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MCAT Content / AAMC MCAT Biology Question Pack 2 Solutions

### AAMC QPack Bio2 [Web]

Biology Question Pack Volume 2: Passage 1

1) First thing we want to do is look at Figure 1 from the passage and compare the rates of actin subunit addition of both ends.

Figure 1 is above. We want to find the concentration of free actin where the plus end of the microfilament grows faster than the minus end. Concentration of free actin is shown on the X-axis. One thing I want to note here. At very low concentrations, meaning 1 micromolar or below, we have a loss of actin subunits on both the plus and minus end. Only after crossing the 1 micromolar threshold do we start getting addition of actin subunits on the plus end.

The solid line represents the plus end, and it has a much larger slope. That’s because the plus end of the microfilament grows faster. Dotted line represents the minus end and has the smaller slope, meaning it grows more slowly. That means at every concentration above 1 micromolar we have the plus end of the microfilament growing, and growing faster than the minus end.

1. Exactly at 1 µM. This answer choice corresponds to a net change of zero to the plus end. The rate of loss is equal to the rate of addition, so there’s no net change. We’re not seeing the plus end growing.
2. Only between 1 µM and 4 µM. This answer choice is partially true. We do have the plus end of the microfilament growing faster than the minus end between these concentrations. But this answer choice says “Only.” We’re not limited to just these concentrations. Answer choice B is an incomplete answer, but it’s better than option A.
3. At any concentration greater than 1 µM. This answer choice is consistent with our breakdown of the question. We said that at every concentration above 1 micromolar we have the plus end of the microfilament growing, and growing faster than the minus end. This is superior to answer choices A and B.
4. At any concentration. This is not true because at a concentration under 1 micromolar, there is a loss of actin subunits on both the plus and minus ends. This contradicts our breakdown and Figure 1 from the passage. We’re left with our correct answer: answer choice C.

2) I have to make sure to be careful with the wording. I want to explain why microfilament lengths do NOT change when the sarcomere shortens in a muscle contraction. We have our image of the muscle sarcomere in the question stem, with labels. Myosin is also known as thick filament. Thin filament is actin, and also troponin and tropomyosin. The A band spans the length of the thick filament and also contains thin filaments.

We have the Z-line labeled which defines the boundaries of each sarcomere. I-band is the region with only thin filaments, and H-zone contains only thick filaments. When the sarcomere shortens, the H-zone, I-band, the distance between Z-lines all become smaller. The minus sign of the microfilament is anchored in that Z-line. Even though we have different parts of the sarcomere shortening, the microfilament is anchored in the Z-line, and only the distance between microfilaments will decrease.

1. The – ends of the microfilaments are capped by Z lines, and the actin subunit concentration is kept above 1 µM in muscle cells. First part of this answer choice is consistent with our breakdown. Microfilaments are capped by Z-lines on the minus end. But the microfilament length isn’t actually varying on the plus side either. The distance between microfilaments varies we said, not the length of the microfilaments. Both sides are capped to prevent addition or subtraction of actin subunits. An actin subunit concentration above 1 micromolar would normally mean addition to the plus end, but we mentioned there are no actin subunits being added to that end. Second part of the answer choice isn’t a great explanation for why length doesn’t change.
2. The – ends of the microfilaments are capped by Z lines, and the + ends are capped by another protein. This answer choice matches our breakdown, and what we just said when we went through answer choice A. We have both ends of the microfilaments capped, which prevents microfilament length from changing. Answer choice B is superior to answer choice A.
3. The actin subunit concentration is kept above 4 µM in muscle cells. This answer choice is out of scope. It’s not explaining why microfilament lengths don’t change. At a concentration above 4 micromolar, we would expect both the plus and minus end of the microfilament to add actin subunits. Not what’s happening here.
4. The – ends polymerize and the + ends depolymerize at the same rate. This is the opposite of our breakdown. We expect both ends to be capped, and no growth. The only way answer choice D would work is if neither side was capped, and we had treadmilling. Not the case here because the minus end is anchored in the z-line. We’re left with our correct answer, answer choice B. The minus ends of the microfilaments are capped by Z lines, and the plus ends are capped by another protein

3) We can reference the passage to find where the author mentions force generation specifically. The passage says “Researchers suspect that microfilaments can generate force, even in the absence of myosin, by elongating and pushing against a structure such as the plasma membrane.”
Normally microfilaments help generate forces used in cellular contraction and basic cell movements. This is done in association with myosin. In this case, the author mentions force can be generated even in the absence of myosin. That force is generated by elongating and pushing against a structure. Elongation happens with the addition of subunits to the ends of the microfilaments.

1. Amoeboid movement stops upon exposure to cytochalasins. This answer choice can explain the theory of force generation. First thing we’ll note is cytochalasins bind to the positive end of a microfilament and prevents the addition of actin subunits to that end. So, this poison stops amoeboid movement, and stops elongation. This is a good option for now.
2. Amoeboid movement cannot occur if mitosis is blocked. This answer choice insinuates that mitosis leads to cell movement. There’s no mention of force generation and mitosis together in the passage. Answer choice A directly answered the question being asked, and answer choice B isn’t related to the passage.
3. Moving Amoeba cells produce more troponin than do stationary ones. This answer choice is similar to answer choice B. There’s no mention of troponin in the passage. We did just mention in our previous question that thin filament is made of actin, and also troponin and tropomyosin. That’s the only mention we made, and there’s no way to conclude that additional troponin supports the theory.
4. The rate of movement is inversely proportional to the viscosity of the medium in which the Amoeba moves. This is a third straight answer choice that is not discussed in the passage. We didn’t discuss the medium in which microfilaments are found, or discuss viscosity influencing subunit addition rate. Best option here was the first one: answer choice A.

4) We’ll use Figure 1. We want to know about treadmilling, or when the rate of subunit addition at the plus end equals the rate of subunit loss at the minus end.

We have Figure 1 right above again, and we said we want to know when the rate of subunit addition at the plus end, equals the rate of subunit loss at the minus end. The key to this question is looking at both ends at the same time. If we look at a concentration of 1.5 micromolar, we see rate of subunit addition at the plus end is 2 units greater than zero and we see the rate of subunit loss at the minus end is 2 units less than zero. That means there is a net change of zero, or treadmilling. Let’s look for a concentration of 1.5 micromolar in the answer choices.

This is a numerical answer with no rounding or estimating. We solved for a value of 1.5 micromolar. We can compare all of our answer choices at once. Our correct answer is going to be answer choice C.

5) One more time we’ll rely on Figure 1. We want to know the concentration of free actin at which there is exclusively net loss of actin subunits.

Figure 1 is above again, and we said we want to know the concentration of free actin at which there is a net loss of actin subunits. That means we have a value less than zero on the Y-axis for both the plus and minus ends. We see that’s true for the minus end at any concentration below 4 micromolar; for the plus end, that’s true at any concentration below 1 micromolar. The key here is looking at both rates and not just one. That means at concentrations below 1 micromolar, both the plus and minus ends of the microfilament experience a net loss of subunits.

This was another numerical answer with no rounding or estimating. We solved for an exact answer, which was any concentration below 1 micromolar. We can compare all of our answer choices at once. Answer choice A is the only answer that matches that value exactly.

Biology Question Pack Volume 2: Passage 2

6) I am going back to the passage to pinpoint the aspect of experiment 1 that shows cell-to-cell communication determines cell fate. There are a lot of smaller words and numbers we’ll be referencing, so it can be beneficial to have the passage in front of you as you go through the solutions.

I’ve added Figure 1 and Experiment 1 here. We can focus on the 2nd paragraph here. It says “The cultured AB cells produced neurons and skin, but no muscle, whereas the cultured P1 cells gave rise to all of the tissues produced by P1 cells of an intact embryo.

What’s the takeaway here? Cultured AB cells produced only neurons and skin, but no muscle. The cultured P1 cells gave rise to all of the tissues normally produced by P1 cells of an intact embryo. What did this tell us? The P1 cells are acting the same, regardless of the presence of AB. But AB produces a different result when isolated from P1. That likely means AB is typically dependent on communication with P1 to produce normally. P1 is the same, regardless of the presence of AB.

1. The fate of an isolated AB cell differs from that of an AB cell in an intact embryo. This answer choice is consistent with my breakdown of the question. I mentioned AB cells normally produce neurons, skin, and muscle. But the isolated AB cell only produced neurons and skin. I said this is because AB is typically dependent on communication with P1. This is a good answer choice to start.
2. The fate of an isolated P1 cell is indistinguishable from that of a P1 cell in an intact embryo. This answer choice is technically a correct statement, but does it answer the question being asked? Because we can’t distinguish between the isolated P1 cell and the P1 cell in the intact embryo, we can’t support the hypothesis that cell-to-cell communication is involved in the determination of cell fate. This answer choice is out of scope and doesn’t answer the question being asked.
3. At the two-cell stage, isolated blastomeres can divide and differentiate. This answer choice is inconsistent with the breakdown of the question and what we were told in the passage. Isolating blastomeres at the two-cell stage led us to find that AB cells are dependent on communication with P1 to produce normally. We can stick with our superior answer here, answer choice A.
4. Several different blastomeres can produce both neurons and muscle tissue. This answer choice is similar to answer choice B. The statement is technically correct, but it doesn’t actually answer the specific question being asked. There are multiple blastomeres that can produce neurons and muscle tissue, but that doesn’t support the hypothesis that cell-to-cell communication is involved in the determination of cell fate. Once we get rid of option D, we’re left with our correct answer, answer choice A: The fate of an isolated AB cell differs from that of an AB cell in an intact embryo

7) Let’s go back to Experiment 1-this is going to be similar to our previous question.

I’ve added Figure 1 and Experiment 1 here again. We’re going to focus on what we focused on for Question 6.

Cultured AB cells produced only neurons and skin, but no muscle. The cultured P1 cells gave rise to all of the tissues normally produced by P1 cells of an intact embryo.

That means the P1 cells are acting the same, regardless of the presence of AB. But AB produces a different result when isolated from P1. That means AB is dependent on communication with P1 to produce normally. We expect the direction of signaling of a two-cell embryo is P1  AB

1. Answer choice A says AB → P1. This is the opposite of what I said during the breakdown of the question. We expect AB is dependent on communication with P1, not the other way around.
2. P1 → AB. This answer choice matches what I said during the initial breakdown, and also what we covered in Question 6. AB is dependent on communication with P1 to produce normally. We expect the direction of signaling of a two-cell embryo is P1 AB. This answer choice is superior to answer choice A.
3. P1 → P2. This answer choice is out of scope. We don’t cover P2 in Experiment 1. We already have an answer choice that matches the passage and predicted answer exactly, so we’re sticking with answer choice B for now.
4. zygote → AB. Answer choice D is similar to answer choice C. There is no mention of the one-celled zygote in Experiment 1. Instead we focused on separating the cells of a two-cell embryo and focusing on AB cells and P1 cells. We’re left with our correct answer, answer choice B.

8) Previously we used Experiment 1, now we’re using Experiment 2.

We have Figure 1 and Experiment 2 here. From the passage, we know we normally have AB cells produce neurons, skin, and muscle. When these cells were treated with cycloheximide, which is an inhibitor of translation, the cells only produced neurons and skin; no muscle was produced.

When the cells were treated with actinomycin D, which is an inhibitor of transcription, the cells produced neurons, skin, and muscles, which is what’s normally produced by AB cells. Transcription produces mRNA, and translation produces proteins. Our correct answer is going to focus on translation and proteins. Inhibiting protein synthesis affected the normal production of the two-cell embryo.

1. DNA. Neither of our inhibitors directly inhibit the production of DNA. That means if DNA were involved, we’d expect the same results in both sets of AB cells. That’s not the case here. The cells that were incubated in the presence of cycloheximide produced only neurons and skin, and no muscle.
2. Messenger RNA. This would be the correct answer if the results of Experiment 2 were switched. The inhibitor that affected translation was the one that caused no muscle to be produced. It was not the inhibitor that affected transcription. This is a better answer than answer choice A, because if messenger DNA were the correct answer, we’d at least be able to figure it out through experimentation. We can eliminate answer choice A.
3. Ribosomal RNA. This is similar to answer choice A. rRNA acts as the building blocks of ribosomes, but isn’t the end product of transcription or translation. There’s no indication in the experiment that rRNA is involved in signaling interaction at the two-cell stage. Answer choice B is still a superior answer choice.
4. Protein. This answer choice matches our breakdown of the question. When cells are treated with cycloheximide, cells only produced neurons and skin, and no muscle. Cycloheximide inhibits translation, which ultimately produces proteins. Answer choice D, protein, is going to be our best and correct answer.

9) We’ll reference Experiment 3. We’ll look for a pair of cells where we have gut differentiation and specification. Also, note how passage and visual-heavy these questions are. This is 4 straight questions where we’re getting details from the passage. Make sure to understand the passage well in your initial read through.

We have Experiment 3 and Figure 2 shown here. Shaded circles say gut differentiation, and the unshaded circles say no gut differentiation. We see the 3rd set down, the EMS and P2 cell combination. That’s where gut differentiation results at all times. That’s the indicator that communication between EMS and P2 results in gut differentiation.

1. Answer choice A says P2 and EMS. This answer matches our breakdown of the question exactly. We have gut differentiation at every time on the timeline when we have EMS and P2 together. We can keep this answer choice for now, but we’re going to remain unbiased and quickly go through our other answers. There is a good chance this is going to be our correct answer because matches my breakdown exactly.
2. AB1 and EMS. This answer choice contradicts what we saw in Figure 2, and our breakdown. If we look at the timeline, we don’t have the same differentiation we see with P2 and EMS. We can eliminate answer choice B.
3. P1 and AB1. This answer choice contradicts what we saw in Figure 2, and our breakdown. Similar to answer choice B, we don’t have gut differentiation the same way we do for EMS and P2. We can eliminate answer choice C.
4. AB1 and AB2. This answer choice is similar to answer choices B and C. It contradicts what we saw in Figure 2 and the breakdown. We can stick to our correct answer, answer choice A: P2 and EMS

10) We’re tracking the sequence by which gut differentiation happens. From zygote to P1, and ultimately to gut.

Looking at the Figure, the predicted sequence is Zygote P1 EMS E

1. AB. This doesn’t match what I said in breaking down the question. There is no sequence that goes Zygote P1 AB. This contradicts Figure 1.
2. EMS to E. This matches our predicted sequence exactly. I said the predicted sequence is Zygote P1 EMS E. We can keep answer choice B-it’s superior to answer choice A.
3. P2 to EMS to E. This answer choice has two of the correct components, but also mentions P2. That’s not the correct sequence based on the Figure.
4. both P2 and EMS. Another answer choice that doesn’t match the Figure or the breakdown of the sequence. We’re left with our correct answer, answer choice B. The sequence is Zygote P1 EMS  E.

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