Blueprint Diagnostic Chem/Phys Discrete 1

Session 263

Today we review Blueprint’s MCAT Diagnostic, C/P Discrete 1. Tune in to discover why you got it right… or wrong.

Alex is back from Blueprint MCAT. We are covering the diagnostic from Blueprint MCAT which everyone gets for free along with so many other free things including the study planner tool, the flashcards, the half-length diagnostic, and the full-length test. If you would like to follow along on YouTube, go to premed.tv.

Listen to this podcast episode with the player above, or keep reading for the highlights and takeaway points.

[03:39] Question 11

Which of the following explains why the pitch of person’s resonant voice harmonics rises when they inhale helium instead of air?

A.The wavelength increases due to the change in temperature from the helium atoms.

B.The change in timbre results in a higher perceived frequency of the voice.

C.The frequency decreases due to the change in air density from helium atoms.

D.The speed of the sound wave remains constant regardless of changes to the medium through which the sound wave propagates.

Through Process:

The question is asking: why does the pitch of someone’s voice rise? Why does their voice sound higher? Alex hints that only one of them actually relates to the question being asked, which is B.

Correct Answer: B

[07:56] Question 12

The figure shown below shows the pressure and volume changes of the left ventricle during the cardiac cycle for two different patients.

Which of the following is true?

A.The change in systolic pressure in patient 1 is greater.

B.The change in diastolic pressure in patient 1 is greater.

C.The work done by patient 2’s left ventricle is greater.

D.The work done by patient 1’s left ventricle is greater.

Through Process:

We can leave A and B out because they look almost identical.

C – If the relative pressures are the same, but it’s moving more blood then it’s doing more work. That’s the case of Patient 2 and we can go with this answer without even looking at D.

Correct Answer: C

[14:46] Question 13

In the figure below, the black line represents an enzyme-catalyzed reaction and the grey line represents that same reaction after adding 0.05 mM of factor X. Factor X can best be described as:

A.a noncompetitive inhibitor.

B.an allosteric activator.

C.a competitive inhibitor.

D.a chelating agent.

Through Process:

'If you're not sure what the words mean and there's one that seems pretty good for you, don't waste loads of time puzzling it out.'Click To Tweet

This is an enzyme kinetics question. There are two quantities that we want to pay attention to when we look at an enzyme activity versus concentration graph. In this case, the concentration of the substrate is on the x-axis and the activity of the enzyme is on the y-axis. The black line represents it by itself and the maximum value is often described as V max.

The second quantity is called KM, which is defined as the substrate concentration required for this enzyme to achieve half of its V max or half of its maximum activity.

In this case, all of the different inhibitor types for enzymes will affect these two quantities in different ways depending on how they work.

A – This would lower the V max, so it lowers the maximum number of reactions per second that this enzyme can catalyze. But it wouldn’t affect the KM at all. Non-competitive inhibitors bind an enzyme at a site that isn’t its active site.

B – If it’s an activator, it would increase the enzyme’s activity. And we can see here the gray line lagging below the normal enzyme line. The gray line is definitely an inhibitor of some description.

C – This binds the enzyme at the active site. A classic feature of competitive inhibitors is if you add enough substrate, you can eventually get the enzyme to where it would have been normally anyway, which is what’s happening here. Competitive inhibitor increases the KM because the KM is the concentration required to get to half of its activity. So if you impartially inhibit it, you need a higher concentration to get that same level of activity. But it leaves the V max unchanged because if you just drown the end enzyme and its substrate, it’ll get up, it’ll get up to its original activity level eventually.

D – This is irrelevant in this case.

Correct Answer: C