Session 218

Today, Madeline and I discuss when/if a student should void the MCAT. We jump into the 7th Bio/Biochem passage with questions on amino acids, muscles, and more!

This podcast is in collaboration with Blueprint MCAT. 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.

[02:15] General MCAT Tips

If a student is feeling just absolutely destroyed passage after passage discrete section after discrete section, your subjective judgment in the heat of the moment is probably wrong. And you’re probably doing better than you think you are. The truth is like we are bad at judging how we’ve done.

So never avoid an exam just because of your subjective evaluation of how you’re doing because it’s not going to be accurate.

[05:47] Passage 7 (Questions 35 – 39)

Paragraph 1

A blood sample was taken from a patient with normal kidney function who displayed symptoms of a heart attack. In order to confirm the suspected diagnosis, a sample of the patient’s blood was collected and analyzed for the presence of creatine kinase (CK) isoenzymes.

Note: We’re highlighting in bold the things we think you should be highlighting. We’re looking at these two different things, maybe we will be comparing or contrasting them. And what might be able to do that is the presence of the(CK) isoenzymes.

[06:22] Paragraph 2

Isoenzymes are enzymes with different amino acid sequences that perform similar biological functions. Increased concentrations of CK isoenzymes in the blood can indicate tissue damage in the heart, skeletal muscle or kidneys. CK has three cytosolic isoenzymes. CK1 is found exclusively in the brain, CK2 is found exclusively in cardiac muscle, and CK3 is present in both skeletal and cardiac muscle. Each isoenzyme is a dimer containing B subunits, M subunits or both B and M subunits. The cytosolic CK B subunit contains a smaller proportion of hydrophobic residues and a greater proportion of acidic residues with low pI values than does the cytosolic CK M subunit, which is enriched in asparagine and lysine. CK M subunits are slowly converted into two modified subunits, M1 and M2, by plasma carboxypeptidases via successive cleavage of C-terminal amino acid residues.

Note: As long as we’ve highlighted those names, we can always come back and find that information there.

So even though we only highlighted a lot of little segments, this gives us a really good idea that we’re looking at these three isoenzymes designed, this is what they’re made up, this is what those residues are, they can be broken down, and this is how. It’s a great summary of what’s happening.

[08:55] Paragraph 3

The activity levels of each of the three isoenzymes in the patient’s serum, drawn and tested approximately 12 hours after the onset of his symptoms via a colorimetric method, are given in Figure 1.

Figure 1 Serum creatine kinase (CK) isoenzyme activity levels for the patient (top table) and a normal reference range (bottom table)

Note: We’re highlighting in bold the things we think you should be highlighting.

[09:57] Paragraph 4

The results of an agarose gel electrophoresis of the isolated isoenzymes, each loaded into an immobilized pH-gradient-polyacrylamide gel at pH 7, are shown in Figure 2. During a heart attack, CK isoenzymes appear approximately 4-8 hours after the onset of chest pain and peak within 24 hours. They return to baseline levels within 48-72 hours of the initial onset of symptoms.

Note: We’re highlighting in bold the things we think you should be highlighting.

Figure 2 Electrophoretic mobility of isolated creatine kinase (CK) protein fractions

[11:28] Question 35

Given the information presented in the passage and the results of the serum electrophoresis, what is the most likely subunit composition of the CK 2 isoenzyme?

1. MM
2. MB
3. BB
4. The subunit composition cannot be determined by the given information.

Thought Process:

Paragraph 2 says “Each isoenzyme is a dimer containing B subunits, M subunits or both B and M subunits.” And so, we know that it has to have a combination of BB and MM or MB. And the three different types, which are CK3, CK2, and CK1 in that second figure pinpointed where we should go.

We need to have an MM, and we said it has to be more basic, and BB is going to be more acidic. Well, there’s only one more option that’s in the middle, which is your CK2, which would be reasonable to speak that it’s a combination of both.

D – Although we feel like it’s true when they say there’s not enough information. But it’s really important to be able to evaluate this question independently of that.

And so, if we can try to evaluate it apart from that answer choice, then we may be able to come up with the more correct answer choice.

A – This is going to be the one that’s the most basic, which is going to be your CK3.

B – This is going to be the middle one, which is going to have your acidic and your basic elements to it. So it’s going to be more around that middle part, which will be where your CK2 is. So that will make sense.

C – This is going to be super acidic, close to that pH equals zero, which is your CK1. So that’s going to be pretty much parallel to that.

So from that information, we could isolate that CK2 is your MB sub-unit type.

Correct Answer: B

[18:38] Question 36

CK isoforms containing M1 and M2 subunits migrate nearer the anode in gel electrophoresis than do CK isoforms containing the unmodified M subunit. The region near the C-terminus of the unmodified M subunit is enriched in which of the following amino acids?

1. Lys, Asn, Arg
2. Leu, Cys, Pro
3. Lys, Ala, Asp
4. Gln, Asp, Glu

Thought Process:

This is an example of the need to rephrase this question because there are a lot of words there in just two sentences. So it’s a good time to say what is important and what is not important. 

They’re saying when you have your modified M1 and M2 subunits, they go near to the anode. So if we look at our M1 and M2 subunits in our passage, they say you’re going to have this cleavage of the C terminal amino acid residues. So something there is cleaved that makes them want to go closer to the anode.

Now, the first thing you have to know is what an anode is. And in gel electrophoresis, the anode is a positively charged side. 

If the anode is positive, and they’re going nearer to the anode, maybe they’re taking away some type of positive charge or they’re adding in some type of negative charge. But we’re not adding anything so we have to be taking something away that allows them to be closer to something positive.

It also makes sense that we’re taking away something that’s also positive because two positives would repel. So if we take away one of the positives, you’re going to be able to get closer to that original positive.

Therefore, we’re only going to be looking for the positive ones. That’s the important thing here. Once you can call through that question stem and understand you’re looking for the positive amino acids to be taken off, it makes things easier. Because then, you’re just looking for something positive.

A – So here we have a which is Lysine, Asparagine, and Arginine. Lysine and arginine are both positively charged amino acids. So that would make sense.

B – Leucine, cysteine, and prolene. Well, we have some polarity there, but no actual charges. So we can cross that one out.

C – Lysine, alanine, and aspartate. Well, lysine is positive, but aspartate is negative, so that basically renders it neutral. We can cross this out because it doesn’t have that overall positivity we’re looking for.

D – Glutamine, aspartate, and glutamate. So again, we have two negatives, which are exactly the opposite of what we want to be looking for.

Correct Answer: A

[22:27] Question 37

Troponin isoenzymes are used as an alternative biomarker in the diagnosis of heart attacks. In which of the following muscle types does the troponin complex function in contraction?

1. Skeletal muscle
2. Smooth muscle

III. Cardiac muscle

1. I only
2. I and II only
3. I and III only
4. I, II, and III

Thought Process:

Troponin is obviously a big part of muscle contraction. The question is, where does it work?

Skeletal muscle and smooth muscle are really different because they have completely different functions. Therefore, they probably don’t have the same mechanistic properties.

And so, if it is skeletal then it’s probably not smooth. And we know it’s skeletal because it’s in every single one of the answer choices. Since they’re opposites, we can take out II, which means the only answer choice that would make sense is actually C.

Correct Answer: C

[26:33] Question 38

Considering the results of the laboratory testing given in the passage, which of the following diagnoses concerning the patient is most likely correct?

1. The patient is unlikely to be suffering a heart attack, because the elevated CK isoenzymes are more closely associated with damage to the brain than with damage to cardiac muscle.
2. Additional testing is necessary, because the CK isoenzymes that are elevated are not specific for cardiac muscle damage, and may also indicate damage to skeletal muscle.
3. Additional testing is necessary, because CK isoenzymes that are markers of cardiac damage appeared at normal levels, but may have been elevated prior to the time at which the samples were drawn.
4. The patient is likely suffering a heart attack, because his CK isoenzyme levels are indicative of recent cardiac muscle damage in a patient with no signs of kidney disease.

Thought Process:

We do know that it has no signs of kidney disease because that’s given in the first paragraph. So that part checks out.

They have CK3 elevated levels, but more importantly, their CK2, which is specific to your cardiac muscle that is indicative of your heart attack. So D would be the correct answer here.

Correct Answer: D

[31:53] Question 39

Which of the following statements is true regarding peptide bonds found in the CK subunits analyzed?

A.They possess partial double bond character.

B.They are ionized at physiological pH.

C.They occur most commonly in the cis configuration.

D.They are cleaved by high urea concentrations.

Thought Process:

This is actually a discrete question so you can stop looking at the passage, and they’re asking you specifically about peptide bonds. So whether or not you draw it on your paper, you imagine it in your mind, drawing out a peptide bond could be really useful in this scenario, or just remembering what they are.

Peptide bonds are going to be amide bonds. It’s going to have a carbonyl, a C=O bond, which is connected to nitrogen. And nitrogen has a lone pair and is best able to basically resonate in to create a double bond with your C. Then kick some electrons up to your oxygen to create a resonance structure wherein you’re delocalizing your electrons.

Basically, it just means that you can move those electrons within those double bonds in an amide bond. So they’re asking us which of these choices describes a peptide bond.

A – We did just say those electrons can toggle between your CO bond and your CN bond. So it makes sense it would have without one character in each of those bonds. So that’s a possibility.

B – This is specifically asking about the peptide bond. It’s not asking about amino acids. And that’s what’s really important. So ionization does not happen to the peptide bond. It’s happening to your side chains, your amine, and your carboxylic acid. It’s not going to be happening within that peptide bond itself. So that’s why B is wrong.

C – The peptide bond is actually in the trans configuration. So this is just a fault description of that peptide bond because it’s actually in its opposite, which is the trans configuration. So this is wrong.

D –  A lot of people really want to claim this because high urea concentrations is a type of denaturation technique used for proteins. However, denaturation means splitting up the different levels of protein structure such as your quaternary, tertiary, and secondary. And peptide bonds are part of that primary structure, not secondary, tertiary, or quaternary. So high urea concentrations would not affect the peptide bonds there.

Correct Answer: A

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