Biochemistry on the MCAT – Breaking Down a Passage

Session 31

This week, we discuss a biochemistry passage where Bryan breaks down how to read the passage and look through the answers. Biochemistry is Bryan’s favorite part of the MCAT and I’m not sure if the same holds true for you. They’re all probably bad and biochemistry is probably the “lesser of two evils” but we’re here to make it less bad for you.

[02:30] Biochemistry Passage:

Potentiometric titration is a useful means of characterizing an acid. No indicator is used. Instead, the cell potential is measured across the analyte solution. When cell potential is plotted against titrant volume added, the equivalence point is the cell potential at the inflection point, the midpoint of the steep segment of the titration curve.

For polyprotic acids, an acidic hydrogen will produce an inflection point only if it is not very weakly acidic and if its ionization constant differs from that of any other acidic hydrogen of the acid by at least a factor of 104.

Captopril (molecular weight: 217.29 g/mol), shown in Figure 1, is a competitive inhibitor of angiotensin-converting enzyme (ACE). In the figure for Captopril, we’re shown two acidic Hydrogens, one on a Sulfur that has a pKa of 9.8 and another on a carboxylic acid that has a pKa of 3.7. So Captopril has two acidic protons, one at 9.8 and one at 3.7 for the pKa’s.

Students studying captopril were provided the following in vivo IC50 values (the minimum plasma concentration needed to inhibit 50% of target enzyme activity). The pH range shows that below 3.7, the IC50 is 0.058. Between 3.8 – 9.5, the IC50 drops from 0.05 all the way down to 0.01 so it’s much lower. Once the pH gets above 9.6, the IC50 goes back up to over 0.06. So a notable lower IC50 in the middle of the pH range.

So we’ve got an organic molecule Captopril with two acidic protons at a pKa of 9.8 to 3.7. The IC50, the amount of this inhibitor you need to get to 50% inhibition is really low, in between those two numbers. So it’s much more powerful inhibition in that part of their pH range.

[05:00] Question #24:

If students performed an enzyme-inhibition assay using Captopril, which of the following changes in the kinetic parameters of ACE should be expected?

Answer choices: The Vmax goes up or down or is unchanged and the Km is increased or decreased or unchanged.

Bryan’s Insights:

Changes in Vmax or changes in Km depend on what kind of an inhibitor it is. Going back to the passage, we know that Captopril is a competitive inhibitor. What you’re expected to walk into the test is that as a competitive inhibitor, it doesn’t change the Vmax. So it’s unchanged. That then narrows us down to choices (C) and (D).

In order to get to the Vmax, you have to add a lot more substrate and flood out the inhibition with tons of the natural substrate. This means that the Km (the substrate concentration required to reach 1/2 Vmax) goes up.

So Captopril as a competitive inhibitor, Vmax is unchanged, Km is increased. And that leads us to answer choice (D).

[06:36] Pure Outside Knowledge

This is a question where about 70% of the students get right. So this is somewhere between easy to medium level of difficulty. But it’s important that for questions like this, make sure you’re keeping up with the crowd. This means you need to find places where you can do better than other premeds. At the bare minimum, on the more straightforward questions like this, you really need to be keeping up with the pack. So competitive inhibitor, Vmax unchanged, Km increased.

This is strictly a matter of knowledge. It’s a common style of questioning on the MCAT. Every passage is basically going to have one or two questions like this where you have to look up a single fact from the passage, in this case, the Captopril was a competitive inhibitor. Other than that, it is just pure outside knowledge.

[07:50] Question #25:

Which of the following protonation states of the Captopril thiol and Carboxyl groups is required to maximize Captropril’s inhibition of ACE?

We have to decide whether to have the thiol group be protonated or deprotonated and have the carboxylic acid group either protonated or deprotonated.

Bryan’s Insights:

Going back to the table, the extra little step of reasoning we have to apply here is that if you want to maximize the inhibition, you want to minimize the IC50 (the amount of inhibitor needed to get to 50% inhibition). So if you’re a really strong inhibitor doing your job really well, you don’t need to add much. A little dab will get you to half inhibition. Based on the table, it was the pH range in the middle, between 3.8 and 9.5. Unsurprisingly, this drug does its best work at a normal physiological pH (around 7 to 7.3).

Looking at the Figure for Captopril, if your pH is in the middle of the pH range, the carboxylic acid is going to be deprotonated, which means you have to strip the proton off that since the pKa for that is only 3.7. If the pH is over the pKa (the pH is really high and you’re in a really basic environment relative to that group or molecule), you deprotonate it. So you deprotonate the carboxyl group.

The hydrogen and the sulfur to the thiol group had a pKa of 9.8 so we’re below that. Relative to 9.8, being a neutral pH is actually relatively acidic from the Sulfur’s point of view. The sulfur would then be protonated.

Knowing that the thiol is going to be protonated and the carboxylic group is going to be deprotonated, this would then lead us to answer choice (B).

[10:18] The Paradox in Biochemistry

It’s a common style of reasoning in biochemistry where they something like, “This is variable is the amount of x needed to get to y.” Seeing that, you’d say, “Lower amounts of x, it’s a powerful molecule so you don’t need as much x to get the job done.” It’s this weird paradoxical thing in biochemistry where smaller numbers have more oomph to whatever you’re looking at.

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