MCAT Biochemistry Questions 101-200ANS Reading Graphs MCAT Research Analysis
MCAT Biochemistry Questions 101-200ANS Reading Graphs MCAT Research Analysis
Explanations for 101-200
101. The correct answer is d. ATP inhibits PFK, allosterically, to prevent glycolysis from running when there is
an abundance of cellular energy. Likewise, AMP stimulates the enzyme. Fructose-6-phosphate has to be
available for PFK to convert it to Fructose-1, 6-bisphosphate. It does so by activating Fructose-2, 6-
phosphatase (in the liver). Feedback regulation occurs at an early step of a pathway, so pyruvate kinase is not a
likely candidate to function in attenuating glycolysis.
102. The correct answer is b. Citrate is also one of the feedback regulators of glycolysis. It inhibits PFK-1.
103. The correct answer is d. In aerobic conditions, Pyruvate is converted to Acetyl-CoA so that it can enter
the TCA cycle. In anaerobic/hypoxic conditions, lactic acid or alcohol (in yeast) is formed. This is why you get
sore after strenuous exercise.
104. The correct answer is d. To find the number of stereoisomers on a carbohydrate, you need to know how
many chiral centers there are. A Ketohexose has 3 chiral centers and 2n= 8 different stereoisomers.
105. The correct answer is a. Lactose is a disaccharide, comprised of a glucose and a galactose linked by an
106. The correct answer is d. In aqueous solution, the most stable conformation is the chair conformation.
107. The correct answer is a. The OH group on carbons 1 & 4 of two glucose monomers form an acetyl
linkage to form the majority of bonds in starch. The branch points are between carbons 1 & 6.
108. The correct answer is b. The organic chemistry definition of an acetal is a functional group with the
following connectivity R2C(OR')2, where both R' groups are organic fragments. The central carbon atom has
four bonds to it, and is therefore saturated and has tetrahedral geometry. The two R'O groups may be
equivalent to each other or not.
109. The correct answer is b. The C5 hydroxyl group attacks the C1 carbon, which leaves the sixth carbon
(CH2OH) moiety “hanging off ” the glucose molecule.
110. The correct answer is b. This describes the majority of glycosidic linkages in starch.
111. The correct answer is b. There are three chiral centers on an aldopentose. 2n= 8.
112. The correct answer is d. 2n= 8.
113. The correct answer is d. The utility of FRAP is that it allows one to gain an understanding of the mobility
of both phospholipids and proteins within the lipid bilayer.
114. The correct answer is c. FRAP can be used to investigate 2-D & 3-D diffusion, and has a very high
115. The correct answer is d. Peripheral proteins are usually found on one side or the other, depending on their
function. Increasing the average length of the fatty acid tails of membrane lipids will increase the transition
temperature of the membrane. Transverse diffusion through a bilayer is not thermodynamically favorable
without an enzyme, but lateral diffusion readily occurs.
116. The correct answer is a. This is how cells adapt to warmer climates.
117. The correct answer is a. The NET entropy of the entire system is what determines the thermodynamic
favorability of bilayer formation.
118. The correct answer is d. Only tryptophan is nonpolar. The others are polar charged, or polar uncharged
119. The correct answer is d. As discussed in question 117, the increase in the entropy of water is what drives
120. The correct answer is d. Aquaporins transport water across membranes via facilitated diffusion, which
does not require ATP hydrolysis.
121. The correct answer is a. The four regions in part B are those that are hydrophobic in nature, so they are
the ones that are able to transverse the bilayer.
122. The correct answer is b. The nonpolar amino acid residues are: G A F V W I L M P.
123. The correct answer is b. Because it contains transmembrane segments, it is likely involved in some type of
membrane transport. The hydrophilic regions inside of the protein indicate that it is likely involved in
facilitated diffusion, with those residues functioning to “carry” the water molecules through the protein. The
absence of any information pointing to a residue that could interact with the phosphate groups on ATP
precludes active transport as being a likely choice.
124. The correct answer is a. The type of amino acid residue is indicated by the numbers. To interact with
water, the residue needs to be polar.
125. The correct answer is c. An enzyme is capable of converting on type of amino acid into another, not an
ion such as mercury. Answer choice b looks as if it could be possible, but considering how many water
molecules in solution there are it’s unlikely that inhibition of the transport requires this many mercury
molecules. Answer choice c makes sense, because it is capable of binding to cysteine’s side chain, and given the
figure, this is a plausible mechanism of inhibition.
126. The correct answer is d. Aquaporins have polar interiors that become narrow enough at the end for one
molecule of water to “drip” into the other side.
127. The correct answer is a. This protein is used by the bacterium to regulate its internal pH.
128. The correct answer is c. Two proteins function in intestinal glucose transport. Na+/glucose symport (both
being transported in the same direction) and a glucose uniport.
129. The correct answer is c. This process is described in questions 120, 123, 126 with re. to aquaporins.
130. The correct answer is a. It does so by importing protons from the outside, and pushing sodium out.
131. The correct answer is d. A clue is given in part B, where transmembrane proteins are depicted. A
Ramachandran diagram demonstrates rotation.
132. The correct answer is b. The negative values on the y-axis indicate polar amino acid residues, whereas
positive values indicate those that are nonpolar. A clue to this can be found in the diagram in question 121.
Obviously, a residue that interacts with water must be polar.
133. The correct answer is d. Described above.
134. The correct answer is c. Beta sheets, as well as helices are stabilized by hydrogen bonds between every 4th
and n-1 amino acid. An oxygen atom cannot donate hydrogen. It has electrons to exchange in order to accept a
135. The correct answer is d. It makes sense that the NMR and XRD information should give similar structural
information. Proteins often are dynamic structures, not static ones. This is key to their function.
136. The correct answer is a. The side chains of amino acids are not involved in stabilizing secondary structure.
137. The correct answer is a. This is because the peptide bond has partial double bond character through
138. The correct answer is c. This reagent prevents the reformation of disulfide bonds.
139. The correct answer is d. In other words, there is restricted rotation about a peptide bond.
140. The correct answer is d. Conversely, joining of amino acids to form a peptide requires energy and is an
141. The correct answer is a. Doubling the amount of enzyme would double the rate, because there’s more
enzyme’s there to do more work. Doubling the amount of enzyme wouldn’t affect the affinity of those
enzymes for their substrate.
142. The correct answer is d. These assumptions hold true for all Michaelis-Menten-type enzymes.
143. The correct answer is d. Although the Km is not always the best indicator for the affinity of an enzyme for
its substrate, it is safe to make this general assumption for the majority of enzymes. A low Km (generally)
indicates that the enzyme has a high affinity for substrate.
144. The correct answer is c. The Enzyme-Substrate (ES) complex is present at approximately constant
concentration. Michaelis-Menten-type enzymes are those with hyperbolic curves.
145. The correct answer is b. See above.
146. The correct answer is d. Transition state analogs bind to the enzyme’s active site with a higher affinity than
its normal substrate. For this reason they are often used as antigens to produce catalytic antibodies.
147. The correct answer is d. Conversely, the more positive the value of ΔG0’ is for a reaction, the slower a
reaction will be.
148. The correct answer is c. This is because it lowers the hill, or energy of activation, for the formation of the
transition state. This increases the rate of both the forward and reverse reactions.
149. The correct answer is b. See above.
150. The correct answer is a. In other words, a proton is pulled away from water, making it OH-. A nucleophile
attacks positive charges.
151. The correct answer is b. Creatine Kinase isn’t even presented on the graph. The spike of MB indicates that
it is likely the best choice, because BB’s spike is very small when comparing day 1 v. day 2.
152. The correct answer is a. His enzymes are elevated when he gets to the hospital, indicating that the
infarction occurred prior to his trip there. On day 2 he experienced cardiac arrhythmia, not an infarction
(information from the passage). The passage also states that after cardioconversion his symptoms got better,
153. The correct answer is a. This is presented in the graph.
154. The correct answer is d. A competitive inhibitor binds only to the enzyme’s active site, and decreases the
maximum velocity of the reaction. It does not change the enzyme’s affinity for its substrate.
155. The correct answer is d. The binding of most allosteric modulators causes a conformational change in the
active site. This either makes catalysis more, or less effective. This is because the conformational change puts
key catalytic residues closer or further away.
156. The correct answer is d. Remember that a Lineweaver-Burk Plot is a double reciprocal plot.
157. The correct answer is c. Another way to relieve competitive inhibition is to increase the concentration of
158. The correct answer is c. ATC is a cooperative enzyme, and has a sigmoidal rate v. substrate concentration
profile. Michaelis-Menten-type enzymes have a hyperbolic saturation curve.
159. The correct answer is d. By process of elimination: ATP is a positive effector of the enzyme, enhancing
the enzyme when there is energy available for anabolism. It does so by binding to the R (relaxed) state of the
enzyme. CTP, the negative effector, binds to a similar type of regulatory subunit and binds to the T (tight) state
160. The correct answer is b. PFK-1 is the pace-setting enzyme of glycolysis. Hexokinase is the committed step
of the pathway.
161. The correct answer is b. Hexokinase and phosphofructokinase require ATP to catalyze their reactions, so
their reactions are not readily reversible. These two steps also require bypass reactions for the opposing
162. The correct answer is b. One of magnesium’s physiological functions is to stabilize the phosphate groups
163. The correct answer is a. In anaerobic conditions, glucose is catabolized through glycolysis. The subsequent
pyruvate formation cannot be oxidized by the oxygen-requiring pyruvate dehydrogenase; so in order to
regenerate re-oxidize the NADH formed and create cellular energy pyruvate is instead converted to lactate-bylactate
164. The correct answer is c. PFK-1 is the rate limit of glycolysis, and is strongly inhibited by both ATP and
citrate. Fructose-6-phosphate is its substrate, and the enzyme forms fructose-1,6-bisphosphate. The enzyme
also requires ATP to catalyzes its highly exergonic reaction.
165. The correct answer is d. The main regulation of glycolysis occurs at the PFK-1 step, an enzyme that is
under the control by several levels of modulation (hormonal, substrate availability, and by feedback
166. The correct answer is d. Glycolysis occurs in the cytosol, and is an anaerobic pathway.
167. The correct answer is a. See the explanation for question 163.
168. The correct answer is d. Different types of carbohydrates are able to enter glycolysis by their conversion
to either glucose or fructose.
169. The correct answer is a. Answer choices b & c list substrates, not enzymes. HK, PFK-1 and PK are all
kinase enzymes, involved in phosphorylating substrates. Isomerases are not involved with the transfer of
energy, so they are not points of regulation.
170. The correct answer is d. Both alpha helices and beta sheets are outlined as ribbons on this figure. The
picture also depicts the rearrangements occurring between the various residues. This is key to the enzyme’s
ability to convert its substrate into product; placing key residues in closer proximity to do (in this case, acid
171. The correct answer is d. Remember that the R state is the active/high affinity state of the enzyme. This
also holds true for the oxygen transport protein hemoglobin. The T (tight) state of the enzyme is the inactive/
low affinity state. Because the aforementioned residues are in contact with each other in the R state, its safe to
assume that this is not random.
172. The correct answer is c. Proline residues are helix-breakers, and are not found in high abundance.
173. The correct answer is d. Clearly answer choice a is incorrect. Peptide bonds most commonly occur in the
trans conformation, which prevents steric hindering between adjacent R groups.
174. The correct answer is c. The shielding of nonpolar groups from contact with the aqueous environment is
what drives the tertiary structure of proteins.
175. The correct answer is b. The hydrogen bonds in anti-parallel beta-sheets are stronger than those in parallel
beta-sheets. The nth residue in beta-sheets forms a hydrogen bond with the n minus 4th residue.
176. The correct answer is d. Only XRD is high resolution enough to give information re. the precise spacing
177. The correct answer is d. The C-N peptide bond is shorter than the C-N bonds in aliphatic amines because
of resonance. For the same reason, the C-O bond is longer than that found in “normal” C-O bonds found in
carboxylic acids. All four substituents lie in the same plane.
178. The correct answer is a. Valine is a nonpolar residue. The hydrophobic effect is what drives proteins into
their native structures. Re. c & d: X-ray crystals are much more high resolution than NMR data; and the
resolution goes higher as the Angstrom number gets smaller.
179. The correct answer is a. Pepsin is a protease that functions in the stomach. In this highly acidic
environment (≈ pH 2) trypsin would not function.
180. The correct answer is b. The Michaelis constant, or Km, is equal to (k-1 + K2)/k1. k1 is the rate of
formation of the enzyme –substrate complex (ES complex), and k2 is the rate of product formation from the
181. The correct answer a. The specificity pocket of chymotrypsin must be able to accommodate a
hydrophobic aryl side chain, because it cleaves on the C-side of large, hydrophobic amino acids. Organophosphates
bind to the serine component of the triad. Aspartate is a part of the catalytic triad, but functions to
make histidine a better at deprotonating serine.
182. The correct answer is d. Serine proteases cleave at the C-side of their targets, and the first product is the –
NH-group of the following amino acid.
183. The correct answer is a. The primary structure of a protein is the amino acid sequence. The 3-D structure
(tertiary structure) of this class of enzymes places the catalytic residues in close proximity to each other, but
this is accomplished with the residues being in different positions depending on the specific enzyme. If they
had the same primary structure, they would be the same enzyme. In other words, there’s more than one
primary structure arrangement that leads to these proteins folding into a conformation that puts the catalytic
residues in position to cleave peptides.
184. The correct answer is d. A zymogen is a protein that requires modification before it becomes an active
enzyme. The zymogen form of chymotrypsin does not possess the oxyanion hole required for catalysis.
185. The correct answer is c. The serine of the catalytic triad is the site of formation of a covalent enzyme/
substrate intermediate. Aspartate makes histidine better at abstracting a proton from serine as discussed in the
explanation for question 181.
186. The correct answer is d. 2-mercaptoethanol is used to cleave disulfide bonds. Cleavage with the reagent cut
the 60-kDa subunit’s disulfide bond, yielding two 30-kDa fragments. Because the 20-kDa fragment subunits
were not cleaved, non-covalent bonds must link them.
187. The correct answer is a. Another way to state this is that larger molecules move further/faster during gel
filtration chromatography, and elute from the column quicker than smaller molecules do. Blue dextran is a large
molecule that serves as a standard used with this technique, and represents the smallest elution volume.
188. The correct answer is c. Membrane ultrafiltration is good for buffer exchange and concentrating protein
samples, because you can choose a membrane with a specific pore size.
189. The correct answer is a. See the explanation for question 187.
190. The correct answer is b. The boundaries of the shaded region corresponds to 0.1 & 0.9 equivalents of
added base, or 10% & 90% of the weak acid’s titration.
191. The correct answer is a. All weak acids display similar relationships on a pH v. base titration curve. Their
buffering regions will be 1 pH unit above and below the pKa.
192. The correct answer is d. From looking at the titration curve you can see that the endpoint occurs at
around pH 7 on the y-axis. At this point, the weak acid has donated its protons to the base that was added
193. The correct answer is b. The visible spectrum falls between 400nm and 700nm. Amino acids absorb in the
194. The correct answer is b. Histidine is a polar charged amino acid, with a pKR of about 6. It would not be
found on the interior of the enzyme, but in a location where it could interact with either a substrate or the
aqueous environment. The side chain would be positively charged at pH values below the pKR, and uncharged
above it. Lysozyme utilizes aspartate and glutamate to do acid base chemistry, not histidine.
195. The correct answer is b. Valine is a nonpolar amino acid residue. Nonpolar amino acids are found on the
inside of proteins because of their tendency to group in a manner that excludes water. If it were to exist on the
exterior of a protein, it would need to be in contact with something nonpolar. The core of the bilayer consists
of the nonpolar tails of phospholipids. Serine is a polar uncharged amino acid residue, and would be found in a
location where it can associate with the aqueous environment or a substrate of some sort.
196. The correct answer is d. The first step to solving this problem is to align the residues in the order they are
listed at the end of the question. At pH 7.9, the N-terminus of the peptide (alanine’s 9.8) has a positive charge.
The C-terminus of the peptide (valine’s 2.3) has a negative charge. Histidine’s side chain would be deprotonated
because it has a pKa of 6. Leucine does not have an ionizable side chain and its amino and carboxyl groups are
tied up in the peptide. Similarly, the C-terminus and N-terminus of respective alanine and valine can be
ignored. +1 + -1=0.
197. The correct answer is b. Remember that the thermodynamics of protein folding takes into account the
system, not just the protein. There are many more water molecules in the environment than there are residues
in a protein. The water molecules around the peptide (in contact) will be ordered, but the majority will be
unordered. Entropy is a measure of disorder, and the second law of thermodynamics states that the entropy of
a system is always increasing. The hydrophobic effect plays a very large role in biochemistry, because cells exist
in aqueous environments.
198. The correct answer is a. Methionine has a thioether group. Cysteine has a thiol group.
199. The correct answer is b. Glutamine, threonine, and asparagine are all polar uncharged amino acids. Alanine
is a nonpolar amino acid. Aspartate, histidine, and glutamate are polar charged. Serine is a polar uncharged
amino acid, but it is included with histidine, which makes it a wrong answer choice.
200. The correct answer is d. Tryptophan, proline, and leucine are all nonpolar amino acids, so they would
likely be buried inside the protein. Lysine is a polar charged amino acid and is commonly found in the active
site’s of enzymes, where is often plays a role in acid-base catalysis.