MCAT Question of the Day

Correct Answer: D. Both A and B

Many cellular enzymes exhibit decreasing activity with increasing levels of reaction product. This feedback inhibition help regulate the activity of enzymatic processes. Often this process occurs through non-competitive inhibition. The product of the chemical reaction binds to the enzyme in a location other than the active site, which causes a conformational change in the enzyme and decreases enzymatic activity. Allosteric activation describes a process in which enzymatic activity is increased because of binding of some molecule to a site other than the active site.

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Correct Answer: A. 1, 4

Rotation of plane polarized light is indicative of a chiral molecule. If not, the compounds must be meso or achiral. Compounds 1 and 4 are meso compounds, which, despite having two or more stereocenters, are not chiral (superposable on its mirror image). Compounds 2 and 3 are chiral molecules.

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Correct Answer: C. Delta(Grxn) < 0 at only low temperatures

On the MCAT, numbers can be intimidating, but remember that if the answer choices have no numbers in them, we should think conceptually instead of empirically. Gibbs equation helps us determine if a reaction is spontaneous or non-spontaneous. It is given as:
G = H – TS

In this case, the enthalpy change H is a negative number (exothermic). We know that the entropy change (S) is also negative (the products are more ordered than the reactants) because the reactants are 2 moles of gas and the products are only 1 mole of gas (a synthesis of phosphorus pentachloride). Our equation now be read differently as:
G = (negative number) – temperature * (negative number)

Temperature is in Kelvin and thus always a positive number, but we must consider two possibilities regarding temperature: (1) If this positive temperature is relatively low, the second part of the equation is a small positive number. Thus, if you add a small positive number to a negative number, it will probably still be negative (G<0). (2) If the positive temperature is relatively high, the second part of the equation is a large positive number. If you add a large positive number to a negative number, it will probably become positive (G>0).

In both these cases, ambiguity is important because we don’t know the temperature or S. The only answer that appropriately describes the case we have here is C.

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Correct Answer: A. Cyanide blocks the passage of electrons from cytochrome oxidase a3 to oxygen

Cyanide blocks the passage of electrons from cytochrome oxidase a3 to oxygen since it has a very high binding affinity for the enzyme. The cell cannot use oxygen so even if the person can breathe, the cells cannot use the oxygen that is delivered to tissues. Oligomycin inhibits the electron transport chain by directly affecting ATP synthase. A class of poisons called uncouplers makes the lipid bilayer permeable to hydrogen ions, which eliminates the electrochemical gradient.

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Correct Answer: A. Gastrula

The correct answer is A. The oocyte and morula are both precursors to the blastula stage. The embryoblast and the trophoblast are both components of the blastula, but are not the next stage in embryonic development. Once the blastula undergoes gastrulation, it will become a gastrula with three distinct layers of tissue: endoderm, mesoderm, and ectoderm.

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Correct Answer: A. Irreversible competition.

Penicillin covalently binds to the active site of the bacterial enzyme and permanently disables its action. This mode of action is inherent to irreversible inhibitors. Notice this is similar to competitive inhibitors, but competitive inhibitors are reversible and adding more substrate can overcome the competitive inhibitor. Here is an excellent explanation on enzyme mechanics.

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Correct Answer: D. losing an existing allele due to random sampling.

Genetic drift represents the change in an allele frequency in a population due to random sampling. Influence from genetic drift is particularly important in small populations. The random changes in allele frequencies could result in the elimination of an existing allele from a population.

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Correct Answer: B. tert-butyl carbocation

The most stable fragment likely formed from 2,2-dimethylpentane fragmentation is t-butyl species.

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Correct Answer: C. N2(g) 0.75 and H2(g) 2.25 mol.L^-1.s^-1

The rate of consumption of the reactants will be the product of the rate of formation of the product multiplied by the ratio of the (co-efficient of the reactant) / (co-efficient of the product).

Rate N2(g) = 1.50 x 1/2 = 0.75 mole.L^-1.s^-1 and H2(g) 1.50 x 3/2 = 2.25 mole.L^-1.s^-1.

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Correct Answer: D. Benzoic acid, an organic compound, is insoluble in water. Extraction into water can be achieved with aqueous sodium hydroxide (deprotonation -> polar sodium benzoate, now soluble in water). Transfer of the benzoate back into organic solution is accomplished by treatment with hydrochloric acid (protonation -> less polar benzoic acid).

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