MCAT Biochemistry PDF 1-100A Research Analysis by Axilogy.com
1. The correct answer is a. To find this answer you need to use the Michaelis-Menten formula: V0=(Vm)([S])/ Km + [S]. For the Vm value, you can roughly estimate by drawing a line from the top of Cut-Ting’s curve over to the y-axis (≈94). Next you should find the Km value; which can be accomplished by drawing a line from 1⁄2 the Vm to Cut-Ting’s curve, and then down to the x-axis (≈0.8). When these numbers are plugged into the formula, only answer choice A is close.
2. The correct answer is d. All of the statements are true. Note that it’s a perfectly good strategy (for the purposes of answering a multiple choice question) to simply round decimal numbers to make computations simple.
3. The correct answer is B. To find this answer, you need to plug your numbers into the Michaelis-Menten formula (above) and compare the two values (64 and 75, respectively). Next, you will divide 75 by 64, and then multiply by 100. The answer you get is the only one that makes sense. Remember, for the purposes of this exercise it’s absolutely fine to estimate. The goal of this exercise if for you to practice using the Michaelis- Menten formula.
4. The correct answer is d. With the exception of the initial stage of an enzymatic reaction, which is usually completed within milliseconds of mixing enzyme and substrate, the [ES] remains approximately constant until the substrate is nearly depleted. So, the rate of synthesis of ES complex must equal its rate of consumption over most of the course of the reaction. In other words, The ES complex maintains a steady state and the [ES] can be treated as having a constant value.
5. The correct answer is d. Each of the elementary reactions that make up a Michaelis-type enzymatic reaction is characterized by a rate constant: k1 & k -1 are the forward & reverse rate constants for formation of the ES complex (the first reaction), and k2 is the rate constant for the decomposition of ES to P (the second reaction). So we can assume for the sake of mathematical simplicity, that the second reaction is irreversible; that is, no product is converted back to substrate.
6. The correct answer is a. Allosteric effectors bind to a site other than the enzyme’s active site, and either decrease or increase the ability of the enzyme to bind its substrate. This is because the binding of the effector causes a shift in the enzyme’s active site, which will make it better, or worse, at catalyzing its reaction.
7. The correct answer is c. Transition-state analogs are often highly effective inhibitors of enzyme- catalyzed reactions because they bind to the enzyme active site with much higher affinity than do the substrates. For this reason, they are often referred to as “suicide substrates”. This feature makes them of great importance in rational drug design.
8. The correct answer is b. Both dialysis and increasing the concentration of substrate with relieve competitive enzyme inhibition. Increasing the concentration of substrate will allow it “outcompete” the inhibitor for access to the enzyme’s active site.
9. The correct answer is a. ATCase is allosterically inhibited by cytidine triphosphate (CTP), a pyrimidine nucleotide, and is allosterically activated by adenosine triphosphate (ATP), a purine nucleotide. CTP, which is a product of the pyrimidine biosynthetic pathway, is an example of a feedback inhibitor, since it inhibits an earlier step in its own biosynthesis. Thus, when CTP levels are high, CTP binds to ATCase, thereby reducing the rate of CTP synthesis. Conversely, when cellular [CTP] decreases, CTP dissociates from ATCase and CTP synthesis accelerates.
10. The correct answer is d.
11. The correct answer is b. Hydropathy plots are quantitative measures of hydrophobicity (on the y-axis) vs. primary structure (on the x-axis).
a. Is wrong because it describes a Ramachandran plot.
c. Is wrong because a hydropathy plot measures hydrophobicity. Because of this fact, it is primarily used to make structural predictions re. integral (transmembrane) proteins.
d. Is wrong because arginine is a polar charged amino acid. The nonpolar amino acids are: GAFVWILMP. 12. The correct answer is d. Hydropathy plots measure hydrophobic indices of the R-groups as a function of their position in a proteins primary structure (which is the sequence of amino acids in a peptide chain). Additionally, this technique can be used for both soluble proteins and membrane-associated proteins (Which must contain both hydrophilic and hydrophobic residues in order for the protein’s association to be thermodynamically favorable).
b. Is wrong because this answer choice describes what a Ramachandran plot indicates.
13. The correct answer is d. Hydropathy plots can be used to predict the region of a soluble protein that is likely to be buried in the interior. Remember that a soluble protein is soluble in water; which means that the exterior of the protein is hydrophilic in nature. The most likely amino acid residues to reside on the exterior of a protein are the polar uncharged & polar charged ones; TSCQYN & DEHRK, respectively..
a. Is wrong because the y-axis of a hydropathy plot-plots the hydrophobic index; and doesn’t give any information re. the acid-base chemistry of the R-group.
b. Is wrong because leucine is a nonpolar amino acid, and lysine is a polar charged amino acid.
c. Is wrong because this technique wouldn’t give you enough information to fully understand the overall tertiary structure of a protein. Hydropathy plots are mainly used to gain an idea as to where the transmembrane segments of the peptide are most likely to be.
14. The correct answer is b. Proteins are more positively charged when they are in a solution below their isoelectric points than when they are in solutions that are above their isoelectric point. Recall that the isoelectric point (pI) is the point at which a molecule has no net charge. For example, Arginine (Arg, R) has three groups with three different pK’s (α-NH3: 8.99, α-COOH: 1.89, and side chain guanidino group: 12.48). The isoelectric point of arginine is (8.99+12.48)/2= 10.7. In a pH of 7, “the proton” would be on the R-group, giving it a plus charge.
a. Is wrong because a protein (or amino acid) would be negatively charged in a solution that’s above its pI. For example, glutamate (Glu, E) is a an amino acid residue that possesses three ionizable groups, with three different pK’s (α-NH3: 9.47, α-COOH: 2.10, and side chain γ-COOH group: 4.07). The isoelectric point of glutamate can be found by adding the two most-similar groups and then dividing by two, to get 3.1. Because the values of pK groups are referenced to a pH if 7.0, this is what you would make your comparison to. The R group (γ-COOH) would be negatively charged in a pH of 7, because it is more acidic than the solution.
c. Is wrong because proteins are LEAST soluble at their isoelectric point. This is the basis of isoelectric focusing.
d. Is wrong because SDS is a detergent that imparts a large negative charge on what it binds to. SDS is commonly used in techniques that separate molecules by their sizes.
15. The correct answer is d. In a group of weak acids, the strongest will have the smallest pKa. It will also have the largest Ka; pKa=-log Ka.
a. Is wrong because arginine’s side chain has a pKa of 12.48, which is a strong base than aspartate’s (3.90).
b. Is wrong because, intuitively, increasing the [base] in a solution is going to increase the solution; not decrease it.
c. Is wrong for the same reason answer choice b is wrong. Adding acid is going to decrease the pH of the solution, not increase it.
16. The correct answer is d. Serine is a polar charged amino acid, and valine is nonpolar
a. Is wrong because the group can function as both a hydrogen bond donor and acceptor.
b. Is wrong because the hydrophobic core of the bilayer is comprised of phospholipid tails, which are hydrophobic (nonpolar). Serine is a polar uncharged amino acid. The residues that would reside in this region as part of an integral protein would have to be similar in polarity (Nonpolar amino acids: GAFVWILMP).
c. Is wrong for the same reason that answer choice a is incorrect.
17. The correct answer is a. Histidine is likely to be found on the exterior of a soluble protein, because it is a polar charged amino acid.
Answer choices b, c, and d are are incorrect because they are true statements; and the question calls for you to select the false answer choice.
18. The correct answer is b. To find the pI of Glu, you would add the two most similar groups (the R group and the α-COOH group). See the explanation for question 4.
a & d are incorrect for the same reasons. See the explanation for the correct answer choice.
c. Is incorrect because aspartate has an α-COOH group as its side chain; which would be negatively charged if it were ionized.
19. The correct answer is c. To find the answer to these types of questions, follow these steps:
 Draw the peptide from the N-terminus to the C-terminus in the order it is given in the question. (Shown below this explanation)
 Eliminate the numbers that are not relevant (those that lose their ionizability due to their inclusion into a peptide bond). In this case, phenylalanine’s (F) 1.8, arginine’s (R) 1.8 & 9.0, all of proline’s (P) pKa values, and tyrosine’s (Y) 2.0 & 9.4, groups.
 Compare each of the numbers left to the pH of the solution, and assign each group a charge. If the number is lower than that of the solution, then “the proton” will go to the higher number. If the number is higher than that of the solution, the inverse is also true. For example the N-terminus of the peptide has a pKa of 9.2, which in a solution with a pH of 2, will be protonated (as will the side chain groups of R & Y). Note that if the pKa is equal to the pH of the solution, if will possess half of its charge, because 50% of the time “the proton” will be on the ionizable group and 50% of the time “the proton” will be in solution. In this example, the C-terminus (Y’s COOH group) has a -1/2 charge.
 Add the charges. From left to right: +1 + +1 + -1/2 = +1.5.
20. The correct answer is d. Isoleucine is a nonpolar amino acid. The others (arginine, aspartate, and asparagine are polar).
21. The correct answer is d. Kinetic data can be plotted in double-reciprocal form to determine KM and Vmax. The formula changes from the Michaelis-Menten-type to 1/V0=(KM/Vm)(1/[S] + 1/VM). A plot of this equation is linear and has a slope of KM/ VMax, a 1/[S] intercept of -1/KM, and a 1/V0 intercept of 1/VMax.
22. The correct answer is c. The intercept of this axis is -1/KM. 23. The correct answer is a. The intercept of this is axis is 1/VMax.
24. The correct answer is d. Double the amount of enzyme will double the rate of the reaction. The Michaelis constant, KM, is defined as KM=k-1+k2/k1. To find this value on a reaction rate v [S] plot, you need to divide
the VM by 2, course this value over to the curve, and down to the x-axis.
25. The correct answer is d. No matter how many intermediates are introduced, the model will still fit the
Michaelis E + S ⇌ES!E+P form.
26. The correct answer is d. If you have information about a given enzyme’s KM and the concentration of
substrate, you can work backwards to find the VMax, and thus the value for V0.
27. The correct answer is b. Strictly speaking; KM is not a direct measure of an enzyme’s affinity for a substrate. However, it is indirectly related to affinity. If the catalysis step is very fast, then affinity and KM can appropriately be considered to have approximately equal values.
28. The correct answer is c. kcat=VM/[E]T. This quantity is also known as the turnover number, because it is the number of reactions that each active site can catalyze per unit of time.
29. The correct answer is b. The ES complex maintains a steady state and [ES] can be treated as having a constant value: d[ES]/dt=0.
30. The correct answer is c. An enzyme displaying Michaelis kinetics will always have a hyperbolic shape on a V0 v. [S] plot. Sigmoidal curves indicate an allosteric enzyme.
31. The correct answer is a. There is an absence of rotation around a peptide bond due to resonance.
b. is incorrect because the figure shows an arrow around the Cα-N bond.
d. is incorrect because a Ramachandan plot would give information provided by the groups that rotate, and are not a part of the squared region of the peptide in the figure.
32. The correct answer is d. The easiest way to solve this problem is by drawing out a peptide and comparing it to the answer choices. The alpha carbon of an amino acid is connected to four groups: H3N, H, an R group, and COOH. When incorporated into a peptide, carboxylate group is connected to the amino group of the next residue, as well to its amino group. Only answer choice d portrays a proper peptide bond.
33. The correct answer is b. Amino acid R-groups are found on the outside of the helix spiral only. Both right and left handed helices form, with the vast majority occurring in the right-handed conformation. The hydrogen bonds that form the helix are generated by hydrogen bonds between amino and carboxyl groups, not R-groups.
34. The correct answer is b. The tertiary structure (or spacial arrangement in 3-D) is a function of amino acid residues that are generally near each other in sequence, but not necessarily side-by side.
35. The correct answer is d. X-ray diffraction (XRD, or X-ray crystallography) is a technique used to identify the atomic and molecular structure of a crystal. Light and electron microscopy do not provide this level of detail.
36. The correct answer is c. Due to resonance, the C/O bond is significantly longer, and the C-N peptide bond is shorter.
a. is incorrect because the barrier to free rotation around the peptide bond is high, due to its partial double bond character.
d. is incorrect because, for example, there is a charge on the oxygen of the carbonyl group when their is only one bond connecting it to carbon.
37. The correct answer is a.
b. is incorrect, because formation of a peptide bond requires the input of energy (endergonic). It takes energy to build things, and breaking them down into their smaller counterparts releases energy.
c. is incorrect, because a peptide is an example of an amide bond. Ester bonds (RCOOR) are present in fats.
d. is incorrect, because the formation of peptide bond is a condensation reaction. To help yourself remember this, know that during a condensation linking two amino acids, water is released, and condenses on your glassware!
38. The correct answer is b. Antiparallel β-sheets are stronger than parallel ones.
a. is incorrect, because interactions between the N & C groups of amino acids stabilize both β-sheets and α- helices.
c. is incorrect. There are many examples of amino acids that contain both types of secondary structure.
39. The correct answer is b. Leaving the disulfide bonds intact would allow you to see where they reside in the native protein.
a. is incorrect, because carboxypeptidase is specific to the carboxyl-terminus, not the side chain.
c. is incorrect, because cyanogen bromide (CNBr) cleaves proteins at Met residues, but alters the methionine side chain. Unless the next amino acid in the sequence is serine, CNBr will produce homoserine lactone.
d. is incorrect, because some amino acids are so near each other in atomic mass (for instance glutamine: 128.1 and lysine: 128.2) that other techniques must be utilized to distinguish them from each other.
40. The correct answer is d. Cis configurations (Cis=same side) are generally less favorable that trans (opposite sides) due to steric hinderance.
b. is incorrect, because there is restricted rotation about a peptide bond.
c. is incorrect, because peptide bonds do not create the chemical difference between α-helices and β-sheets.
41. The correct answer is d.
42. The correct answer is d.
43. The correct answer is d.
44. The correct answer is c.
45. The correct answer is d.
46. The correct answer is d. (k-1 + k2)[ES] ≈ k1 [E][S]. k-1 is the dissociation of the ES complex. k2 is the formation of product. k1 is the formation of the ES complex from E + S.
47. The correct answer is a. Adding additional intermediates won't change the basic mathematical form of the Michaelis-Menten formula, so the model can be used with enzymes that have more than one intermediate and their reaction mechanism.
c. The Km is not an accurate measure of every enzyme’s affinity for its substrate, although you can generally make this assumption.
48. The correct answer is b. The free energy of reaction depends on the identity of the reactants and products. Remember that the ∆ G is the change in Gibbs’ free energy, which is the energy required or released to take a specific amount of reactant converted to products. The model can only be used for enzymes that have sigmoid curve, not allosteric enzymes.
49. The correct answer is c. The addition of the catalyst alters the ∆ G of the transition state for both the forward and reverse reactions. The enthalpy term, or heat, will affect the entropy term. For example, increasing the enthalpy term to make an endergonic reaction with a negative ∆ S value exergonic.
50. The correct answer is c. The induced fit model describes the interaction between a substrate and an enzyme in which the three dimensional shape of the enzyme changes due to its chemical binding with the substrate. Covalent Bond formation between substrate and enzyme is in fact a feature of some enzyme mechanisms, but there are other types such as acid-base catalysis. Most enzymes are able to distinguish between pro-chiral forms of substrate. The transition state of an enzyme-catalyzed reaction will fall lower than that of the uncatalyzed reaction.
51. The correct answer is c. The ∆G of the transition state is smaller in the forward direction then it is for the
reverse direction for a reaction that is exergonic in the forward direction. Catalysts accelerate the rates of reaction by lowering the ∆G of the transition state. Catalysts do not decrease did ∆G0 of the overall reaction, they simply decrease the ∆G of the transition state.
52. The correct answer is a. The presence of the catalyst alters only the ∆G of the transition state, which is the highest point on a free energy versus reaction cord mid diagram. The presence of a catalyst alters the rate of both the forward and reverse reactions.
53. The correct answer is c. Competitive inhibitors bind to the enzyme via non covalent interactions, therefore they inhibition can be relieved by either dialysis or by increasing the concentration of substrate. A noncompetitive inhibitor find site other than the active site, therefore increasing the concentration of substrate nor dialysis will relieve inhibition.
54. The correct answer is c. k-1 is the mathematical representation of the substrate dissociating from the enzyme. In the Michaelis-Menten model, k-1 and k1 are first-order rate constants.
55. The correct answer is a. In practice it is very difficult to assess Vmax accurately from direct plots of vo versus [S], because, even at substrate concentrations as high as [S] 10 KM, the Michaelis-Menten equation indicates that vo is only 91% of Vmax, so that the value of Vmax will almost certainly be underestimated. A better method for determining the values of Vmax and KM uses the reciprocal of the Michaelis–Menten equation. These values can be plotted on a Lineweaver-Burk plot.
56. The correct answer is d. A competitive inhibitor will increase the Km of an enzyme, but will have no effect on the maximum velocity (Vmax). Additionally, a competitive inhibitor binds to the enzyme at the active site. 57. The correct answer is a. this question needs revision
58. The correct answer is c. Transition states occur at the highest point of an enzyme reaction coordinate diagram. As stated above, catalyst stabilized the transition state for both the forward and reverse directions. Transition state analogs buying two enzymes at their active sites with a higher affinity than their natural substrate. Competitive inhibition occurs at the active site only.
59. The correct answer is d.
60. The correct answer is d. PFK-1 is the major regulatory enzyme in the glycolytic pathway. It is under hormonal control by both insulin and glucagon. Controlling this enzyme prevents what is known as a futile cycle, and prevents the simultaneous occurrence of both glycolysis and gluconeogenesis.
61. The correct answer is a. a high citric concentration “tells” the cell that there is plenty of substrates available further down the pathway for the generation of cellular energy. Therefore, the glycolytic pathway will be inhibited an intermediate will be fluxed towards areas where they can better be utilized.
62. The correct answer is b. F6P activates 6-phosphofructo-2-kinase & inhibits fructose-2,6-bisphosphatase, thus increasing the fructose 2,6-bisphosphate-level. This promotes PFK activity, which will drive carbohydrate degradation. This is also known as feed forward activation.
63. The correct answer is d. Insulin promotes gluconeogenesis, which is the generation of glucose from non- carbohydrate precursors. This is accomplished by the inhibition of PFK-1. conversely, glucagon causes the phosphorylation of PFK; which acts to promote the glycolytic pathway by activation of the enzyme.
64. The correct answer is b. Epinephrine, or adrenaline, is released by your adrenal glands, and acts to speed up your metabolism to provide you with cellular energy in short bursts. The glycolytic pathway represents a means to quickly produce this needed energy.
65. The correct answer is b. One fact re. gluconeogenesis that you should always remember is that it only occurs in the liver(hepatocytes), and to a lesser extent, the kidney. This pathway does not occur and muscle cells, ever.
66. The correct answer is c. As described above in relation to phosphofructokinase, feed–forward activation allows for a later step and metabolism to control a key step in another pathway; in this case glycolysis.
67. The correct answer is d. A futile cycle is prevented by the action of bifunctional enzyme; which controls the rates of both glycolysis and gluconeogenesis.
68. The correct answer is c.
69. The correct answer is a.
70. The correct answer is d.
71. The correct answer is b. 72. The correct answer is d. 73. The correct answer is d. 74. The correct answer is b. 75. The correct answer is c. 76. The correct answer is b. 77. The correct answer is d. 78. The correct answer is b. 79. The correct answer is a. 80. The correct answer is d.
81. The correct answer is d. 82. The correct answer is d. 83. The correct answer is c. 84. The correct answer is d. 85. The correct answer is b. 86. The correct answer is d. 87. The correct answer is c. 88. The correct answer is c 89. The correct answer is b. 90. The correct answer is c.
91. The correct answer is b. 92. The correct answer is a. 93. The correct answer is c. 94. The correct answer is a. 95. The correct answer is b. 96. The correct answer is b. 97. The correct answer is c. 98. The correct answer is d. 99. The correct answer is c. 100. The correct answer is a.