TPR test 1 wrong
A barometer is made by erecting an evacuated glass tube over a dish filled with mercury. Atmospheric pressure at sea level causes the mercury to rise inside the tube. If this barometer were moved from sea level to a mountain several kilometers in altitude, the mercury column would most likely: A. fall due to the decreased atmospheric pressure. B. rise due to the decreased atmospheric pressure. C. fall due to the increased atmospheric pressure. D. rise due to the increased atmospheric pressure.
A. Atmospheric pressure on a mountain several kilometers in altitude would be less than at sea level (eliminate choices C and D). Since the mercury in the dish feels less pressure from the atmosphere, it would not be pushed up as high in the tube (eliminate choice B and choice A is correct).
Which of the following is a true statement? A. A thermodynamic equilibrium between folded and unfolded scr SH3 can be achieved through the application of shearing forces. B. Unfolding of scr SH3 becomes increasingly endergonic with the application of shearing forces. C. The application of different amounts of force has no effect on which portion of the peptide participates in the transition state. D. Unfolding the protein domain scr SH3 along different pathways results in different unfolded peptide structures.
A. Choice B may be eliminated as the difference between the initial free energy (GF) and the free energy of the final unfolded state becomes increasingly negative with increasing force (increasingly exergonic). Choice C may be eliminated by the information in Figure 1c, which indicates that the application of various forces changes the portion of the peptide involved in the transition state. Choice D may be eliminated by the fact that each pathway ends at both the same reaction coordinate and the same free energy in Figure 1b. This indicates each path ends at the same configuration. Choice A is correct, and may be explained by the fact that the thermodynamics of unfolding are slightly endergonic in the absence of force, but with the application of force, the final value of the free energy decreases, as shown in Figure 1b. Somewhere in this process ΔG will be 0 and the system will be at thermodynamic equilibrium.
While completing a ritual, such as repeated locking and unlocking of the front door, an individual with obsessive-compulsive disorder would most likely feel: A. distressed. B. profound relief for the rest of the day. C. content. D. happy.
A. People with obsessive-compulsive disorder often feel the need to perform certain rituals or routines repeatedly. These individuals are often compelled by intrusive thoughts. An individual compelled to perform behavior repeatedly, such as continually locking and unlocking a door, will often feel distressed by these intrusive thoughts and the need to perform these behaviors (choice A is correct). While performing these rituals can sometimes cause a temporary sense of relief, there is no profound relief for the rest of the day, only a transient sense of relief that may come from temporarily satisfying the intrusive thoughts (choice B is wrong). An individual compelled to perform repeated behavior would have no particular reason to feel content or happy (choices C and D are wrong).
Which of the following statements best describes pyruvate kinase? A. It catalyzes the conversion of phosphoenolpyruvate to pyruvate and uses one ADP molecule. B. Its quaternary protein structure is most similar to hemoglobin and myoglobin. Your Answer C. It is involved in gluconeogenesis causing the conversion of pyruvate to glucose during high energy states. D. It has several isoforms due to alternative splicing via the spliceosome and ribosome.
A. Pyruvate kinase catalyzes the last step of glycolysis (eliminate choice C), where phosphoenolpyruvate is converted to pyruvate. Because it is a kinase, pyruvate kinase transfers an inorganic phosphate from phosphoenolpyruvate to ADP to form a molecule of ATP (choice A is correct). Paragraph 3 discusses how PKM2 can exist as a dimer or tetramer, each of which would contain more than one peptide chain and would therefore have quaternary protein structure. Hemoglobin also has quaternary structure (since it contains four peptide chains) but myoglobin does not (eliminate choice B). Since PKM2 is one isoform of pyruvate kinase, there are presumably others. Isoforms are different forms of the same protein and can be due to gene duplication or alternative splicing. However, the ribosome has no function in splicing (eliminate choice D).
Which of the following best theorizes how verbal insight problems most likely activate the wrong set of internal representations? A. Spreading activation model B. Feature-detection theory C. Weber's Law D. Bottom-up processing
A. The passage describes verbal insight problems as those containing words that tend to activate the wrong set of internal representations. The spreading activation model posits that the mind maintains networks of words that are associated with each other to varying degrees. When one word is activated, those words with the strongest associations to the activated word are also activated. The way verbal insight problems use words that lead to incorrect activations is best explained by the spreading activation model (choice A is correct). Feature-detection theory is a theory of visual perception that proposes that different neurons fire in response to detecting different features; in other words, some neurons fire only in response to vertical lines or motion or contrast. Feature-detection theory cannot explain how verbal insight problems most likely activate the wrong set of internal representations because these problems are word-based and not image-based. While feature-detection theory may help explain how we learn to read, it does not help explain how word problems would trick us the way the verbal insight problems do (choice B is wrong). Weber's Law states that two sensory stimuli must differ by a minimum constant proportion (not a constant amount) in order for us to notice the difference; this law does not explain how verbal insight problems activate the wrong set of internal representations (choice C is wrong). Bottom-up processing suggests that we start with the information gathered by sensory receptors and build up to a final representation in our brain; this type of processing tends to be used more with novel stimuli. It is the opposite of bottom-up processing, top-down processing, that helps to, in part, explain whyverbal insight problems (like the example with Mark Antony and Cleopatra) are so tricky (choice D is wrong).
Raclopride-H+ can best be described as: A. an amphiphilic molecule that has two stereoisomers. B. a Lewis base that has two stereoisomers. C. an amphiphilic molecule that has four stereoisomers. D. a Lewis base that has four stereoisomers.
A. This is a two by two question. In the reaction shown in figure 1, raclopride-H+ donates a proton in the first step (acting as a Brønsted-Lowry acid). Lewis bases are electron pair donors, and as H+ has no electrons choices B and D can be eliminated. Raclopride-H+ has a hydrophobic halogenated aromatic portion, and a hydrophilic side chain (protonated amine), making it an amphiphilic or amphipathic molecule. The maximum number of stereoisomers a compound may have is given by the formula 2n, where n is the number of chiral centers present. Raclopride-H+ has one chiral center, so it has two stereoisomers (choice C is wrong and choice A is correct).
Aldosterone's mechanism of action on its target cell is most similar to that of which of the following hormones? A. Testosterone B. Glucagon C. Insulin D. ACTH
A. Two general classes of hormones are those that are small hydrophobic molecules like steroid hormones and those that are peptides. The steroid hormones, which include aldosterone and testosterone, diffuse through the plasma membrane to bind to a receptor which enters the nucleus to regulate transcription of a specific set of genes (choice A is correct). Peptide hormones, such as glucagon, insulin, and ACTH, cannot diffuse into the cell since they are large and hydrophilic, so they bind to cell-surface receptors to transduce a signal into cells (choices B, C, and D are wrong).
It appears that in their original predictions, participant wives who reported steeper drops in marital satisfaction might have used which defense mechanism? A. Displacement B. Denial C. Repression D. Reaction formation
B. According to Freud, people use ego defense mechanisms in order to protect themselves from the effects of anxiety. The results of the current research offer the following about the women who experienced the steepest declines in marital satisfaction: (1) these women reported higher levels of external stress and (2) following this, these women had more positive predictions for their respective marriages. The fact that these women were more optimistic about their futures, despite the presence of significant external stressors, suggests the presence of a cognitive bias (as described in the final paragraph) that offers self-protection. Denial is the defense mechanism that is most closely related to this situation (choice B is correct). Denial is the refusal to accept external realities that are threatening to the self; for example, the refusal to consider the effect that external stressors could have on marital satisfaction. Displacement is the shift of emotions, such as aggressive or sexual impulses, to safer outlets (choice A is wrong). Repression is the psychological attempt to subdue emotionally painful memories (choice C is wrong). Reaction formation is expression of the opposite of one's feelings in order to conceal unacceptable or dangerous emotions (choice D is wrong).
An activated aldosterone receptor most directly regulates activity of which of the following enzymes? A. DNA polymerase B. RNA polymerase II C. Na+/K+ ATPase D. Renin
B. Aldosterone binds to aldosterone receptors to regulate transcription of a specific set of genes. The enzyme that synthesizes mRNA is RNA polymerase II, so it is this enzyme that would be most directly affected by the activated aldosterone receptor (choice B is correct). DNA polymerase is used in replication, not transcription (choice A is incorrect), and while aldosterone may ultimately affect the activity of the Na+/K+ ATPase and renin, these would be indirect effects (choices C and D are incorrect).
The OCA2 gene overlaps with another gene called HERC2, which has two alleles: the wild type allele, and a recessive allele A1 which has been linked to a genetic predisposition to Crohn's disease. A woman with an OCA2305R : HERC2wt chromosome and an OCA2305W : HERC2A1 chromosome mates with a homozygous OCA2305R / 305R : HERC2A1 / A1 man. Their children will most likely be: A. 25% normal with blue eyes, 25% normal with brown eyes, 25% blue eyed with a risk of Crohn's disease and 25% brown eyed with a risk of Crohn's disease. B. 50% normal with blue eyes and 50% brown eyed with a risk of Crohn's disease. C. 50% blue eyed with a risk of Crohn's disease and 50% normal with brown eyes. D. 100% blue eyed with a risk of Crohn's disease.
B. Based on information in the passage, alleles of OCA2 are one of the major determinants of eye color. The woman in the question stem will have brown eyes because she is heterozygous for OCA2 and has an allele that is associated with blue eyes (305R) and an allele associated with brown eyes (305W). Since the question stem says that the OCA2 gene and the HERC2 gene overlap, they must be linked (they are 0 map units apart). This means that crossing over will not occur between these two genes and they will be inherited as a unit. The father in this question is acting like a testcross; that is, he is homozygous recessive for both genes. Therefore, the offspring produced will either get the OCA2305R : HERC2wt chromosome or the OCA2305W : HERC2A1 chromosome from the mother and an OCA2305R : HERC2A1 chromosome from the father. The children will be 50% OCA2305R / 305R : HERC2wt / A1 (blue eyed with no increased risk of Crohn's disease; note that the A1 allele of HERC2 is recessive to the wild type allele) and 50% OCA2305W / 305R : HERC2A1 / A1 (brown eyes with an increased risk of Crohn's disease). Thus, choice B is correct (eliminate choices A, C, and D). Note than choice A is what would occur if the two genes were not linked.
How could the James-Lange theory of emotion best be explained by the dynamical systems theory concept of emergence? A. Perception of the emotional stimulus is determinative for the emotional experience. B. Different collections of physiological markers combine to represent distinct emotions. C. Convergence of perceptive stimuli onto the thalamic and hypothalamic structures creates a global emotional experience. D. Cognition and physiology combine to create a more complex emotional experience than the constituent components.
B. Emergence, according to the passage, describes the way that a system composed of many separate elements combines these elements in such a way that the system itself is more complex than the constituent elements themselves. Since the James-Lange theory deals with the way different physiological states represent emotions, this theory of emotion could be explained by the dynamic systems concept of emergence if different collections of physiological markers combine to represent distinct emotions (choice B is correct). Perception of the emotional stimulus being determinative does not bring together different constituent elements, as in the definition of emergence, but is more similar to sensitivity to initial conditions (choice A is wrong). Convergence onto thalamic and hypothalamic structures is prominent in the Cannon-Bard theory, not the James-Lange theory (choice C is wrong). The way that cognition and physiology combine during emotion is a central component of the Schachter-Singer, or two-factor, theory of emotion (choice D is wrong).
How many grams of NaI should be added to 500 mL of a saturated solution of AgI to make a solution that is 1.5 x 10-15 M Ag+? A. 9.0 g B. 4.5 g C. 0.06 g D. 0.03 g
B. Figure 1 indicates that when [AgI] = 1.5 x 10-15 M, the concentration of NaI = 0.06 M. In 500 mL (0.5 L), this solution should contain 0.030 mol NaI. Since the molar mass of NaI is 150 g/mol, this amounts to 4.5 g of NaI.
All of the following conclusions can be drawn from the data in Figure 1 EXCEPT: A. training enabled participants to correctly solve a greater percentage of ANPs than they would have solved without the training. B. training interfered with participants' ability to solve problems other than ANPs. C. training was not helpful for breaking constraints other than ambiguous names. D. those without training generated a greater percentage of hypotheses that failed to break the ANP constraints.
B. Figure 1 shows solution rates and correct hypothesis rates for ANPs and OPs for both groups. For ANPs, the training group outperformed the control group in terms of both the solution rate (~70% vs. ~30%) and the correct hypothesis rate (~85% vs. ~45%). In other words, training enabled participants to correctly solve a greater percentage of ANPs than they would have solved without the training (choice A is true and therefore wrong). Additionally, since the control group produced a smaller percentage of correct hypotheses, by definition, it produced a greater percentage of incorrect hypotheses than did the training group (choice D is true and therefore wrong). The OP sections of the chart indicate that the training group performed similarly to the control group, both in terms of solution rate and correct hypothesis rate, which suggests that training did not help participants to break constraints other than ambiguous names (choice C is true and therefore wrong). Nevertheless, since the training group produced the same percentage of correct solutions as the control group on OPs, it would be inaccurate to conclude that training interfered with participants' ability to solve problems other than ANPs (choice B is false and therefore correct).
A highly proliferating cell would most likely: A. overexpress hexokinase and fructose-1,6-bisphosphatase. B. express high levels of the lactate transporter and the glutamine transporter. C. power cell growth by running the electron transport chain and oxidative phosphorylation. D. upregulate pyruvate dehydrogenase kinase and downregulate phosphofructokinase.
B. Highly proliferating cells would express large amounts of hexokinase (a key glycolysis enzyme) but would not also overexpress fructose-1,6-bisphosphatase because this is an enzyme involved in gluconeogenesis. Even highly proliferative cells will avoid concurrently running reciprocally regulated pathways (eliminate choice A). The passage says that rapidly dividing tumor cells produce and export large amounts of lactate, and import large amounts of the amino acid glutamine (choice B is correct). The focus of this passage is how highly proliferative tumor cells power growth via glycolysis and fermentation, making choice C an unlikely correct answer (eliminate choice C). Based on information in the last paragraph, pyruvate dehydrogenase kinase activity is high in some rapidly growing cells, but a high rate of glycolysis will lead to high PFK expression (eliminate choice D).
If the front mirror in Figure 1, which is made of glass of refractive index 3/2, were repositioned so that the laser beam strikes at an angle of 30° to its normal, what would be the angle of reflection? A. sin-1(1/3) B. 30° C. sin-1(1/31/2) D. 60°
B. If the laser beam strikes the mirror at an angle of 30° relative to the normal, then this is the angle of incidence, and, by the Law of Reflection, it is also the angle of reflection. The refractive index of the mirror is irrelevant.
A Q-switched laser can be used to treat skin blemishes and to remove tattoos. The Q switch momentarily interrupts the inducing light creating a build-up of energy within the crystal. This does not increase the overall energy of the laser, but concentrates it into shorter time periods or pulses. A longer interruption with the Q-switch most likely would increase the: A. total amount of work done by the laser B. power of each laser pulse C. wavelength of the laser light. D. frequency of the laser light.
B. Since the overall energy of the laser does not change, neither will the frequency, wavelength, nor work done by the laser. This leaves choice B: Concentrating the energy into a shorter time period increases the power of each pulse (since power equals energy delivered per unit time, by definition).
During spermatogenesis, spermatids: A. are frozen in meiosis II until after fertilization. B. have already undergone meiotic recombination. C. have four copies of the genome per cell. D. have no nucleus.
B. Spermatids are the sperm precursors that have completed meiosis but have not yet fully matured. A mature sperm has fully completed meiosis, and this is true of spermatids as well; it is the ova that are frozen in meiosis II until after fertilization (eliminate choice A). Recombination occurs in both oogenesis and spermatogenesis during meiotic prophase I, which the spermatid already completed (choice B is correct). The spermatid has passed through two reductive divisions in meiosis to end up with only one copy of the genome (eliminate choice C). The cell does have a nucleus. In fact, the sperm has virtually no cytoplasm, but the genome is still packaged into a nucleus (eliminate choice D).
A researcher attempts to explain the phenomenon, as described in the passage, that couples who reported greater relationship satisfaction had greater physiological co-regulation. The researcher concludes that the cognitive interpretation of the physiological state of both partners led to their reported feelings of satisfaction. This is most similar to which of the following? A. Emotional intelligence B. Schachter-Singer theory C. Cannon-Bard theory D. James-Lange theory
B. The Schachter-Singer theory posits that emotional experience is determined by one's physiological state and the cognitive interpretation of that state. This is most similar to the description in the question stem in which couples experience physiological co-regulation and cognitive interpretation (choice B is correct). Emotional intelligence is the ability to control, interpret, and understand one's own emotions and the emotions of others. This is not as specific as choice B in capturing the information in the question stem (choice A is wrong). The Cannon-Bard theory focuses on the central role of the hypothalamus in regulating emotions and is less specifically concerned with physiological activation and cognition (choice C is wrong). The James-Lange theory asserts that emotional experience is primarily based on physiological arousal, and that each different physiological state is associated with an emotion (choice D is wrong).
Why does the solubility of AgI decrease in an aqueous solution when NaI is added? A. A lower [Ag+] shifts Reaction 1 to the right. B. A higher [I-] shifts Reaction 1 to the left. C. A higher [Na+] shifts Reaction 1 to the right. D. A higher [Ag+] shifts Reaction 1 to the left.
B. The equilibrium described by Reaction 1 will be shifted to the left if more I- ions are present in solution (which would occur if NaI were added), thus precipitating more AgI(s) and decreasing its solubility compared to that in pure water. This phenomenon is known as the common ion effect, and is a consequence of Le Châtelier's Principle. While both choices A and D are true statements, only a shift to the left indicates a decrease in solubility of AgI (eliminate choice A); the addition of NaI also has no effect on the concentration of Ag+ ions in order to cause a shift (eliminate choice D). Since Na+ is not a part of Reaction 1 it cannot affect the equilibrium position of the reaction (eliminate choice C).
Which of the following is the best explanation for why urea, CO(NH2)2, may be used to study the unfolding of scr SH3? A. The entropic loss due to unfolding of scr SH3 is a result of urea increasing the solubility of the protein interior. B. Urea acts to stabilize the denatured state of scr SH3 by associating with backbone polar groups and charged side chains. C. Urea stabilizes the multiple transition states observed for the unfolding of scr SH3. D. The addition of urea reduces S—S bonds in the structure of scr SH3.
B. Using process of elimination can quickly narrow the options given in the answer choices. Choice A presents an inconsistency because unfolding will cause entropy to increase, not decrease, due to the disordering effect of unfolding (eliminate choice A). Based on its structure, urea, CO(NH2)2, will interact with the polar moieties of a protein via hydrogen bonding, which include the peptide backbone and polar residues (charged or uncharged). By disabling intramolecular hydrogen bonding urea is capable of stabilizing the denatured state of a protein (choice B is correct). As stated in the passage, only a single unfolding pathway is followed when a chemical denaturant, such as urea, is applied. Thus, there can only be one transition state (eliminate choice C). Choice D is incorrect because urea is not a reducing agent and will not affect disulfide bonds.
Retention time in gas chromatography is the time it takes a compound to be eluted and detected after injection into the system. Of the following compounds, which will display both a short retention time and a high Rf in analysis by thin-layer chromatography? A. I B. II C. III D. IV
C. A low retention time in gas chromatography is associated with a volatile (low boiling point) compound, while a high Rf in thin-layer chromatography is associated with a nonpolar compound. Thus, the best answer for this question is a nonpolar compound with a low boiling point. Choices A and B, being alcohols, will have high boiling points due to their ability to hydrogen bond, so are incorrect answers. Choices C and D are both nonpolar compounds, but they differ in size by one carbon. The compound with more carbons (and less branching) in choice D, 2,3-dimethylpentane, will have a higher boiling point because it has greater London dispersion forces, so is also incorrect.
If a silent mutation occurs at position 6 in the β chain of hemoglobin, what will be the overall charge of the resulting tryptic peptide in a solution buffered to pH 5? A. -2 B. -1 C. 0 D. +1
C. A silent mutation within the coding region of DNA does not alter the amino acid sequence of a protein, thus the amino acid at position 6 in the tryptic peptide will still be Glu. Thus, the tryptic peptide to evaluate is identical to the HbA fragment shown in the passage. By determining the charge states on each individual amino acid in the tryptic peptide at pH 5, the overall charge on the peptide can be calculated. Recall that when the pH of a solution exceeds the pKa of a functional group, that group is deprotonated. This applies only to the acidic residues in the peptide. So, from left-to-right: the N-terminal Val bears a +1 charge on its α-amino group; His bears a +1 charge on its side chain; Leu, Thr, and Pro are uncharged; each Glu bears a -1 charge on its side chain; and the C-terminal Lys is neutral overall (+1 on its side chain and -1 on its α-carboxyl group). The sum total of these charges [(+1) + (+1) + (-1) + (-1) + (0)] is 0, or answer choice C.
Which of the following represents the best way that researchers could have improved their ability to draw conclusions about observed differences in behavior at marked vs. unmarked crosswalks? A. Informing the public of the study prior to conducting it B. Measuring behavior of pedestrians at marked crosswalks during the day and comparing it to the behavior of pedestrians observed at unmarked crosswalks at night C. Measuring pedestrian behavior at unmarked crosswalks at specific times of day, on specific days of the week, over time, and then adding signs at these same crosswalks, and measuring pedestrian behavior again at the same specific times of day and specific days of the week, over time D. Including more of each type of crosswalk in the study
C. A within-subjects design would have allowed the researchers to make broader inferences about their results. Even though the current results are significant, it can be argued that the difference between crosswalks was largely due to the differences between pedestrians who used each type of crosswalk. A within-subjects design would allow researchers to examine how the same sample of pedestrians responds to marked vs. unmarked crosswalks (choice C is correct). Informing the public about the study prior to conducting it might make pedestrians behave unnaturally, which would be a limitation since researchers would be unable to predict how pedestrians would behave if crosswalk laws were implemented in other states (choice A is wrong). Measuring pedestrian behavior at different times of the day for each crosswalk would be adding another difference to the study, thereby limiting the ability to claim that the signs are what impacted behavior (choice B is wrong). Increasing the number of crosswalks studied might help strengthen inferences, but nothing in the passage suggests that this was an issue; the passage already stated that the study included a representative sample of crosswalks (choice D is wrong).
The substrate in the experiment is: A. catalase. B. potassium permanganate. C. hydrogen peroxide. D. water.
C. Catalase is an enzyme that acts on hydrogen peroxide, reducing it to water and oxygen (choice A is wrong and choice C is correct). In an enzymatic reaction, the substrate is that on which the enzyme acts. Potassium permanganate serves in the titration but is not the substrate (choice B is wrong). Choice D refers to water, a product.
Which of the following can the researchers conclude strictly from the data shown in Figure 2? A. Lipid hydrolysis cannot occur in the absence of raclopride. B. DOPC + 10 mol% of raclopride would have a higher rate of lipid hydrolysis than DOPC + 5 mol% of raclopride. C. Lipid hydrolysis with raclopride is significantly faster than any background degradation process. D. The rate of lipid hydrolysis with DOPC + 5 mol% of raclopride increases linearly with time.
C. Figure 2 shows the rate of lipid hydrolysis of the lipid DOPC in the presence and absence of raclopride by measuring the concentration of the hydrolyzed lipid, lyso-PC, over time. While the data show that the rate of lipid hydrolysis is appreciable in the presence of raclopride and negligible in its absence, as raclopride acts a simple Brønsted acid, it would be incorrect to say that this exact molecule is required for any kind of lipid hydrolysis (eliminate choice A). Figure 2 does show that the rate of background lipid hydrolysis is comparatively much slower than raclopride-mediated lipid hydrolysis (choice C is correct). While a higher concentration of raclopride may increase the rate of lipid hydrolysis, there is no data in Figure 2 to support this statement, and as the raclopride-H+ is regenerated in the mechanism, there is no guarantee that increasing its overall concentration leads to a greater rate (eliminate choice B). The slope of Figure 2 is the rate of lipid hydrolysis (change in concentration over a change in time), and as the slope with DOPC + 5 mol% of raclopride is constant, choice D is incorrect.
A human tibia with a cross-sectional area of 0.0002 m2 undergoes a 1% change in length when compressed by a force of 20,000 N. What is the approximate elastic energy density within the bone while compressed if the bone has not reached its yield point? A. 5 × 103 J/m3 B. 1 × 104 J/m3 C. 5 × 105 J/m3 D. 1 × 106 J/m3
C. From Figure 1, the elastic energy density is equal to the area under the stress vs. strain graph. If the yield point has not been reached, then the region under the graph has the shape of a right triangle, whose area is (1/2) × base × height = (1/2) × strain × stress. In this case, the strain is 1% = 10-2, and the stress is F = 2 × 104 N divided by A = 2 × 10-4m2, which equals 108 N/m2. Thus, the elastic energy density is (1/2)(10-2)(108 N/m2) = 5 × 105 J/m3.
The point mutation denoted as β represents the substitution of an amino acid integral in the transition state for which of the following pathway(s). A. P1 B. P2 C. P3 D. P2 and P3
C. From Figure 1c, the point mutation denoted as β is important in the high-shearing-force regime. With high shearing forces P3 dominates, making choice C the correct answer. The mutation β is specifically not important in the situation corresponding to the shearing forces making P2 dominant, eliminating choice D.
Which of the following best describes the role of fructose-2,6-bisphosphate? A. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating fructose-1,6-bisphosphatase and inhibiting phosphofructokinase. B. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting hexokinase. C. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating phosphofructokinase and inhibiting fructose-1,6-bisphosphatase. D. It exerts reciprocal control on glycolysis and gluconeogenesis by stimulating hexokinase and inhibiting fructose-1,6-bisphosphatase.
C. Fructose-2,6-bisphosphate exerts reciprocal control on glycolysis and gluconeogenesis through phosphofructokinase and fructose-1,6-bisphosphatase, and has no impact on hexokinase activity (eliminate choices B and D). Fructose-2,6-bisphosphate stimulates phosphofructokinase, which is used in glycolysis, and inhibits fructose-1,6-bisphosphatase, which is used in gluconeogenesis (choice C is correct and eliminate choice D).
If pirenzepine is administered to a Zollinger-Ellison syndrome patient, which of the following will occur? A. Luminal secretion of H+ will decrease, because high intracellular Ca2+ will inhibit protein kinases. B. Luminal H+ secretion will increase, because low intracellular cAMP will increase protein kinase activity. C. Luminal H+ secretion will decrease, because low intracellular Ca2+ will lead to lowered protein kinase activity. D. Luminal H+ secretion will increase, because intracellular pH will rise.
C. If acetylcholine is blocked, calcium levels will be low, protein kinase activity low, and H+ secretion low (choice C is correct). Calcium will be low (eliminate choice A) and H+ secretion will decrease, not increase (eliminate choices B and D).
Which of the following could explain the Warburg effect? Metabolic differences allow cancer cells to adapt to hypoxic (oxygen-deficient) conditions inside solid tumors. Cells with low proliferation rates often have a high ratio of glycolysis to mitochondrial respiration. Some oncogenic changes shut down the mitochondria because these organelles are involved in apoptosis, which would result in cell death. A. I only B. III only C. I and III only D. I, II, and III
C. Item I is true: tumors can grow quite quickly and often the inside of tumors don't have sufficient blood supply. This can lead to hypoxic conditions inside the tumor. Since glycolysis and fermentation don't require oxygen, they could facilitate cell growth in anaerobic conditions (choice B can be eliminated). Item II is false: the passage is about highly proliferative cells that use glycolysis and fermentation to power growth so a statement about slow growth does not explain the Warburg effect discussed in the passage (choice D can be eliminated). Item III is true: if a tumor cell is relying on glycolysis and fermentation instead of mitochondrial cell respiration, it will have less use for the mitochondria in general. Since the mitochondria can initiate apoptosis, less reliance on this organelle could lead to apoptosis resistance, conferring a survival advantage to the cancerous cell (choice A can be eliminated and choice C is correct).
Which one of the following statements concerning the SN2 reaction mechanism is true? A. It proceeds best with tertiary substrates and is a two-step mechanism. B. It proceeds with retention of stereochemistry. C. It proceeds best with primary substrates and is a one-step reaction. D. It proceeds through a carbocation intermediate.
C. SN2 cannot operate with a tertiary substrate because the nucleophile must attack at the same time the leaving group leaves, and tertiary substrates are too sterically hindered (eliminate choice A). SN2 goes with inversion of stereochemistry, not retention (eliminate choice B). SN1 reactions have carbocation intermediates (eliminate choice D), while reactions are one-step processes with no intermediates. Choice C is the best answer because unhindered substrates allow the nucleophile to attack while the leaving group leaves.
Perceiving the color of a pH indicator requires what type of visual processing? A. Depolarization of cone cells to trigger hyperpolarization of bipolar neurons. B. Depolarization of cone cells to trigger depolarization of bipolar neurons. C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons. D. Hyperpolarization of cone cells to trigger hyperpolarization of bipolar neurons.
C. The answer choices contain pairs of information so proceed with a 2x2 question elimination strategy. When cone cells are not exposed to light and are thus not processing the visual stimuli of color, they are depolarized. Since the question is asking about perceiving color, the cone cells need to be hyperpolarized (eliminate choices A and B). When cone cells hyperpolarize, the bipolar neurons are no longer being inhibited and can then depolarize in order to transmit signal to the ganglion and eventually the optic nerve (choice C is correct; eliminate choice D).
The change in Q, the magnitude of charge on each surface of the axon membrane that is depicted in the graph above, could be produced by which of the following? A. A decrease in the resistance of the Na+ channels B. A steady decrease in the voltage across the axon membrane C. Inactivity of the Na+/K+ pump D. An increase in the resistance of the leakage channels
C. The graph shows how the charge decreases as a capacitor discharges, which occurs when the voltage between the plates is no longer maintained. If the Na+/K+ pump (the voltage source) stopped working, the voltage between the sides of the membrane would no longer be maintained, so C is the best choice. B is wrong since it says "a steady decrease in the voltage"; if V decreased steadily, then, since Q ∝ V, Q would also decrease steadily and its graph would be a straight line
If the graph in Figure 1 were redrawn for cartilage undergoing tensile stress, then the angle θ would be: A. greater, because cartilage has a smaller Young's modulus than does bone. B. greater, because cartilage has a larger Young's modulus than does bone. C. smaller, because cartilage has a smaller Young's modulus than does bone. D. smaller, because cartilage has a larger Young's modulus than does bone.
C. The passage states that cartilage does not withstand tension as well as bone. Therefore, the Young's modulus of cartilage must be lower than that of bone, eliminating choices B and D. Since stress = E × strain, a lower value of E corresponds to a lower stress for the same strain, so the graph of stress vs. strain in the elastic region would be less steep; that is, the angle θ would be smaller.
Which of the following is the anticodon sequence on the tRNA for the start codon? A. 5'-AUG-3' B. 5'-UAC-3' C. 5'-CAU-3' D. 5'-GUA-3'
C. The start codon on the mRNA would be 5'-AUG-3', therefore, the anticodon on the tRNA would be the complementary sequence: 3'-UAC-5' (or alternatively written, 5'-CAU-3').
Change in the kinetic parameters of Arf1 following treatment with an experimental Arf1 inhibitor is the same as the change measured after replacing the wild type gene encoding Arf1 with the mutant gene encoding [Δ17] Arf1. The experimental Arf1 inhibitor is most likely to be a: A. Non-competitive inhibitor B. Competitive inhibitor C. Mixed Inhibitor D. Uncompetitive inhibitor
Competitive inhibitors increase the apparent Km of inhibited enzyme for substrate by competing with substrate for access to the enzyme active site. Such inhibitors do not affect the catalytic efficiency, and thus kcat, for inhibited enzyme. This is consistent with the differences in the measured parameters of wild type Arf1 and [Δ17] Arf1 shown in Table 1 (choice B is correct). Non-competitive inhibitors decrease Vmax (and thus kcat) while having no effect on Km for the inhibited enzyme (eliminate choice A). Mixed inhibitors decrease Vmax (and thus kcat) while either increasing or decreasing Km for the inhibited enzyme (eliminate choice C). Uncompetitive inhibitors decrease Vmax (and thus kcat) while decreasing Km for the inhibited enzyme (eliminate choice D).
Two nuclides with the same excess number of neutrons over protons (that is, the same value of N-Z) are called isodiapheres. Which of the following is an isodiaphere of 68/30 Zn? A. 67/31 Ga B. 74/32 Ge C. 72/34 Se D. 78/35 Br
D. 68Zn has 30 protons and 38 neutrons, so the excess number of neutrons over protons is N-Z = 38 - 30 = 8. Of the given choices, only choice D, 78Br shares this property; it has 35 protons and 43 neutrons, giving N-Z = 8.
The activation free energy (GTS) is composed of the enthalpy (HTS) and entropy (STS) of activation by the same relation as free energy, enthalpy and entropy are related for energy minima (GF, HF and SF). Which of the following is likely true regarding the transition from the folded state (F) to the transition state (TS)? A. ΔH (F → TS) < 0 B. ΔG (F → TS) < 0 C. ΔS (F → TS) < 0 D. ΔH (F → TS) > 0
D. Choice B can be immediately eliminated, as the free energy of activation will always be higher than the free energy of either minimum it connects. To eliminate choice C it is helpful to remember that proteins are highly ordered macromolecules. In fact, just the term "unfolding" implies that there is an orderly arrangement being undone, and hence it is very unlikely that such unfolding would require a negative change of entropy. This, on its own, means that choice D is correct as ΔG is positive from the folded state to the transition state and ΔS is also positive (ΔG = ΔH - TΔS). Still, reasoned further, the bonding (enthalpic) components that hold a folded protein together must overcome the entropic penalty associated with this ordering. Therefore, ΔH (TS → F) must be very negative, meaning the opposite process is very positive, and choice D is correct.
Figure 1 best supports which of the following conclusions about the participants in the study? A. Participants given situational attribution training were faster at categorizing photos of AAs when they were primed with stereotypic AA trait words compared to those in the control group. B. Participants given the situational attribution training were slower at categorizing photos of AAs when they were given stereotypic AA trait word primes than when they were given non-trait primes. C. Participants given situational attribution training were slower at categorizing photos of people in general compared to those in the control group. D. Participants showed no greater automatic association between people and trait words that are not stereotypic of that group than between people and non-trait words.
D. Figure 1 shows that when participants were categorizing white photos, there was no difference in speed whether they were given a stereotypic AA trait prime or a non-trait prime (choice D is correct). Participants given situational attribution training were slower at categorizing photos of AAs when they were primed with stereotypic AA trait words compared to those in the control group (choice A and choice C are wrong). Trained participants were equally fast in categorizing photos of AAs whether they were given a stereotypic AA trait prime or a non-trait prime (choice B is wrong).
Based on Figure 1, which of the following is least supported? A. The OCA2 proteins in the eight species studies would be considered homologous. B. The OCA2 protein is conserved across eukaryotes. C. Zebrafish and Japanese killifish have a OCA2 common ancestor protein which is more related that that between horse and human. D. Humans and chimpanzees are more closely related than fission yeast and the wild boar.
D. Homologous proteins or genes are those that have evolved from a common ancestor. This matches the information in Figure 1, since the OCA2 protein in all organisms on the figure originated with the ancestor protein represented by the point at the bottom (choice A is supported and can be eliminated). The eight organisms on the figure are all eukaryotes and from diverse families (fungi, mammals, fish; choice B is supported and can be eliminated). Remember that yeast are fungi and therefore eukaryotic. OCA2 in zebrafish and Japanese killifish share a common ancestor which is not far away in evolutionary terms (i.e., is not very far down the diagram). In contrast, the common ancestor protein between human and horse OCA2 protein is the point at the bottom of the diagram; this is farther away in evolutionary terms (choice C is supported and can be eliminated). While choice D may be true based on logic and background information on evolution, it is not supported by Figure 1; this phylogenetic tree contains information on how the OCA2 proteins are evolutionarily related, not how organisms are related (choice D is the least supported and the correct answer choice).
Two blocks are suspended from the ends of a massless meter stick as shown below: How far from the center of the stick must the rope be attached in order to maintain rotational equilibrium? A. 10 cm B. 20 cm C. 25 cm D. 30 cm
D. Intuitively, the rope must be closer to the heavier block to balance the stick. Since the block on the left is 4 times heavier, the rope must be attached at a point which is 4 times closer to the left-hand block than to the right-hand block to balance the torques due to the weights of the blocks. So, letting x denote the distance from the 4 kg block to the rope's suspension point, the distance from the 1 kg block to the suspension point is 4x. Thus, x + 4x = 100 cm, which implies x = 20 cm. The rope is therefore suspended at a point which is 50 - 20 = 30 cm from the center of the stick.
The diaphragm plays an important role in respiration. During inspiration, the diaphragm: A. relaxes, causing alveolar pressure to drop below atmospheric pressure. B. contracts, causing alveolar pressure to rise above atmospheric pressure. C. relaxes, causing alveolar pressure to rise above atmospheric pressure. D. contracts, causing alveolar pressure to drop below atmospheric pressure.
D. Recognize theh opportunity to treat this question as a 2 X 2 elimination. Inspiration is the drawing of air into the lungs. The diaphragm contracts and flattens during inspiration (eliminate choices A and C), expanding the chest cavity; the lungs (which are stuck to the inside wall of the chest cavity) expand as well. The expansion of the lungs decreases the pressure in the alveoli, causing air to move into the lungs from the exterior (eliminate choice B and choice D is correct).
Which indicator would be most useful in signaling the second equivalence point of a typical neutral amino acid in a titration using NaOH? A. Methyl red B. Phenol red C. Phenolphthalein D. Alizarin yellow
D. The most useful indicator for the titration in question will change from its undissociated to dissociated form around the pH of the equivalence point. Since the second equivalence point for the amino acid shown is roughly at a pH of 10.5-11.0 (the steep part of the curve at the far right of the graph), the best indicator will be one color below this range and another color above it. Only alizarin yellow changes at such a high pH range.
For children who exhibit an avoidant attachment style while growing up, parents do not make themselves available emotionally and the child responds by withdrawing. Research shows that children who demonstrate an avoidant style did not connect with a parent emotionally. According to the third paragraph, what could account for this deficit? A. Low level of satisfaction and an "out of sync" relationship B. Inadequate social superstructure C. A poor support network within the family D. Lack of physiological co-regulation in the mother-child dyad
D. The parents in the question stem fail to connect with the child emotionally, leading to the avoidant attachment style. In the passage, physiological co-regulation of emotion is the mechanism through which the mother and child connect emotionally (choice D is correct). Relationship satisfaction and being "in sync" is discussed in the passage regarding romantic relationships, not parent-child interactions (choice A is wrong). In Marxist theory, the superstructure is the collection of social organizations and groups that form the society's structures of power (choice B is wrong). A support network, though important for a child's upbringing, does not specifically deal with emotional interaction between a parent and child as described in the question stem (choice C is wrong).
Socialized medicine is a term used to describe governmental regulation of health care, with this public administration of health services being funded through taxation. This is also often referred to as universal health care. In the United States, there are some socialized insurance programs, such as military medicine. However, despite the implementation of the Affordable Care Act, private companies continue to provide most of the nation's health care. Public opinion has been slow to accept the notion of universal health care, as evidenced by the opposition to presidential reform efforts during the Truman, Clinton, and Obama administrations. This hesitation could be attributed to several factors, such as the public's agreement with conservative critics. and best matches the approach to inequalities described in the theories of: A. Emile Durkheim. B. Karl Marx. C. Ludwig Gumplowicz. D. Max Weber.
D. The theoretical perspective most concerned with social inequalities is conflict theory. Classical sociologists associated with this theory include Karl Marx, Ludwig Gumplowicz, and Max Weber. In contrast, Emile Durkheim is more associated with structural functionalism, which is focused on contributions to social stability (choice A is wrong). The persistent opposition to forms of socialized medicine best reflects the theories of Max Weber. Unlike the other theorists, Weber argued that the presence of inequalities would not necessitate the collapse of capitalism. He suggested that responses to inequalities are moderated through additional social factors, such as agreement with authority figures (e.g., public political figures; choice D is correct). The public's persistent opposition challenges the theories of Karl Marx. Marx argued that social inequalities, and subsequent conflict and internal tensions as a result of power differentials, would lead to the rise of socialism. The Marxist perspective then suggests the rise of socialized medicine, as opposed to the continuation of capitalistic private systems (choice B is wrong). Finally, the theories of Ludwig Gumplowicz focus on cultural and ethnic conflicts that are not relevant to the question (choice C is wrong).
A CO2 laser used as a laser scalpel produces a beam of laser light with a wavelength of 10.6 µm. Compared to the excited electrons in the laser shown in Figure 1, the excited electrons in the CO2 laser most likely have: A. greater mass. B. less mass. C. a greater energy difference between their normal state and their excited state. D. a smaller energy difference between their normal state and their excited state.
D. The wavelength of the CO2 laser is 10 times greater than the wavelength of the laser shown in Figure 1. Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the laser light is the energy released by transitions of electrons dropping to a lower energy level, less emitted energy implies a smaller difference between the energy levels. The mass of an electron is independent of its atomic or molecular energy state, eliminating choices A and B (the mass of the atom or molecule as a system changes with electron excitation according to relativity theory, but not the mass of the electron itself, and anyway that is well beyond the scope of the MCAT).
If a fully saturated solution of AgI, with precipitate present, were treated with NaCl instead of NaI, which of the following observations is likely? A. As NaCl is added, all precipitates are dissolved into the aqueous solution. B. The decrease in [AgI] is even more drastic than with the addition of NaI in Figure 1. C. There is no change in the amount of undissolved AgI. D. The concentration of [I-] increases.
D. Unlike NaI, NaCl does not have a common ion with AgI and will therefore NOT cause a decrease in the solubility for AgI with increasing concentration (eliminate choice B). The following will act as a competing reaction when [Cl-] concentrations become sufficiently large: Ag+ (aq) + Cl- (aq) → AgCl (s) With this in mind, there will be no situations wherein the solution is free of precipitate (eliminate choice A). As the dissolved [NaCl] concentration increases, AgCl will be precipitated from solution, which will enable additional AgI to dissolve (eliminate choice C). The increased dissolution of AgI will cause the increase in [I-], even as [Ag+] levels remain low.
Due to its GTPase activity, Arf1 may best be categorized enzymatically as a(n): A. lyase. B. transferase. C. oxidoreductase. D. hydrolase.
The GTPase activity of Arf1 catalyzes the hydrolysis of GTP. Hydrolases are enzymes that catalyze the cleavage of, among other bond types, the phosphoric anhydride bonds found in GTP (choice D is correct). Lyases are enzymes that cleave bonds without the addition of water (non-hydrolytically) (eliminate choice A). Transferases are a class of enzymes involved in the transfer of a functional group from a donor molecule to an acceptor molecule (eliminate choice B). Oxidoreductase catalyze oxidation-reduction reactions (eliminate choice C).
The results of the experiment suggest that: A. the ability of ArfGAP1 to interact with [Δ17] Arf1 depends on an alpha-helical structure. B. [Δ17] Arf1 and wild type Arf1 differ at their CTA1 binding site. C. ArfGap1 decreases the GTP binding affinity of Arf1. D. the maximal rate of GTP hydrolysis by [Δ17] Arf1 is less than that of wild-type Arf1.
The kinetic parameters of wild type Arf1 without ArfGAP1 and of [Δ17] Arf1 are nearly identical. Combined with passage information that the structure of [Δ17] Arf1 differs from that of wild type Arf1 at the GTP binding site, and that ArfGAP1 stabilizes the Arf1 GTP binding site, the most likely conclusion is that [Δ17] Arf1 is unable to interact with ArfGAP1 at the binding site because of the structural disruption of an alpha helix at the GTP binding site (choice A is correct). The experiment does not test the ability of either wild type Arf1 or [Δ17] Arf1 to bind CTA1 (eliminate choice B). The Km of wild type Arf1 is smaller, rather than greater, in the presence of ArfGAP1, suggesting that ArfGAP1 increases, rather than decreases, the GTP binding affinity of Arf1 (eliminate choice C). The maximal rate of the maximal rate of GTP hydrolysis by [Δ17] Arf1 and wild-type Arf1 depends on kcat, a measure of the catalytic efficiency of the enzymes. Table 1 shows kcat, to be nearly equal, but slightly higher, for [Δ17] Arf1 (eliminate choice D).