Princeton Review MCAT Practice Test Questions and Answers (Latest Update 2023) Verified An, Exams of Nursing

Download Princeton Review MCAT Practice Test Questions and Answers (Latest Update 2023) Verified An and more Exams Nursing in PDF only on Docsity!

Answers (Latest Update 2023) Verified Answers

Sickle-cell anemia (HbS) results from a substitution to valine from glutamic acid at position 6 of the β chain of hemoglobin. Which of the following best explains why the isoelectric point of HbS is higher than that of HbA? A. The side chain of glutamic acid is less acidic than that of valine. B. Glutamic acid is isoelectric at a lower pH than is valine. C. Valine is isoelectric at a lower pH than is glutamic acid. D. Glutamic acid has a net charge of 0 at its isoelectric point. - Correct Answer✅B. Glutamic acid is isoelectric at a lower pH than is valine. The isoelectric point is the pH at which an amphoteric molecule has a net electric charge of zero. Glutamic acid is more acidic, not less acidic than valine (A is wrong). Glutamate would require more acidic pH to neutralize its charge (C is wrong). D is a true, but irrelevant statement. The best answer is B. The negative charge of glutamic acid would require a more acidic pH to neutralize, making the isoelectric point of HbA (with glutamic acid) lower than that of HbS (with valine). A mixture of aspartate and phenylalanine is separated into its component molecules by thin layer chromatography on a silica plate eluted with benzene. Which of the following best explains why the separation occurs? A. Aspartate will move farther with the mobile phase, because it has a polar side chain.

Answers (Latest Update 2023) Verified Answers

B. Aspartate will move farther with the mobile phase, because it has a nonpolar side chain. C. Phenylalanine will move farther with the mobile phase, because it has a polar side chain. D. Phenylalanine will move farther with the mobile phase, because it has a nonpolar side chain. - Correct Answer✅D. Phenylalanine will move farther with the mobile phase, because it has a nonpolar side chain. The side chain on aspartate is -CH2COO-, which is very polar, while the side chain on phenylalanine is -CH2Ph, which is nonpolar. Since "like dissolves like," and the mobile phase is nonpolar, the molecule which is less polar (phenylalanine, in this case) will move farther with the mobile phase. SEE OTHERSIDE FIRST D. Thickness of the membrane Correct Answer "Charge stored per unit voltage," Q / V, is the definition of capacitance, C. The equation for the capacitance of a parallel-plate capacitor is C = κε0A / d, where κ is the dielectric constant, ε0 is a universal constant (the permittivity of free space), A is the area of each plate, and d is the distance between the plates. Of the choices given, only D, decreasing the thickness of the membrane (that is, decreasing d), would increase the capacitance, C. - Correct Answer✅Ion flow in neurons can be characterized as an electrical circuit for both the resting neuron (Figure 1) and the active neuron (Figure 2). The membrane is a capacitor, slow leakage channels are a 25 MΩ resistor, and the Na+/K+ pump is a voltage generator. In a resting axon, there is no net transfer of charge across the axon membrane. Figure 2 includes the

Answers (Latest Update 2023) Verified Answers

additional Na+ influx (a 4 kΩ resistor) of an action potential. Other ion fluxes are ignored. According to the figures, decreasing which of the following would create the greatest increase in charge stored per unit voltage on an axon membrane in its rest state? A. Leakage channel resistance B. Na+ channel resistance C. Area of the membrane surfaces D. Thickness of the membrane Across the membrane of an axon in its rest state: A. the potential is higher on the outside of the cell than the inside, and electric field lines run from the outside to the inside. B. the potential is higher on the outside of the cell than the inside, and electric field lines run from the inside to the outside. C. the potential is lower on the outside of the cell than the inside, and electric field lines run from the outside to the inside. D. the potential is lower on the outside of the cell than the inside, and electric field lines run from the inside to the outside. - Correct Answer✅A. the potential is higher on the outside of the cell than the inside, and electric field lines run from the outside to the inside. The voltage across the membrane is -70 mV from the exterior to the interior of the cell; thus, the potential is higher on the exterior of the cell (C and D are wrong). Since the potential is higher on the exterior than in the interior,

Answers (Latest Update 2023) Verified Answers

the electric field lines must point from the outside to the inside (choice A). IN PHYSICS, FIELD LINES GO TOWARD THE MORE NEGATIVE AREA. Which of the following statements regarding RNA molecules is NOT true? A. RNAs can act as enzymes to catalyze reactions. B. Some RNAs have more than four different types of bases. C. Some RNAs are synthesized in the nucleolus. D. RNAs are insusceptible to alkaline hydrolysis. - Correct Answer✅D. RNAs are insusceptible to alkaline hydrolysis. RNA molecules have decreased stability compared to DNA in part because of their susceptibility to alkaline hydrolysis due to the presence of hydroxyl group at 2'-C position (choice D is not true of RNA and is the correct answer choice). Some RNAs have enzymatic function (such as in telomerase) and they are termed ribozymes (A is true). tRNA has unique and modified bases apart from the traditional four bases A,U,C, and G (such as inosine, B is true). rRNA is synthesized in the nucleolus (C is true). Two blocks are suspended from the ends of a massless meter stick; a 4kg weight from one side and a 1kg weight from the opposite side. How far from the center of the stick must the rope be attached in order to maintain rotational equilibrium? A. 10 cm B. 20 cm

Answers (Latest Update 2023) Verified Answers

C. 25 cm D. 30 cm - Correct Answer✅D. 30 cm 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. - Correct Answer✅B. power of each laser pulse. Since the overall energy of the laser does not change, neither will the frequency, wavelength, nor work done by the laser. This leaves choice B: Concentrating the energy into a shorter time period increases the power of each pulse (since power equals energy delivered per unit time, by definition). A CO2 laser used as a laser scalpel produces a beam of laser light with a wavelength of 10.6 μm. Compared to the excited electrons in a laser with a wavelength of 1.06um, the excited electrons in the CO2 laser most likely have: A. greater mass.

Answers (Latest Update 2023) Verified Answers

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. - Correct Answer✅D. a smaller energy difference between their normal state and their excited state. The wavelength of the CO2 laser is 10 times greater than the 1. wavelength laser. Therefore, the frequency and the energy of the CO2 laser are 10 times lower. Since the laser light is the energy released by transitions of electrons dropping to a lower energy level, less emitted energy implies a smaller difference between the energy levels. The mass of an electron is independent of its atomic or molecular energy state (A and B are wrong). If the front mirror in a laser, which is made of glass of refractive index 3/2, were repositioned so that the laser beam strikes at an angle of 30° to its normal, what would be the angle of reflection? A. sin-1(1/3) B. 30° C. sin-1(1/31/2) D. 60° - Correct Answer✅B. 30°

Answers (Latest Update 2023) Verified Answers

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

Answers (Latest Update 2023) Verified Answers

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

Answers (Latest Update 2023) Verified Answers

B. a Lewis base that has two stereoisomers. C. an amphiphilic molecule that has four stereoisomers. D. a Lewis base that has four stereoisomers. SEE OTHER SIDE FIRST Protonation of the carbonyl group at the oxygen atom in the first step of the reaction in Figure 1 makes the carbonyl-carbon a good electrophile. As the protonated carbonyl was an ester, a carboxylic acid derivative, it will undergo an addition-elimination reaction. Water acts as the nucleophile, attacking the electrophilic carbonyl at the carbon atom as shown below (B is correct, eliminate choice C). The carboxylate is then eliminated as the leaving group (eliminate choice D). A ligand is a molecule that binds to a central metal atom to form a coordination complex. There are no metal atoms involved in the reaction (eliminate A). - Correct Answer✅In the second step of the reaction shown in Figure 1, the role of water can best be described as a(n): A. ligand. B. nucleophile. C. electrophile. D. leaving group. SEE OTHER SIDE FIRST Item I is false. There is no change in the oxidation states in any of the individual atoms in the reaction shown in Figure 1 (eliminate A and C). Item

Answers (Latest Update 2023) Verified Answers

II is true. During the reaction, an ester is converted into a carboxylic acid. Carboxylic acid derivatives undergo an addition reaction followed by an elimination reaction. Item III is false (eliminate choice D). Saponification is the base-mediated hydrolysis of an ester. While an ester is hydrolyzed in the reaction, it is achieved under acidic conditions. The answer is B. - Correct Answer✅Which of the following occur(s) in the reaction shown in Figure 1? I. Oxidation II. Elimination III. Saponification A. I only B. II only C. I and II only D. II and III only SEE OTHER SIDE FIRST Figure 2 shows the rate of lipid hydrolysis of the lipid DOPC in the presence and absence of raclopride by measuring the concentration of the hydrolyzed lipid, lyso-PC, over time. While the data show that the rate of lipid hydrolysis is appreciable in the presence of raclopride and negligible in its absence, as raclopride acts a simple Brønsted acid, it would be incorrect to say that this exact molecule is required for any kind of lipid hydrolysis (eliminate A). Figure 2 does show that the rate of background lipid hydrolysis is

Answers (Latest Update 2023) Verified Answers

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

Answers (Latest Update 2023) Verified Answers

D. Closer, because blue light is refracted less due to its longer wavelength - Correct Answer✅B. Closer, because blue light is refracted more due to its shorter wavelength Correct Answer Choices C and D can be eliminated immediately since blue light has a shorter wavelength than red light. The refractive index of a transparent medium increases with increasing frequency of the transmitted light. Thus, the refractive index of the lens for blue light would be slightly higher than for red light. This would imply that blue light experiences more refraction than red light. Since the lens is a converging one (because diverging lenses cannot form real images), the blue light would be bent more sharply toward the axis than the red light, so the image of the blue object would be formed closer to the lens. SEE OTHER SIDE The pictured epoxide has two chiral centers, but it also has a plane of symmetry. For this reason, it is a meso compound, and its two chiral centers must have opposite configurations (one is R while the other must be S). - Correct Answer✅A. It is meso. B. Its absolute configuration is 7R, 8R C. Its absolute configuration is 7S, 8R. D. Its absolute configuration is 7S, 8S. Which one of the following statements concerning the SN2 reaction mechanism is true?

Answers (Latest Update 2023) Verified Answers

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. - Correct Answer✅C. It proceeds best with primary substrates and is a one-step reaction. SN2 cannot operate with a tertiary substrate because the nucleophile must attack at the same time the leaving group leaves, and tertiary substrates are too sterically hindered (eliminate choice A). SN2 goes with inversion of stereochemistry, not retention (eliminate choice B). SN1 reactions have carbocation intermediates (eliminate choice D), while reactions are one-step processes with no intermediates. Choice C is the best answer because unhindered substrates allow the nucleophile to attack while the leaving group leaves. A cyclist collides with a tree, and the collision creates an intense compressive stress on the cyclist's left humerus (upper arm bone), causing a fracture. Which of the following changes in the initial conditions most likely could have prevented the fracture? A. A decrease in his velocity by a factor of two B. A decrease in his mass by a factor of two C. An increase in his mass by a factor of two D. A decrease in the cross-sectional area of his humerus by a factor of two - Correct Answer✅A. A decrease in his velocity by a factor of two

Answers (Latest Update 2023) Verified Answers

The cyclist's kinetic energy just before the impact will be converted into elastic and inelastic energy of deformation upon impact to cause damage to the tree and to the cyclist. Decreasing this kinetic energy will result in a less violent collision. Because KE = (1/2)mv^2, a decrease in the cyclist's velocity would have the greatest effect (since v is squared) in lowering his KE. SEE OTHER SIDE FIRST The passage states that cartilage does not withstand tension as well as bone. Therefore, the Young's modulus (E) of cartilage must be lower than that of bone, eliminating choices B and D. Since stress = E × strain, a lower value of E corresponds to a lower stress for the same strain, so the graph of stress vs. strain in the elastic region would be less steep; that is, the angle θ would be smaller. Thus the answer is "C. smaller, because cartilage has a smaller Young's modulus than does bone." - Correct Answer✅If the graph were redrawn for cartilage undergoing tensile stress, then the angle θ would be: A. greater, because cartilage has a smaller Young's modulus than does bone. B. greater, because cartilage has a larger Young's modulus than does bone. C. smaller, because cartilage has a smaller Young's modulus than does bone. D. smaller, because cartilage has a larger Young's modulus than does bone. Paget's disease is characterized by massive new bone formation resulting in softening of the bone. A patient with this disease may exhibit bone deformities such as bowing of the tibia or humerus. Compared to normal bone, a sample of bone from an individual with Paget's disease would most likely have:

Answers (Latest Update 2023) Verified Answers

A. a smaller Young's modulus and a lower yield point. B. a smaller Young's modulus and a higher yield point. C. a larger Young's modulus and a lower yield point. D. the same Young's modulus and a lower yield point. - Correct Answer✅A. a smaller Young's modulus and a lower yield point. Paget's disease weakens the bone and thus decreases its Young's modulus. This eliminates choices C and D. The bowing of the tibia and humerus implies that the bone has gone past the yield point, implying that this disease also lowers the yield point of the patient's bones. SEE OTHER SIDE FIRST From Figure 1, the elastic energy density is equal to the area under the stress vs. strain graph. If the yield point has not been reached, then the region under the graph has the shape of a right triangle, whose area is (1/2) × base × height = (1/2) × strain × stress. In this case, the strain is 1% = 10^-2, and the stress is F = 2 × 10^4 N divided by A = 2 × 10^-4m2, which equals 10^8 N/m2. Thus, the elastic energy density is (1/2)(10-2)(108 N/m2) = 5 × 10^5 J/m3. - Correct Answer✅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/m B. 1 × 104 J/m

Answers (Latest Update 2023) Verified Answers

C. 5 × 105 J/m D. 1 × 106 J/m Which one of the following statements is true regarding PCl3? A. The phosphorus has sp2 hybridization, and the molecule has a trigonal pyramid shape. B. The phosphorus has sp3 hybridization, and the molecule has a tetrahedral shape. C. The phosphorus has sp2 hybridization, and the molecule has a trigonal planar shape. D. The phosphorus has sp3 hybridization, and the molecule has a trigonal pyramid shape. - Correct Answer✅D. The phosphorus has sp3 hybridization, and the molecule has a trigonal pyramid shape. Attack this 2x2 style question first by hybridization, and then by molecular shape. The phosphorous atom has three bonds to chlorine and a lone electron pair, for a total of four electron groups. As such it must be sp hybridized (eliminate A and C). Because the fourth electron group is a lone pair, the shape cannot be tetrahedral (eliminate B). Which one of the following compounds would be most soluble in ethanol? A. Methane, CH B. Formic acid, HCOOH C. Stearic acid, CH3(CH2)16COOH

Answers (Latest Update 2023) Verified Answers

D. 1-Chlorooctodecane, CH2Cl(CH2)16CH3 - Correct Answer✅B. Formic acid, HCOOH B. Choices A and D cannot hydrogen bond with ethanol, so should be quickly eliminated. Formic acid is a highly polar molecule, making it soluble in the polar alcoholic solvent. However, stearic acid has a long (18-C) nonpolar chain. Therefore, it will be less soluble in ethanol. Formic acid, HCOOH (Ka = 1.8 × 10-4), dissociates in water according to Reaction 1. What is the hydrogen ion concentration in a 0.05 M aqueous solution of HCOOH? A. 9.0 × 10^-6 M B. 3.6 × 10^-4 M C. 3.0 × 10^-3 M D. 6.0 × 10^-2 M - Correct Answer✅ SEE OTHER SIDE FIRST The most useful indicator for the titration in question will change from its undissociated to dissociated form around the pH of the equivalence point. Since the second equivalence point for the amino acid shown is roughly at a pH of 10.5-11.0 (the steep part of the curve at the far right of the graph), the best indicator will be one color below this range and another color above it. Only alizarin yellow changes at such a high pH range. - Correct Answer✅Which indicator would be most useful in signaling the second equivalence point of a typical neutral amino acid in a titration using NaOH?

Answers (Latest Update 2023) Verified Answers

A. Methyl red B. Phenol red C. Phenolphthalein D. Alizarin yellow SEE OTHER SIDE FIRST Since bromothymol blue changes from yellow to blue over the range of pH 6.0-7.6 (while the pH increases, or the [H+] decreases), at a pH below 6.0 the solution will be yellow, above 7.6 it will be blue, and between 6.0-7.6 it will be green. Therefore, the transition point in question occurs at a pH of 7. (eliminate C). Since pH + pOH = 14, that means the pOH of the solution should be 7.6 + pOH = 14, or 6.4 (eliminate A and D). - Correct Answer✅If bromothymol blue changes from yellow to blue as the pH of the solution increases, at what value of pOH will a solution containing this indicator just begin to turn from blue to green? A. 6. B. 6. C. 7. D. 8. Perceiving the color of a pH indicator requires what type of visual processing?

Answers (Latest Update 2023) Verified Answers

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. - Correct Answer✅C. Hyperpolarization of cone cells to trigger depolarization of bipolar neurons. The answer choices contain pairs of information so proceed with a 2x question elimination strategy. When cone cells are not exposed to light and are thus not processing the visual stimuli of color, they are depolarized. Since the question is asking about perceiving color, the cone cells need to be hyperpolarized (eliminate choices A and B). When cone cells hyperpolarize, the bipolar neurons are no longer being inhibited and can then depolarize in order to transmit signal to the ganglion and eventually the optic nerve (C is correct; eliminate D). Which of the following is the best explanation for why urea, CO(NH2)2, may be used to study the unfolding of the protein scr SH3? A. The entropic loss due to unfolding of scr SH3 is a result of urea increasing the solubility of the protein interior. B. Urea acts to stabilize the denatured state of scr SH3 by associating with backbone polar groups and charged side chains. C. Urea stabilizes the multiple transition states observed for the unfolding of scr SH3.

Answers (Latest Update 2023) Verified Answers

D. The addition of urea reduces S—S bonds in the structure of scr SH3. - Correct Answer✅B. Urea acts to stabilize the denatured state of scr SH3 by associating with backbone polar groups and charged side chains. Using process of elimination can quickly narrow the options given in the answer choices. A is inconsistent because unfolding will cause entropy to increase, not decrease, due to the disordering effect of unfolding (eliminate A). Based on its structure, urea, CO(NH2)2, will interact with the polar moieties of a protein via hydrogen bonding, which include the peptide backbone and polar residues (charged or uncharged). By disabling intramolecular hydrogen bonding urea is capable of stabilizing the denatured state of a protein (choice B is correct). The passage stated only a single unfolding pathway is followed when a chemical denaturant, such as urea, is applied. Thus, there can only be one transition state (eliminate choice C). Choice D is incorrect because urea is not a reducing agent and will not affect disulfide bonds. The transition state's activation free energy (G^TS) is composed of the enthalpy (H^TS) and entropy (S^TS) of activation by the same relation as free energy, enthalpy and entropy are related for energy minima (G^F, H^F and S^F). Which of the following is likely true regarding the transition from the folded state (F) to the transition state (TS)? A. ΔH (F → TS) < 0 B. ΔG (F → TS) < 0 C. ΔS (F → TS) < 0 D. ΔH (F → TS) > 0 - Correct Answer✅D. ΔH (F → TS) > 0

Answers (Latest Update 2023) Verified Answers

B can be immediately eliminated, as the free energy of activation will always be higher than the free energy of either minimum it connects (think of the reaction coordinate graph). To eliminate C it is helpful to remember that proteins are highly ordered macromolecules. In fact, just the term "unfolding" implies that there is an orderly arrangement being undone, and hence it is very unlikely that such unfolding would require a negative change of entropy. This, on its own, means that D is correct as ΔG is positive from the folded state to the transition state and ΔS is also positive (ΔG = ΔH - TΔS). Still, reasoned further, the bonding (enthalpic) components that hold a folded protein together must overcome the entropic penalty associated with this ordering. Therefore, ΔH (TS → F) must be very negative, meaning the opposite process is very positive, and choice D is correct. 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' - Correct Answer✅C. 5'-CAU-3' The start codon on the mRNA would be 5'-AUG-3', therefore, the anticodon on the tRNA would be the complementary sequence: 3'-UAC-5' (or alternatively written, 5'-CAU-3'). SEE OTHER SIDE FIRST

Answers (Latest Update 2023) Verified Answers

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. I and II, being alcohols, will have high boiling points due to their ability to hydrogen bond, so they are incorrect answers. III and IV are both nonpolar compounds, but they differ in size by one carbon. The compound with more carbons (and less branching) in choice IV, 2,3-dimethylpentane, will have a higher boiling point because it has greater London dispersion forces, so is also incorrect. III is the best answer. - Correct Answer✅ SEE OTHER SIDE FIRST 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. - Correct Answer✅ SEE OTHER SIDE FIRST D. The charge of the droplets differs because the amount of ionization due to the x-rays is random.

Answers (Latest Update 2023) Verified Answers

A can be eliminated because, though it is true that drag for these droplets is proportionate to diameter, that fact is irrelevant because there is no drag for something that isn't moving, like a droplet in static equilibrium. Moreover, the fact that drag wasn't mentioned in the passage should make one suspicious of this choice. B is directly contradicted by the data: compare drops #1 and #2, for example. C is also inconsistent with the given information, namely that density of the oil is a constant 920 kg/m3. D is correct: there is no way to control the exact number of excess electrons in the ionized air that will cling to a falling oil droplet, but the net charge must be a whole number multiple of the elementary charge. - Correct Answer✅What best explains the variation among the values of the floating voltage for the different droplets? A. As the diameter of the droplets changes, the drag force each experiences as it falls changes proportionately. B. Larger volume droplets require greater voltage to achieve static equilibrium. C. Smaller droplets have greater density, which in turn affects the voltage needed to keep them afloat. D. The charge of the droplets differs because the amount of ionization due to the x-rays is random. SEE OTHER SIDE FIRST The passage states that no significant difference is observed between the obese and lean mice. While this may seem to contradict the leptin resistance hypothesis, leptin is a hormone which will be expected to affect target tissues differently. In the passage, the leptin resistance hypothesis

Answers (Latest Update 2023) Verified Answers

specifically relates to satiety and, while it is known that leptin activates the sympathetic nervous system, it is unclear how renal sympathetic nerve activity (RSNA) is related to the sensation of satiety. Therefore the lack of a significant difference between obese and lean mice does not provide significant evidence to strengthen or weaken the leptin resistance hypothesis (choice D is correct). The lack of difference between RSNA in lean and obese mice is not supportive of the hypothesis that leptin resistance inhibits leptin function (answer choice A is wrong). As mentioned above, leptin resistance is primarily associated with mitigating a sense of satiety when full; it is not clear whether an increase in renal sympathetic activity should be included within the leptin resistance syndrome (answer choice B is wrong). Given the dramatic increase in nerve activity with increasing - Correct Answer✅While it has been shown that leptin can activate the sympathetic nervous system, the most classic characteristic of leptin resistance is a decrease in the sense of satiety after feeding. In order to test whether leptin resistance in obese participants affected all functions of leptin, an experiment was performed in which yellow obese agouti mice and normal lean mice were injected intravenously with three different leptin concentrations while renal sympathetic nerve activity (renal SNA) was measured. Surgery was performed to expose the appropriate nerve fascicles near the kidney and platinum electrodes recorded the multi-fiber renal SNA (RSNA) in anesthetized mice. Both groups showed an enhancement of renal sympathetic activity correlating to the increased leptin injection. No significant difference in RSNA activity was detected between lean and obese mice as depicted in Figure 1. This research suggests that leptin resistance in obese individuals may be a selective phenomenon in the sense that not all functions of the hormone are affected by leptin resistance.

Answers (Latest Update 2023) Verified Answers

What impact do the results shown in Figure 1 have on the hypothesis that leptin resistance secondary to obesity reduces all effects of leptin? A. Supports the hypothesis due to the increase in neural activity with injection of leptin in obese mice B. Weakens the hypothesis due to the increase in neural activity with injection of leptin in obese mice C. Weakens the hypothesis due to the lack of impact of leptin on renal sympathetic nerve activity. D. Neither supports nor weakens the hypothesis due to the differential effect of hormones on differing tissues. A researcher proposed to block the leptin receptor in order to treat hypertension in obese patients. Which of the following detection methods would work best to identify the receptor in the different tissues affected by leptin? I. Northern blot II. Western blot III.Southern blot A. I only B. II only C. I and III only