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• Question 151:

  Auxin stimulates stem elongation and is involved in the process of phototropism. If plants bend toward a light source, in which region of the plant is auxin most likely to be found?

  A. The shaded side of a stem

  B. The top of a shoot

  C. The bottom of a shoot

  D. The sunny side of a stem

  E. The top side of leaves

  Correct Answer: A

  Auxin is found in higher concentrations on the shaded side than the sunny side of a stem. More elongation on the shaded side causes the stem to bend toward the light.

• Question 152:

  In photosynthesis, high-energy electrons in Photosystem II are transferred along an electron transport chain and eventually end up in high-energy molecules used in the Calvin Cycle. Which molecule provides electrons to replace those lost by Photosystem II after light stimulation?

  A. CO2

  B. FADH2

  C. NADPH

  D. ATP E. H2O

  Correct Answer: E

  Water and carbon dioxide are the two essential consumable molecules in photosynthesis. First, water is split into oxygen, protons, and electrons, and then carbon dioxide is used in the Calvin cycle to create glucose. The electrons from splitting water are used in photosystem II, the protons are used to create NADPH, and oxygen is a waste product of the splitting of water.

• Question 153:

  Which of the following is a characteristic of an enzyme cofactor?

  A. It binds to an enzyme's active site.

  B. It is covalently bound to the enzyme.

  C. It is consumed in the enzymatic reaction.

  D. It inhibits the enzymatic reaction.

  E. It binds to an allosteric site.

  Correct Answer: B

  A cofactor binds to the active site along with the substrate in order to catalyze an enzymatic reaction. Like the enzyme, it is not consumed by the reaction. Allosteric effectors bind to a second binding site on the enzyme, not the active site.

• Question 154:

  

  The diagram below depicts a metabolic pathway. When product D accumulates, the production of product C decreases, D is an inhibitor of which enzyme?

  A. C'

  B. B'

  C. A'

  D. D'

  E. E'

  Correct Answer: B

  This is an example of negative feedback, a process whereby an increase in an outcome causes a decrease or slowing in the pathways that led to the outcome.

• Question 155:

  All of the following molecules are soluble in water except:

  A. Polysaccharides

  B. Triglycerides

  C. Hydroxyl groups

  D. Polypeptides

  E. Carboxylic acids

  Correct Answer: B

  Triglycerides are hydrophobic. They consist of three fatty acids joined to a glycerol molecule, and because of their long hydrocarbon chains, they are not soluble in water.

• Question 156:

  Which of the following carbohydrate polymers serves as an energy storage molecule in plants?

  A. Chitin

  B. Cellulose

  C. Starch

  D. Glycogen

  E. Phospholipids

  Correct Answer: C

  Plants have cellulose as the major structural component of their cell walls. However, plants store energy as starch, not cellulose. Starch is a polymer of -glucose molecules, whereas cellulose is a polymer of -glucose molecules. The different chemical bonds between glucose molecules in starch and cellulose make the difference in whether or not the polymer is digestible in plants and animals.

• Question 157:

  Human predation has caused the population of cheetahs to decline dramatically. Changes in allele frequencies in the remaining population of cheetahs would most likely be due to:

  A. Natural selection

  B. Gene flow

  C. The founder effect

  D. The bottleneck effect

  E. Mutation

  Correct Answer: D

  The bottleneck effect occurs when populations undergo a dramatic decrease in size. It could be due to natural or artificial causes.

• Question 158:

  (R)-1-Fluoro-1-iodopropane is reacted with NaN3 in HMPA. Which of the following is the major product of this reaction?

  

  A. (R)-1-Azido-1-fluoropropane

  B. (R)-1-Azido-1-iodopropane

  C. (S)-1-Azido-1-fluoropropane

  D. (S)-1-Azido-1-iodopropane

  Correct Answer: A

  Although the reactants may look complicated, this reaction is actually just a classic SN2 reaction. Azide (N3-) is the nucleophile, and iodide (I-) is the leaving group. Note that, although fluorine is a halogen, fluoride (F-) does not follow the group trend and is not a good leaving group due to its small size.

  However, SN2 reactions are stereoselective with an inversion of stereochemistry and therefore only create one of the two possible R-S isomers. Since the leaving group and the nucleophile end up having different priorities, the absolute configuration (R vs. S) must be recalculated. Start by drawing out the new molecule (remembering to invert the stereochemistry, like turning an umbrella inside out) and ranking the attached atoms in order of decreasing atom mass: F > N > C > H. From there, connect 1 ?gt; 2 ?gt; 3 with an arrow. The arrow is counterclockwise, initially indicating an S configuration. Finally, check atom 4: If it is into the page, leave the configuration how it is; if it is coming out of the page, reverse the configuration. In this case, H is coming out of the page, so the final configuration is R.

  

• Question 159:

  Given the following reaction conditions, which statement is most accurate?

  

  A. The reaction follows first-order kinetics and is a concerted reaction.

  B. The reaction follows first-order kinetics and involves formation of a carbocation.

  C. The reaction follows second-order kinetics and is concerted.

  D. The reaction follows second-order kinetics and involves formation of a carbocation.

  Correct Answer: C

  This substitution reaction shows chloride (Cl-) being replaced by hydroxide (OH-). Substitution reactions occur mainly via one of two mechanisms: SN1 (unimolecular kinetics and two steps with a carbocation intermediate) or SN2 (biomolecular kinetics and one step). In this situation, the mechanism isn't shown but can be inferred based on the reaction conditions. First, the reactant with the carbons (the substrate) has a strong leaving group; the Cl- that detaches is relatively stable as an ion in solution by itself (think of how table salt, NaCl, is able to readily dissolve into Na+ and Cl- in water). Second, the substrate has primary substitution, meaning the carbon attached to the leaving group is only attached to one other carbon, which in turn means that there is little steric hindrance but also that the carbon doesn't have many other carbons to stabilize it if it were to gain a charge. Third, the other reactant, OH-, is a strong base and strong nucleophile, indicating that it can readily attack the substrate on its own. Finally, the solvent DMSO (dimethyl sulfoxide) is polar aprotic so can stabilize the leaving group without deactivating the nucleophile. All of these factors point toward an SN2 reaction. Since an SN2 reactions is always concerted, occurring in one step without forming discrete intermediates.

• Question 160:

  What is the oxidation state of each nickel on the reactant side of the following reaction?

  2 NiO(OH) + Cd + 2 H2O ?gt; 2 Ni(OH)2 + Cd(OH)2

  A. -2

  B. -1

  C. +1

  D. +3

  Correct Answer: D

  Oxidation numbers provide a way to keep track of the movement of electrons in a reaction. Several rules govern how oxidation numbers are calculated, but in this situation, it's only important to recognize the oxidation numbers of the common components attached to nickel [Ni] on the reactant side and to remember that the sum of the internal oxidation numbers of a molecule equals that molecule's net charge. In this case, nickel oxide [NiO(OH)] is neutral, so the sum of its internal oxidation numbers must be 0. Hydroxide [OH-] always has an oxidation number of -1, and oxygen [O] almost always has an oxidation number of -2. To make these charges cancel out to equal 0, the oxidation number of nickel [Ni] must be +3: 3 - 1 - 2 = 0.