This week, Madeline and I discuss ways to mentally calm down for the final two sections after the lunch break. We dive back into passages and questions.
Listen to this podcast episode with the player above, or keep reading for the highlights and takeaway points.
[02:18] The Right Mindset for Taking the Bio/Biochem Section
As students go back to their seats for this section, Madeline recommends that students really take their break and take some breaths. Understand that what happened before that’s done is already done. The only thing you can change is the next process. You have a clean slate now. Go in and you can still absolutely crush it.
And even if you feel you might have bombed it, you have no idea. There have been so many students who came out of tests who said that was the worst test of their life. And they got their scores back and it was their best test. Of course, that’s not always the case. But just because you feel a certain way does not mean that you did badly.
[04:36] Passage 5 (Questions 23 – 26)
The flux of water across biological membranes is facilitated by transmembrane protein channels called aquaporins (AQPs). AQPs passively transport water in response to osmotic gradients while excluding the movement of ions, including protons; thus, they are highly important for cell volume regulation. One particular aquaporin, AQP5, has been a recent focus of study due to its involvement in cystic fibrosis and several other diseases.
Note: Notice we’re highlighting the keywords in bold.
[05:41] Paragraph 2
To better understand this topic, researchers developed a mutant construct in human embryonic kidney (HEK) cells in which the S156 residue was replaced with glutamate. This mutation had been previously observed to have a phosphomimetic (phosphorylation-mimicking) effect on the residue. Membrane expression of AQP5 in these S156E cells was compared to expression in wild-type HEK cells. Researchers also assessed the effect of myristoylated PKI 14–22 amide, a known PKA inhibitor, on both S156E and WT cells. Figure 1 shows the measurements of expression performed after 30 minutes of inhibitor exposure.
Note: Notice we’re highlighting the keywords in bold. Basically, we’re just highlighting what it is and what has been done on that human construct to make it the S156. T cells.
Figure 1 Membrane expression of AQP5 in wild-type and mutant HEK cells with or without PKA inhibition
[08:37] Paragraph 3
The proper regulation of membrane protein abundance requires a delicate balance between two opposing processes: translocation to and internalization from the membrane. Translocation and internalization of AQP5 have been shown to be regulated by at least two factors: the phosphorylation state of the S156 residue on the AQP5 protein and the signaling activity of both cyclic AMP (cAMP) and protein kinase A (PKA). Interestingly, increased cAMP levels were observed to have a biphasic effect on the sub-cellular distribution of AQP5, with decreased expression in the plasma membrane in the short term (< 1 hour), followed by an increase in membrane abundance in the long term (> 1 hour). Dysfunctional trafficking of AQP5 has been implicated in several human disease states, including Sjögren’s autoimmune syndrome, bronchitis and cystic fibrosis.
Note: Notice we’re highlighting the keywords in bold. We’re highlighting the short term and the long term because that shows us the effect and the timeline.
Some students have this tendency to hyperlight, which means they highlight everything. But the skill of reading a paragraph, pausing, reflecting, and then highlighting avoids that because then you’re thinking about what’s actually important versus what’s not. It’s based on your reflection. And so, that’s a helpful way to avoid that hyper-highlighting.
[12:33] Question 23
Researchers later isolate another residue that tends to be phosphorylated at position 259 of the AQP5 protein. If they desire to replicate the results described in the passage with a phosphomimetic mutant construct, this construct is most likely to be:
First off, if you don’t understand anything here and you can’t find those things mentioned anywhere in the passage, skipping the question could be a great strategy.
In Paragraph 2, we highlighted phosphorylation-mimicking. It makes sense that we’re going to be looking at the same part of the passage where we highlighted as it’s being referenced in a question.
Then it specifically says that this type of residue is going to have the mutant of S156E cells. Now here, this is where a little bit of content comes in. Because this is a really common way to show a specific amino acid switch at a certain number of residues within a primary amino acid chain.
Your primary amino acids are set up like a beaded necklace or a beaded chain. Each one of those is given a position like 123, all the way up to, 156 here. And each one of those has a specific amino acid that’s going to be known as serine, cysteine, alanine, etc. And that’s going to be shown as its one-letter code.
S156E is a serine residue at a 156 position. The E is what they’re switching to from Serine. So the E is now the glutamic acid. It’s one of your acidic amino acids, which means it has a negative charge.
We’re going from something that’s polar with an OH group for your serine to something that has a negative charge to your glutamine.
We just want to see something that’s parallel to that same change in our amino acids. So the 259 in our answer choices don’t matter because those are all the same.
We just need to see if there’s something that looks like serine. And we look for something that looks like a negative, or an acidic amino acid that’s being changed to.
If it’s talking about a protein and about residues on a protein, usually residues are amino acids. And then we know amino acids have these one-letter codes. So maybe even if you didn’t quite know that that’s like a convention of showing when you’re switching an amino acid on a residue, you might still be able to get there using those tips and tricks. You know what residues are in context so you know what’s happening here.
A – T is threonine, which has an OH group and D is aspartic acid which has a negative. So this one may already seem good.
B – Y is tyrosine, which does have an OH group, but it’s going to A which is an alanine, which is a nonpolar. So we cross out B.
C – F is a phenylalanine, which does not have an OH group. So we could cross that.
D – G is glycine going to a W, which is tryptophan. Glycine is nonpolar, so we can also cross that one out. So by the process of elimination, it’s A.
Correct Answer: A
[19:24] Question 24
The results in Figure 1 and the information in the passage most strongly support which of the following conclusions?
- Phosphorylation of S156 by protein kinase A promotes the immediate localization of AQP5 to the plasma membrane.
- Phosphorylation of S156 promotes the internalization of AQP5 in the short term.
- Protein kinase A promotes the internalization of AQP5 in the short term.
- 30 minutes of exposure to protein kinase A stimulates the internalization of AQP5, a process that is upregulated when S156 is phosphorylated.
The first thing here is they tell you where to look, which is wonderful. They say the results in Figure 1. So in this case, we go to Figure 1.
Because there can be so many conclusions to draw from it, this is a great way to use a strategy called process of elimination or POE. So we can go into each answer choice and see if it fits with Figure 1. If not, then we can cross it out and keep going.
Going back to the figure, try to understand what we’re looking at. The graph has the x axis, we have AQP5 and our phosphomimetic version. And then on our y axis, we have relative membrane expression.
It means that you’re getting a higher localization to that membrane if you have a higher number on that y axis.
We also have two different options – the uninhibited control, and then the PKA inhibition, which is the protein kinase A. And that’s what’s happening here for inhibiting your PKA.
Madeline just made a correction she said earlier ago about skipping the figures. And she clarified that you do want to know what you’re looking at before you do jump into the answer choices. So don’t make the mistake she just made.
A – This is a graph that just says something has been inhibited and something has been phosphorylated. We don’t have a mechanism through which that has happened. There’s a keyword here where it’s phosphorylation by protein kinase A. Nothing has told us that protein kinase A is actually the kinase or the protein that phosphorylates S156. So in that case, that’s actually going beyond what can be given by the graph. Because it’s talking about a mechanistic factor that is not involved here. So we cross this out because it’s saying that PKA is doing the phosphorylation.
B – So if we’re looking at our graph, and we see phosphorylation of S156, which is that second column, and we see an increased number after that PKA inhibition of your relative membrane expression of your AQP5. Well, if you’re getting more on the membrane, you’re not going to have more internalized. And so because of that reason, B is going to be wrong. Because it’s saying internalization, but we’re actually getting an increase of localization to the membrane.
C – It said 30 minutes and now we’re looking at the part where it says PKA promotes internalization. Here, we just said that PKA inhibition is when you take out PKA and you’re going to have things go into the membrane or localize. So if you put PKA back in, then it would make sense because you’re going to have more internalization. Protein kinase A would then promote the internalization because we take out protein kinase. So now, your AQP5 is going to the membrane and not being internalized. This is a little bit tricky because it’s almost an opposite process of what’s being shown in the graph.
D – Figure 1 shows the 30 minutes of exposure to the PKA inhibitor promotes the number and expression in both the wild type and the mutant strains. So because of that, PKA does stimulate internalization. But the important thing here is that the graph does not provide sufficient information to clearly predict what would happen after 30 minutes of exposure to just PKA. So here, we don’t have something with just PKA in it. It’s not isolating the factors in order to draw this conclusion. We only have AQP5 wild type with this phosphomimetic version of our AQP5, but we don’t have anything where it’s only with PKA and not PKA.
Correct Answer: C
[26:52] Question 25
Which of the following enzymes should the researchers add to the cell samples if they want to reverse the general catalytic effects of protein kinase A?
- Glycogen phosphorylase
- Protein phosphatase 1
- Lactate dehydrogenase
Reframing the question: It’s a pseudo-discrete question of how does a protein kinase A work and what enzymes potentially would block that?
Tip: If you’re ever looking at a question you don’t understand, rephrase the question. If you’re able to rephrase it, then you probably understand what it’s asking. If you’re not able to rephrase it, then maybe you need to look at it a little bit longer. Maybe reword it and do a little mental gymnastics to understand what’s happening.
Going through the answer choices:
This is protein kinase A. And there are two ways to go about this. The first way is knowing what a kinase does. A kinase is a specific enzyme that adds a phosphate onto a protein specifically. And if a phosphate is added on by a kinase, the opposite would be to take away a phosphate group.
And so here, what would take away a phosphate group would probably be something with “phospho” in it. So we take away D from that because that’s lactate dehydrogenase. It doesn’t have anything with phospho in it.
So now, we’re left with A, B, and C.
We’re talking about something that phosphorylates a protein. Then which of those three would make sense that you’re taking a phosphate group off of a protein?
And that’s Protein phosphatase 1, which is C, because that’s the only one that actually tells you that it’s being taken from a protein.
A – We’re not talking about glycogen here.
B – The type of enzyme we’re talking about here isn’t the “phosphogluco” part, it’s actually the mutase. So we’re talking about a mutation. It mutates something or changes its shape. And that doesn’t have to do with phosphate groups necessarily.
But if you didn’t know anything about the function of enzymes, you might be able to get it just by saying protein kinase is going to do something to a protein and C is the only thing protein there.
Correct Answer: C
[30:52] Question 26
A rapid mechanism is thought to govern the localization of AQP5 in response to changes in extracellular osmolarity. If this mechanism is independent of both PKA activity and S156 phosphorylation, which of the following will most likely be observed?
- HEK cells exposed to the most hypotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow out of the cells.
- HEK cells exposed to the most hypotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow into the cells.
- HEK cells exposed to isotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow out of the cells.
- HEK cells exposed to isotonic conditions will display the greatest degree of AQP5 membrane localization, allowing water to flow into the cells.
Reframing the question: It’s saying that there’s a localization that changes based on extracellular osmolarity. It’s a mechanism that’s independent of everything in the passage, which can be observed. So aquaporins going to the membrane are dependent on osmolarity. What can we expect to happen when we have more aquaporins in our membrane, dependending on osmolarity?
Prediction: So we’re talking about aquaporins. It’s something that allows for passive diffusion. It’s actually mentioned in that first paragraph. Aquaporins allow for passive transport of water that goes on to the membrane and has to do with osmolarity. So maybe we’re talking about osmosis, maybe some hypertonic/hypertonic solutions.
Going through the answer choices:
A – They’re saying hypotonic conditions will allow water to flow out of the cells. If we think about osmosis, it doesn’t make sense because hypotonic would be flowing into the cells.
B – This one makes sense because we did say hypotonic would flow into the cells.
C – Isotonic means you have the same tenacity of both conditions. So it doesn’t make sense if there’s any water flowing. So no to this.
D – This has the same reasoning as answer choice C, so this is out as well.
Correct Answer: B
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