MCAT Full length TPR

Fat molecules in adipose tissue are important energy reservoirs in mammals. After eating, excess dietary non-esterified fatty acids (NEFAs) are esterified into triacylglycerols (TAGs), which are subsequently stored in cytosolic lipid droplets of adipocytes. When required, TAG stores are mobilized via hydrolytic cleavage and the resulting NEFAs are delivered via the circulation to peripheral tissues for β-oxidation and ATP production. This breakdown of lipids into glycerol and free fatty acids is called lipolysis. The relative balance of TAG hydrolysis and NEFA esterification controls the cellular concentration of NEFAs, and this process is tightly regulated. Many of the enzymes used in this process are under hormonal control, and most are linked to upstream signaling pathways. Lipolysis involves hydrolyzing ester linkages between the glycerol backbone and long chain fatty acids. This multi-step process is controlled by three specific hydrolases or lipases. First, adipose triglyceride lipase (ATGL) selectively performs the first and rate-limiting step, hydrolyzing a molecule of TAG to generate one diacylglycerol (DAG) and one NEFA. Next, hormone-sensitive lipase (HSL) hydrolyzes DAG into monoacylglycerol (MAG) and one NEFA (note that although it preferentially catalyzes this step, HSL is actually capable of hydrolyzing TAGs, DAGs, and MAGs at all steps of the reaction). Finally, monoglyceride lipase (MGL) efficiently cleaves MAG into glycerol and one NEFA. G0S2 (G0/G1 Switch Protein 2) directly interacts with the N-terminal domain of ATGL, and expression of G0S2 correlates with insulin levels. In order to determine G0S2 expression patterns, several tissues samples were collected from a healthy lab rat. The samples were lysed and three different macromolecules were isolated from each. PCR was performed on one, reverse-transcriptase PCR was performed on another, and the final was used for western blot analysis (Figure 1). Based on information in the passage, which of the following is true? A. Compared to fatty acids, molecules of triacylglycerol are relatively inert. B. A competitive inhibitor for ATGL would decrease DAG catabolism. C. HSL is responsible for generating glycerol and MAG, but not DAG. D. TAGs contain two stable ester linkages.

A A. Part of the reason fatty acids are stored as triacylglycerol is because TAGs are much more stable than free fatty acids (choice A is correct). The passage says that ATGL hydrolyzes TAG into DAG and one free fatty acid. It does not hydrolyze or catabolize DAG (choice B is wrong). The passage also says that HSL can hydrolyze TAG, DAG and MAG. It therefore generates DAG, MAG, glycerol and free fatty acids (choice C is wrong). TAGs contain a glycerol backbone and three fatty acids. They therefore contain three ester linkages (choice D is wrong).

Which of the following is likely true of the pentose phosphate pathway? A. It is more active in adipose tissue than in muscle tissue. B. NADPH is produced and can be used in the electron transport chain and oxidative phosphorylation to generate ATP. C. It is a series of isomerizations of six-carbon sugars. D. The inhibition of carbamoyl phosphate synthase by purine nucleotides helps to regulate the production of UTP and CTP.

A A. The pentose phosphate pathway replenishes NADPH, which the passage states is used by the cell in biosynthetic pathways such as fatty acid biosynthesis. This is more likely to happen in adipose (fat) tissue than in muscle tissue (choice A is correct). NADPH cannot, however, be used to generate ATP via the electron transport chain and oxidative phosphorylation; NADH is needed for that pathway (choice B is wrong). The pathway produces ribose, a 5-carbon sugar from glucose, a 6-carbon sugar; this involves a decarboxylation, not just an isomerization (atomic rearrangement, choice C is wrong). Lastly, the passage states the carbamoyl phosphate is inhibited by UTP, which is a pyrimidine, not a purine (choice D is wrong). **I chose D because I figured PPP would be more likely in muscle cells since NADPH acts as an antioxidant and muscle cells do a lot of cell resp. but passage says that NADPH is used for fatty acid synthesis which happens in fat cells.

Which of the following presents a difficulty for electronic stethoscope use? A. The amplification system may enhance the intensity of ambient sounds near the system that are not of medical interest. B. Low resistance in the wires connecting the chestpiece to the earpieces is needed for high sound fidelity but may cause currents to reach dangerously high levels, increasing the risk of electric shock for the listener. C. Low frequency sounds from the bell require more amplification to be audible, and electronic amplification increases as a function of frequency. D. Standing waves in the tube part of the stethoscope can destructively interfere with the electric signal, reducing sound volume.

A Choice A seems possible, because the second paragraph of the passage indicates that all sounds picked up by the electronic stethoscope will be amplified, and this may make it hard to distinguish important sounds from unimportant ones. B is wrong because electric shock would not be a risk unless listener held live wire. C is wrong because electronic amp is talked about in passage as having to increase with frequency. D is wrong because standing waves are sound waves and electrical signals are electromagnetic so they can't interfere destructively.

Taurine is a derivative of the amino acid cysteine, shown below. Which of the following reactions is a necessary part of the biochemical conversion of cysteine to taurine in the body? (look up pic of taurine) A. Decarboxylation of the carboxyl group B. Protonation of the basic amine C. Reduction of the thiol group D. Conversion of the primary amine to an imine

A Compare the structure of cysteine given in the question to the structure of taurine, which can be deduced from the structure of taurocholic acid given in the passage. The amino group is still present in taurine and has been coupled to the carboxyl group of the bile acid to form a new amide bond (eliminate choices B and D). The thiol group in cysteine cannot be reduced anymore, but has instead been highly oxidized with the addition of five bonds to oxygen in order to form taurine (eliminate choice C). By process of elimination, decarboxylation must have taken place. The COOH group of the amino acid has been lost since no carbonyl group is present in the new substituent of Compound 4 in the passage.

Reaction 1 can also be run using peptide derivatives that contain Se in place of S. Compared to the original reaction, the replacement of S with Se results in: (S acts as nucleophile in reaction with polar protic solvent) A. a faster reaction, since Se is a better nucleophile than S. B. a slower reaction, since Se is a weaker nucleophile than S. C. a faster reaction, since Se is a better electrophile than S. D. a slower reaction, since Se is a better electrophile than S.

A Nucleophilicity increases DOWN the PT and since Se is larger, it is more polarizable, especially since it's a polar protic solvent

Which of the following statements is/are true with regard to the mechanics of skeletal muscle contraction? I. Reduced ATP levels could result in high intracellular calcium and persistent contraction II. Calcium binds to calmodulin and allows actin and myosin to interact III.The shorter the sarcomere length, the stronger the contraction A. I only B. I and II only C. II and III only D. I, II, and III

A. Item I is true: in the absence of ATP, not only will the myosin and actin filaments be unable to disconnect, Ca2+ active transporters will not be able to return Ca2+ to the sarcoplasmic reticulum (choice C can be eliminated). Item II is false: calcium binds to troponin (not calmodulin) in skeletal muscle (choices B and D can be eliminated and choice A is correct). Note that Item III is false: the length-tension relationship states that there is an optimum sarcomere length to ensure the most forceful contraction. While it is true that shorter sarcomere length translates to a stronger contraction up to a point, too much overlap causes the actin and myosin filaments to bump together and reduces the strength of contraction.

According to Skinner, in terms of behavior modification, punishment: A. is typically only effective as long as the punishment is present. B. is the single most effective form of behavior modification. C. is more effective when applied in an inconsistent manner. D. will cause an increase in likelihood of future misbehavior.

A. Skinner asserted that punishment is not an ideal form of behavior modification because it is only effective so long as the punishment, or threat of punishment, is present. For example, every time your cat jumps onto the counter, you squirt it with water and it jumps down. But when the threat of being sprayed is not present (e.g. you are in another room or not home), your cat is free to jump on the counters, and thus is learning to manipulate the misbehavior in order to avoid the punishment (choice A is correct). For those very reasons, punishment is not the single most effective form of behavior modification (choice B is wrong). Punishment is most effective when applied consistently, not inconsistently (choice C is wrong). Finally, there has been no definitive evidence that the use of punishment causes misbehavior, though it can lead to sneakier forms of the misbehaviors it is being used to extinguish (choice D is wrong).

Social identity theorists think that the greater the identification with the in-group: A. the more likely in-group bias is to occur. B. the less likely in-group bias is to occur. C. the less likely one is to seek out membership in other in-groups. D. the more likely one is to seek out membership in other in-groups.

A. Social Identity research indicates that strong in-group identification often leads to stronger in-group bias (the preference for other in-group members) and consequent out-group derogation, defined as discrimination against those not in the in-group due to perceived threat (choice A is correct and choice B is wrong). Social identity theorists hold that people's strength of identity with their in-groups significantly impacts their attitudes toward incompatible out-groups. However, individuals could potentially be members of multiple, non-contradictory in-groups (alumni groups, sports fandoms, etc.), and their decisions to join such groups are not necessarily influenced by the strength of identification with groups to which they already belong (choices C and D are wrong).

Which of the following statements can be inferred from information in the passage? A. Increased urinary cAMP is associated with decreased renal phosphate transport B. cAMP-dependent Protein Kinase A (PKA) leads to activation of NPT2a C. NPT2a is a G-protein coupled receptor (GPCR) D. A novel drug that inhibited NHERF1 activity would lead to decreased risk of both calcium renal stones and uric acid renal stones

A. The passage states in the last paragraph that certain mutations in NHERF1 lead to impaired TmP/GFR values, as well as increased urinary cAMP levels. Therefore increased urinary cAMP levels are associated with decreased renal phosphate transport (as quantified by TmP/GFR values) (choice A is correct). Since increased cAMP is associated with decreased renal phosphate transport, it is unlikely that activation of PKA by cAMP would result in activation of the phosphate transport protein NPT2a (choice B is wrong). The passage states that NPT2a is a sodium-phosphate transporter and not a GPCR (choice C is wrong). The passage also states that NHERF1 knockout mice have increased uric acid excretion, and that this is associated with renal stones, thus a drug that inhibits NHERF1 is unlikely to decrease the risk of these stones (choice D is wrong).

"Due to the significance of If in controlling heart rate, numerous pharmacological agents have been developed to control heart rate. HCN channel blockers, such as ivabradine, have been used to decrease heart rate in patients who have cardiac ischemia (insufficient blood supply to cardiac tissue). In order to better understand the dose-response relationship of ivabradine on HCN channels, the fractional inhibition of If current was measured at various ivabradine concentrations." Which of the following best describes why ivabradine is used in patients with cardiac ischemia? A. Ivabradine decreases heart rate, reducing the severity of hypoxia. B. Ivabradine decreases If, increasing intracellular Ca2+ stores for myocyte contraction. C. Ivabradine blocks HCN channels, decreasing cAMP concentrations and activating the parasympathetic nervous system. D. Parasympathetic stimulation causes epinephrine release, activating G-protein coupled receptors that result in elevated cAMP levels.

A. The passage states that ivabradine blocks HCN channels, causing a decrease in heart rate. A decrease in heart rate will then decrease oxygen use by the oxygen-starved cardiac tissue (choice A is correct). There is no information in the passage to support that ivabradine is involved with intracellular Ca2+ stores (choice B is wrong). Inactivation of HCN channels will not have any direct effect on cAMP concentrations or activate the parasympathetic nervous system; in fact it is cAMP that activates the channels, not the other way around (choice C is wrong). It is the sympathetic nervous system that causes the release of epinephrine, not the parasympathetic nervous system (choice D is wrong).

In the electrolysis of water shown below, a current of 2 amps is applied to 180 mL of H2O(l) for 6 hours and 42 minutes. How many grams of H2(g) are formed? (Faraday's constant = 96,500 C/mol) 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) A. 0.25 g B. 0.5 g C. 5.0 g D. 10.0 g

B **start with time for these questions, convert 6 hrs and 42 mins to 24000 sec: (24000s)x(2C/1s)x(1 mol e-/ 96500C)x (1 mol H2/2mol e-)x (2gH2/ 1 mol H2)= 0.5g

If a mutation rendered creatine phosphokinase non-functional, which of the following is most consistent with feedback inhibition of this biosynthetic pathway? A. Increased creatine levels B. Decreased guanidoacetate levels C. Increased creatine phosphate levels D. Decreased creatine phosphate levels

B B. Increased levels of creatine and decreased levels of creatine phosphate would result from creatine phosphokinase inactivity, regardless of feedback inhibition (note that there cannot be two correct answer choices, so choices A and D must both be wrong). High levels of creatine inhibiting the enzyme responsible for a precursor to creatine (guanidoacetate) is an example of feedback inhibition (choice B is correct). In addition, with enzyme dysfunction, levels of creatine phosphate would not be expected to increase (choice C is wrong).

"A recent study measured the tubular maximal reabsorption of phosphate (TmP) normalized for the glomerular filtration rate (GFR) - the TmP/GFR value - in 207 subjects with known calcium nephrolithiasis. The study found that 20% of these subjects with stone formation and normal parathyroid function had decreased TmP/GFR values." A decreased TmP/GFR value would result in which of the following? Hypophosphatemia Increased 1,25-dihydroxy vitamin D production Hyperphosphatemia A. I only B. I and II only C. III only D. I, II and III

B B. Item I is true: A decrease in the TmP/GFR value (the maximal reabsorption of phosphate divided by the glomerular filtration rate) means that less phosphate is reabsorbed from the filtrate back to the blood. This would lead to hypophosphatemia, i.e., decreased phosphate levels in the blood (choice C can be eliminated). Item II is true: The decrease in serum phosphate levels would lead to increased 1,25-dihydroxy vitamin D production, since vitamin D functions to increase serum calcium and phosphate levels when they are low (choice A can be eliminated). Item III is false: As described above, a decrease in the TmP/GFR value means that less phosphate is reabsorbed from the filtrate back to the blood, leading to hypophosphatemia not hyperphophatemia.

Two resistors R1 and R2 are connected in series to a battery with no internal resistance. The current through each is measured. A third resistor, R3, is then connected in parallel to R2. How does this affect the currents throughR1 and R2? A. I1 and I2 both increase. B. I1 increases and I2 decreases. C. I1 decreases and I2 increases. D. I1 and I2 both decrease.

B B. The combination of R2 and R3 has less resistance than R2 alone. This is because, in parallel, 1 / Rparallel = 1 / R2 + 1 / R3 > 1 / R2 and therefore, Rparallel < R2. The equivalent resistance of the entire circuit is R1 + Rparallel, which has now decreased. Since the current supplied by the battery is V / Req, where V is the battery's voltage, decreasing Req increases the total current. The current flows through R1; therefore, I1 must also increase. This eliminates choices C and D. Increasing the current through R1 would also increase the voltage across it. This means that there would be less voltage available to the parallel portion of the circuit. Less voltage across R2 means less current through it.

In one performance, the dancer spins at a constant speed while holding a basket of flowers in her hand. While spinning, the dancer holds her arms out to the side, so her arms make a 90° angle with her body. Which of the following is true for the circular motion of the basket of flowers? The spinning causes a centripetal force, which pulls the basket out from the center of the circle. The dancer's arm provides the centripetal force, which pulls the basket in towards the center of the circle. The basket has a constant velocity which is equal to the tangential velocity of the dancer. A. I only B. II only C. II and III only D. I, II, and III

B Centripetal force is a force pointing to the center of the circle so I is wrong. III is wrong because velocity's vector is constantly changing since she's spinning

Which of the following statements is correct with regard to the role of creatine phosphokinase? A. It is involved in the committed step for creatine phosphate formation. B. It stabilizes the transition state in the formation of creatine from creatine phosphate. C. It lowers ΔG, making the reaction more spontaneous. D. It raises ΔG, making the reaction more spontaneous

B Creatine phosphokinase is the enzyme used to catalyze the reversible reaction between creatine and creatine phosphate. Catalysts increase the reaction rate by lowering the activation energy (stabilizing the transition state) of the reaction (choice B is correct). Note that this enzyme will stabilize the transition state between creatine and creatine-phosphate in BOTH directions. Because the specific reaction is reversible, it cannot be a "committed step" (choice A is wrong). Furthermore, catalysts do not affect ΔG and spontaneity (choices C and D are wrong). **committed step means irreversible

The dissociation constant for fourth NAD+ to GAPDH is 1.5x10^5 What is the equilibrium constant for the binding of the fourth NAD+ to GAPDH? A. 1.5 × 10-5 B. 6.7 × 10-6 C. 1.5 × 105 D. 6.7 × 106

B Do 1/1.5x10^5 to get 6.7 × 10-6

When GAPDH binds the first molecule of NAD+, a series of oxidation-reduction reactions occur in which glyceraldehyde-3-phosphate is phosphorylated and brought to a higher energy level. Which of the following is true concerning these processes? (Dissociation constant is .15x10^5) A. The binding of NAD+ increases the spontaneity of the overall reaction. B. The association of NAD+ with the enzyme is an endergonic process. C. Glyceraldehyde-3-phosphate stabilizes the transition state in the binding of NAD+ and increases the reaction rate. D. Binding of NAD+ results in a more stable enzyme-NAD+ complex.

B Given the binding constant for NAD+ is the reciprocal of its dissociation constant, the equilibrium constant for the binding of NAD+ with the enzyme is exceedingly small (<10-4) which corresponds with a positive ΔG (remember that ΔG° = -RT ln K). The binding of NAD+ to the enzyme is therefore an endergonic process (choice B is correct and choices A and D are wrong). Note that choices A and D are identical as making the product more stable will increase its stability. There is no evidence in the passage to indicate that glyceraldehyde-3-phosphate stabilizes a transition state. In fact, the enzyme is responsible for increasing the rate of the reaction, not glyceraldehyde-3-phosphate (choice C is wrong).

Electronegativity, electron affinity, and ionization energy all increase across a periodic table row and decrease down a periodic table group. Why does acidity not follow this trend? A. Acidity relates to the ability of the nuclear protons to attract hydrogen atoms and not valence electrons. B. Acidity relates to the stability of the conjugate base, and larger atoms form more stable conjugate bases even when they are less electronegative than smaller atoms. C. Acidity is not influenced by nuclear shielding and effective nuclear charge, while the other trends are. D. Acidity increases across a periodic table row like the other trends, but it increases down a group because the electronegativity of the elements approaches that of hydrogen.

B Going across a period, acidity increases just like the other trends mentioned, but going down a column, acidity increases unlike the other trends because the attraction b/t H and X decreases so dissociation is more likely. Larger atomic anions are more stable conjugate bases because they can distribute charge better so size overcomes high values of electronegativity. Choice A is eliminated because acidity relates to the ability of an atom to attract valence electrons away from a hydrogen ion, thereby making proton donation (loss of a hydrogen ion) easier. Choice C is eliminated because acidity does increase with increasing effective nuclear charge across the periodic table. Choice D is eliminated because the electronegativity of elements only approaches that of hydrogen when going down the columns on the right side of the periodic table. In addition, this choice does not adequately explain the role of atomic radius in determining acidity.

The 10 kg dancer leaps into the air with an initial velocity of 5 m/s at angle of 45° from the floor. How far will she travel in the air horizontally before she lands on the ground again? A. 1.25 √2 m B. 2.5 m C. 2.5 √2 m D. 5 m

B Have to find time in air, so first use a=v/t>> -10=-5(sin45)/t>> t=√2/4. but this is the time to the top of the parabola, so have to multiply by 2 to get √2/2. Then, plug this in to x=vt>> (5cos45)(√2/2)=2.5m.

Two circular pipes both have radii of 3.5 meters. They both steadily eject water horizontally, and the water from each pours onto a measuring grid 5 meters below. The stream from the first pipe lands on the measuring grid 20 cm away from a position directly below the pipe mouth, and the stream from the second pipe lands 5 cm away from a position directly below its pipe mouth. How long would it take the second pipe to pour out as much water as the first pours out in 1 minute? A. 2 minutes B. 4 minutes C. 8 minutes D. 16 minutes

B In answering this question, it is essential to take advantage of the fact that almost everything is the same between the two flow streams. The given difference is how far they travel horizontally, and the question asks about a difference in time. The task is to determine the relationship between these variables. First, no matter how far the streams fall (5 m or any other distance), they both fall the same distance, so they both take the same amount of time to get to the measuring grid. In the projectile motion equation R = v0xt, if t is constant, then horizontal displacement is proportional to horizontal velocity. The flow speed coming out the second pipe is therefore one-fourth the flow speed coming out the first pipe, since it only travels 5 cm instead of 20 cm. Furthermore, according to f = Av, flow rate is directly proportional to flow speed if area is held constant. Since the two pipes have the same radius, they have the same area. Thus, the flow rate of the fluid coming out of the second pipe is one-fourth what it is in the first pipe. Next, since flow rate is also equal to volume over time, and since they are ultimately going to pour out equal amounts of water (equal volumes), flow rate is inversely proportional to time. The second pipe has one-fourth the flow rate, so it should take four times as much time. Four times as long as 1 minute is 4 minutes.

The strength of the bond between Lewis acid and Lewis base pairs is a function of the strength of their acidity and basicity, respectively. The B—N bond in Compound 1 below will be the strongest when what is true about X and Y? A. X = R1, Y = R3 B. X = R2, Y = R1 C. X = R3, Y = R3 D. X = R1, Y = R2 R1: trifluorophenyl group R2: trimethylphenyl group R3: tribromophenyl group

B In this the Boron is the acid and the Nitrogen is the base. To make nitrogen a better Lewis base, it should be able to give electrons so R2 has electron donating species (methyls) which delocalizes the positive charge when bond is formed. To make boron a better Lewis acid, it needs to want electrons more, so 3 fluorine groups are very electron accepting species.

"Cyclic nucleotide-gated (CNG) channels are non-selective ion channels that are activated by cyclic nucleotides, such as cGMP and cAMP. Upon binding cyclic nucleotides, CNG channels become permeable to a variety of alkali ions, resulting in either cellular depolarization or hyperpolarization. CNG channels are composed of four protein subunits with a central pore. Each subunit consists of six transmembrane segments (S1-S6) and a P-loop (a motif known for binding phosphate). Both the S6 segment and P-loop play an important in ion selectivity in the pore, and the C-terminus contains a cyclic nucleotide-binding domain (CNBD). CNG channels are of particular importance in the visual and olfactory systems, nephron collecting duct cells, and pacemaker cells of the heart." CNG channels are likely utilized in which of the following regulatory pathway(s): ADH regulation of water retention. aldosterone regulation of sodium and water retention. regulation of olfactory neuronal depolarization. A. I only B. II and III only C. I and II only D. I, II, and III

B Since Item III is found in exactly two answer choices, start analyzing it first; whether it is true or false, you'll be able to eliminate half the answer choices. Item III is true: the passage mentions that HCN channels are involved in the olfactory system and olfactory neurons are involved in relaying sensory information about smell in the nose, where HCN activation may be involved in the generation of action potentials (choices A and C can be eliminated). Since both remaining choices include Item II it must be true and you can focus on Item I. Item I is false: the passage states that CNG channels play an important role in sensory systems, the renal collecting ducts, and cardiac pacemaker cells. ADH acts at the collecting duct to control reabsorption of water, however CNG channels are selective for ions, and are therefore unlikely to be useful in the movement of water across the cell membrane of collecting duct cells (choice D can be eliminated and choice B is correct). Note that Item II is in fact true: aldosterone regulates Na+ reabsorption by acting on the distal convoluted tubule and the collecting duct. Since CNG channels are selective for cations, they could be used in Na+ reabsorption.

The dancer is wearing a necklace which consists of a chain and a pendant of mass, m. When the dancer is spinning at a constant speed, the pendant pulls away from her body, and the chain makes an angle, θ, with the dancer's body. If the pendant is spinning at a constant speed, v, and the chain has a tension, T, what is the radius of the circle created by the pendant? A. mv2 / T B. mv2 / T sin θ C. mv2 / T cos θ D. T sin θ / mv2

B The question describes the pendant travelling in a circle making an angle, θ, with the dancer's body. Notice that the component of tension directed into the center of the circle is T sin θ, not the full tension in the chain. It is just this component of tension that provides the centripetal force. Centripetal force is always equal to mv2 /r so in this case T sin θ = mv2 / r. Solving for the radius of the circle shows r = mv2 / T sin θ. The correct answer is choice B.

If I represents the intensity of the sound wave from an earpiece, A represents the area of the eardrum to which the sound is delivered, and t represents the time spent listening to the sound, which of the following expressions gives the energy received by the eardrum? A 1/IAt B. IAt C. I/At D. At/I

B The relationship between intensity and power is I = P/A, so the power received by the ear must be P = IA. Next, the relationship between power and energy is P = W/t, so the energy received by the ear (equivalently, the work done on the ear) must be W = Pt = IAt.

When given a new diagnosis of AD, studies suggest that people respond in many different ways, often employing various defense mechanisms. Which of the following responses would be considered intellectualization? A. Denying the diagnosis by questioning the credentials of the physician and seeking a second opinion. B. Focusing on the details of the diagnosis and seeking out as much information as possible about the disease. C. Physically lashing out at family and loved ones. D. Trying hard to forget the diagnosis ever occurred by focusing on other things.

B. According to psychoanalytic theory, intellectualization is a defense mechanism that involves an individual dealing with her or his emotions by focusing on the intellectual aspects of the problem or issue (choice B is correct). Denial is a defense mechanism that involves an individual behaving as though the problem or issue doesn't exist (choice A is wrong); displacement is a defense mechanism that involves an individual taking out their frustration on others (choice C is wrong); and repression is a defense mechanism that involves an individual trying to push the problem or issue into her or his subconscious, pretending it never happened (choice D is wrong).

An individual with factitious limb disorder pushes her way to the front of the line at a ticket window, claiming that she cannot stand for long periods of time. A bystander, who has been waiting in line patiently for 20 minutes, is disgusted. She looks at her friend and mutters, "How rude!" Which of the following could best explain the bystander's reaction? A. The bystander effect B. Attribution theory C. Conflict theory D. Functionalism

B. Attribution theory attempts to explain why individuals interpret the actions of others in the ways they do. In this case, the bystander is making the assumption that the individual with factitious limb disorder is rude. Specifically, this is an example of an internal attribution; the bystander blames the behavior of the individual with factitious disorder on an internal personality characteristic (i.e., she is rude; choice B is correct). The bystander effect occurs when someone needs help and there are many people around, but no one helps the person in need; this phenomenon does not best explain the bystander's reaction (choice A is wrong). Conflict theory is a sociological theory that attempts to explain how power relates to social order and related conflict; there is no mention of power in the question stem (choice C is wrong). Functionalism is a sociological theory that refers to the interdependence that exists among institutions within society; this theory does not explain the bystander's reaction (choice D is wrong).

Based on the design of the study described in the passage, what limits the researchers' ability to draw conclusions about the causal relationship between conception risk and race bias? A. The sample contained only White participants. B. The study did not employ random assignment, since participants could not be randomly assigned to a particular phase of their menstrual cycle. C. The study looked only at race relations between Black and White individuals. D. The study did not take into account non-oral contraceptives such as NuvaRing, Mirena, or IUDs.

B. Causation is extremely difficult to determine when experimenting with humans, particularly because all of the variables in a given experiment must by controlled by the experimenter, and subjects must be randomly assignment to experimental and control groups. Therefore, random assignment of subjects to a group (in this case, phases of the menstrual cycle) is one of the many variables that should have been controlled for in order to determine a causal relationship between conception risk and race bias (choice B is correct). While the variety of race relations and effects of non-oral contraceptives are important variables to consider, neither specifically limits the researchers' abilities to draw conclusions about the causal relationship between conception risk and race bias (choices C and D are wrong). The fact that the sample only contained White participants limits the researchers' ability to draw conclusions about how their results might apply to the general population, not about causality (choice A is wrong).

Viral infection would lead to an increase in which of the following cellular pathways? Aerobic respiration Pentose phosphate pathway cAMP mediated protein kinase activation A. I only B. I and II only C. II and III only D. I, II, and III

B. Item I is true: viral infected cells are undergoing a fair amount of cellular activity, including DNA replication, RNA transcription, and protein synthesis, all of which contribute to an increased need for ATP, and thus oxidative respiration (choice C can be eliminated). Item II is true: the pentose phosphate pathway produces essential precursors for nucleotide synthesis, necessary for both DNA replication and RNA transcription. It's likely that the activity of this pathway would be increased during viral infection (choice A can be eliminated). Item III is false: cAMP mediated protein kinase activity is increased when G protein coupled receptors are bound and activated; there is no reason to assume this in viral infection (choice D can be eliminated and choice B is correct).

Myasthenia gravis is an autoimmune neuromuscular disorder caused by the production of antibodies against acetylcholine receptors. Which of the following explains why individuals who are affected with this disease experience muscle weakness? A. The antibodies bind to acetylcholine receptors and stimulate a downstream response. B. The antibodies bind to acetylcholine receptors and prevent endogenous acetylcholine from binding. C. The antibodies bind to acetylcholine receptors and cause helper T cells to destroy the receptors. D. The antibodies bind to acetylcholine receptors and directly degrade the receptors

B. Myasthenia gravis manifests as muscle weakness because when the antibodies are bound to ACh receptors, endogenous acetylcholine cannot bind. Thus, muscles cannot contact despite the signal from the motor neuron (choice B is correct). If the antibodies stimulated a downstream response, the muscles would contract rather than feel weak (choice A is wrong). Antibodies can mark antigens for destruction by macrophages and other phagocytes, but not by helper T cells (choice C is wrong), and antibodies are not enzymes; they cannot directly degrade their antigens (choice D is wrong).

Palmitic acid has the chemical formula, C16H32O2. Which of the following best describes the appropriate product distribution after 4 rounds of beta-oxidation? A. 1 C12H24O2; 4 NADH, 4 acetyl-CoA; 4 FADH2. B. 1 C8H16O2; 4 NADH; 4 acetyl-CoA; 4 FADH2. C. 1 C8H16OSCoA; 4 NADH; 4 acetic acid; 4 FADH2. D. 1 C6H12O2; 5 NADH; 5 acetyl-CoA; 5 FADH2

B. Recall that each round of beta-oxidation produces 1 acetyl-CoA, 1 NADH, 1 FADH2, decreasing the fatty acid chain length by two carbons. After 4 rounds of beta-oxidation, there will be 4 acetyl-CoA produced (choices C and D are wrong), 4 NADH, 4 FADH2 and a fatty acid that is reduced by 8 carbons, C8H16O2 (choice A is incorrect; choice B is correct)

If the investigators in the study wanted to allow for the maximum chance that negative stereotypes would not interfere with the interactions between the adolescents and older people in the experimental groups, they might pair adolescents with older people: A. whom the adolescents did not know at all. B. whom the adolescents knew very well. C. of the opposite gender as the adolescents. D. who lived in a different city than the adolescents.

B. Stereotypes are weakest when individuals or groups are familiar with each other (choice B is correct), and they are strongest when individuals or groups are unfamiliar with each other or when they belong to different groups (choices A, C, and D are wrong).

The pentose phosphate pathway (PPP) produces ribose-5-phosphate from glucose-6-phosphate, and generates NADPH, which is used by the cell in biosynthetic pathways (such as fatty acid biosynthesis) as a reducing agent. Ribose-5-phosphate is converted to 5-phosphoribosyl-1-pyrophosphate (PRPP) by the enzyme ribose phosphate pyrophosphokinase. PRPP is an essential precursor in the biosynthesis of all nucleotides. Ribose phosphate pyrophosphokinase is inhibited by both ADP and GDP nucleotides. The committed step in purine nucleotide synthesis is catalyzed by the enzyme amidophosphoribosyl transferase, which uses glutamine and PRPP as substrates. This enzyme is inhibited by AMP and GMP and is activated by high concentrations of PRPP. An intermediate in purine biosynthesis is inosine monophosphate (IMP). The conversion of IMP to AMP is inhibited by AMP, and the conversion of IMP to GMP is inhibited by GMP. An essential precursor in pyrimidine biosynthesis is carbamoyl phosphate, which is generated by the enzyme carbamoyl phosphate synthase. This enzyme is inhibited by UTP and activated by ATP and PRPP. The production of CTP from UTP is inhibited by CTP. These reactions are summarized in Figure 1. Synthesis of which of the following are subject to end-product inhibition? A. AMP, GMP, and CTP only B. AMP, GMP, CTP, and UTP C. AMP and GMP only D. UTP and CTP only

B. The diagram shows that AMP, GMP, and CTP all exert end-product (feedback) inhibition, and the passage states that UTP inhibits carbamoyl phosphate synthase, a necessary enzyme in the pathway to UTP synthesis.

Sodium laurel sulfate, a common ingredient in hand soap, functions due to the combination of its hydrophilic head (pKa = 1.9) and hydrophobic tail. This amphipathic nature allows for the removal of hydrophobic substances with water. Which of the following would most likely increase the effectiveness of sodium laurel sulfate? A. Benzene B. Dilute aqueous sodium bicarbonate C. Aqueous phosphoric acid (pH = 1.9) D. Lemon juice

B. The head of sodium laurel sulfate (SLS) is a weak acid. It is most hydrophilic when deprotonated and charged. This head can be protonated in lower pH solutions and become less effective. Therefore, addition of a hydrophilic weak base like sodium bicarbonate (NaHCO3) would favor the deprotonated, active form of SLS. Choices C and D are both acidic and eliminated. Benzene is a hydrophobic compound with essentially non-acidic protons (pKa of about 43), eliminating choice A.

Compared to TTXS Na+Cs, TTXR Na+Cs have which of the following characteristics? A. Open at a lower membrane potential threshold value B. A lower micromolar affinity for tetrodotoxin C. Expression in dorsal root ganglion neurons of older mammals D. More substantive control of the gastrointestinal system

B. The passage says that tetrodotoxin binds to TTXS Na+Cs with a binding affinity of 5 to 15 nM. These channels are sensitive to TTX. The TTX-resistant channels will not be affected by TTX, and will probably have a lower affinity for this neurotoxin (choice B is correct). Opening at a lower membrane potential threshold value would mean sodium channels are more sensitive and would open earlier than in the normal situation. There is no information in the passage to support that this would occur (choice A is wrong). If you compare the two rows of data on dorsal root ganglion neurons in Table 1, you will see that young rats (3 days old) express more TTX-resistant channels and older rats (10 days) express more sensitive channels (choice C is wrong). The table also tells you that the enteric nerve sensory fibers (which control the digestive system) express equal amounts of the TTX-sensitive and the TTX-resistant sodium channels (choice D is wrong).

"Tetrodotoxin binds to the fast voltage-gated sodium channel at the extracellular pore opening of the ion channel." Tetrodotoxin most likely has which of the following effects on the neuron? A. A lower than normal resting membrane potential B. Limited ability to propagate an action potential C. Introduction of a plateau period after depolarization D. A shorter hyperpolarization phase due to less efficient repolarization

B. The passage says that tetrodotoxin binds to the fast voltage-gated sodium channel at the extracellular pore opening of the ion channel. The voltage-gated sodium channel is important in depolarization (choice B is correct). The sodium-potassium ATPase and potassium leak channels maintain the resting membrane potential (choice A is wrong), and voltage-gated potassium channels repolarize the cell after depolarization. If they are working properly and are not affected by TTX, there is no reason that a plateau period would be observed (choice C is wrong) and no reason for repolarization to be altered (choice D is wrong).

Which of the following offers the best explanation for why Line 1-2 has a steeper slope than Line 4-5? (Temp vs volume graph with 1-2 being in liquid phase and line 2-4 being flat and 4-5 being less steep and in the vapor phase). A. The heat of fusion for water is greater than the heat of vaporization. B. The entropy for liquid water is greater than the entropy for water vapor. C. Liquid phases are less compressible than gas phases. D. The universal constant R is smaller for liquids, thus resulting in a higher value of T/V in the ideal gas law.

Both have change of temp w respect to volume in liquid water and water vapor. A is wrong bc 1-2 and 4-5 are not phase changes (2-4 is a phase change). B is wrong because entropy is greater. D is wrong because R is only used for gases. C is correct because liquids do not exhibit a large change in volume when temp changes but gases do.

The hepatitis B virus (HBV) is a member of the hepadnavirus family and is found worldwide as a cause of chronic liver disease and hepatocellular cancer. The fully infectious virus, also referred to as the Dane particle, has a partially double-stranded DNA genome attached at one end to a viral polymerase. It is contained in an icosahedral nucleocapsid core tucked in an outer lipid envelope containing multiple membrane surface proteins. The virion carries four genes denoted by C, X, P, and S. Gene C encodes the core protein used to form the viral structure. Gene P encodes the polymerase. Gene S encodes the surface antigen which is used to recognize and enter liver cells. It has three in-frame start codons referred to as pre-S1, pre-S2, and S. Gene X has an unknown function. Presence of the surface antigen indicates that an individual is capable of transmitting the disease. The immune response to HBV is complex and can cause confusion for the clinician. Upon viral entry to the body by blood or mucosal contact, the first antigen recognized is the surface antigen (HBsAg) which can trigger a humoral response (production of anti-sAg). As hepatocyte infection occurs, viral replication leads to the eventual production of the corresponding antibody (anti-core). IgM is the transient first-line antibody, replaced by IgG for lasting defense (as in long-term exposure or vaccination). A third antigen, referred to as the e antigen (HBeAg), is a splice variant of the core antigen. It serves as a marker of active replication and infectivity and, like other antigens, provokes a humoral response (anti-eAg). Typically, acute HBV is cleared from the body within 6 months from infection. As a defense to host immunity, hepatitis B produces non-infectious spherical and filamentous bodies lacking core and genetic components. They are composed only of lipid and protein and mimic the outermost layer of the Dane particle. These are produced in excess and can reach concentrations of up to 1013 particles/mL (300 mg/mL) which is far in excess of corresponding antibody concentrations. In long term HBV infection, failure of an effective humoral response against HBsAg results in ongoing, often asymptomatic infection. However, these individuals remain capable of transmitting the disease and are known as "chronic carriers." An interesting phenomenon occurs with a related (-)RNA virus, hepatitis D (HDV). It is missing the HBV equivalent of gene S. By itself, HDV is harmless, but when paired with HBV, sharing of the surface antigen occurs and a superinfection results, with much more extensive disease than with HBV alone. If a filamentous body were isolated from an infected individual and injected into an uninfected healthy individual what antibodies would be formed? A. Anti-core IgM B. Anti-eAg IgG C. Anti-sAg IgM D. Anti-sAg IgG

C A filamentous body is described in the passage as lacking core and genetic elements, thus anti-core antibodies would not be formed (choice A is wrong). The passage further states that IgM is the first line antibody and is replaced by IgG for long term protection. Since the filamentous bodies are being injected into an uninfected individual, the first antibody to be made will be IgM (choices B and D are wrong and choice C is correct).

While dancing, a ballet dancer uses many principles of physics in order to balance, demonstrate flexibility and strength, and perform advanced turns. One of the primary skills of a ballet dancer is superior balance. Dancers must subtly shift their center of mass in order to maintain perfect and seemingly effortless balance. Female dancers often dance en pointe, which means raising the body and balancing on the tips of the toes in specially designed pointe shoes. When dancing en pointe, the dancer is required to balance her entire body weight on a space that is usually less than 15 cm2, while a flat foot is typically about 180 cm2. In order to demonstrate flexibility and strength, dancers often extend their legs into the air in a développé. A développé involves standing on one leg (the standing leg) while the other leg (the working leg) first bends at the knee to raise the foot along the standing leg, then extends straight into the air. The final position requires holding the working leg straight and as high into the air as possible while the standing leg remains straight and the dancer maintains balance. To perform a développé well, the dancer must have flexibility in the working leg so the angle between the legs is as large as possible, and strength in both hips in order to hold the working leg still in the air. To maintain equilibrium, the hips must be strong enough to counter the torque created from the force of gravity. Figure 1 Final body position for développé Another difficult skill performed by ballet dancers is a pirouette, which is a turn performed on one leg. While standing on one leg (the standing leg), the dancer bends the knee of the other leg (the rotating leg) until the foot reaches the knee of the standing leg. Using the muscles in the rotating leg, the dancer creates torque in the rotating leg's hip joint, which causes the dancer to start spinning. To stop spinning, the dancer uses the muscles in the standing leg to create torque in the standing leg's hip joint, which causes the dancer to stop spinning. Often, a pirouette is performed while remaining balanced en pointe. In order to make it easier to learn to pirouette, dancers almost always hold their arms in the shape of a circle in front of their bodies, but more experienced dancers can pirouette with their arms in any position. Which best describes why it is easier to balance on a flat foot than en pointe? A. Since the area in contact with the floor is greater on a flat foot, there is more torque in the ankle. B. Since the area in contact with the floor is smaller en pointe, there is more torque in the ankle. C. Since the area in contact with the floor is greater on a flat foot, the center of mass can be in a wider range of positions. D. Since the area in contact with the floor is smaller en pointe, the center of mass can be in a wider range of positions.

C According to the passage, dancers manipulate their center of mass to balance. The torque in the ankle does not contribute to balance, eliminating choices A and B. The location of the center of mass of the dancer is the location where her mass acts as a single point. If this location is above an area that is in contact with the ground, she will be balanced. Since there is a significantly greater area in contact with the ground on a flat foot, there are a greater number of locations where the center of mass can be located and the dancer will still be balanced. Choice D is false. The correct answer is choice C.

GAPDH converts an aldehyde into a carboxylic acid in order to change NAD+ to NADH. Which of the following is true regarding this reaction in humans? A. The aldehyde is reduced. B. NAD+ is oxidized. C. NAD+ gains 2 e- and one proton. D. The carboxylic acid (pKa = 2.19) is mostly in its protonated form.

C C. An aldehyde (—CHO) gains an O when it is converted to a carboxylic acid (-COOH). By definition, gaining an O is oxidation which makes choice A incorrect. Choice B is incorrect because NAD+ gains elections in order to become a neutral molecule. Gaining electrons is reduction, not oxidation. Choice D is incorrect because according to the Henderson-Hasselbalch equation (pKa - pH = -log([A-] / [HA])), the conjugate base (deprotonated form) is favored if pH > pKa. Since the human body (with the exception of the gastrointestinal tract) has a pH higher than 2.19, the deprotonated form of the carboxylic acid is favored. Recalling that an H+ is equivalent to a proton, it is apparent that an H+ is added when converting NAD+ → NADH. Since NADH is a neutral molecule, the positive +1 charge from the proton and the +1 charge from the NAD+ must be balanced by two electrons.

Cooperative binding is a special case of allosteric regulation. It occurs when the affinity for a ligand changes depending on the amount of ligand already bound. This stipulation requires the molecule to have multiple binding sites and quaternary structure. Depending on the interactions between the active sites, the macromolecule could exhibit positive cooperativity, negative cooperativity, or non-cooperativity. Positive cooperativity occurs when the binding of one ligand increases the affinity for other ligands. For instance, hemoglobin (Hb) is made up of four subunits. Each deoxyhemoglobin unit has a rather low affinity for oxygen, but once one of the subunits binds with molecular oxygen, the hemoglobin changes from a "tense" to "relaxed" conformation. In the latter state, hemoglobin has a higher affinity for oxygen. Certain conditions such as lowered pH alter this equilibrium in the body by promoting the stability of the tense state of hemoglobin. Negative cooperativity occurs when the binding of one ligand decreases the affinity for other ligands. The enzyme glyceraldehyde-3-phosphate dehydrogenase (GAPDH) and its substrate nicotinamide adenine dinucleotide (NAD+) exhibit this type of relationship. Involved in the sixth step of glycolysis, GAPDH has four possible active sites. The dissociation constants (K′) for each site are listed below in Table 1. Which of the following could affect any of the dissociation constants for hemoglobin? Decreasing the temperature Adding NH3 Increasing the [O2] A. I only B. III only C. I and II only D. I, II, and III

C C. K′ is constant for any equilibrium at a certain temperature, and changing the temperature altersK′. Therefore, Item I is correct and choice B can be eliminated. According to the passage, lowering the pH alters the dissociation equilibrium of hemoglobin, preferentially stabilizes its tense form, and lowers its affinity for O2. This indicates that the hemoglobin-oxygen dissociation equilibrium as a whole and its dissociation constant are pH dependent. Ammonia, NH3, is a weak base and will affect the pH of the system and alter K′. Item II is correct and choice A can be eliminated. Adding [O2] will temporarily disturb the system's equilibrium, but the system will return to equilibrium according to Le Châtelier's principle until Q = K′. K′ will remain constant with the addition of [O2] which makes Item III false and choice C the correct answer.

What would be the effect over-expressing G0S2 in an adipocyte? A. Elevated ATGL activity B. Limited gluconeogenesis C. Massive lipid accumulation D. Stimulation of protein catabolism

C C. The passage states that G0S2 directly interacts with ATGL, and that expression of G0S2 correlates with insulin levels. Insulin is released when serum glucose levels are high, or when humans are in a recently fed state; lipolysis would be low in this situation, meaning G0S2 likely inhibits ATGL and lipolysis (choice A is wrong). Over-expression of G0S2 would therefore inhibit lipid breakdown and cause massive lipid accumulation (choice C is correct). There is no information in the passage to support the fact that G0S2 affects gluconeogenesis or protein catabolism (choices B and D are wrong). *since AGTL breaks down lipids, it's activity would not be elevated.

Which of the following statements regarding creatinine is LEAST likely to be true? A. The amount of creatinine produced per day will be relatively high in bodybuilders. B. Allosteric regulation does not affect the formation of creatinine to creatine. C. Vegetarians will produce less creatinine than meat-eaters. D. In general, males will produce more creatinine than females.

C C. The passage states that the amount of creatinine produced is related to muscle mass. Creatinine results from the dephosphorylation of creatine phosphate to regenerate ATP; it is safe to assume that individuals with more muscle require more ATP, and are thus likely to produce more creatinine. Bodybuilders typically have more muscle than the average person (choice A is likely to be true and can be eliminated), and males typically have greater muscle mass than females (choice D is likely to be true and can be eliminated). Formation of creatinine occurs non-enzymatically, so it will not be affected by allosteric regulation (choice B is true and can be eliminated). However, although the passage states that vegetarians rely solely on internal synthesis for the formation of creatine, it does not make any statements or assumptions regarding the formation of creatinine. Further, it cannot be assumed that the levels of creatine (and thus indirectly creatinine) are low just because vegetarians rely on internal synthesis (choice C is least likely to be true and is the correct answer choice).

At pH values just below 4, Mito-pH was no longer found to specifically accumulate in the mitochondria. Which of the following is the most likely cause for this phenomenon? A. At pH levels below 4 the amide bond in the organic linker is deprotonated, giving excess negative charge to the compound. B. At pH levels below 4 the hydroxyl groups on the spirolactone moiety are protonated. C. The cyanine group can be protonated in solutions with pH less than 4. D. At pH values below 4 the spirolactone moiety ring-opens to the carboxylate form.

C Choice A is false, since deprotonation occurs at high pH (more basic) rather than low pH. The passage states that the cyanine portion of the molecule, rather than the spirolactone portion, is responsible for accumulation in the mitochondria (eliminate choice B). Beyond this, the pKa of a protonated phenol is -6.7, far below a pH of 4, meaning no protonation would be expected at this pH. Finally, the spirolactone opens at elevated pH levels (eliminate choice D). The passage states that the specific accumulation of Mito-pH is a result of the cyanine group. This moiety contains a tertiary amine group that may be protonated around pH 4, making choice C the correct answer.

Creatine is one of the eight naturally occurring guanidine-derived compounds, and is synthesized from three amino acids: arginine, glycine, and methionine (Figure 1). The best natural sources of creatine are meat, fish, and other animal products, and the average creatine intake from dietary sources is estimated to be 1 gram per day. As plants are very low in creatine, vegetarians rely solely on internal creatine synthesis. In humans, over 95% of the total creatine content is located in skeletal muscle; approximately one-third of this is free creatine, and the remainder is phosphorylated. Creatinine is formed in muscle from creatine phosphate by a nonenzymatic dehydration and loss of phosphate. The amount produced is related to muscle mass and remains remarkably constant from day to day. Figure 1 Biosynthesis of creatine Creatine supplementation has become mainstream in athletic training and competition. During exercise, ATP is hydrolyzed to form ADP. As energy demands increase and ATP is depleted, the body begins to utilize creatine-phosphate stores; these are broken down to produce creatine and a high-energy phosphate group, which is used to reform ATP. If creatine stores are low, fatigue sets in much faster and exercise intensity is reduced. Creatine is rephosphorylated during periods of recovery. This is dependent on oxygen supplies, and muscles can typically restore 50% of their creatine phosphate levels in about one minute. There are three types of creatine supplements: creatine monohydrate, creatine phosphate, and creatine citrate. Creatine monohydrate is the most common and is simply creatine bound with water. Each molecule of creatine monohydrate is made up of 88% creatine and 12% water by mass. Creatine phosphate is more expensive than creatine monohydrate and has never shown to be more effective. In addition, creatine phosphate is only 62.3% creatine and 37.7% phosphate. Creatine citrate became popular because it is more water soluble than other forms of creatine and dissolves better, but it is only 40% creatine. When taking creatine supplements, one must also take into consideration the absorption of creatine into the muscle cells. Studies have shown that consumption of simple carbohydrates at the time of supplementation improves uptake, whereas consumption of protein has the opposite effect. A study was conducted on six subjects to determine the effects of creatine supplementation on muscle creatine levels. Quadriceps muscle biopsies were taken before and after three months of supplementation (with 20 grams of the respective creatine types), and the creatine concentrations were analyzed. The following table summarizes the results from the study: Question 40 Crown ethers (18-crown-6 shown below) are a family of compounds often used to aid in the crystallization of compounds bearing amine entities from cold, ether-type solvents. How do these compounds aid crystallization? A. Crown ethers greatly decrease the polarity of the solvent, causing crystallization of the polar amine. B. Crown ethers are not miscible in ether solvents and act as inhomogeneous sites for crystal formation. C. The crown ether saturates the hydrogen bonding capabilities of the amine, decreasing its interaction with the solvent. D. The crown ether hydrogen bonds with solvent molecules, preventing them from interacting with amines in solution.

C Crown ether molecules are ether-type compounds. As such they don't significantly alter the polarity of an ether solvent, will be miscible with it, and contain no protic hydrogens to hydrogen bond with the solvent, eliminating choices A, B, and D. However, since one molecule can efficiently hydrogen bond all the protic hydrogens of an amine, crown ethers shield the amine from hydrogen bonding interactions with the solvent, decreasing the amine's solubility.

One common medical tool is the stethoscope, often used to listen to sounds made inside the body. Most stethoscopes consist of a two-sided chestpiece, a hollow tube, and earpieces. The chestpiece usually has a diaphragm and a bell, the latter for transmitting low frequency sounds (20 to 80 Hz) and the former for high frequency (80 to 200 Hz). When the listener wishes to hear sounds of a given frequency range, he or she presses the appropriate side of the chestpiece to the part of the body that is creating the sound. Traditional acoustic stethoscopes produce only low levels of sound in the earpieces. As a result, many current stethoscopes are now electronic, rather than acoustic: they use electronic technology to amplify all the sound picked up by the stethoscope. Such stethoscopes allow quiet sounds in the body to be heard more clearly than otherwise would be possible. Many medically relevant sounds in the body can be detected by a doctor with a stethoscope. For example, the bell side of the stethoscope can allow a doctor to listen for heart murmurs, caused by abnormally turbulent blood flow that creates audible vibrations in the heart or arteries. Also, using the diaphragm side of a stethoscope to listen to breathing can allow a doctor to detect crackling or popping in the lungs, which can be caused by fluid in the lungs and may be an indicator of pneumonia. Why would a doctor use the bell to detect heart murmurs but the diaphragm to detect pneumonia? A. Artery vibration is usually much quieter than breathing abnormalities, and the bell helps detect the lower-energy sounds. B. Artery vibration is usually much louder than breathing abnormalities, and the diaphragm helps detect the lower-energy sounds. C. Turbulent blood flow usually creates low frequency sounds, but crackling in the lungs is usually high frequency. D. Turbulent blood flow usually creates high frequency sounds, but crackling in the lungs is usually low frequency.

C Loudness does not correspond w energy and frequency so A and B can be eliminated.

In modern radiotherapy, superconducting wire coils are used to create magnetic fields that focus particle beams in order to deliver the maximum dose of ionizing radiation to a tumor site while minimizing the destruction of healthy tissue. For which of the following types of radiation would this technique NOT be useful? A. Alpha particles B. Proton beams C. Gamma photons D. Beta particles

C Magnetic fields exert effects upon moving charged particles and photons have no charge so are unaffected by magnetic fields.

Bile is a greenish fluid produced in the liver that helps dispose of the liver's waste products and aids in the digestion of fats. It is stored in the gallbladder and emptied into the small intestine via the common bile duct when needed. The principle components of bile are bile acids, cholesterol, and bilirubin. The liver enzymatically converts cholesterol (1) into one of two primary bile acids, cholic acid (2) or chenodeoxycholic acid (3). These acids are then coupled with glycine, an amino acid, or taurine, one of the few known naturally occurring sulfonic acids. Taurocholic acid (4) is readily converted into its salt in the duodenum upon mixing with pancreatic secretions containing lipase and bicarbonate. Bile acids are amphipathic. It is this property that allows them to emulsify fat globules into microscopic micelles, increasing fat surface area and aiding digestion by lipase. Bacteria in the colon convert Compounds 2 and 3 into their respective secondary bile acids 5 and 6, shown below. Bile salts are not passively absorbed in the small intestine, but secondary bile acids are actively absorbed in the colon where the pH is less basic. Up to 95% of the bile acids produced by the liver are reabsorbed and can be used in the digestive process up to twenty times. Which of the following contributes to the fact that the small intestine cannot passively absorb bile salts? I. Taurine and glycine have low pKa values. II. Bile salts are hydrophobic. III. Pancreatic secretions increase intestinal pH. A. I only B. I and II only C. I and III only D. II and III only

C Salts are charged, making them water soluble or hydrophilic; Item II is false (choices B and D can be eliminated). By looking at the remaining choices, Item I must be true, so focus on Item III to answer the question quickly. The passage states that pancreatic secretions contain bicarbonate, a base, which raises the pH of the small intestine. High pH means a high concentration of OH- is present to deprotonate the bile acid, yielding its charged conjugate base. Since Item III is a true statement, choice C is the best answer. To address Item I, the pKa of the carboxyl group of an amino acid like glycine is about 2 (and the sulfonic acid of taurine is even more acidic), while the pH of the small intestine is greater than 7. When the pKa of a functional group is smaller than the environmental pH, the group will exist in its deprotonated or charged form. This confirms Item I as true. Because passage says "these acids are coupled with glycine or taurine... taurocholic acid is readily converted into its salt in the duodenum upon mixing with pancreatic secretions..." so these are negatively charged in basic environment because they have low pKa meaning are able to become a bile salt.

At a pH of 9.5, tyrosine is strongly attracted to the positive end of an electrophoresis gel, whereas lysine moves a very small distance from the origin. What is the best explanation for this observation? A. The pI of lysine is greater than tyrosine because the ammonium side chain in lysine is more acidic. B. At high pH, the R group in tyrosine is uncharged whereas the R group in lysine is negatively charged. C. The phenol side chain in tyrosine is acidic, whereas the amino side chain in lysine is basic. D. The oxygen in tyrosine's side chain is more electronegative than the nitrogen in lysine's side chain.

C Since the pKa values of all the groups are nearly identical, it's the different side chains that determine the differences in charge between the molecules. At pH = 9.5 tyrosine's side chain is uncharged while lysine's is positively charged (eliminate choice B). Choice A is also false; even though lysine has the greater pI near 9.5, it can't be due to a more acidic side chain as a higher pKa means a less acidic group. Phenols are weak acids, as indicated by a pKa of 10.1. However, the pKa of lysine's side chain is for the conjugate acid of the basic free amino group, making choice C the best answer. Electronegativity and polarity are not deciding factors in the migrating abilities of amino acids in an electrophoresis experiment (eliminate choice D).

Within a single biological system, the pH within certain organs can vary by a few pH units, or orders of magnitude in terms of actual hydrogen ion concentration. Even within the cell, individual organelles maintain different pH levels, optimized for or created by their functional roles in the cell. Mitochondria are known to function optimally at a slightly basic pH level (~8.0). Deviations from this value may be indicative of malfunctions within the organelle, and strategies have been sought to identify a biologically safe indicator capable of preferential accumulation in the mitochondria and presenting spectroscopic markers that change according to mitochondrial pH within the range of interest. One such recently reported compound, dubbed Mito-pH, is the product of linking two photoactive fluorophores, one of which is pH sensitive and the other pH insensitive, into one compound. Both Mito-pH and the reaction responsible for its pH sensitivity are shown in Figure 1. The mitochondrial targeting ability of the molecule is driven by the cyanine moiety, while the pH sensitivity results from the reversible ring opening of the spirolactone moiety. The ring-closed spirolactone moiety shows no absorbance or emission; however, when opened the unit absorbs light with a wavelength of 490 nm and emits at 520 nm. The cyanine moiety absorbs at 560 nm and emits at 600 nm. The changes observed in the respective emission with changing pH are shown in Figure 2a and 2b. Figure 1 Structure and pH sensitivity of Mito-pH Since the cyanine group is largely unaltered by pH changes in the range studied (4.85 - 9.56), the ratio of ring-closed vs. ring-opened Mito-pH can be determined by scaling the normalized ratios of the emissions at 520 nm and 600 nm at any given pH. This value (normalized Em(520)/Em(600)) represents the ratio of ring-opened to ring-closed forms of Mito-pH. Figure 2c shows a plot of this ratio against pH. (Figure c shows pI at ~7) Figure 2 a) Plot of the emission intensity of a 10 μM solution of Mito-pH at 520 nm from pH 4.85 to 9.65 b) Plot of the emission intensity of a 10 μM solution of Mito-pH at 600 nm from pH 4.85 to 9.65 c) Plot of the ratios of the normalized emission values of Mito-pH at 520 and 600 nm across a range of pH values (see text for description) Which of the following values best approximates the pKa of Mito-pH? A. 5.0 B. 6.1 C. 7.3 D. 8.7

C The pKa of any system is equal to the pH where exactly half of the compound in solution is in the protonated form and half is in the deprotonated form. In the case of Mito-pH, this is when half is ring-opened and half is ring-closed. Figure 2c indicates that the normalized ratio of Em(520):Em(600), which represents the ratio of ring-opened to ring closed, is 0.5 at 7.3, making choice C correct.

Acetaminophen is one of the most commonly used drugs in the United States. It inhibits both isoforms of the cyclooxygenase enzyme (COX-1 and COX-2) and is used primarily for its analgesic and antipyretic properties. In massive overdoses, acetaminophen can cause fulminant hepatitis as well as necrosis of hepatocytes. With normal dosing, however, most of the drug undergoes direct glucuronidation and sulfation reactions, which produce metabolites that can be excreted in the urine. Approximately 10% of acetaminophen is oxidized by P-450 isoenzymes 2E1 and 1A2 in hepatocytes to form N-acetyl-p-benzoquinone imine (NAPQI). Once generated, NAPQI can bind vital proteins in the cell leading to cell death. Alternatively, NAPQI can bond to glutathione (GSH) to form a mercapturic acid, another stable metabolite excreted in the urine. Figure 1 Acetaminophen metabolism When glutathione in the cell is depleted, however, NAPQI forms additional toxic protein adducts, accelerating the rate of cell death. One common cause of GSH depletion is chronic alcohol consumption, which not only induces the action of CYP2E1 but also utilizes GSH, as shown in Figure 2. Though glutathione reductase can convert glutathione disulfide (GSSG) back into GSH, the rate at which it does so may be insufficient to keep up with the rate of acetaminophen metabolism. Figure 2 The role of glutathione in ethanol metabolism (MEOS = microsomal ethanol oxidizing system; SOD = superoxide dismutase) Acetaminophen is commonly administered through three different routes: intravenous (IV), oral, and rectal. Figure 3 indicates how the bioavailability and rate of absorption of the drug differs for the three routes over time. Based on information provided in the passage, which of the following statements describes a result of administration of acetaminophen to a patient with a history of chronic alcohol consumption? A. Plasma NAPQI increases and GSH decreases. B. Plasma NAPQI increases and GSH increases. C. Hepatocyte NAPQI increases and GSH decreases. D. Hepatocyte NAPQI increases and GSH increases

C The passage states that these reactions occur in hepatocytes (choices A and B can be eliminated). Chronic alcohol consumption induces CYPE21 (increasing the rate of NAPQI generation with administration of acetaminophen) while also depleting GSH stores. Higher levels of NAPQI will, in turn, consume even more GSH, depleting levels further (choice C is correct and choice D is wrong).

A woman suffers a hemorrhage into her pituitary gland, destroying only the portion of the anterior pituitary responsible for producing luteinizing hormone. What will be the effect of this destruction on her menstrual cycle? A. Her menstrual cycle will not be affected by the destruction. B. She will still ovulate and release estrogen, but not progesterone. C. She will become anovulatory. D. She will still ovulate and release progesterone, but not estrogen.

C. If the portion of the anterior pituitary that produces LH is destroyed, then LH levels will not surge during the menstrual cycle and ovulation would not occur (choice C is correct and choices B and D are wrong). In the absence of ovulation, the corpus luteum would not develop either, although if FSH is unaffected, the follicle would most likely develop and release estrogen. Regardless, the menstrual cycle would be affected (choice A is wrong).

Certain lung cancers can secrete PTHrP (parathyroid hormone related-protein), which, while distinct from PTH, binds and activates PTH receptors. Which of the following laboratory findings would be consistent with a patient that had a PTHrP-secreting tumor? A. ↓ Serum [calcium], ↑ PTH, ↓ risk of nephrolithiasis B. ↑ Serum [calcium], ↑ PTH, ↑ risk of nephrolithiasis C. ↑ Serum [calcium], ↓ PTH, ↑ risk of nephrolithiasis D. ↓ Serum [calcium], ↓ PTH, ↑ risk of nephrolithiasis

C. Parathyroid hormone (PTH) is usually secreted by the four small parathyroid glands located posterior to the thyroid and functions to increase serum calcium levels. PTHrP activates PTH receptors and therefore would also cause increased serum calcium levels (choices A and D are can be eliminated). The increase in serum calcium would feedback to the parathyroid gland to reduce the secretion of PTH (choice B can be eliminated and choice C is correct). Note that these patients would be at risk for nephrolithiasis; the increased serum calcium would lead to a greater excretion of calcium in the urine (hypercalciuria).

What famous social psychology experiment was designed to study in-group favoritism? A. Solomon Asch's conformity experiments B. Stanley Milgram's shock experiments C. The Robbers Cave experiment D. The Tuskegee syphilis experiment

C. The Robber's Cave experiment showed that even arbitrary group distinctions (camp teams) can cause bitter rivalry and discrimination, thus demonstrating in-group/out-group biases (choice C is correct). Solomon Asch's conformity experiments looked at the likelihood of conformity in a group setting (choice A is wrong) and Stanley Milgram's shock experiments analyzed obedience and the power of authority figures (choice B is wrong). The Tuskegee syphilis experiment was an extremely unethical medical experiment (not social psychology experiment) that demonstrated a fair amount of race bias on the part of the experimenters toward the participants, but was not attempting to study in-group/out-group bias (choice D is wrong).

Suppose the researchers want to examine the effect of motivation on older adult's formation of new LTM. To measure this, they follow up with the older adults two weeks after the study and ask them to recall details from a text read during the initial phase of the experiment (the text was standardized for all groups in the experiment, and was unrelated to the life and media problems). If the researchers subscribe to the depth-of-processing model, they would hypothesize that: A. presentation of a media problem, then analysis of the text will yield the best performance on the recall task. B. rehearsal of the text, then presentation of a life problem will yield the best performance on the recall task C. presentation of a life problem, then analysis of the text will yield the best performance on the recall task. D. presentation of a life problem, then rehearsal of the text will yield the best performance on the recall task.

C. The depth-of-processing model holds that information is transferred from short-term memory (STM) to long-term memory (LTM) when it is processed at deeper levels of analysis. The dual-store model, on the other hand, holds that information is moved from STM to LTM when it is maintained in STM for a sufficiently long period of time via rehearsal. Therefore, proponents of the depth-of-processing model would utilize an analytical task rather than mere rehearsal (choices B and D are wrong). Logically, for the motivational factor to enhance any cognitive capacity, the motivational trigger would have to precede the cognitive task. Accordingly, the life problem would need to precede an analytical task to produce the greatest effect (choice C is correct; choice A is wrong).

"To understand changes in enzyme levels in patients with elevated iron levels, gene expression in Hfe-/- mice was monitored by RT-PCR. Carnitine palmitoyl transferase was significantly elevated, as were levels of PDH kinase 4 (Pdk4), which phosphorylates and inhibits PDH. PDH levels were also significantly decreased. Lastly, fatty acid oxidation was measured in mice that were fed high fat diets, by the method described previously. Oxygen consumption and heat production were also monitored in the same mouse models to further confirm changes in beta-oxidation. Multiple strains (129 and C57) were tested for control purposes. The results of these experiments are shown in Figure 2." Given the role of Pdk4, is its elevation in the mice induced to express HH symptoms consistent with the decreased glucose oxidation seen in diabetic patients? A. No; PDH is needed for gluconeogenesis, and Pdk4 levels should be decreased in diabetic patients. B. No; PDH activity should be decreased in diabetic patients due to decreased intracellular glucose concentration. C. Yes; high Pdk4 levels will limit glucose oxidation in diabetic patients, who have decreased rates of cellular glucose uptake. D. Yes; elevated Pdk4 levels will increase acetyl-CoA production, leading to increased production of ketone bodies, as seen in diabetic patients.

C. The passage states that Pdk4 inhibits PDH, which converts pyruvate to acetyl-CoA following glycolysis. Elevated levels of Pdk4 will thereby decrease PDH activity, resulting in decreased acetyl-CoA production. Decreased acetyl-CoA production will decrease the rate of glucose oxidation in the Hfe-/- knockout mice. Similarly glucose oxidation rates are low in diabetic patients due to limited entry of glucose into body cells (choice C is correct). PDH is used to convert pyruvate into acetyl-CoA, and not utilized in gluconeogenesis (choice A is wrong). PDH activity is expected to be low in diabeticpatients, which is consistent with the elevated Pdk4 levels in the Hfe-/- knockout mice (choice B is wrong). Increased Pdk4 levels will decrease, not increase acetyl-CoA production (choice D is wrong).

The process by which the autonomic nervous system activates HCN channels is best described by which of the following? A. Sympathetic stimulation causes acetylcholine release, thereby activating cardiac tyrosine kinase receptors, resulting in elevated cAMP levels. B. Parasympathetic stimulation causes acetylcholine release, thereby activating tyrosine kinase receptors, resulting in elevated cAMP levels. C. Sympathetic stimulation causes epinephrine release, thereby activating G-protein coupled receptors, resulting in elevated cAMP levels. D. Parasympathetic stimulation causes epinephrine release, thereby activating G-protein coupled receptors, resulting in elevated cAMP levels.

C. The passage states that cAMP binding opens and activates HCN channels, causing increased depolarization of the heart, and therefore should increase heart rate. The sympathetic nervous system is responsible for increasing heart rate, while the parasympathetic nervous system is responsible for slowing it (choices B and D are wrong). Activation of the sympathetic nervous system results in the release of norepinephrine from post-ganglionic neurons and epinephrine from the adrenal medulla (choice A is wrong). Epinephrine acts on G-protein coupled receptors, which activate adenylate cyclase to convert ATP to cAMP (choice C is correct).

Which of the following ranks the redox-active species of the electron transport chain in order of decreasing electron affinity? A. O2 > FAD > CoQ > NAD+ B. NAD+ > FAD > CoQ > O2 C. O2 > CoQ > NAD+ > FAD D. O2 > CoQ > FAD > NAD+

D Because O2 is the terminal species in the electron transport chain (ETC), it must have the highest electron affinity and thus be first in the list (eliminating choice B). Since NADH donates its electrons to the first electron carrier in the ETC, the product of its oxidation, NAD+, would have the lowest electron affinity and would be last in the list (eliminating choice C). FADH2 enters the ETC by donating its electrons directly to ubiquinone (CoQ) so the oxidized FAD will have a lower electron affinity than CoQ (eliminating choice A).

An object is placed in front of a concave mirror between the focal point and the mirror. If the object is then moved closer to the mirror, the image would be: A. larger and farther from the mirror. B. larger and closer to the mirror. C. smaller and farther from the mirror. D. smaller and closer to the mirror.

D Conceptually, an object placed directly at the focal point would produce no image (i.e., an image infinitely far away). Therefore it makes sense that moving away from the focal point would produce an image closer to the mirror. This would eliminate choices A and B. Also, when an object is placed between the focal point and the mirror, the image is virtual and larger than the object. As the object moves closer to the mirror, the image would get smaller and eventually become the same size as the object when the object is touching the mirror. The answer is therefore choice D. Using equations, recall that 1 / o + 1 / i = 1 / f, where f is positive for a concave mirror and remains constant. If o decreases, the (1 / o) term must increase, which means the (1 / i) term must decrease to compensate. Since i is negative for virtual images, decreasing the (1 / i) term means becoming less negative, which means that the magnitude of i decreases, which corresponds to the image moving closer to the mirror. As far as magnification goes, if we solve the above equation for i, we get i = of / (o - f). Magnification = -i / o = f / (f - o). Since o < f, then decreasing o would increase the denominator (f - o), and thus decrease the magnification. Another option would be to choose two object distances (say o = f / 2 and o = f / 4) and calculate the image distances and magnifications.

Which of the following is a true statement about the diaphragm? A. It contains both skeletal and smooth muscle cells. B. Its effector neurotransmitters are norepinephrine and acetylcholine. C. It is innervated by the phrenic nerve and autonomic nervous system. D. It receives neural signals from the cerebral cortex and the brain stem.

D Diaphragm is purely skeletal muscle so A is wrong. And because it's only skeletal, only Ach is used as neurotransmitter so B is wrong. It is innervated only by the phrenic nerve not the autonomic nerves so C is wrong. The phrenic nerve originates in cerebral cortex for voluntary breathing and in brain stem for involuntary control.

Lipolysis is followed by β-oxidation, in order to ensure the cell can harvest ATP from lipid molecules. Which of the following is true of this process? A. An isomerase and a reductase are required for complete oxidation of a monounsaturated fatty acid. B. Lipolysis of a DAG will generate twice the amount of fatty acids and glycerol, compared to lipolysis of a MAG. C. Unlike cell respiration, fatty acid catabolism starts in the mitochondria and finishes in the cytosol. D. The cell requires more than twice the number of NAD+ electron carriers compared to FAD electron carriers, in order to harvest ATP from fatty acids.

D It is true that cells require more than twice the number of NAD+ electron carriers compared to FAD electron carriers, in order to harvest ATP from fatty acids (choice D is correct). As an example, catabolism of a ten-carbon saturated fatty acid will require four rounds of β-oxidation. This will generate four FADH2, four NADH, and five acetyl-CoA. The acetyl-CoAs will enter the Krebs cycle and will generate 15 more NADH and 5 more FADH2 (as well as 5 GTP). In total then, 19 NADH are generated (requiring 19 NAD+) and 9 FADH2 are generated (requiring 9 FAD). β-oxidation of a monounsaturated fatty acid requires only an isomerase, not a reductase (choice A is wrong). Lipolysis of a DAG will generate two fatty acids and one molecule of glycerol. Lipolysis of a MAG will generate one fatty acid and one molecule of glycerol, so the number of fatty acids generated is doubled, but not the amount of glycerol generated (choice B is wrong). Fatty acid catabolism starts in the cytosol, and is mostly performed in the mitochondrial matrix (choice C is wrong).

N-acetylcysteine (NAC) is often administered for the treatment of acetaminophen overdose to help prevent the toxic metabolite, NAPQI, from accumulating in hepatocytes. This is most likely because NAC: A. increases the production of NAPQI. B. acts as an inhibitor of glutathione reductase. C. acts to stimulate the action of CYP1A2. D. possesses a, sterically unhindered functional group equivalent to glutathione

D NAPQI is designated ready for excretion with the binding of glutathione, which contains a sulfhydryl (-SH) functional group. Since NAC is a derivative of cysteine, it also has a sulfhydryl group in its side chain; this can act as a precursor to glutathione synthesis, thus providing more substrate for the detoxification of the reactive NAPQI (choice D is correct). The accumulation of NAPQI is the opposite of the treatment goal of reducing NAPQI (choice A is wrong). An inhibitor of glutathione reductase would prevent the reduction of GSSG into GSH, which as stated previously, is necessary for the elimination of NAPQI; this would also lead to its accumulation (choice B is wrong). Stimulating CYP1A2 would also lead to increased NAPQI (choice C is wrong).

From one point in space, Point S, to another, Point T, electric potential increases continuously from 100 V to 200 V. Which of the following must be true of the electric field near these points? A. Field lines point away from both S and T. B. Field lines point towards bothS and T. C. Field lines point from S to T. D. Field lines point from T to S

D T is at a higher potential so field lines would point from high potential to low potential since they are the direction a positive charge would move.

"Another study showed that mutations in the sodium-hydrogen exchanger regulatory factor 1 (NHERF1) also leads to phosphaturia. NHERF1 interacts with the C-terminal tail of NPT2a and NPT2c. This interaction leads to regulation of NPT2a trafficking and transcription. Patients with mutations in NHERF1 were found to have lower TmP/GFR values. They were also found to have increased urinary cyclic AMP (cAMP) concentrations. Furthermore, it has been found that NHERF1 also interacts with urate transporter 1 which regulates uric acid transport in the proximal convoluted tubule and, indeed, NHERF1-knockout mice have increased uric acid excretion." Which of the following results would NOT support the hypothesis that disregulation of the NPT2a and/or NPT2c transporters may lead to nephrolithiasis? A. A missense mutation in NPT2a in mice has been shown to impair renal phosphate reabsorption. B. Subjects with osteoporosis associated with hypophosphatemia who had impaired TmP/GFR values were found to have heterozygous mutations in the NPT2a gene. C. Hereditary hypophosphatemic rickets with hypercalciuria has been found to be caused by mutations of NPT2c. D. A mutation in NHERF1 is found to affect a novel phosphate transporter, leading to phosphaturia

D. As stated in the passage, phosphaturia (phosphate in the urine) is associated with nephrolithiasis. A defect in the NPT2a transporter resulting in impaired renal phosphate transport would lead to phosphaturia (since if the phosphate is not reabsorbed it will be in the urine), which may then cause nephrolithiasis (choice A would support the hypothesis and can be eliminated). Likewise, patients with impaired TmP/GFR values and hypophosphatemia would have phosphaturia (since, once again, the phosphate is not being reabsorbed and is therefore in the urine), possibly leading to nephrolithiasis. If these patients are found to have mutations in the NPT2a gene, this supports the hypothesis (choice B can be eliminated). If a mutation in NPT2c has been shown to cause hypophosphatemia and hypercalciuria, this could lead to nephrolithiasis (choice C would support the hypothesis and can be eliminated). As mentioned in the passage, NHERF1 has been shown to affect NPT2a and NPT2c and its level of activity can be associated with renal stone formation. However, if a mutation in NHERF1 has been found to affect a novel phosphate transporter that could lead to phosphaturia, this would NOT support the hypothesis that disregulation of NPT2a and/or NPT2c leads to nephrolithiasis (choice D is the correct answer choice).

In a culture of mammalian skeletal muscle cells, the consumption of oxygen and glucose is measured. Which of the following would occur in response to inhibition of electron transport? A. Oxygen consumption will increase, and glucose consumption will decrease. B. Oxygen consumption will increase, and glucose consumption will increase. C. Oxygen consumption will decrease, and glucose consumption will decrease. D. Oxygen consumption will decrease, and glucose consumption will increase.

D. In the absence of electron transport, oxygen is not needed (it is the final electron acceptor in the transport chain) so the consumption of oxygen would decrease (choices A and B are wrong). Since the electron transport chain is not available to make energy (ATP) the cells will rely solely on anaerobic respiration - glycolysis. Further, since the energy demands of the cells have not changed, and since glycolysis makes fewer ATP than electron transport, the rate of glycolysis will have to increase to keep the level of ATP normal. Thus, glucose consumption would increase (choice D is correct and C is wrong).

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

D. Item I is true: the passage and the diagram both indicate that NADPH is generated in this pathway (choice B can be eliminated). Item II is true: the diagram indicates that ribose-5-phosphate can be converted to fructose-6-phosphate, which is an intermediate in the glycolytic pathway (choices A and C can be eliminated and choice D is correct). Note that Item III is also true: ribose-5-phosphate is indicated by both passage and diagram as a product.

Normative social influence describes: A. the process by which humans learn the norms of their society. B. a process whereby an individual conforms their behavior to match everyone else's behavior because he or she wants to do the right thing and feels that everyone else knows something he or she doesn't know. C. the likelihood that an individual's behavior will be considered "normal" in any given situation. D. a situation where a person's behavior conforms or changes because she or he wants to be liked by others

D. Normative social influence describes when someone changes her or his behavior (conforms) because she or he has a desire for the approval of others and wants to avoid rejection (choice D is correct). Socialization is the term generally used to describe the process by which humans learn the norms of their society (choice A is wrong). Informational social influence describes when an individual's behavior conforms or changes because he or she wants to do the "right thing" (choice B is wrong). The likelihood that an individual's behavior will be considered "normal" in any given situation is not described by normative social influence (choice C is wrong).

"Tetrodotoxin (TTX, Figure 1) is a potent neurotoxin produced by bacteria, often those living in a symbiotic relationship with another organism. TTX has been detected in pufferfish, newts, red alga, crabs, starfish, octopus and many types of worms. Depending on the species, TTX can be used as either a predatory venom or as a defensive biotoxin to ward off predation. Despite its high toxicity, some species have developed resistance to TTX, a phenomenon that is currently under investigation. The toxin can enter the body of a victim by inhalation, ingestion, injection, or through scraped skin. Injection is the most dangerous." TTX-producing bacteria have been found in symbiotic relationships with: A. animals only. B. animals and worms. C. plants, worms and fish. D. animals and at least one other kingdom to eukaryotes.

D. The passage says that TTX-producing bacteria have been found associated with fish, newts, crabs, starfish, octopus and worms; these are all animals (choice B is wrong). Paragraph 1 also lists red alga, which are protists, or unicellular eukaryotes (choices A and C are wrong; choice D is correct).

What does a larger molecule do to gas chromatography retention time?

Increases the time because they are less volatile than smaller molecules.

Does frequency change when wave travels from one medium to another

No.

Could you detect a nondisjunction in all of mom's oocytes?

No. nondisjunction takes place on a cell-to-cell basis during anaphase of meiosis. Primary oocytes are halted in prophase I (before anaphase has taken place) making the observation of nondisjunction in the mother's oocytes highly unlikely

What is fluorescent in situ hybridization of chromosomes? Could it be used to detect a translocation of chromo?

Uses fluorescent DNA probes to bind to chromosomes. If done on chromos, could detect a translocation by detecting a gene on an unexpected chromo.

The Beck Depression Inventory (BDI) is a survey consisting of 21 multiple-choice questions that is the most widely used self-report test for depression. Questions about depressive symptoms such as hopelessness, sleeplessness, fatigue, etc. are scored from 0 (no symptom of depression) to 3 (severe symptom of depression) and totaled. Higher scores indicate more severe depression. If the BDI were administered to the participants in the second study, how would the BDI scores mostly likely correspond with the cortisol levels? A. BDI scores would increase as cortisol levels declined. B. Low BDI scores would correspond to high cortisol levels. C. BDI scores and cortisol levels would be directly correlated. D. BDI scores would be independent of cortisol levels

`C. The Beck Depression Inventory (BDI) is scored such that higher scores correspond to more severe depression. Elevated cortisol levels are directly correlated to decreased brain serotonin and an increase in depression; therefore BDI scores and cortisol levels are directly correlated (as one increases, so does the other; choice C is correct; choices A and B are wrong). Depression and cortisol have been shown by numerous studies to be linked, so BDI scores (a measure of depression) will not be independent of cortisol levels (choice D is wrong).

ostrich effect

an individual avoiding an apparently risky financial situation by pretending it does not exist

unreliable witness effect

associated with eyewitnesses in court cases being unable to accurately and reliably recall the events they witnessed

Catecholamines are:

epinephrine, norepinephrine, dopamine

Harlow experiment findings

extreme social isolation causes long-term and irreversible negative impacts; baby monkeys prefer a "cloth mother" to a "nutrient mother"; and securely attached monkeys are better adjusted than insecurely attached monkeys

Ennui

mental state characterized by lethargy and apathy, often associated with depression

Social reciprocity

responding to a kind or generous action with another kind or generous action

If at same volume, but a higher temp... A. higher pressure. B. lower pressure. C. higher vapor mass. D. lower vapor mass.

A PV=nRT or can think that with more temp, more avg KE, more force against container so more pressure

Besides detoxification of drugs such as acetaminophen, the liver is involved in and regulates several different biochemical pathways. Which of the following is NOT a biochemical activity of the liver? A. Regulation of carbohydrate metabolism such as glycogenolysis, glycogenesis, and gluconeogenesis B. Production of lipases and bile for fat digestion C. Deamination of amino acid and conversion of the resulting ammonia to urea D. Lipid metabolism, including cholesterol and lipoprotein synthesis

B The pancreas makes lipases and bile does not do fat digestion, instead it emulsifies the fat to make it easier to re-absorb

Which of the following is the correct relationship for the heat required when going from Point 1 to Point 3? A. q=mc/\T B. q=mc/\T+m/\H(vap) C. mc/\T<q<mc/\T+m/\H(vap) D. mc/\T>q>mc/\T+m/\H(vap)

C Because point 4 is end of phase change so doesn't go to completion so would be more than q=mcT but less than q=mcT+mL

Which of the following would be detected in the blood of a person chronically infected with hepatitis B? A. Anti-core IgG, anti-sAg B. HBsAg, anti-core IgM, anti-core IgG C. HBsAg, anti-core IgG D. HBsAg, anti-core IgG, anti-sAg

C The passage states that chronic infection involves the "failure of an effective humoral immune response against HBsAg," i.e., the absence of anti-sAg (choices A and D are wrong). The passage also states that IgM type antibodies are the first response, and that over time these are replaced with IgG; thus in chronic infection, only anti-core IgG would be present (choice B can be eliminated and choice C is correct).

A 10 kg dancer raises her leg in the air in a développé. Her 2 kg leg is 1 m long and can be assumed to have uniform mass throughout its length. What angle, θ, between the working leg and the standing leg will create the most torque in the dancer's hip? A. 30° B. 45° C. 90° D. 135°

C Torque is calculated with T=Frsin(theta) so to maximize this, you would want sine to be closest to 1, which 90° makes 1.

Which has higher oxidation potential? Fe or Ca

Ca because transition metals are less reactive than alkali earth metals.

"As was hypothesized, success at the object permanence task was positively correlated with certain areas of mental development, specifically orientation/engagement and emotional regulation.... Object permanence tasks can also be used as working memory assessments for infants and young children." Suppose Baby X is a male born at 29 weeks gestational age at 990 g. According to the study, if Baby X is tested when he is 20 months old and demonstrates good emotional regulation, then he will: A. likely also demonstrate a strong ability to regulate his attention. B. demonstrate good orientation and engagement as well. C. perform better on other measures of object permanence than 20-month-old female children. D. probably also score high on early working memory tasks.

D. The passage notes that success at the object permanence task is positively correlated with emotional regulation. Earlier on, the passage explains that object permanence is a measure of working memory in young children. If Baby X demonstrates good emotional regulation, then he will probably also score high on early working memory tasks, since object permanence is a measure of working memory (choice D is correct). The passage makes no mention of attention (choice A is wrong). Orientation/engagement is significantly correlated with object permanence, but is not necessarily significantly correlated with emotional regulation (choice B is wrong). The study does not indicate a correlation between gender and object permanence, so it is not possible to conclude that Baby X will perform better on other measures of object permanence than similarly-aged female children (choice C is wrong).

Integrative reminiscence

the process by which older people may take stock of their lives and come to terms with previously unresolved conflicts