Phospholipids are abundant components of the plasma membrane and contain esters of many different fatty acids (C is correct).

A. Cholesterol B. Triglycerides C. Phospholipids D. Peptidoglycan layer

SEE OTHER SIDE FIRST The polarity of a bond depends on the difference in electronegativity between the two bonding atoms: The larger the difference, the more unequal the sharing of electrons, and the more polar the bond. Thus, the Cl—Cl bond is completely nonpolar, eliminating choices C and D. To find the final answer, look at the other extreme of the ranking problem for the remaining answer choices. Carbon is more electronegative than Si, therefore Cl, which is the most electronegative element depicted, will show the greatest electronegativity difference with Si. Therefore, of the four bonds, the Si—Cl bond should be the most polar (eliminate B).

A. III < I < IV < II B) III < IV < II < I C) I < II < IV < III D) II < IV < I < III

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

A. Linear 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 (A is correct and eliminate 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 (eliminate B and C).

Since constitutively active B-Raf mutants commonly cause cancer, inhibitors of their kinase domain have been developed. One strategy involves high-affinity binding by inhibitors to the nonpolar ATP binding pocket of the B-Raf kinase domain. Two these inhibitors are PLX 4720 and 4032. A. V and L B. D and N C. D and K D. K and S

A. V and L The passage states that the substances are inhibitors of the hydrophobic B-Raf ATP binding site. Given this, PLX 4720 and PLX 4032 would be expected to bind most strongly to active sites enriched in hydrophobic residues, such as valine (V) and leucine (L). Aspartic acid (D) is an acidic amino acid and will be negatively charged in the body, so choices B and C can be eliminated. Lysine (K) is a basic amino acid and will be positively charged, making choice D incorrect.

SEE OTHER SIDE FIRST The difference in pressure between two points in a static fluid is given by the expression ρ(fluid)gy, where y is the vertical distance between the two points. (Note that A and C can be eliminated, because the units are wrong: density times acceleration times velocity does not give pressure.)

Which of the following expressions is equal to the pressure difference between Point c and Point d? A. ρ(gas)gv B. ρ(gas)gy C. ρ(liquid)gv D. ρ(liquid)gy

SEE OTHER SIDE FIRST At equilibrium ΔG = 0. Since reaction 2 has a negative value of ΔG°, it must be true that ΔG > ΔG°.

Which of the following is true for Reaction 2, once it is allowed sufficient time to achieve equilibrium? A. ΔG < ΔG° B. ΔG = ΔG° C. ΔG > ΔG° D. ΔG = RT

SEE OTHER SIDE The pictured epoxide has two chiral centers, but it also has a plane of symmetry. For this reason, it is a meso compound, and its two chiral centers must have opposite configurations (one is R while the other must be S).

A. It is meso. B. Its absolute configuration is 7R, 8R C. Its absolute configuration is 7S, 8R. D. Its absolute configuration is 7S, 8S.

Paget's disease is characterized by massive new bone formation resulting in softening of the bone. A patient with this disease may exhibit bone deformities such as bowing of the tibia or humerus. Compared to normal bone, a sample of bone from an individual with Paget's disease would most likely have: A. a smaller Young's modulus and a lower yield point. B. a smaller Young's modulus and a higher yield point. C. a larger Young's modulus and a lower yield point. D. the same Young's modulus and a lower yield point.

A. a smaller Young's modulus and a lower yield point. Paget's disease weakens the bone and thus decreases its Young's modulus. This eliminates choices C and D. The bowing of the tibia and humerus implies that the bone has gone past the yield point, implying that this disease also lowers the yield point of the patient's bones.

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 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. an A-not-B error. In an A-not-B error, the individual perseveres in looking for an object in a location in which 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 (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 (B is wrong). Trust-versus-mistrust is the first stage of Erik Erikson's theory of psychosocial development (C is wrong). Conservation, which is a logical thinking ability attained during Piaget's concrete operational stage of development (around roughly ages 7-11), refers to a child's capacity to determine that a certain quantity remains the same despite adjustment of the container, shape, or apparent size (D is wrong).

Each of the following is a personality trait included in McCrae and Costa's five-factor model of personality (the "Big Five"), EXCEPT: A. optimism. B. agreeableness. C. extraversion. D. neuroticism.

A. optimism. Optimism is not one of the "Big Five" personality traits (A is correct). The five-factor model of personality traits (the "Big Five") includes: openness to experience, conscientiousness, agreeableness (B can be eliminated), extraversion (C can be eliminated), and neuroticism (D can be eliminated).

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 potential is higher on the outside of the cell than the inside, and electric field lines run from the outside to the inside. The voltage across the membrane is -70 mV from the exterior to the interior of the cell; thus, the potential is higher on the exterior of the cell (C and D are wrong). 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). IN PHYSICS, FIELD LINES GO TOWARD THE MORE NEGATIVE AREA.

A. ME0322's trivalency allows it to associate more effectively with HAd37's fiber protein. The first paragraph states that HAd37's fiber protein is mainly responsible for recognition of host cells, and that the fiber protein has a trimeric structure. Figure 2 reveals that ME0322 is a trivalent molecule with sialic acid molecules attached to it via aliphatic linkers. It thus follows that ME0322's trivalency might allow it to associate more effectively with HAd37's trimerically-arranged binding pockets than sialic acid alone (A is correct). Ligands are bound in specific positions by proteins and do not rotate once in place (B is wrong). Although anti-GD1a antibodies might recognize ME0322, antibodies do not play a role in the experiment in question (C is wrong). Figure 3 reveals that HAd37 is more effectively inhibited by ME0322 than sialic acid (choice D is wrong).

Adenovirus capsids are made up of two types of protein, hexameric or pentameric capsomere proteins, which form an icosahedral shell, and trimeric fiber proteins, which extend from the vertices of the capsomere proteins. While the capsomere proteins' main function is to protect the adenovirus genome, the fiber proteins have an important role in determining viral tropism by recognizing and binding to certain motifs on the surfaces of target host cells. In Experiment 2, why might HAd37 infection be more effectively inhibited by ME0322 than sialic acid? A. ME0322's trivalency allows it to associate more effectively with HAd37's fiber protein. B. The aliphatic linker in ME0322 allows more rotation in HAd37's binding pocket. C. ME0322's trivalency renders it unrecognizable by antibodies. D. None of the above; HAd37 is more effectively inhibited by sialic acid.

Which one of the following compounds would be most soluble in ethanol? A. Methane, CH4 B. Formic acid, HCOOH C. Stearic acid, CH3(CH2)16COOH D. 1-Chlorooctodecane, CH2Cl(CH2)16CH3

B. Formic acid, HCOOH B. Choices A and D cannot hydrogen bond with ethanol, so should be quickly eliminated. Formic acid is a highly polar molecule, making it soluble in the polar alcoholic solvent. However, stearic acid has a long (18-C) nonpolar chain. Therefore, it will be less soluble in ethanol.

Which of the following is the best explanation for why urea, CO(NH2)2, may be used to study the unfolding of the protein scr SH3? A. The entropic loss due to unfolding of scr SH3 is a result of urea increasing the solubility of the protein interior. B. Urea acts to stabilize the denatured state of scr SH3 by associating with backbone polar groups and charged side chains. C. Urea stabilizes the multiple transition states observed for the unfolding of scr SH3. D. The addition of urea reduces S—S bonds in the structure of scr SH3.

B. Urea acts to stabilize the denatured state of scr SH3 by associating with backbone polar groups and charged side chains. Using process of elimination can quickly narrow the options given in the answer choices. A is inconsistent because unfolding will cause entropy to increase, not decrease, due to the disordering effect of unfolding (eliminate A). Based on its structure, urea, CO(NH2)2, will interact with the polar moieties of a protein via hydrogen bonding, which include the peptide backbone and polar residues (charged or uncharged). By disabling intramolecular hydrogen bonding urea is capable of stabilizing the denatured state of a protein (choice B is correct). The passage stated only a single unfolding pathway is followed when a chemical denaturant, such as urea, is applied. Thus, there can only be one transition state (eliminate choice C). Choice D is incorrect because urea is not a reducing agent and will not affect disulfide bonds.

A Q-switched laser can be used to treat skin blemishes and to remove tattoos. 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. power of each laser pulse. 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).

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. 5'-CAU-3' 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').

Fluorescence in situ hybridization (FISH) is a powerful molecular genetic laboratory technique, where fluorescently labeled probes are hybridized to fixed chromosome clusters. A FISH probe should be made of: A. an antibody that binds to a DNA epitope, covalently linked to a fluorescent segment of DNA. B. a segment of double-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorescent protein. C. a segment of single-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorochrome. D. a single-stranded piece of RNA.

C. a segment of single-stranded DNA which is complementary to the gene being studied, covalently linked to a fluorochrome. The passage states that during FISH, fluorescently labeled probes are hybridized to chromosome clusters. This is similar to how probes are used in Southern and Northern blots. The probe must be single-stranded DNA or RNA if it is going to hybridize or bind to the chromosomes (which are denatured first, also similar to the process in Southern or Northern blotting), and should be covalently linked to the fluorescent molecule (C is correct). An antibody is not typically used to bind DNA, and in any case, there is no easy way to generate a "fluorescent segment of DNA." It is much easier to use fluorescent molecules or dyes (A and D are wrong). A double-stranded probe will not bind the chromosomes because it will be complementary to itself (B is wrong).

Stilbestrol is a crystalline non-steroid with estrogenic effects often superior to those of the estrogen, estradiol. If radiolabeled stilbestrol were administered to experimental chicks, given its effect on the oviduct, the 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. in the nuclei of oviduct tissue. When stimulated by the addition of a ligand such as stilbestrol, estrogen receptors will localize within the nucleus, where it regulates genes by binding to enhancers and promoters. Radiolabeled stilbestrol would localize with estrogen receptors in the nucleus (C is correct). There are no estrogen receptors in the plasma membrane or mitochondria (eliminate A and D). Some estrogen receptors 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 (eliminate B).

SEE OTHER SIDE FIRST D. hydride shift. The hydroxide ion can deprotonate both the 3-pentanone and the 2-methyl-3-pentanone (eliminate A). Aldol condensation can occur between the enolates formed from the deprotonations described above and any ketones in solution (eliminate B). Nucleophilic substitutions can occur between any enolate and methyl iodide (eliminate C). Hydride shifts cannot occur from an enolate, but only from a carbocation intermediate, which cannot form under the basic conditions of the reaction (D is correct).

Competing reactions in Experiment 1 are responsible for the product mixture. All of the following reactions contribute to the product mixture EXCEPT: A. deprotonation. B. aldol reaction. C. nucleophilic substitution. D. hydride shift.

Because of their high glycolytic rates, tumor cells uptake large amounts of 2-18F-2-deoxyglucose (FDG), a modified radioactive substrate of hexokinase. This uptake can be imaged via positron emission tomography (PET), which allows diagnosis of cancer and also monitoring of how tumors and metastases are responding to treatment. Which of the following is a true statement? A. If anticancer treatment is working well on a patient, more FDG will be detected on a positron emission tomography scan. B. Expression of telomerase, a ribonucleoprotein complex, is typically higher in normal somatic cells than in tumor cells. C. Glutamine is a hydrophobic amino acid with one chiral center, a basic amino group and an acidic carboxyl group. D. Tumor cells can use cap-independent translation to make proteins when growth conditions are not ideal.

D. Tumor cells can use cap-independent translation to make proteins when growth conditions are not ideal. Cap-independent translation uses internal ribosomal entry sites (IRESs) and allows a cell to translate proteins during sub-optimal growth conditions (because less regulation is necessary). Since tumors typically grow quickly, they are often short of oxygen and nutrients and must deal with acidic and CO2-rich growth conditions. Activating cap-independent translation would allow the tumor cells to continue proliferating even under these less-than-optimal conditions (D is correct). If an anticancer treatment is working, less FDG should be detected (A is wrong). Normal somatic cells do not express telomerase. This enzyme is only expressed in the germ line, by some white blood cells, and in some tumor cells (B is wrong). Glutamine is a hydrophilic polar amino acid (C is wrong). Note that explicit knowledge of cap-independent translation was not necessary, only the ability to recognize the other three statements as false.

Recently a married couple experienced mortgage foreclosure, which has resulted in negative thoughts and behaviors on the part of the husband who feels responsible, thus creating significant strain in the relationship. The pair decided to pursue professional counseling to address this strain. The therapist approached sessions with the intention of helping the husband relearn these undesired thoughts and behaviors through reconditioning and other general methods of practice. This technique is described as a(n): A. psychoanalytic approach. B. exposure and response prevention approach. C. humanistic approach. D. cognitive behavioral approach.

D. cognitive behavioral approach. The problematic behavior described includes negative self-defeating thoughts. This problem is best addressed through cognitive behavioral therapy, or CBT (D is correct). Through techniques such as those described in the question stem, CBT can help individuals learn healthier methods of thinking. Psychoanalytic therapy addresses problems that are often unconscious and/or stem from childhood experiences, with the goal of reducing anxiety through self-insight; this is accomplished by the therapist's analyzing the patient's past (A is wrong). Exposure and response prevention, a specific method of CBT, addresses problematic behaviors, such as obsessive-compulsive behaviors, with the goal of helping individuals discontinue them; this is accomplished by exposing clients to their fears and preventing stressful responses (B is wrong). Humanistic therapy addresses problems that prevent self-understanding with the goal of personal growth through self-insight; this is accomplished by establishing a relationship between the therapist and the client that is based on trust (C is wrong).

The theoretical paradigm LEAST suited to microsociological studies is the: A. constructionist perspective. B. symbolic interactionist perspective. C. dramaturgical perspective. D. conflict theory perspective.

D. conflict theory perspective. Theories with a micro level orientation are concerned with the nature of social interactions on the scale of the individual; theories with macro level orientation, in contrast, are concerned with the broader social structure. The main microsociological perspective is symbolic interactionism, which includes the dramaturgical approach (B and C are suited to microsociological studies and can be eliminated). Social constructionism has both macro and micro versions, which reflects the fact that social realities are constructed both through greater social structures and through face-to-face social interactions (A is suited to microsociological studies and can be eliminated). Conflict theory, however, is a macrosociological perspective concerned with social inequalities as a result of the greater social structure, which is of particular concern in capitalistic societies (D is not suited to microsociological studies and is the correct answer).

The transition state's activation free energy (G^TS) is composed of the enthalpy (H^TS) and entropy (S^TS) of activation by the same relation as free energy, enthalpy and entropy are related for energy minima (G^F, H^F and S^F). Which of the following is likely true regarding the transition from the folded state (F) to the transition state (TS)? A. ΔH (F → TS) < 0 B. ΔG (F → TS) < 0 C. ΔS (F → TS) < 0 D. ΔH (F → TS) > 0

D. ΔH (F → TS) > 0 B can be immediately eliminated, as the free energy of activation will always be higher than the free energy of either minimum it connects (think of the reaction coordinate graph). To eliminate C it is helpful to remember that proteins are highly ordered macromolecules. In fact, just the term "unfolding" implies that there is an orderly arrangement being undone, and hence it is very unlikely that such unfolding would require a negative change of entropy. This, on its own, means that D is correct as ΔG is positive from the folded state to the transition state and ΔS is also positive (ΔG = ΔH - TΔS). Still, reasoned further, the bonding (enthalpic) components that hold a folded protein together must overcome the entropic penalty associated with this ordering. Therefore, ΔH (TS → F) must be very negative, meaning the opposite process is very positive, and choice D is correct.

Ghrelin, the "hunger hormone," is released when the stomach is empty. It increases hunger and promotes secretion and motility in the digestive tract. The receptor for ghrelin is found on the same cells in the brain as the receptor for leptin. Which of the following is a true statement? A. The interaction of leptin and ghrelin can be considered competitive inhibition because they produce antagonistic effects. B. The interaction of leptin and ghrelin is considered to be allosteric inhibition. C. The interaction of leptin and ghrelin is considered uncompetitive because ghrelin only binds when leptin is already bound. D. There is no interaction between leptin and ghrelin because they operate on separate receptors.

Even though they bind to the same cells in the brain, the fact that leptin and ghrelin bind to separate receptors means that there cannot be any kind of inhibition or interaction between them (D is correct). For competitive inhibition to occur, two things must be trying to bind to one site on the same receptor. The fact that they have separate receptors prevents this (A is wrong). Likewise, allosteric inhibition requires that the inhibitor bind to a separate site on the same enzyme or receptor (B is wrong), and there is nothing in the question stem or passage to suggest that ghrelin can only bind when leptin is bound (C is wrong).

SEE OTHER SIDE FIRST The passage states that cartilage does not withstand tension as well as bone. Therefore, the Young's modulus (E) of cartilage must be lower than that of bone, eliminating choices B and D. Since stress = E × strain, a lower value of E corresponds to a lower stress for the same strain, so the graph of stress vs. strain in the elastic region would be less steep; that is, the angle θ would be smaller. Thus the answer is "C. smaller, because cartilage has a smaller Young's modulus than does bone."

If the graph were redrawn for cartilage undergoing tensile stress, then the angle θ would be: A. greater, because cartilage has a smaller Young's modulus than does bone. B. greater, because cartilage has a larger Young's modulus than does bone. C. smaller, because cartilage has a smaller Young's modulus than does bone. D. smaller, because cartilage has a larger Young's modulus than does bone.

SEE OTHER SIDE FIRST Figure 1 shows that when participants were categorizing white photos, there was no difference in speed whether they were given a stereotypic AA trait prime or a non-trait prime (D is correct). Participants given situational attribution training were slower at categorizing photos of AAs when they were primed with stereotypic AA trait words compared to those in the control group (A and C are wrong). Trained participants were equally fast in categorizing photos of AAs whether they were given a stereotypic AA trait prime or a non-trait prime (B is wrong).

Figure 1 best supports which of the following conclusions about the participants in the study? A. Participants given situational attribution training were faster at categorizing photos of AAs when they were primed with stereotypic AA trait words compared to those in the control group. B. Participants given the situational attribution training were slower at categorizing photos of AAs when they were given stereotypic AA trait word primes than when they were given non-trait primes. C. Participants given situational attribution training were slower at categorizing photos of people in general compared to those in the control group. D. Participants showed no greater automatic association between people and trait words that are not stereotypic of that group than between people and non-trait words.

SEE OTHER SIDE FIRST Since bromothymol blue changes from yellow to blue over the range of pH 6.0-7.6 (while the pH increases, or the [H+] decreases), at a pH below 6.0 the solution will be yellow, above 7.6 it will be blue, and between 6.0-7.6 it will be green. Therefore, the transition point in question occurs at a pH of 7.6 (eliminate C). Since pH + pOH = 14, that means the pOH of the solution should be 7.6 + pOH = 14, or 6.4 (eliminate A and D).

If bromothymol blue changes from yellow to blue as the pH of the solution increases, at what value of pOH will a solution containing this indicator just begin to turn from blue to green? A. 6.0 B. 6.4 C. 7.6 D. 8.0

Brown fat will increase energy expenditure in the form of heat better than any other adipose tissue in the body. In brown fat, leptin indirectly activates uncoupling protein 1 (UCP-1) located in the inner mitochondrial membrane. UCP-1 activation will dissipate the proton gradient across the inner mitochondrial membrane and STOP which of the following reactions in mitochondria? A. Electron transport chain B. Krebs cycle C. Oxidative phosphorylation D. NAD+ reduction

In the absence of the proton gradient across the inner mitochondrial membrane no electrochemical gradient has been built and oxidative phosphorylation, i.e. the formation of ATP by ATP synthase, will not proceed (C is correct). The electron transport chain will continue and pass electrons along the transport chain, but the pumped protons will return to the matrix of the mitochondria with the help of UCP-1 activation (A is wrong). The Krebs cycle provides NADH and FADH2 as substrates to the electron transport chain and receives NAD+ and FAD for continuation of its chemical cycle. This action can proceed unhindered during UCP-1 activation and choice B can be eliminated. Reduction of NAD+ in the mitochondria takes place in the Krebs cycle and the pyruvate dehydrogenase complex reaction and occurs prior to the action of events induced by UCP-1 (D is wrong).

SEE OTHER SIDE FIRST The carbon β to the nitrogen acts as a nucleophile, because it reacts with methyl iodide, which is an electrophile; this eliminates choices C and D. The partial negative charge on the β carbon is due to resonance, as can be seen most easily by analogy to an enolate ion. A is the best answer.

In the alkylation step of Experiment 2, the carbon β to the nitrogen acts as: A. a nucleophile due to partial negative charge from resonance. B. a nucleophile due to partial negative charge from induction. C. an electrophile due to SN2-type reaction. D. an electrophile due to an inductive effect.

SEE OTHER SIDE FIRST Protonation of the carbonyl group at the oxygen atom in the first step of the reaction in Figure 1 makes the carbonyl-carbon a good electrophile. As the protonated carbonyl was an ester, a carboxylic acid derivative, it will undergo an addition-elimination reaction. Water acts as the nucleophile, attacking the electrophilic carbonyl at the carbon atom as shown below (B is correct, eliminate choice C). The carboxylate is then eliminated as the leaving group (eliminate choice D). A ligand is a molecule that binds to a central metal atom to form a coordination complex. There are no metal atoms involved in the reaction (eliminate A).

In the second step of the reaction shown in Figure 1, the role of water can best be described as a(n): A. ligand. B. nucleophile. C. electrophile. D. leaving group.

SEE OTHERSIDE FIRST This is a two by two question. In the reaction shown in figure 1, raclopride-H+ donates a proton in the first step (acting as a Brønsted-Lowry acid). Lewis bases are electron pair donors, and as H+ has no electrons choices B and D can be eliminated. Raclopride-H+ has a hydrophobic halogenated aromatic portion, and a hydrophilic side chain (protonated amine), making it an amphiphilic or amphipathic molecule. The maximum number of stereoisomers a compound may have is given by the formula 2n, where n is the number of chiral centers present. Raclopride-H+ has one chiral center, so it has two stereoisomers (C is wrong and A is correct).

Raclopride-H+ can best be described as: A. an amphiphilic molecule that has two stereoisomers. B. a Lewis base that has two stereoisomers. C. an amphiphilic molecule that has four stereoisomers. D. a Lewis base that has four stereoisomers.

What forms of the amino acid GABA predominate at any given pH?

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.

SEE OTHER SIDE FIRST D. The charge of the droplets differs because the amount of ionization due to the x-rays is random. A can be eliminated because, though it is true that drag for these droplets is proportionate to diameter, that fact is irrelevant because there is no drag for something that isn't moving, like a droplet in static equilibrium. Moreover, the fact that drag wasn't mentioned in the passage should make one suspicious of this choice. B is directly contradicted by the data: compare drops #1 and #2, for example. C is also inconsistent with the given information, namely that density of the oil is a constant 920 kg/m3. D is correct: there is no way to control the exact number of excess electrons in the ionized air that will cling to a falling oil droplet, but the net charge must be a whole number multiple of the elementary charge.

What best explains the variation among the values of the floating voltage for the different droplets? A. As the diameter of the droplets changes, the drag force each experiences as it falls changes proportionately. B. Larger volume droplets require greater voltage to achieve static equilibrium. C. Smaller droplets have greater density, which in turn affects the voltage needed to keep them afloat. D. The charge of the droplets differs because the amount of ionization due to the x-rays is random.

SEE OTHER SIDE FIRST The most useful indicator for the titration in question will change from its undissociated to dissociated form around the pH of the equivalence point. Since the second equivalence point for the amino acid shown is roughly at a pH of 10.5-11.0 (the steep part of the curve at the far right of the graph), the best indicator will be one color below this range and another color above it. Only alizarin yellow changes at such a high pH range.

Which indicator would be most useful in signaling the second equivalence point of a typical neutral amino acid in a titration using NaOH? A. Methyl red B. Phenol red C. Phenolphthalein D. Alizarin yellow

SEE OTHER SIDE FIRST Figure 2 shows the rate of lipid hydrolysis of the lipid DOPC in the presence and absence of raclopride by measuring the concentration of the hydrolyzed lipid, lyso-PC, over time. While the data show that the rate of lipid hydrolysis is appreciable in the presence of raclopride and negligible in its absence, as raclopride acts a simple Brønsted acid, it would be incorrect to say that this exact molecule is required for any kind of lipid hydrolysis (eliminate A). Figure 2 does show that the rate of background lipid hydrolysis is comparatively much slower than raclopride-mediated lipid hydrolysis (choice C is correct). While a higher concentration of raclopride may increase the rate of lipid hydrolysis, there is no data in Figure 2 to support this statement, and as the raclopride-H+ is regenerated in the mechanism, there is no guarantee that increasing its overall concentration leads to a greater rate (eliminate choice B). The slope of Figure 2 is the rate of lipid hydrolysis (change in concentration over a change in time), and as the slope with DOPC + 5 mol% of raclopride is constant, choice D is incorrect.

Which of the following can the researchers conclude strictly from the data shown in Figure 2? A. Lipid hydrolysis cannot occur in the absence of raclopride. B. DOPC + 10 mol% of raclopride would have a higher rate of lipid hydrolysis than DOPC + 5 mol% of raclopride. C. Lipid hydrolysis with raclopride is significantly faster than any background degradation process. D. The rate of lipid hydrolysis with DOPC + 5 mol% of raclopride increases linearly with time.

SEE OTHER SIDE FIRST Item I is false. There is no change in the oxidation states in any of the individual atoms in the reaction shown in Figure 1 (eliminate A and C). Item II is true. During the reaction, an ester is converted into a carboxylic acid. Carboxylic acid derivatives undergo an addition reaction followed by an elimination reaction. Item III is false (eliminate choice D). Saponification is the base-mediated hydrolysis of an ester. While an ester is hydrolyzed in the reaction, it is achieved under acidic conditions. The answer is B.

Which of the following occur(s) in the reaction shown in Figure 1? I. Oxidation II. Elimination III. Saponification A. I only B. II only C. I and II only D. II and III only

SEE OTHER SIDE FIRST From Figure 1, the elastic energy density is equal to the area under the stress vs. strain graph. If the yield point has not been reached, then the region under the graph has the shape of a right triangle, whose area is (1/2) × base × height = (1/2) × strain × stress. In this case, the strain is 1% = 10^-2, and the stress is F = 2 × 10^4 N divided by A = 2 × 10^-4m2, which equals 10^8 N/m2. Thus, the elastic energy density is (1/2)(10-2)(108 N/m2) = 5 × 10^5 J/m3.

A human tibia with a cross-sectional area of 0.0002 m2 undergoes a 1% change in length when compressed by a force of 20,000 N. What is the approximate elastic energy density within the bone while compressed if the bone has not reached its yield point? A. 5 × 103 J/m3 B. 1 × 104 J/m3 C. 5 × 105 J/m3 D. 1 × 106 J/m3

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. A decrease in his velocity by a factor of two The cyclist's kinetic energy just before the impact will be converted into elastic and inelastic energy of deformation upon impact 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)mv^2, a decrease in the cyclist's velocity would have the greatest effect (since v is squared) in lowering his KE.

Which of the following statements about groupthink is/are true? I. It can stifle individuality. II. It is always harmful. III. It can only occur within small groups of people. A. I only B. III only C. I and II only D. II and III only

A. I only: It can stifle individuality. A. Item I is true: When it comes to groupthink, individuals often avoid thinking independently because they do not want to upset the status quo. In succumbing to groupthink, group members frequently stifle their individuality as they try to appease others (B can be eliminated). Item II is false: Despite its bad reputation, groupthink can sometimes be beneficial; for instance, it can lead to job security for those who choose to conform, as well as productivity within the workplace. Regardless, it is not accurate to state that it is always harmful ( C and D can be eliminated). Item III is false: Groupthink is often associated with small cohesive groups of individuals; however, it may occur within larger groups as well (A is correct).

One of the ways glycolytic flux is controlled in tumor cells is via the M2 splice isoform of pyruvate kinase (PKM2). This enzyme can exist as a dimer or a tetramer. Upstream signaling events cause phosphorylation of the less active dimeric form, which promotes the conversion of pyruvate to lactate. The more active tetrameric form promotes the conversion of pyruvate into acetyl-CoA. Which of the following statements best describes pyruvate kinase? A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. B. Its quaternary protein structure is most similar to hemoglobin and myoglobin. C. It is involved in gluconeogenesis causing the conversion of pyruvate to glucose during high energy states. D. It has several isoforms due to alternative splicing via the spliceosome and ribosome.

A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. Pyruvate kinase catalyzes the last step of glycolysis (eliminate 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 (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 D).

SEE OTHER SIDE FIRST 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 (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 (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 (D is wrong). There is no Gestalt law of solids (C is wrong).

A. Law of continuity B. Law of proximity C. Law of solids D. Law of similarity

Which of the following best theorizes how verbal insight problems most likely activate the wrong set of internal representations? A. Spreading activation model B. Feature-detection theory C. Weber's Law D. Bottom-up processing

A. Spreading activation model The passage describes verbal insight problems as those containing words that tend to activate the wrong set of internal representations. The spreading activation model posits that the mind maintains networks of words that are associated with each other to varying degrees. When one word is activated, those words with the strongest associations to the activated word are also activated. (A is correct). Feature-detection theory is a theory of visual perception that proposes that different neurons fire in response to detecting different features; in other words, some neurons fire only in response to vertical lines or motion or contrast (B is wrong). Weber's Law states that two sensory stimuli must differ by a minimum constant proportion (not a constant amount) in order for us to notice the difference (C is wrong). Bottom-up processing suggests that we start with the information gathered by sensory receptors and build up to a final representation in our brain; this type of processing tends to be used more with novel stimuli. It is the opposite of bottom-up processing (top-down processing) that helps to, in part, explain why verbal insight problems are so tricky (D is wrong).

The rules that 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. display rules. The rules governing emotional displays within a particular culture are called display rules (A is correct). Emotional expression is what is being regulated by the display rules (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 (C is wrong). Emotional responses refers to the experience of feeling emotions, which is thought to be innate; there is no such thing as "cultural emotions," per se (D is wrong).

If the front mirror in a laser, which is made of glass of refractive index 3/2, were repositioned so that the laser beam strikes at an angle of 30° to its normal, what would be the angle of reflection? A. sin-1(1/3) B. 30° C. sin-1(1/31/2) D. 60°

B. 30° If the laser beam strikes the mirror at an angle of 30° relative to the normal, then this is the angle of incidence, and, by the Law of Reflection, it is also the angle of reflection. The refractive index of the mirror is irrelevant.

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 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. 50% normal with blue eyes and 50% brown eyed with a risk of Crohn's disease. Correct Answer 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.

Object permanence is the awareness that physical objects exist even when they are not seen or experienced through any other physical sense. Jean Piaget's theory of cognitive development proposed that infants progress through a series of steps in understanding object permanence during the sensorimotor stage, with complete understanding occurring around 18-24 months. Complete understanding of object permanence occurs during which of Freud's psychosexual stages? A. Oral stage B. Anal stage C. Phallic stage D. Latency stage

B. Anal stage According to the first paragraph, object permanence is attained by roughly 18-24 months. 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 (B is correct). Freud's oral stage occurs from birth to roughly age 1, which means that this stage would occur before object permanence is attained at roughly 18-24 months (A is wrong). Freud's phallic stage (from about 3-6 years) and latency stage (from about 6-12 years) both occur long after the attainment of object permanence (C and D are wrong).

Some of the characteristic clinical findings of Down syndrome include mental disability, hypotonia, blind-ending jejunum, and heart malformations, including a common arterial trunk coming off both the left and right ventricles. There is also an increased risk of developing leukemia and early-onset Alzheimer's dementia. 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. Aortic arterial blood oxygen saturation is lower than normal. 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, 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 (B is correct) and the pulmonary arterial oxygen concentration will be higher than it is normally (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 (C is wrong).

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. Closer, because blue light is refracted more due to its shorter wavelength Correct Answer 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.

Wives with more positive predictions (e.g., expected improvements in relationship quality) reported higher levels of external stress and physical aggression. Therefore, it was suggested that these predictions might be affected by individual perceptions of risk factors in the relationship. For example, when women are facing stressful circumstances, positively-biased cognitive processes might be used to regulate contextual risks. Furthermore, results indicated that women with higher levels of external stress also reported significantly greater declines in relationship satisfaction in the measured timeframe following marriage. It appears that in their original predictions, participant wives who reported steeper drops in marital satisfaction might have used which defense mechanism? A. Displacement B. Denial C. Repression D. Reaction formation

B. Denial 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 respective marriages. 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 the defense mechanism that is most closely 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 that external stressors could have on marital satisfaction. Displacement is the shift of emotions, such as aggressive or sexual impulses, to safer outlets (A is wrong). Repression is the psychological attempt to subdue emotionally painful memories (C is wrong). Reaction formation is expression of the opposite of one's feelings in order to conceal unacceptable or dangerous emotions (D is wrong).

Dynamical systems theory is concerned with finding qualitative solutions to mathematical questions involving complex dynamic systems, which involve a fixed or iterative time-dependent equation that describes the evolution of the system over time. Two fundamental principles contribute to the difficulty of predicting the evolution of complex systems: emergence and sensitivity to initial conditions. Emergence describes the way that a system composed of many separate elements combines these elements in such a way that the system itself is more complex than the constituent elements themselves. Sensitivity to initial conditions, popularly known as the butterfly effect, postulates that even seemingly slight changes to early states of a complex system can lead to radically different qualitative states as the system evolves over time. How could the James-Lange theory of emotion best be explained by the dynamical systems theory concept of emergence? A. Perception of the emotional stimulus is determinative for the emotional experience. B. Different collections of physiological markers combine to represent distinct emotions. C. Convergence of perceptive stimuli onto the thalamic and hypothalamic structures creates a global emotional experience. D. Cognition and physiology combine to create a more complex emotional experience than the constituent components.

B. Different collections of physiological markers combine to represent distinct emotions. Emergence, according to the passage, describes the way that a system composed of many separate elements combines these elements in such a way that the system itself is more complex than the constituent elements themselves. Since the James-Lange theory deals with the way different physiological states represent emotions, this theory of emotion could be explained by the dynamic systems concept of emergence if different collections of physiological markers combine to represent distinct emotions (choice B is correct). Perception of the emotional stimulus being determinative does not bring together different constituent elements, as in the definition of emergence, but is more similar to sensitivity to initial conditions (choice A is wrong). Convergence onto thalamic and hypothalamic structures is prominent in the Cannon-Bard theory, not the James-Lange theory (choice C is wrong). The way that cognition and physiology combine during emotion is a central component of the Schachter-Singer, or two-factor, theory of emotion (choice D is wrong).

It has been suggested that one reason tumor cells have novel metabolism is to generate the additional biomolecules required to increase biomass, which is needed to support high rate of proliferation. Which is the best explanation of how this could 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. 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. 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 (B is correct). The pentose phosphate pathway generates ribose-5-phosphate (not ribose-2-phosphate) and this allows nucleotide anabolism (not catabolism, C is wrong). Paragraph 2 describes how glutamine is first converted into glutamate, not the other way around (D is wrong).

Sickle-cell anemia (HbS) results from a substitution to valine from glutamic acid at position 6 of the β chain of hemoglobin. Which of the following best explains why the isoelectric point of HbS is higher than that of HbA? A. The side chain of glutamic acid is less acidic than that of valine. B. Glutamic acid is isoelectric at a lower pH than is valine. C. Valine is isoelectric at a lower pH than is glutamic acid. D. Glutamic acid has a net charge of 0 at its isoelectric point.

B. Glutamic acid is isoelectric at a lower pH than is valine. The isoelectric point is the pH at which an amphoteric molecule has a net electric charge of zero. Glutamic acid is more acidic, not less acidic than valine (A is wrong). Glutamate would require more acidic pH to neutralize its charge (C is wrong). D is a true, but irrelevant statement. The best answer is B. The negative charge of glutamic acid would require a more acidic pH to neutralize, making the isoelectric point of HbA (with glutamic acid) lower than that of HbS (with valine).

A researcher proposed to block the leptin receptor in order to treat hypertension in obese patients. Which of the following detection methods would work best to identify the receptor in the different tissues affected by leptin? I. Northern blot II. Western blot III.Southern blot A. I only B. II only C. I and III only D. I, II, and III

B. II only Correct Answer Item II is true: The leptin receptor is a protein and located in the cell membrane of different tissue types. Detection of a protein can be done using a Western blot. Items I and III are false: a Northern blot detects RNA sequences, and a Southern blot detects DNA sequences. Neither one of these two methods will assure that the final protein product will be formed in the cells tested.

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. No, since the phenotype ratio is close to the expected ratio for this cross. 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 (B is correct). Recombination can occur between linked genes; it just happens less frequently (eliminate A). 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% (eliminate C). 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 (eliminate D).

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. An 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. RNA polymerase II Aldosterone binds 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 (B is correct). DNA polymerase is used in replication, not transcription (A is incorrect), and while aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin, these would be indirect effects (C and D are incorrect).

A researcher attempts to explain the phenomenon that couples who reported greater relationship satisfaction had greater physiological co-regulation. The researcher 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. Schachter-Singer theory B. The Schachter-Singer theory posits that emotional experience is determined by one's 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 (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 (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 (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 (D is wrong).

If the children in the experiment had been offered no material incentive (i.e., no reward chips), how might this have affected the results? A. The children would not have given up on the harder problems. B. The experimental results may have been confounded by motivational variables. C. The children with unsolvable problems would have given up less frequently. D. Children with an internal locus of control would have found the matching patterns less intrinsically rewarding than they did when the chips were provided.

B. The experimental results may have been confounded by motivational variables. The students were given chips that could be redeemed for toys, presumably to ensure that the children put effort into the pattern-matching tasks. Had no incentive been provided to the children, the experimental results may well have been confounded by motivational variables; the students may have been far less motivated and have exerted less effort, especially on the unsolvable pattern tasks (B is correct). Without a reward, the children would likely give up on the harder (unsolvable pattern) tasks more readily (A and C are wrong). There is nothing in the passage to indicate that children with an internal locus of control would be less likely to find the matching patterns intrinsically rewarding if no reward of a material nature was provided (D is wrong).

A viroid is a very small, circular, segment of single-stranded RNA that is capable of causing infection. Unlike true viruses, viroids do not have capsids of their own and do not produce proteins. These viroids can bind to complementary sequences of plant mRNA, silencing translation of host mRNA. A botanist studying viroids has isolated a new viroid capable of causing disease in potato plants. The genome was isolated and studied. In order to fit the definition of a viroid as defined in the passage, all of the following must be true EXCEPT: A. a lack of thymine. B. a 1:1 ratio of cytosine to guanine. C. the absence of coding regions for capsid proteins. D. ribose instead of deoxyribose.

B. a 1:1 ratio of cytosine to guanine. The passage states that viroids are single stranded RNA molecules. RNA lacks thymine (A is true and can be eliminated) and uses ribose instead of deoxyribose (D is true and can be eliminated). The passage states that viroids lack capsids, so would not have coding regions for these proteins; in fact, the passage states that viroids do not produce proteins at all (this is what makes HBV unusual and not a true viroid, C is true and can be eliminated). However, the ratio of cytosine to guanine only has to be 1:1 if the nucleic acid is double stranded. Since viroids are single stranded, this ratio could be different (B is false and the correct answer choice).

SEE OTHER SIDE FIRST 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.

Based on the data collected from the medical students' study, what percentage of the isolates were methicillin-resistant Staphylococcus aureus? A. 115/120 B. 115/160 C. 115/345 D. 115/785

Rapidly dividing tumor cells produce and export large amounts of lactate, and import large amounts of the amino acid glutamine. A highly proliferating cell 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. express high levels of the lactate transporter and the glutamine transporter. 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 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 (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).

During spermatogenesis, spermatids: A. are frozen in meiosis II until after fertilization. B. have already undergone meiotic recombination. C. have four copies of the genome per cell. D. have no nucleus.

B. have already undergone meiotic recombination. Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (eliminate A). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (B is correct). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (eliminate C). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (eliminate D).

Wives with more positive predictions (e.g., expected improvements in relationship quality) reported higher levels of external stress and physical aggression. Therefore, it was suggested that these predictions might be affected by individual perceptions of risk factors in the relationship. For example, when women are facing stressful circumstances, positively-biased cognitive processes might be used to regulate contextual risks. Furthermore, results indicated that women with higher levels of external stress also reported significantly greater declines in relationship satisfaction in the measured timeframe following marriage. For both correlations, there were no significant findings among the male participants. Attribution is a concept that addresses how individuals view behavior, both our own behavior and that of others. In the process of reasoning about behavior, there are often errors. The findings discussed in the passage suggest a(n): A. self-serving bias. B. optimism bias. C. just-world phenomenon. D. actor-observer bias.

B. optimism bias. According to the passage (paragraph 4), the participants who had more positive predictions for the future of their relationship also reported higher levels of physical aggression. The fact that these participants maintain positive expectations for their relationship despite the presence of problematic behaviors suggests the presence of an attribution bias. The optimism bias describes the belief that bad things will not happen to oneself in the future. This can be applied to the finding that people continue to predict positive changes in their relationship, as opposed to divorce, even in the presence of, for example, aggression (B is correct). The self-serving bias describes the attribution of positive experiences to one's own character and actions but negative experiences to factors external to one's own character and actions (A is wrong). The just-world phenomenon describes the belief that the world is fair; success or failure is the result of positive or negative behavior, respectively (C is wrong). The actor-observer bias describes the belief that one's actions are situational while the actions of others are the direct result of their personalities (D is wrong)

The Cannon-Bard theory of emotion: A. emphasizes the role that cognitive interpretation plays in our experience of emotion. B. suggests that our physiological arousal and experience of emotion happen simultaneously and independently. C. suggests that the amygdala is responsible for our experiences of fear and anger. D. states that physiological arousal is the most important facet of the emotional experience.

B. suggests that our physiological arousal and experience of emotion happen simultaneously and independently. The Cannon-Bard theory does propose that the experience of the emotion and the physiological arousal that accompanies that emotion are experienced simultaneously and independently of each other; this was supported by Cannon and Bard's research using cats with severed vagus nerves--the cats still experienced both physiological arousal and fear, even though physiological information from their bodies was no longer being conveyed to their brains (B is correct). The Schachter-Singer theory of emotion emphasizes the role that cognitive interpretation plays in our experience of emotion, not the Cannon-Bard theory (A is wrong). While the amygdala is largely involved in our experiences of fear and anger, that is not what the Cannon-Bard theory of emotion suggests (C is wrong). Nor does the Cannon-Bard theory state that physiological arousal is the most important facet of the emotional experience, since this theory suggests that the cognitive aspect of emotion can still occur without physiological input (D is wrong).

SEE OTHER SIDE FIRST Homologous proteins or genes are those that have evolved from a common ancestor. This matches the information in Figure 1, since the OCA2 protein in all organisms on the figure originated with the ancestor protein represented by the point at the bottom (choice A is supported and can be eliminated). The eight organisms on the figure are all eukaryotes and from diverse families (fungi, mammals, fish; choice B is supported and can be eliminated). Remember that yeast are fungi and therefore eukaryotic. OCA2 in zebrafish and Japanese killifish share a common ancestor which is not far away in evolutionary terms (i.e., is not very far down the diagram). In contrast, the common ancestor protein between human and horse OCA2 protein is the point at the bottom of the diagram; this is farther away in evolutionary terms (choice C is supported and can be eliminated). While choice D may be true based on logic and background information on evolution, it is not supported by Figure 1; this phylogenetic tree contains information on how the OCA2 proteins are evolutionarily related, not how organisms are related (choice D is the least supported and the correct answer choice).

Based on Figure 1, which of the following is least supported? A. The OCA2 proteins in the eight species studies would be considered homologous. B. The OCA2 protein is conserved across eukaryotes. C. Zebrafish and Japanese killifish have a OCA2 common ancestor protein which is more related that that between horse and human. D. Humans and chimpanzees are more closely related than fission yeast and the wild boar.

SEE OTHER SIDE FIRST Figure 1 shows that the patient was more likely to identify an image as being one of himself when it was presented to his right hemisphere. Accordingly, one can reasonably infer that the right hemisphere was more active in the process of self-identification. Answer choices that include the left hemisphere as the location of the bias can therefore be eliminated (B and D are wrong). In order to produce a deficit in self-recognition, which is related to self-identification, the stroke must occur in the same hemisphere wherein the patient exhibited the self-identification bias (C is correct and A is wrong).

Based on the information in Figure 1, which of the following would most likely cause a deficit in self-recognition, assuming that the subject of the case study exhibits a typical pattern of response? A. A stroke in the left hemisphere, because the patient had a self-identification bias in the right hemisphere. B. A stroke in the left hemisphere, because the patient had a self-identification bias in the left hemisphere. C. A stroke in the right hemisphere, because the patient had a self-identification bias in the right hemisphere. D. A stroke in the right hemisphere, because the patient had a self-identification bias in the left hemisphere. C. A stroke in the right hemisphere, because the patient had a self-identification bias in the right hemisphere.

SEE OTHER SIDE FIRST This is a bookend question, as either the biggest or smallest value for Kd is the most likely answer (eliminate choices A and C). The passage states that the substances were tested for their ability to bind the ATP binding pocket of the B-Raf ATP kinase domain with high-affinity, and that the dissociation constant for the bound inhibitor, Kd, quantifies the dissociation of the ligand-protein complex. Therefore, the substance that binds most strongly will have the smallest value for Kd. According to Table 1, PLX 3479 possesses the greatest binding affinity as it has the smallest dissociation constant, and would hence be the most likely choice for further testing (eliminate choice B).

Based upon the data reported in Table 1, researchers chose one compound for further testing as a potential B-Raf inhibitor. Which compound was most likely to have been chosen? A. PLX 4720 B. PLX 4032 C. PLX 2021 D. PLX 3479

A sample of 222Rn is collected in a sealed glass container and allowed to stand for 11.5 days. After this time, approximately what percentage of the original 222Rn remains if its half-life is 3.82 days? A. 3% B. 6% C. 12% D. 25%

C. 12% Since the passage states that the half-life of 222Rn is a little less than 4 days, a time period of 11.5 days is approximately 3 half-lives. After 3 half-lives have elapsed, the amount of radioisotope remaining is (1/2)^3 = 1/8 = 12.5%, so choice C is best.

The following measurements were obtained by an investigator who observed a light ray that propagated across a boundary between air and Material X: 1. speed of light in air = 3.0 × 108 m/s 2. frequency of light in air = 5.0 × 1014 Hz 3. angle of incidence = 30.0° 4. angle of refraction = 23.5° 5. density of Material X = 5.0 g/cm3 At what speed does light propagate through Material X? (Note: sin 23.5° = 0.4) A. 1.4 × 108 m/s B. 1.7 × 108 m/s C. 2.4 × 108 m/s D. 3.7 × 108 m/s D. 3.7 × 108 m/s

C. 2.4 × 108 m/s Snell's Law gives n1 sin θ1 = nX. sin θX → (1) sin30° = n sin(23.5°) → 0.5 = nX(0.4) → n = 5/4. Since the index of refraction of Material X is 5/4, then, by definition, the speed of light is reduced by a factor of 5/4 from its speed in vacuum (or air). Thus, vX = (4/5)c = (4/5)(3 × 10^8 m/s) = 2.4 × 10^8 m/s. (Note that D can be eliminated immediately since it is greater than c = 3 × 10^8 m/s.)

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. 590 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 A, B, and D are eliminated.

Researchers determined that, at 25°C, the rate constant for the binding of PLX 2021 to the B-Raf kinase domain ATP binding pocket is 2 x 10^-2 M^-1 s^-1. The PLX 2021-B-Raf's Kd is 3.5 x 10^-5. Given this, the rate constant for the dissociation of PLX 2021 from the B-Raf kinase domain ATP binding pocket is nearest what value? A. 5 x 10^-3 s-1 B. 5 x 10^-4 s-1 C. 7 x 10^-7 s-1 D. 6 x 10^2 s-1

C. 7 x 10-7 s-1 The equilibrium constant, Kd, for the dissociation of PLX 2021-B-Raf kinase domain is equal to the ratio of the rate constants in the formula to the left. At 25°C, Kd = 3.5 x 10^-5 M. Thus 3.5x10^-5 = (ligand dissociation from protein)/(2x10^-2). (2 x 10-2 M-1s-1)(3.5 x 10-5 M) = 7 x 10^-7 s-1.

The sound level of a sound wave that is striking an eardrum is increased from 35 dB to 55 dB. By what factor has the intensity of the sound increased? A. 2 B. 20 C. 100 D. 200

For each increase by 10 in the decibel level, the intensity of the sound increases by a factor of 10. In this case, the decibel level has increased by two units of 10 (since 55 - 35 = 20 = 10 + 10), so the sound intensity increases by two factors of 10, that is, by a factor of 10^2 = 100.

Which of the following correctly pairs the type of stressor with a scenario describing that stressor? A. Catastrophes; sudden divorce filing B. Significant life changes; wartime displacement C. Daily hassles; mortgage deadlines D. Ambient stressors; moving to another residence

C. Daily hassles; mortgage deadlines There are three main forms of stressors: catastrophes, significant life changes, and daily hassles. Catastrophes are unpredictable large-scale stressors, such as natural disasters, that impact many people to a significant extent. A sudden divorce filing is an example of a significant life change, not a catastrophe (A is wrong). Significant life changes are high stress events that occur with greater frequency than catastrophes, such as marriage, divorce, or loss of a loved one. Wartime displacement is an example of a catastrophe, not a significant life change (B is wrong). Daily hassles are common irritations in life, such as scheduling errors. A mortgage deadline is an example of a daily hassle, as it impacts individuals regularly (C is correct). Ambient stressors are common irritations that are present in one's environment, such as noise or traffic, and can be related to daily hassles. However, moving to another residence is an example of a significant life change, not an ambient stressor (D is wrong).

Which Eriksonian stage have those bicultural individuals who are able to successfully integrate both cultures into their personality successfully navigated? A. Hope B. Purpose C. Fidelity D. Care

C. Fidelity As seen in Figure 1, the bicultural individuals who are able to successfully integrate both aspects of their identity probably do so because they have successfully resolved the crisis in the Fidelity stage, which involves the crisis of Identity vs. Identity Confusion (C is correct). The primary crisis in the Hope stage is between trust and mistrust, determining whether other people are safe and reliable caregivers, not identity issues (A is wrong). The Purpose stage centers on the Initiative vs. Guilt crisis in which children must learn to feel empowered by taking initiative, or guilty for doing so. This is not related to identity (B is wrong). The crisis which must be resolved in the Care stage is between generativity, or continuing to grow as a person, and stagnation, rather than identity (D is wrong).

Perceiving the color of a pH indicator requires what type of visual processing? A. Depolarization of cone cells to trigger hyperpolarization of bipolar neurons. B. Depolarization of cone cells to trigger depolarization of bipolar neurons. C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons. D. Hyperpolarization of cone cells to trigger hyperpolarization of bipolar neurons.

C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons. The answer choices contain pairs of information so proceed with a 2x2 question elimination strategy. When cone cells are not exposed to light and are thus not processing the visual stimuli of color, they are depolarized. Since the question is asking about perceiving color, the cone cells need to be hyperpolarized (eliminate choices A and B). When cone cells hyperpolarize, the bipolar neurons are no longer being inhibited and can then depolarize in order to transmit signal to the ganglion and eventually the optic nerve (C is correct; eliminate D).

Which of the following are products of the pentose phosphate pathway? I. NADH II. NADPH III. Ribose-5-phosphate A. I only B. III only C. II and III only D. I, II, and III

C. II and III only Correct Answer Item I is false: NADH is an electron-carrier produced in catabolic reactions, like cellular respiration (choices A and D can be eliminated). Note that since both remaining choices include Item III it must be true and we can focus on Item II, which is true: NADPH is produced by the pentose phosphate pathway (choice B can be eliminated and choice C is correct). Item III is in fact true: ribose-5-phosphate is a primary product of the pentose phosphate pathway.

Suppose it were shown that when the students were given solvable puzzles by an experimenter who had previously only given them unsolvable ones, both experimenters were noticeably paying more attention to the students' responses. Under these circumstances, which of the following might confound the results? I. The bystander effect II. Social loafing III. Social facilitation A. I only B. II only C. III only D. I and III only

C. III only Item I is false: the bystander effect occurs when people fail to help strangers in apparent distress because they assume that another onlooker will do so (A and D can be eliminated). Item II is false: social loafing occurs when individuals work in a group and put forth less effort than they would have had they been working alone (B can be eliminated). Item III is true: social facilitation occurs when an individual's performance on a simple task is enhanced by the presence of others. If the researchers are noticeably paying more attention to the children while the latter are trying to complete solvable puzzles, it is possible that this added attention could improve performance through social facilitation. This effect could confound the results of the experiment (C is correct).

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. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase 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 (eliminate B and D). Fructose-2,6-bisphosphate stimulates phosphofructokinase, which is used in glycolysis, and inhibits fructose-1,6-bisphosphatase, which is used in gluconeogenesis (C is correct and eliminate D).

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. It proceeds best with primary substrates and is a one-step reaction. SN2 cannot operate with a tertiary substrate 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.

The Implicit Association Test (IAT) was developed at Harvard University to measure unconscious biases that people may have. The subject is presented with words or photos, one at a time, and asked to categorize the stimuli as either "good" or "bad" via the computer keyboard. The associations between the stimulus and the categories of good and bad are measured via the response time; when the requested categorization (e.g., a particular face with distinctive racial features to be categorized as "good") is congruent with the subject's pre-existing bias (if any), the subject responds faster than when the associations do not match. The discrepancies in response time indicate the bias a person has; longer response times indicate more difficulty in matching the pairings. What might the IAT of a bicultural Asian American with a low BII show if she were asked to pair the word "self" with either "American" or "Asian"? A. She would show a bias during American/self pairings. B. She would not show any bias. C. It would depend on the context that was primed prior to her beginning the IAT. D. She would show a bias during Asian/self pairings.

C. It would depend on the context that was primed prior to her beginning the IAT. Without knowing what priming condition the respondent had been exposed to prior to taking the IAT, it is impossible to assess what the subject's performance would likely be. Had the person been exposed to an American priming condition prior to the IAT, she would likely be faster with pairing the self with the Asian prime, and vice versa, as those with low BII tend to have the opposite cultural identity primed when presented with a cultural prime (C is correct). Since we do not know what the priming condition was, we cannot tell if she would have a bias during the American/self or Asian/self pairings (A and D are wrong). Since she has a low BII we know that she would show a bias, but that the bias would differ depending on which cultural identity was primed before the IAT (B is wrong).

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. No; bacteria are asexual organisms, and in the absence of mutation, all progeny of any one cell are genetically identical to the parent cell. 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 correct). 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 (eliminate A). 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 (eliminate B). They also do not perform meiosis (eliminate D).

A study is being conducted in which participants are given written anonymous name problems and subsequently encouraged to verbalize their thought processes as they try to find at least one solution to the problem. 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 saying them aloud.

C. reading the problems aloud rather than providing the participants with the text. The lateral geniculate nucleus (LGN) is the area behind the retina that serves as an intermediate station for electrochemical signals passed from the retina to the visual cortex of the brain. Accordingly, the vision of a participant with damage to the LGN would be impaired, requiring that the problems be read aloud rather than presented in textual form (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 (B and D are wrong). Even in its original form, the experiment entailed a verbal problem, not an auditory problem (A is wrong).

Two blocks are suspended from the ends of a massless meter stick; a 4kg weight from one side and a 1kg weight from the opposite side. How far from the center of the stick must the rope be attached in order to maintain rotational equilibrium? A. 10 cm B. 20 cm C. 25 cm D. 30 cm

D. 30 cm

Suppose that 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? I. Schizophrenia II. Antisocial personality disorder III. Alzheimer's disease A. I only B. I and II only C. II and III only D. I, II, and III

D. I, II, and III 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 were suffering from schizophrenia it is highly likely 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).

Humans are exposed to radiation from naturally-occurring radioactive elements. The most abundant radioactive nuclide on earth is 238U, which undergoes a series of decays ending in 206Pb, a stable isotope of lead. These decays include the emission of alpha particles (α), which are helium nuclei (4He), beta particles (β-), which are energetic electrons, and gamma rays, which consist of high-energy, massless photons. Though abundant, 238U represents a limited risk to human health as it is naturally only found in solid minerals. However, one of its daughter products, the inert gas 222Rn (t1/2 = 3.82 days), poses a much more substantial risk to public health as it tends to accumulate in living spaces with poor ventilation. Exposure to 222Rn has been found to constitute up to 60% of the standard annual environmental radiation dose in humans, and has been cited as the primary cause of 14% of diagnosed lung cancers. Within the series of consective α-decay steps leading to 222Rn, intermediates have half-lives ranging from thousands to billions of years. (There are also two early β- decays with very short half-lives.) For daughter nuclei of 222Rn, half-lives range from about one-sixth of a microsecond (for 214Po) to 22.3 years for the radioactive isotope 210Pb. One of the interesting features of this naturally-occurring radioactive decay series is the branching that can spontaneously occur at 214Bi. This nucleus can either undergo a β-decay, followed by an α-decay to reach 210Pb (as shown in Figure 1) or the other way around: first an α-decay, then a β-decay. Which of the following is the most likely explanation for the fact that 222Rn is a greater biohazard than the other elements in the 238U decay series? A. It is the alpha emitter with the shortest half-life. B. It is the alpha emitter with the longest half-life. C. It has the greatest mass number in the series. D. It is a gas.

D. It is a gas. Correct Answer Explanation D. Choices A, B, and C are all false, so the answer must be D. Choice A is false because 214Po is an alpha emitter with a shorter half-life than 222Rn, and B is false since 210Pb is an alpha emitter with a longer half-life. Choice C is false because 238U clearly has the greatest mass number in the series. Radon is an odorless, colorless gas that results from the decay of radium (after which it was named) found in soil and rocks. Being a gas, radon can easily seep from the ground (and even the building materials) into homes, where it is inhaled and can lodge in the lungs. Although radon is chemically inert, the alpha particle it emits upon decaying is damaging; furthermore, its daughter nucleus, 218Po, is not chemically inert and also radioactive. Radon is the largest source of radiation encountered by humans (accounting for over half of the total average radiation exposure to the U.S. population).

Socialized medicine is a term used to describe governmental regulation of health care, with this public administration of health services being 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, as evidenced by the opposition to presidential reform efforts during the Truman, Clinton, and Obama administrations. This hesitation could be attributed to several factors, such as the public's agreement with conservative critics. The persistent opposition to forms of socialized medicine matches the approach to inequalities described in the theories of: A. Emile Durkheim. B. Karl Marx. C. Ludwig Gumplowicz. D. Max Weber.

D. Max Weber. The theoretical perspective most concerned with social inequalities is conflict theory. Classical sociologists associated with this theory include Karl Marx, Ludwig Gumplowicz, and Max Weber. In contrast, Emile Durkheim is more associated with structural functionalism, which is focused on contributions to social stability (A is wrong). The persistent opposition to forms of socialized medicine best reflects the theories of Max Weber. Unlike 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; D is correct). The public's persistent opposition challenges the theories of Karl Marx. 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 (B is wrong). Finally, the theories of Ludwig Gumplowicz focus on cultural and ethnic conflicts that are not relevant to the question (C is wrong).

Trisomy 13, also known as Patau syndrome, occurs with a frequency of 1 in 10,000 births. In a given population, what is the frequency of carriers of Patau syndrome? A. 1/100 B. 198/10,000 C. 9,801/10,000 D. None of the above

D. None of the above The Hardy-Weinberg equation can only be used to describe the frequencies of autosomal recessive or dominant traits or conditions, not changes in chromosome number. Since trisomies are not autosomal recessive conditions, the Hardy-Weinberg equation cannot be used, and one cannot predict the occurrence of the carriers in a population (D is correct and A, B, and C are eliminated). Also, because trisomies are not based on dominant or recessive expressions, there is no "carrier" state.

A mixture of aspartate and phenylalanine is separated into its component molecules by thin layer chromatography on a silica plate eluted with 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. Phenylalanine will move farther with the mobile phase, because it has a nonpolar side chain. 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.

Which of the following statements regarding RNA molecules is NOT true? A. RNAs can act as enzymes to catalyze reactions. B. Some RNAs have more than four different types of bases. C. Some RNAs are synthesized in the nucleolus. D. RNAs are insusceptible to alkaline hydrolysis.

D. RNAs are insusceptible to alkaline hydrolysis. RNA molecules have decreased stability compared to DNA in part because of their susceptibility to alkaline hydrolysis due to the presence of hydroxyl group at 2'-C position (choice D is not true of RNA and is the correct answer choice). Some RNAs have enzymatic function (such as in telomerase) and they are termed ribozymes (A is true). tRNA has unique and modified bases apart from the traditional four bases A,U,C, and G (such as inosine, B is true). rRNA is synthesized in the nucleolus (C is true).

When a person belonging to a minority group is primed to recall his or her minority status while engaged in a task that is traditionally thought of as not typical for that group, the individual tends to perform worse than if he or she had not been reminded of the prejudice associated with that status. What is this phenomenon called? A. Racism B. Unstable self-esteem C. Self-verification D. Stereotype threat

D. Stereotype threat When people are primed with a negative stereotype about their group's expected performance in a particular subject, they tend to do worse than if there had been no priming (a phenomenon known as stereotype threat). For example, when female participants are led to believe that the difficult math test they are taking is one on which women do worse than men, they tend to perform in line with the stereotype that "women are bad at math." However, when the second half of the test is presented to the same group without the gender priming, there are no gender differences in outcome. The priming can be as simple as having men outnumber women in a room (D is correct). Racism includes the actions, beliefs, or social systems that place different races in a hierarchy based on stereotyped expectations of people based on their racial characteristics; the concept does not deal directly with performance (A is wrong). The concept of self-esteem is not used to refer to group stereotypes, but is the worth a person places on him or her self (B is wrong). Self-verification posits that people wish others to perceive them as they perceive themselves; a positive view of a person that is not in line with his or her self-concept is likely to be rejected (C is wrong).

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. The concentration of [I-] increases. 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 (eliminate 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 (eliminate choice A). As the dissolved [NaCl] concentration increases, AgCl will be precipitated from solution, which will enable additional AgI to dissolve (eliminate choice C). The increased dissolution of AgI will cause the increase in [I-], even as [Ag+] levels remain low.

Which one of the following statements is true regarding PCl3? A. The phosphorus has sp2 hybridization, and the molecule has a trigonal pyramid shape. B. The phosphorus has sp3 hybridization, and the molecule has a tetrahedral shape. C. The phosphorus has sp2 hybridization, and the molecule has a trigonal planar shape. D. The phosphorus has sp3 hybridization, and the molecule has a trigonal pyramid shape.

D. The phosphorus has sp3 hybridization, and the molecule has a trigonal pyramid shape. Attack this 2x2 style question first by hybridization, and then by molecular shape. The phosphorous atom has three bonds to chlorine and a lone electron pair, for a total of four electron groups. As such it must be sp3 hybridized (eliminate A and C). Because the fourth electron group is a lone pair, the shape cannot be tetrahedral (eliminate B).

A CO2 laser used as a laser scalpel produces a beam of laser light with a wavelength of 10.6 µm. Compared to the excited electrons in a laser with a wavelength of 1.06um, the excited electrons in the CO2 laser most likely have: A. greater mass. B. less mass. C. a greater energy difference between their normal state and their excited state. D. a smaller energy difference between their normal state and their excited state.

D. a smaller energy difference between their normal state and their excited state. The wavelength of the CO2 laser is 10 times greater than the 1.06 wavelength laser. Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the laser light is the energy released by transitions of electrons dropping to a lower energy level, less emitted energy implies a smaller difference between the energy levels. The mass of an electron is independent of its atomic or molecular energy state (A and B are wrong).

Hypothetically, if Reaction I was found to be spontaneous at standard state and higher temperatures, but nonspontaneous at low temperatures, the reaction must result in: A. decreasing entropy and decreasing enthalpy. B. decreasing entropy and increasing enthalpy. C. increasing entropy and decreasing enthalpy.

D. increasing entropy and increasing enthalpy. From the equation ΔG = ΔH - TΔS, we can eliminate B since ΔH > 0 and ΔS < 0 would always give ΔG > 0 (nonspontaneous) no matter what the temperature, and we can eliminate C since ΔH < 0 and ΔS > 0 would give ΔG < 0 (spontaneous) regardless of temperature. Choice A is false since at sufficiently high temperatures, the T ΔS term would become more negative than ΔH, leading to a positive ΔG, contradicting the statement in the question that the reaction is spontaneous at high temperatures. This leaves choice D.

The removal of the adrenal glands will result in a reduction of all of the following EXCEPT: A. K+ secretion in the nephron. B. Na+ reabsorption in the nephron. C. water reabsorption in the nephron. D. renin secretion by juxtaglomerular cells.

D. renin secretion by juxtaglomerular cells. The hormones secreted by the adrenal glands include epinephrine from the medulla, and cortisol, aldosterone, and low levels of sex steroids from the adrenal cortex. Aldosterone increases potassium secretion and also increases sodium reabsorption (A and B are true and thus eliminated). The increased Na+ reabsorption leads to increased water reabsorption, thereby increasing the blood volume (C is true and thus eliminated). The loss of aldosterone would cause water and sodium loss, decreased blood volume and decreased blood pressure. In response to decreased blood pressure, renin secretion would increase, not decrease (choice D is false and thus the correct answer choice).

SEE OTHERSIDE FIRST The figure indicates that when [AgI] = 1.5 x 10-15 M, the concentration of NaI = 0.06 M. In 500 mL (0.5 L), this solution should contain 0.030 mol NaI. Since the molar mass of NaI is 150 g/mol, this amounts to 4.5 g of NaI.

How many grams of NaI should be added to 500 mL of a saturated solution of AgI to make a solution that is 1.5 x 10-15 M Ag+? A. 9.0 g B. 4.5 g C. 0.06 g D. 0.03 g

SEE OTHER SIDE FIRST Enamines form from the combination of an aldehyde or ketone with a secondary amine (like piperidine). Since propylamine is a primary amine (eliminate choice A), an imine is formed instead (choice C is correct). Amides are carboxylic acid derivatives and form from the reaction of an amine with any other derivative (eliminate choice B). A lactam is a cyclic amide, so choices B and D are effectively the same answer (eliminate choice D).

If the students had used propylamine in Experiment 2 in place of piperidine, what functional group would they synthesize in Step 1? A. Enamine B. Amide C. Imine D. Lactam

SEE OTHER SIDE FIRST Intermediate X is an enolate ion which is stabilized by resonance. It is formed by deprotonation of the ketone by hydroxide.

Intermediate X formed in Experiment 1 is: A. an alcohol. B. an alkene. C. resonance stabilized. D. a carbocation.

SEE OTHERSIDE FIRST D. Thickness of the membrane Correct Answer "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 = κε0A / d, where κ is the dielectric constant, ε0 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), would increase the capacitance, C.

Ion flow in neurons can be characterized as an electrical circuit for both the resting neuron (Figure 1) and the active neuron (Figure 2). The membrane is a capacitor, slow leakage channels are a 25 MΩ resistor, and the Na+/K+ pump is a voltage generator. 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. 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

SEE OTHER SIDE FIRST The passage states that no significant difference is observed between the obese and lean mice. While this may seem to contradict the leptin resistance hypothesis, leptin is a hormone which will be expected to affect target tissues differently. In the passage, the leptin resistance hypothesis specifically relates to satiety and, while it is known that leptin activates the sympathetic nervous system, it is unclear how renal sympathetic nerve activity (RSNA) is related to the sensation of satiety. Therefore the lack of a significant difference between obese and lean mice does not provide significant evidence to strengthen or weaken the leptin resistance hypothesis (choice D is correct). The lack of difference between RSNA in lean and obese mice is not supportive of the hypothesis that leptin resistance inhibits leptin function (answer choice A is wrong). As mentioned above, leptin resistance is primarily associated with mitigating a sense of satiety when full; it is not clear whether an increase in renal sympathetic activity should be included within the leptin resistance syndrome (answer choice B is wrong). Given the dramatic increase in nerve activity with increasing leptin dosage, leptin does have an impact on RSNA (choice C is wrong).

While it has been shown that leptin can activate the sympathetic nervous system, the most classic characteristic of leptin resistance is a decrease in the sense of satiety after feeding. In order to test whether leptin resistance in obese participants affected all functions of leptin, an experiment was performed in which yellow obese agouti mice and normal lean mice were injected intravenously with three different leptin concentrations while renal sympathetic nerve activity (renal SNA) was measured. Surgery was performed to expose the appropriate nerve fascicles near the kidney and platinum electrodes recorded the multi-fiber renal SNA (RSNA) in anesthetized mice. Both groups showed an enhancement of renal sympathetic activity correlating to the increased leptin injection. No significant difference in RSNA activity was detected between lean and obese mice as depicted in Figure 1. This research suggests that leptin resistance in obese individuals may be a selective phenomenon in the sense that not all functions of the hormone are affected by leptin resistance. What impact do the results shown in Figure 1 have on the hypothesis that leptin resistance secondary to obesity reduces all effects of leptin? A. Supports the hypothesis due to the increase in neural activity with injection of leptin in obese mice B. Weakens the hypothesis due to the increase in neural activity with injection of leptin in obese mice C. Weakens the hypothesis due to the lack of impact of leptin on renal sympathetic nerve activity. D. Neither supports nor weakens the hypothesis due to the differential effect of hormones on differing tissues.