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MCAT Content / AAMC Section Bank Bb Solutions

AAMC Section Bank BB [Web]

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 1

1) First thing we want to note is the test-maker says “based on the information in the passage.” That doesn’t always mean we have to flip back to the passage. You want to read the passage properly and thoroughly for a reason, and that’s to make sure you fully grasp the big picture. For this specific question, we can flip back to the passage and see what the author mentions about the cross-bridge cycle and possible suppression.

We have part of our passage here where the author talks about the CBC. When you go back to the passage like this, I don’t want you re-reading the entire passage or even entire paragraphs. I’m going to walk you through the important parts here, just to be thorough. When you’re actually practicing or taking the exam, going back to the passage (unless it’s for specific details you don’t want to memorize) can be a waste of time. The author says, “calmodulin….activates an enzyme that phosphorylates amino acid residue 19 of the myosin light chain (LC20). Phosphorylation of LC20 is required to activate the myosin head, which binds to actin. This myosin–actin interaction forms the basis of the cross-bridge cycle (CBC).” There’s phosphorylation of LC20. Phosphorylation is done by a protein kinase; we’re adding a phosphate group to a molecule. How would we suppress CBC? By inhibiting, preventing, or reversing this phosphorylation. Answering this question boils down to knowing our enzyme types. A kinase phosphorylates a molecule, while a phosphatase dephosphorylates a molecule. That means our ideal answer is going to be the opposite of a kinase, or a phosphatase.

  1. Phosphorylase. Phosphorylase enzymes catalyze the addition of a phosphate group from an inorganic phosphate. We want an answer that focuses instead on dephosphorylation. This answer choice contradicts our breakdown, and what the author told us in the passage.
  2. Kinase. This is the opposite of our breakdown of the question, and would actually increase phosphorylation of LC20. Increasing phosphorylation would increase CBC, not suppress CBC. Again, another contradiction. The author mentions phosphorylation actually activates the myosin head, which binds to actin. That interaction forms the basis of the CBC, not suppresses CBC. Both answer choices A and B are, for our purposes, equally incorrect.
  3. Phosphatase. This answer choice matches our breakdown. We said a phosphatase dephosphorylates a molecule. That dephosphorylation wouldn’t allow for the activation of the myosin head, or the binding to actin. The enzyme that most likely would suppress CBC is a phosphatase, that matches our prediction. We can eliminate answer choices A and B because both directly contradicted the passage and our question breakdown.
  4. Synthase. Synthase enzymes are a type of lyase: There’s a breakdown of a molecule to form two different molecules. This could theoretically prevent activation of the myosin head, but this doesn’t directly address the reasoning given in the passage. Our passage specifically mentions phosphorylation of LC20 is required to activate the myosin head, which binds to actin. By finding an enzyme that focuses on dephosphorylation specifically, we have our best answer. We can eliminate answer choice D, and keep our superior answer, answer choice C. The type of enzyme most likely to suppress CBC is phosphatase. 

2) No tricky wording here, we’re simply relating the results in Table 1 with VSM function and cytoskeletal dynamics. 

We have three different artificially adjusted internal pressures: 10, 80, and 120 millimeters of mercury. We have dependent variables listed on the left side: arterial diameter, percent phosphorylated LC20 and calponin, and G-actin amounts.

Arterial diameter decreases with increased pressure. That corresponds to a decrease in resistance, meaning flow stays relatively consistent.

Percent phosphorylated LC20 and calponin. We see a significant increase in both variables as we go from internal pressure of 10 to 80 millimeters of mercury. When we get to 120, there’s a small increase in LC20 percentage, but calponin percentage stays the same. 

G-actin amount. We have a general decrease from lower internal pressure to higher internal pressure. But there’s reversible polymerization of G-Actin into F-actin filaments. Meaning, that decrease in G-actin amount likely means we have an increase in F-actin amount.

  1. Vasoconstriction is associated with an increase in the ratio of F-actin to G-actin. Vasoconstriction is a decrease in arterial diameter, or going from a lower internal pressure to higher internal pressure. That also corresponds to a decrease in G-actin amount, but we said increase in F-actin amount. That means the ratio of F-actin to G-actin is higher when we have vasoconstriction. This answer choice sounds good for now, but we always want to be thorough and look at the other answer choices.
  2. F-actin levels decrease as phosphorylation of LC20 increases. Again, we expect F-actin amounts to increase going from left to right in our table, even though it doesn’t explicitly say so. We have an increase of phosphorylation of LC20 from left to right, we also have an increase in F-actin levels from left to right. This answer choice contradicts Table 1 from the passage, so we can eliminate answer choice B.
  3. Vasodilation does not affect phosphorylation levels of calponin. Vasodilation happens from right to left in the table, as we decrease internal pressure. As there’s vasodilation, we actually see a decrease in the phosphorylation levels of calponin. We can’t establish the phosphorylation levels are unaffected. We can eliminate answer choice C as well because it contradicts Table 1. 
  4. Arterial diameter reduction is always dependent upon increased calponin phosphorylation. Arterial diameter varies with changes in internal pressure. But not always with calponin phosphorylation. Look at internal pressure of 80 millimeters of mercury to 120 millimeters of mercury. We have no change in calponin phosphorylation, but we do have arterial diameter reduction. We can say arterial diameter reduction isn’t always dependent upon increased calponin phosphorylation. This answer choice is too extreme, so we can eliminate answer choice D. We’re left with our correct answer, answer choice A: Vasoconstriction is associated with an increase in the ratio of F-actin to G-actin.

3) This is another answer that is going to come from information in the passage. We’ll note what the author says about latrunculin B and how those effects will affect data in Table 1.

We have a part of our 3rd paragraph from the passage here. The author says Actin depolymerization can be induced by the drug latrunculin B. Depolymerization would be the breakdown of polymerized actin into monomers of globular actin. Essentially more G-actin, and less F-actin. Not much else we need to dissect there, let’s look at Table 1 below.

Question stem asks specifically about 120 millimeters of mercury. What do we expect to happen at that internal pressure specifically? We’d expect the direct results to be a G-actin amount that is greater. We have depolymerization, like we just talked about, which means polymerized actin is broken down into monomers of globular actin, or G-actin. So, our answer is going to be something consistent with increased G-actin levels.

  1. increased phosphorylated LC20 levels. This answer choice is not something we talked about during the breakdown. Phosphorylated LC20 levels depend more on calcium ions and calmodulin, at least for the sake of this passage. Also, as we see increased G-actin levels, we eventually have lower phosphorylated LC20 levels. Not exactly an inverse relationship, but something to consider.
  2. decreased arterial diameter. This answer choice contradicts our breakdown. From a general trend perspective, we expect an increased G-actin amount to correspond to increased arterial diameter, not decreased. We can eliminate answer choice B for being a direct contradiction. Answer choice A presented a relationship that we didn’t explicitly identify, but we can still hold on to answer choice A because it didn’t directly contradict our prediction like option B.  
  3. decreased F-actin levels. This answer choice is consistent with our breakdown. Our prediction was focused on G-actin amounts, not F-actin levels, but what did we say during the breakdown? Depolymerization would be the breakdown of polymerized actin into monomers of globular actin. That means more G-actin, and decreased F-actin levels. We can now eliminate answer choice A because answer choice C is a superior and more direct answer choice.
  4. increased phosphorylated calponin levels. This answer choice is similar to answer choice A. When we have increased G-actin levels, we have decreased phosphorylated calponin levels. That’s the opposite of this answer choice. We can eliminate answer choice D. We’re left with our correct answer, answer choice C: decreased F-actin levels.

4) Essentially, we’re relating the myogenic response with sympathetic stimulation. We’ll do an overview of stimulation of the sympathetic nervous system, and relate to myogenic response. Meaning we’ll focus on blood flow, and varying vessel diameter.

The sympathetic nervous system controls the body’s automatic response to danger. Normally that entails increased heart rate, slowing digestion, and something we’re interested in today: moving blood flow to the heart, muscles, and also the brain. In the muscles we have vasodilation so we can maximize the blood making it to the muscles in our fight or flight response. 

But, if we have increased heart rate and blood is moving toward the brain, we have to be careful. That’s where the myogenic response comes in. We don’t want blood flow to be too high, so we compensate. We can increase resistance to account for the higher pressure. That ensures the flow remains in a safe, steady range-and doesn’t get too high.

  1. equalizes blood pressure in cerebral resistance arteries and the aorta. Besides the heart, we expect the aorta to have the highest blood pressure of anywhere in the body. Cerebral resistance arteries likely can’t handle that same blood pressure, so equalizing blood pressure is unreasonable. 
  2. moderates blood flow to the brain under high pressure. This answer choice sounds like our prediction and what we’ve covered in the passage. The brain can’t have the same increased blood flow coming in as the muscles and other organs. That’s why the myogenic response works to increase resistance of blood flowing to the brain, despite the higher blood pressure. We can hold on to this answer choice and eliminate answer choice A for being unreasonable.
  3. enables cerebral resistance arteries to locally vasodilate. This answer choice is the opposite of our breakdown. We said we expect the arteries to vasoconstrict to add resistance, not vasodilate. If these arteries vasodilated, we’d put the brain at risk because of this sudden, increased blood flow.
  4. redirects blood circulating in the brain to organs in the abdominal cavity. This answer choice also contradicts our breakdown. Stimulation of the sympathetic nervous system means slowing digestion and decreasing the amount of blood directed toward organs in the abdominal cavity. We can also eliminate answer choice D, so we’re left with our correct answer, answer choice B: The myogenic response moderates blood flow to the brain under high pressure.

5) To attack this question, we’ll focus on G-actin and F-actin, and actin polymerization. The only mention the author made in the passage about polymerization is actin polymerization: monomers of G-actin polymerize into F-actin. What structural component are we dealing with? F-actin can actually be described as microfilaments, or actin filaments. Essentially this is a standalone, content question that’s asking us to identify microfilaments are made of actin proteins.

  1. Microtubules. This answer choice seems like a tool the author uses to trick us. Microtubules sound similar to our breakdown, and have some functions similar to microfilaments. But Microtubules are hollow tubes composed of tubulin proteins, not actin. They are longer and wider than microfilaments. They help the cell transport materials within itself and resist shape changes. Not quite what we’re looking for, but we can keep comparing with the remaining answer choices.
  2. Microfilaments. We mentioned microfilaments are thin protein fibers made up of actin proteins; their fundamental role is to absorb tension. They’re the thinnest part of the cytoskeleton. This is what the author was talking about in the passage, so we’re going to keep this answer choice for now. We can eliminate answer choice A because it was a distractor. Answer choice A was only presented because it sounds like microfilaments.
  3. Intermediate filaments. Intermediate filaments have multiple different proteins, and not predominantly actin like our microfilaments. Keratin is an example of a protein that makes up intermediate filaments. These intermediate filaments contribute to cellular structural elements. Answer choice B is still our superior, and most direct answer choice, so we can eliminate answer choice C.
  4. Thick filaments. Thick filaments deal with muscle contraction, and are associated with myosin, not actin. The two work together in contraction, but we’re looking for an answer choice that focuses on actin specifically. We can eliminate this answer choice as well, so the correct answer, and the most direct answer, is answer choice B: Microfilaments.

6) Let’s flip back and see what the author mentions about the amino acid. We’ll use our general knowledge to identify the amino acid, and match it to its structure.

The passage says Calcium ions bind the cytosolic protein calmodulin, which activates an enzyme that phosphorylates amino acid residue 19 of the myosin light chain (LC20). Key thing we want to point out is we have phosphorylation. Phosphorylation is a form of protein modification and regulation. Which amino acids are phosphorylated? For the sake of the MCAT: serine and threonine. Sometimes we include tyrosine. We’re thinking of amino acids with hydroxyl groups in their side chains, so our answer will likely be serine or threonine. This comes down to knowing your amino acid structures, if you’re not familiar with these structures, you may want to spend some time reviewing. 

Answer choice A shows (Glycine)

Answer choice B shows (Valine)

Answer choice C shows (Isoleucine)

Answer choice D shows (Serine)

Only one answer choice matched either serine or threonine, and it was the only one answer choice had that hydroxyl group in its side chain. We can eliminate answer choices A-C and we’re left with our correct answer, answer choice D: Serine.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 2

7) To answer this question, we can revisit Figure 1 in the passage and see what body weight trends we should be expecting. Germ free (GF) condition mean the gut doesn’t become colonized, and no normal production of short chain fatty acids. In other words, no typical activation of GPCRs, and specifically GPCR43.

We have Figure 1 here, and we want to break down the results of the experiment. We can make a broad conclusion from the passage that regardless of which diet we look at in the experiment, the GPCR43 deficient mouse is heavier than its wildtype counterpart. Said differently, the wildtype mouse weighs less than its GPCR43 deficient counterpart that ate the same diet. Meaning we’re expecting the wildtype mice to be lighter, given all other conditions are the same.

One more wrinkle we have to introduce. The question stem mentioned we’re also looking at a germ-free condition where the gut doesn’t become colonized. That means no activation of GPCR43-so even the wildtype genotype that’s in the germ-free condition, should yield the same results as the GPCR43 deficient mice. We’re expecting the wildtype genotype in the conventional condition to be the lowest body weight.

  1. Answer choice A looks off right away. We have a higher body weight in the wildtype. We said the wildtype mouse weighs less than its GPCR43 deficient counterpart that ate the same diet. Let’s keep comparing to the remaining options.
  2. Answer choice B is consistent with our breakdown. Wildtype has a lower body weight, and we mentioned something about this experiment specifically. The question stem mentioned we’re also looking at a germ-free condition where the gut doesn’t become colonized. That means no activation of GPCR43, and even the wildtype in the germ-free condition will have a higher body weight. We can keep this answer choice and eliminate answer choice A because it contradicted Figure 1 and our breakdown. 
  3. Answer choice C shows the wildtype genotype corresponding to a lower body weight. But we’re going to focus on the germ-free condition specifically. We said germ-free means no activation of GPCR43 in the wildtype. That body weight shouldn’t be that low. We eliminate answer choice C as well. It contradicts the results in the passage, and what we were told in the question stem.
  4. Answer choice D is similar to answer choice A. Wildtype typically has a lower body weight. This is showing us the opposite. This answer choice also contradicts our breakdown of the question and passage, so we can eliminate answer choice D. We’re left with our correct answer, answer choice B.

8) To answer this question, we’re going to cover the conditions in the transmembrane domain, and review the properties of amino acid side chains. We want an amino acid that LEAST likely is found in the transmembrane domains; always be careful with the verbiage.

Transmembrane domains are the regions of a protein that are hydrophobic. This becomes just a content question. The amino acids least likely to be found in one of the transmembrane domains is any hydrophilic amino acid. If we can’t narrow it down to a single answer choice, we can try and find 3 that are hydrophobic, and the remaining answer choice is the odd-one out. We can eliminate answer choices for being hydrophobic. 

  1. Aspartic acid. Aspartic acid is a very hydrophilic amino acid. We want an amino acid that’s least likely found in the transmembrane domains, which are hydrophobic. This answer choice matches what we’re looking for already. Let’s quickly go through the other options.
  2. glycine. Glycine is fairly neutral, so we’re still sticking with our hydrophilic, best answer: answer choice A. We can eliminate answer choice B for being inferior to answer choice A.
  3. tryptophan. Tryptophan is slightly hydrophobic, so it would fit in nicely in the transmembrane domains. This contradicts our prediction, and doesn’t correctly answer the question. We can eliminate answer choice C. 
  4. Phenylalanine. Phenylalanine is also slightly hydrophobic, so we have another answer that doesn’t correctly answer our question. We can eliminate answer choice D and stick with the only amino acid listed that’s hydrophilic, answer choice A: Aspartic acid.

9) Let’s recall some broad details from the passage. The leanest mice were the wild type mice that expressed GPCR43. Heaviest mice were the GPCR43 deficient mice. In general, the mice that were fed the high fat diet all got heavier. All of the mice here are being fed a high fat diet, so we’re predicting anything close to wildtype would be the leanest.

  1. Mice treated with antibiotics. This implies we’re killing the microorganisms in the gut. In the passage we found that catabolism of dietary fiber by gut microbiota produces short chain fatty acids. That eventually leads to activation of GPCRs. If we don’t have activation of GPCR43, we expect the mice to be heavier. 
  2. Mice treated with a GPCR43 antagonist. An antagonist would oppose the effects of GPCR43. That means we have a similar answer to answer choice A. Both correspond to the opposite of our wildtype, and correspond to heavier mice. 
  3. Mice in which GPCR43 is overexpressed in WAT. Overexpression of GPCR43 would be exactly what we’re looking for. Overexpression of GPCR43 would account for the high fat diet. GPCR43 deficient mice are heavier according to the passage. Wildtype mouse weighs less than its GPCR43 deficient counterpart that ate the same diet. Meaning presence of GPCR43 equals less body weight. We can eliminate answer choices A and B because both contradict our breakdown and the answer we’re looking for.
  4. Mice treated with a drug that inhibits the generation of intracellular second messengers. Without the generation of intracellular second messengers, we don’t see the effects of GPCR43. That’s similar to answer choice B, so we can also eliminate answer choice D. We’re left with our correct answer, answer choice C: Mice in which GPCR43 is overexpressed in WAT

10) To answer this question, let’s pull up Figure 2 and review what we saw in the passage, then we’ll go through our answer choices.

We have Figure 2 here and we’re going to rely on what you saw in your readthrough of the passage. In the presence of insulin, acetate decreases glucose uptake in wildtype mice, but we don’t see that same suppression in the GPCR43 deficient mice. In the absence of insulin, we don’t see any differences in the groups. The big difference is on the right side where we have the presence of both insulin and acetate. 

  1. Acetate suppresses insulin-mediated glucose uptake in WT adipocytes, but not Gpcr43–/– adipocytes. This answer choice is consistent with our breakdown. We’re focused on insulin-mediated glucose uptake, which is on the right side of our figure. Note the suppressed uptake in the wildtype adipocytes compared to the GPCR43 deficient adipocytes in the presence of acetate. We can hold on to this answer choice for now and see if any other answer choices are superior.
  2. Acetate stimulates insulin-mediated glucose uptake independently of GPCR43 expression in adipocytes. This is the opposite of our breakdown. Acetate actually suppresses insulin-mediated glucose uptake. We can see that clearly on the right side of Figure 2. Glucose uptake is less in the wildtype adipocytes vs. the GPCR43 deficient adipocytes in the presence of acetate. We can eliminate answer choice B for contradicting what we see in Figure 2.
  3. Insulin suppresses glucose uptake in the presence of acetate in WT adipocytes, but not Gpcr43–/– adipocytes. Insulin isn’t suppressing glucose uptake. Note the 4 bars on the left side of the figure and the 4 on the right side. Difference there is insulin. Insulin increases glucose uptake. We can eliminate answer choice C also because it contradicts the figure.
  4. Insulin stimulates glucose uptake in WT adipocytes, but not Gpcr43–/– adipocytes. We see the same amount of glucose uptake in 3 of the 4 adipocytes on the right side of figure 2. The only outlier is the wildtype in the presence of both insulin and acetate. This answer choice also contradicts our figure so we can eliminate answer choice D. We’re left with our correct answer, answer choice A.

11) To answer this question, we’ll have to remember the results in Figure 2, and where GPCR43 is expressed in the body. First thing we want to establish is GPCR43 is expressed in white adipose tissue, and not in muscle. Next thing we want to remember are the results from Figure 2: in the presence of insulin, acetate decreases glucose uptake in wildtype mice, but we don’t see that same suppression in the GPCR43 deficient mice. Also note, the muscle cells are going to be GPCR43 deficient-we only have GPCR43 expressed in white adipose tissue. That means our ideal answer: In white adipose tissue, we’ll have less Akt phosphorylation in the presence of acetate.

We can compare all of our answer choices at once. We expect Akt phosphorylation levels to be equal in muscle. That means we can eliminate answer choices C and D right away. 

We also expect less Akt phosphorylation in the presence of acetate. We can eliminate answer choice A. We’re left with the only answer choice that’s consistent with our breakdown, answer choice B.

12). To answer this question, we’ll note what antibiotics would do to these mice, and what that does to our experimental results. Antibiotics mean no gut microorganisms. That also means no activation of GPCRs. The author explained that relationship earlier in the passage. So essentially, we’re going to see the same results in the wildtype mice as we’re used to seeing in the GPCR43 deficient mice. That typically means increased body weight, and no effects from the presence of acetate. 

  1. increased plasma butyrate levels. This is the opposite of what we said in the breakdown of the question. We said antibiotics means no effects of gut microorganisms. That means no catabolism of dietary fiber, or production of butyrate and acetate. This contradicts our breakdown and the passage. 
  2. decreased body mass. Another answer choice that contradicts our breakdown. We said we’re seeing the same results as we’re used to seeing in the GPCR43 deficient mice, which is an increase in body weight, not a decrease. Neither answer sticks out at this point, so we’ll keep answer choices A and B for now.
  3. Answer choice C says increased volume of adipocytes. This answer choice is consistent with our breakdown. Antibiotics means no activation of GPCRs, and ultimately, we said that leads to increased body weight. Increased body weight is the result of the increased volume of adipocytes. We can eliminate answer choices A and B now-both contradicted our breakdown and the passage. 
  4. decreased insulin sensitivity in adipocytes. We just mentioned in our breakdown of answer choice A: antibiotics means no effects of gut microorganisms. That means no catabolism of dietary fiber, or production of butyrate and acetate. No acetate means insulin sensitivity is normal, and not decreased. We saw that in Figure 2 in the passage. We can eliminate answer choice D and we’re left with our correct answer, answer choice C: increased volume of adipocytes. 

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Questions 13-17

13) This question comes down to knowing content. Specifically, we’ll need to know which amino acids are branched chain amino acids and which are unbranched. I always use a mnemonic to help me here. When I think of the word “branched” I think trees, and trees are a LIVing thing. The reason I capitalized LIV is because the three branched chain amino acids are Leucine, Isoleucine, and Valine. Reading the question stem again, we want an answer choice that does not include one of these three amino acids.

  1. Ala. Alanine is not a branched chain amino acid. You can look at the image above and see alanine only has a methyl side chain. 
  2. Ile. Isoleucine is one of the branched chain amino acids. 
  3. Leu. Leucine is one of the branched chain amino acids.
  4. Val. Valine is one of the branched chain amino acids. Answer choice A remains the best option here.

14) This is a standalone question that relies on knowing the difference between α and β designators. The designation is traditionally used to distinguish molecules with multiple chiral centers. Think of two carbohydrates that are virtually identical, except for their configuration on a single carbon. How do we distinguish between the two? Is there even a reason to distinguish between the two, or are they the same thing for all intents and purposes?

There is certainly a reason! Epimers are stereoisomers that differ in the configuration of atoms attached to a chiral carbon. A big one we see is glucose and its epimers. The difference in the position of the hydroxyl at a chiral carbon creates these epimers. When glucose is in its cyclic form, a chiral center is generated at C-1. That carbon atom that forms the new chiral center (C-1) is the anomeric carbon.  We can see two anomers below and how they differ in position at the anomeric carbon:

  1. enantiomers at an epimeric carbon atom. Be careful with the verbiage here! We do need to distinguish between enantiomers, but that’s with the R and S designations. This does not encompass the α and β designators we talked about in our breakdown of the question.
  2. enantiomers at an anomeric carbon atom. Similar to answer choice A, but we at least mention the anomeric carbon in this case. This is better than answer choice A, but still not 100% consistent with the verbiage in our breakdown.
  3. epimers at an anomeric carbon atom. This is consistent with what we came up with in the breakdown of the question. α and β designators are used to distinguish between epimers at an anomeric carbon atom. The visual above really helps demonstrate the difference between the two if it’s unclear.
  4. epimers at a non-anomeric carbon atom. This answer choice starts out strong until we get to “non-anomeric” which is not true about this question. Answer choice C matched our breakdown exactly, so we can stick with C as our correct answer.

15) This is a standalone question so what I like to do is give a brief overview of the content being tested. In this case, we’re dealing with osmotic pressure. Osmotic pressure is a measure of the tendency of water to move into one solution from another by osmosis. The higher the osmotic pressure of a solution the more water wants to go into the solution. We want to pick the solution with the greatest concentration of solute particles. Why is that? Because of the definition we just covered. There will be greater movement of water in that situation. Quick glance at our answer choices shows multiple molarities, but also compounds that can dissociate into ions. Much like these compounds can be broken down, let’s break down each answer choice.

  1. 0.1 M MgCl2. At first glance, one might incorrectly glance at this answer choice and see 0.1 M and move on. However, this solution can dissociate into multiple ions. MgCl2 can dissociate into three ions, which also means we have 0.3 M in solute particles. We know this question entails knowing more than just which number in the answer choices is the largest, so the fact that we multiplied this molarity by three ions makes this a good start. Not always the case, but we know AAMC typically has something up their sleeve, so we like answer choice A for now.
  2. 0.2 M NaCl. Similar to answer choice A, we’re not taking this answer choice at face value. We know sodium chloride can dissociate into two ions. What does that do to our molarity? It can be multiplied by 2 to give us 0.4 M. This is now our best answer. 
  3. 0.2 M CaCl2. Similar methodology here, but we expect calcium chloride to dissociate into three ions. What does that do to molarity? We multiply by 3 to give us 0.6 M. This is our best answer so far. We can eliminate answer choices A and B. 
  4. 0.5 M Glucose. This is tricky because glucose does not ionize into multiple molecules in solution like we expect from answer choices A-C. While this is our second-best answer, we still have a superior answer with answer choice C.

16) Similar to most standalone question, let’s do a quick overview of the content being tested, then we’ll jump deeper into the specific question and the answer choices. 

Gated ion channels bind a ligand and open a channel through the membrane that allows specific ions to pass through. In contrast, while ligand-gated ion channels rely on ligand binding to open, voltage-gated ion channels open in response to a change in membrane potential. We’re focused on the former (ligand-gated), and want to make sure to not get confused by the latter (voltage-gated). Typically, we see ligand-gated ion channels bind a ligand like a neurotransmitter, and open a channel through the membrane that allows specific ions to pass through. A big one we see is in electrically excitable cells like neurons, which need to react very quickly to a stimulus. When a gated ion channel opens, it rapidly lets ions move through the channel, either into or out of the cell depending on the concentration gradient of ions across the plasma membrane.

  1. Release of Ca2+ from the sarcoplasmic reticulum of a muscle fiber to initiate muscle contraction. This is one distinction we were careful to make in the breakdown of the question. This is mediated by a voltage-gated ion channel, not ligand-gated. 
  2. Influx of Na+ across the axon membrane of a somatic neuron during action potential propagation. Similar to answer choice A, this is mediated by a voltage-gated ion channel, not ligand-gated. Recall that ligand-gated ion channels rely on ligand binding to open, but we don’t have that binding here.
  3. Influx of Na+ across the motor end plate resulting in the depolarization of the muscle fiber membrane. In the previous answer choice, I mentioned ligand-gated ion channels rely on ligand binding to open. Depolarization of the muscle fiber membrane is the result of sodium ion channels binding acetylcholine. I also mentioned that typically, we see ligand-gated ion channels bind a ligand like a neurotransmitter, and open a channel through the membrane that allows specific ions to pass through. We like this answer choice better than answer choices A and B.
  4. Re-entry of Ca2+ back into the sarcoplasmic reticulum of a muscle fiber to end muscle contraction. This process is done by a pump and is powered by ATP. The pump functions to get calcium ions back into the sarcoplasmic reticulum to allow muscle to relax. We’re not looking for an active pump or transport. Answer choice C remains our best answer. 

17) Let’s break this down a bit at a time. A point mutation is a single nucleotide change in the sequence of DNA. Southern blotting is used to detect a specific DNA sequence in a sample. It involves the transfer of DNA to a nylon membrane and, like I mentioned, probing for the presence of certain sequences. To detect point mutations in Southern blotting, researchers will use Restriction Fragment Length Polymorphism, which we typically see as RFLP. RFLP allows us to identify differences in DNA sequences based on the length of fragments caused by restriction enzymes. A restriction enzyme recognizes and cuts specific 4-6 nucleotide-long palindromic sequences in the DNA and cleaves both strands. To find the correct answer, we need to find a palindromic sequence (complementary strand is identical) with a mutation in the bottom, mutant sequence.

a. 

Focus near the middle of the WT here. We have our 4-6 nucleotide-long palindromic sequence 5’-AAGCTT-3’. It’s palindromic because it’s complementary strand will be identical (5’-AAGCTT-3’). In the mutant, however, we see the 2nd A replaced with a T. That’s no longer palindromic, so this checks all the boxes we need to for an answer choice.

b.

The mutation in this answer choice is near the end of the sequence. A C is replaced with an A. While this is a point mutation, it doesn’t disrupt a palindromic sequence like we saw in option A. We can stick with answer choice A for now.

c.

We have our mutation closer to the left side of our strand here, but reasoning here is going to remain the same as answer choice B. Answer choice A remains superior.

d.

Same reasoning here as we used for answer choice B-D. Big takeaway here is recognizing the test-maker mentioned Southern blot for a reason, and recognizing endonucleases cleave at palindromic sequences. 

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 3

18) No tricky verbiage here, we can answer this question just by knowing about the bond that’s cleaved. We’ll revisit the passage, but a quick glance at our answer choices shows we’re going to have phosphorus bound to another atom. We’ll get details about the cleavage taking place, and then we’ll jump into specifics using our general knowledge.

We have part of the passage here. I want to focus on the middle of the second paragraph. It says Upon binding, IN catalyzes cleavage of a GT dinucleotide from each 3PRIME vDNA terminus. This is the cleavage we’re focused on. The cleavage of the GT dinucleotide. How do two nucleotides bond in the first place? DNA is polymerized when the 3′ hydroxyl group of the nucleotide, attacks a 5′ phosphate group to create a phosphodiester bond. When we look at a phosphate group, the only thing the phosphorus is attached to is 4 oxygen atoms. That means our ideal answer is going to be phosphorus bonded to oxygen.

In our breakdown we went through the formation of the dinucleotide and the bonds formed by phosphorus in the phosphate group. Phosphorus only binds oxygen, not carbon, hydrogen, or another phosphorus. That makes it easy to eliminate answer choices A-C. We’re left with our best answer, the only answer that matches our breakdown, answer choice D: the bond that is cleaved is between phosphorus and oxygen.

19) To answer this question, we’re finding the segment of the original viral genome that encoded this protein. We have to be very careful with verbiage and make sure we have the proper segment’s nucleotide sequence. Note, the question stem says “based on the passage,” but that doesn’t always mean we automatically go back to the passage. This is almost like a standalone question, so we’re going to reason this out slowly.

We have the hypothetical mRNA sequence, so let’s work backwards to get to the RNA of the initial HIV: the retrovirus. mRNA is synthesized from DNA in the nucleus. It’s synthesized from a DNA template, but it’s not an exact copy, we have a complementary strand. So that DNA is going to be CCGTTGACTGAT, or in other words, complementary bases to that hypothetical mRNA sequence. But we want the original viral genome. RNA is converted to DNA through reverse transcriptase, but again, we don’t have an exact copy. We have complementary RNA. We’re going to have uracil again instead of thymine, but what that means is, we’re going to have the same nucleotide sequence as the hypothetical mRNA.

  1. 5‘–GGCAACUGACUA–3’. This matches our prediction, and the sequence in the question stem. We like this answer, but we still want to compare with the remaining options.
  2. 5’–TAGTCAGTTGCC–3’. We have thymine here, not uracil, which we know we’d have in in our initial RNA. This answer choice is the DNA strand complement to our correct answer. That’s not what we’re looking for, so we can eliminate this answer choice.
  3. 5’–CCGTTGACTGAT–3’ This answer choice also has thymine instead of uracil. Again, this could be correct if we were asking about the viral DNA. We wanted the sequence of the initial retrovirus RNA. We can eliminate this answer choice.
  4. 5’–UAGUCAGUUGCC–3’. This is the same as answer choice B, only we replaced thymine with uracil. We can eliminate this answer choice and we’re left with our correct answer, answer choice A. 

20) To answer this question, we have to remember a few key points from the passage. Specifically, that oligodeoxynucleotides mimic LTRs and function as competitive inhibitors. We’re focused on our knowledge of competitive inhibition. Let’s go through how we can distinguish between different inhibitor types.

We have some characteristics of enzyme inhibitors and we’re focused on this highlighted line. Competitive inhibition means oligodeoxynucleotides bind integrase, instead of vDNA binding integrase.

The way we test for competitive inhibition, is by keeping the integrase concentration the same. Both the oligodeoxynucleotides and vDNA are binding to that same enzyme active site. We want to run the experiment in the presence and absence of oligodeoxynucleotides. Why do we want to do that? So that we can see the effect on Km and Vmax with and without that competitive inhibitor. We also want to increase vDNA concentration. That’s going to help us find Km (that’s the Michaelis constant and tells us the affinity of an enzyme for substrate) and Vmax (which we can call the maximal velocity at saturating substrate concentrations). We want an answer choice that allows us to take advantage of the properties of competitive inhibitors specifically.

  1. Keep IN concentration constant and measure the initial velocity of the reaction at increasing vDNA concentrations in the presence and absence of the ODN. This answer choice matches our breakdown. We conduct this experiment and see the results. In theory, we should be able to determine Vmax and Km, and find an increase in the apparent Km, but no change on Vmax. This experiment allows us to test that effectively.
  2. Keep vDNA concentration constant and measure the initial velocity of the reaction at increasing concentrations of IN in the presence and absence of the ODN. This answer choice is inconsistent with our breakdown. Increasing the concentration of the enzyme, but not changing the concentration of substrate is going to make it impossible to get information about our Vmax. We want to keep enzyme concentration the same, so we can eliminate this answer choice.
  3. Keep ODN and IN concentration constant and measure initial velocity of the reaction at increasing concentrations of vDNA. This answer choice starts out better since we’re keeping integrase concentration constant. But we need to see the effects of the experiment with, and without oligodeoxynucleotides. We don’t want an experiment where oligodeoxynucleotide concentration is constant, so we can eliminate this answer choice.
  4. Keep IN and vDNA concentrations constant and measure initial velocity of the reaction at increasing concentrations the ODN. This could theoretically work, but this would only work if we know Vmax for certain. We’d have to have very specific enzyme and substrate concentrations. Ideally, we would just run the experiment in the presence and absence of the oligodeoxynucleotide. This is likely our 2nd best answer choice, but our best answer still remains answer choice A, Keep IN concentration constant and measure the initial velocity of the reaction at increasing vDNA concentrations in the presence and absence of the ODN. 

21) In other words, which amino acid is most similar to the amino acid at position 64? The passage said the catalytic core domain has three high conserved residues: D64, D116, and E152. We’re focusing on D64. We have aspartate at positions 64 and 116. Glutamic acid at position 152. Mutation of any of these three residues abolishes integrase activity. If we had a mutation, but the amino acid is similar enough in properties aspartate, we are less likely to affect enzymatic function of integrase. That being said, we do have some clues. We have three highly conserved residues, but the amino acids are all electrically charged amino acids. All acidic, negatively charged side chains. Meaning if we had a substitution, we’d want to substitute with another negative charged amino acid. That means we’d want to substitute with glutamate, if possible. We’d want to avoid positively charged amino acids like lysine, arginine, and histidine. 

  1. Asparagine. This isn’t a negatively charged amino acid. Rather we have a polar, uncharged R-group. Let’s keep comparing to the other options and see what else we can come up with.
  2. Glutamate. This answer choice is the best-case scenario. Glutamate is also negatively charged, just like aspartate, meaning a substitution would be the least likely to affect enzymatic function. We still have that negative charge, unlike answer choice A. We can eliminate answer choice A and keep our superior answer choice, answer choice B: glutamate.
  3. Lysine. Lysine was one of the few positively charged amino acids we absolutely wanted to avoid. We can eliminate answer choice C. 
  4. Valine. Valine also isn’t negatively charged. It has a nonpolar, uncharged R-group. We can eliminate answer choice D. We’ll still stick with our superior answer, the only amino acid listed with the negatively charged R-group, answer choice B: glutamate. 

22) To answer this question, we can quickly reference some details in the passage. We’re going to see the number of amino acid residues in integrase. We’ll account for the fact that we’re looking for the number for a tetramer, not a monomer. And quick preliminary glance at our answer choices to get the proper units: we’re looking for kilodaltons.

We have part of our passage here. I want to focus on the number given in the middle of the paragraph. The author says IN is a 288-residue protein. First things first, that’s going to be for a monomer. The question stem asks about a tetramer, meaning we’ll have: 288 residues multiplied by 4, or 1,152 residues. We want a molecular weight, in kilodaltons. From our general knowledge, average molecular weight of an amino acid is 110 Daltons.

Multiplying 1152 residues by 110 Daltons gives us a total of 126,720 Daltons. Divide by 1000 to give us 126.72 Kilodaltons.

This is a math problem, we did no rounding, we did no estimating. We solved for an exact value. Let’s see if we have an answer choice that matches our prediction of 126.72 Kilodaltons. And just a heads up, the MCAT isn’t a math test. The test-maker likely isn’t going to give you an answer choice to two decimal points. Rather they’ll give you 4 answers that are far apart from one another, and you’ll only get the incorrect answers by not solving the problem correctly. In this case, even though we don’t have an exact match, answer choice D is very close. Note our question stem asks for an approximate molecular weight, which is exactly what we solved for. We can eliminate answer choices A-C, those are all incorrect values. We’re left with our correct answer, answer choice D: 128 kilodaltons.

23) To answer this question, we’ll revisit the mode of action of the inhibitor. We’ll follow that up with general knowledge where we’ll break down the condition needed for the gel electrophoresis. We’re deciding on denaturing, reducing, and native as the 3 roman numeral options.

We have part of our passage here. We’re told One class of these inhibitors are LEDGINs, which bind IN and shift its oligomerization equilibrium toward an inactive tetramer that cannot bind vDNA. That means if the inhibitor is working properly, we see an integrase tetramer instead of an integrase dimer. What does that mean for us? We want to be able to distinguish between the dimer and tetramer. We don’t want the polymer to be denatured, because then we’d only see integrase monomer. We wouldn’t be able to see if tetramers formed. So, we want to view these proteins in their native form. Meaning we’ll eliminate option 1, and keep option 3. Let’s think about option 2 now. What would reducing accomplish? That would remove all disulfide bonds between cysteine molecules. That’s not relevant in this situation, so we’re going to eliminate option 2 also. Let’s look at the answer choices given what we just said. 

We confidently eliminated option 1, which means we can eliminate answer choices A and D right away. We’re deciding between reducing and native, but we said native was the better answer. That means the best answer here is answer choice C: which is option 3 only: Native.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 4

24) Right away, we know we’re not specifically talking about any of the experiments in our passage. We didn’t deal with bovine serum albumin, but we did focus on retinol uptake. Keep reading, it says: Uptake of retinol bound to RBP was also measured in cells that were cotransfected with a small inhibitor RNA (siRNA) targeting STRA6. Which graphic shows the expected result of this experiment?

We’re looking at graphs showing retinol uptake activity. We want to find a graph that corresponds to the information given in the question stem and the passage. 

First thing we want to note, is the author mentioned STRA6-mediated retinol uptake is specific to RBP. What does that mean? We shouldn’t see any retinol uptake in our BSA group.

As we’re making these observations, we can also analyze our answer choices so that’ll make it easier to compare.

Based on what we said, right away we can eliminate answer choices A and B. Next, we have to consider the effects of STRA6 transfection. Those are the gray bars. But we also want to note something mentioned in the question stem. Some cells were cotransfected with an inhibitor RNA targeting STRA6. That’s shown by the + on the right of our graphs along the X-axis.

Looking at answer choice C, we see the same level of retinol uptake in the untransfected and transfected group on the left. We’re not expecting that to be the case. That transfected group has a membrane receptor for RBP, and we saw from Figure 1 in the passage we expect more retinol uptake activity in that group. Answer choice D fixes that problem for us. Very left side we see the STRA6-transfected group have higher retinol uptake activity, just like we expect.

Let’s quickly glance at the group all the way to the right for answer choices C and D. We’re not expecting any change in retinol uptake in the untransfected group in either case. And the inhibitor RNA that’s targeting STRA6 decreases retinol uptake versus the group where the inhibitor isn’t present. Ultimately, answer choice D hits all of our criteria.

25) We can answer this question by flipping back to Figure 1. This question is very broad so we’ll go back to the figure while also comparing all of the 4 answer choices at the same time.

  1. STRA6 activity is associated with the CRBP-I intracellular protein, but not the LRAT intracellular protein. We can see the combination of STRA6+LRAT actually shows higher relative retinol uptake activity. That implies that STRA6 activity is associated with both, but again, we have more uptake in the STRA6 + LRAT combination.
  2. STRA6-mediated retinol uptake in cells is only dependent on the ability to bind retinol. If this statement were true, theoretically the STRA6 line would show the same activity as the LRAT and CRBP-1 lines. Retinol uptake depends on more than just the ability to bind retinol.
  3. Neither LRAT nor CRBP-I are absolutely required for intracellular retinol uptake. Looking at the STRA6 line, we see that there is some retinol uptake activity. That means neither LRAT nor CRBP-1 are absolutely necessary. Both clearly help, but the verbiage is extreme here. This is our best answer choice so far because A and B contradict what we see in Figure 1. 
  4. Release of retinol from holo-RBP is most effective in the STRA6 only condition. Release of retinol implies there is more retinol uptake. That happens the most in the STRA6+LRAT group and in the STRA6+CRBP-1 group. STRA6 uptake is lower. We can also eliminate answer choice D. We’ll stick with our best answer and the only true answer that’s consistent with the figure: answer choice C. 

26) To answer this question, we can look back at Figure 2 from the passage.

X-axis on the graph is usually an independent variable, in this case we have time, given in minutes. Y-axis is usually a dependent variable, and in this case, it represents what we’re testing for during this experiment. We have relative retinol uptake activity, or relative retinol fluorescence. 

  1. STRA6 or STRA6/LRAT condition. These are actually the conditions being tested. The researchers control these testing conditions, they’re not dependent on other variables.
  2. Retinol fluorescence. This is consistent with our breakdown. We conduct the experiment and retinol fluorescence is dependent on other variables, and the experimental setup in general. It’s not controlled, and it matches our breakdown. That means we can now eliminate answer choice A. Those weren’t dependent variables. 
  3. Time duration of assay. This is the independent variable shown on the graph along the X-axis. This directly contradicts our breakdown so we can eliminate answer choice C. The only reason you’d pick answer choice C is if you confuse dependent and independent variables.
  4. Untransfected cell membranes. This is the control group, not the dependent variable. Again, something that’s independent of the other variables in the experiment. We can now eliminate answer choice D as well. We’re left with our correct answer, answer choice B: Retinol fluorescence.

27) To answer this question, we’ll revisit the passage and specifically the experimental data. For now, we’ll pull up Figure 1 to make answering the question easier. 

Our control group is untransfected cells. That means no STRA6. We have almost minimal retinol uptake activity in that group. 

Next group, we have the cells transfected with only STRA6. We see slightly more uptake activity, but still relatively low retinol uptake.

3rd group, we have STRA6 and CRBP-1, which was the group we related to retinol storage. We have an increase in retinol uptake activity in this group compared to the two we’ve covered.

Lastly, we have the cells transfected with STRA6 & LRAT. We have much more retinol uptake activity here. 

  1. Retinol uptake activity by STRA6 is limited without the presence of intracellular binding proteins. This answer choice sounds promising. When we looked at Figure 1, we noticed the STRA6 condition alone showed lower relative retinol uptake activity. As soon as CRBP-1 or LRAT were introduced, retinol uptake activity increased significantly. That shows STRA6 retinol uptake activity is limited without the intracellular binding proteins. As soon as we had those binding proteins, activity shot up. 
  2. The expression of STRA6 is likely to be higher in the liver than in target organs. Let’s not get tricked by this answer choice STRA6 is found on cells in target organs, not in the liver. The liver is where we found the holo-RBP complex. We expect the opposite of this answer choice to be true so we can eliminate answer choice B.
  3. Retinol uptake activity by STRA6 is accelerated when retinol is stored as intracellular retinol rather than as retinyl esters. This answer choice is comparing STRA6+CRBP-1 (which stores retinol) and STRA6+LRAT (which converts retinol to retinyl esters). The retinol uptake in Figure 1 is higher in the conversion group. This answer choice contradicts Figure 1 and our breakdown, so we can eliminate answer choice C.
  4. Binding of holo-RBP to STRA6 is likely to have a Hill coefficient greater than 1. Hill coefficient greater than 1 indicates positive, cooperative binding. That would mean We start with a flatter slope that gets steeper as we have more binding. We’d have an S-curve with a much steeper increase but that’s not the case, nor did the author suggest positive cooperativity. We can eliminate answer choice D. We’re left with our best answer, our correct answer, answer choice A: Retinol uptake activity by STRA6 is limited without the presence of intracellular binding proteins.

28) To answer this question, we’ll do the same thing we did in our previous question. I’ll pull up Figure 2 here. We’ll make some general observations, then we’ll follow that up by going through our answer choices.

Retinol fluorescence is enhanced when retinol is bound to RBP. Our control group, which isn’t exposed to membranes expressing STRA6, has nearly full fluorescence. 

STRA6 and STRA6+LRAT with a concentration of 4 micromolar holo-RBP: Relative fluorescence is much higher over a longer period in the STRA6 line. STRA6+LRAT we see a steady decrease, and a larger decrease overall.

STRA6 and STRA6+LRAT membranes with a lower concentration of holo-RBP. Both have steep decline initially. Over the course of the 3 hours, we see roughly the same trend between the two lines. Sharp decrease, following by a slower decrease and eventual plateau. 

  1. STRA6-catalyzed retinol release increases as the holo-RBP to STRA6 molar ratio increases. This is the opposite of our breakdown. When we have a higher molar ratio, STRA6-catalyzed retinol release is much lower. Note the relative retinol fluorescence. It’s much higher at the higher molar ratio. That means less retinol was released, and more retinol remained bound to RBP. 
  2. Retinol release in the STRA6 alone condition is not affected when the holo-RBP to STRA6 molar ratio changes. This answer choice also contradicts Figure 2. Looking at the two STRA6 lines, we see a stark difference in relative retinol fluorescence. That means there was a difference in retinol release in the two conditions.
  3. STRA6 becomes more dependent on LRAT for retinol release when the holo-RBP to STRA6 molar ratio increases. This answer choice is consistent with our breakdown. Note the STRA6 and LRAT levels were the same in each reaction.
    At lower molar ratios, we have a relatively similar line for both STRA6 and STRA6+LRAT groups. When that molar ratio increases. The retinol release in the STRA6+LRAT condition was much higher. This shows that when that molar ratio increases. STRA6 is more dependent on LRAT for retinol release. We can keep this answer choice. We can eliminate answer choices A and B because those both contradicted Figure 2.
  4. Retinol release is only dependent on the presence of holo-RBP. This answer choice contradicts what we saw in the passage. We’re also seeing the presence of LRAT influencing retinol release. That release isn’t only dependent on the presence of holo-RBP. We can eliminate answer choice D as well. We’re left with our correct answer, answer choice C: STRA6 becomes more dependent on LRAT for retinol release when the holo-RBP to STRA6 molar ratio increases.

29) From the passage, we know the complex forms in the liver. To answer this question, we’re going to go through the functions of the liver. The liver plays a vital role in the digestion of fats, and detoxifying the blood. It produces bile that’s required for the breakdown of fatty components of the food in the duodenum. The liver is mostly composed of hepatocytes which are involved in the synthesis of cholesterol, bile salts, and phospholipids as well. When there is excess glucose in the blood, it’s stored in the liver as glycogen.

  1. store bile. The liver produces bile, but bile is stored in the gallbladder. Not ideal, but we’ll still keep this answer choice for now.
  2. secrete glucagon. This is a function of the pancreas. Glucagon is a peptide hormone secreted from the alpha cells of the pancreas.
  3. produce hydrochloric acid. This is done by the parietal cells of the stomach. Another answer choice that doesn’t match the function of the liver. 
  4. detoxify drugs. This answer choice matches our breakdown of liver functions. This is actually one of the primary functions of the liver: it detoxifies the blood. We can eliminate answer choices A-C. All 3 listed functions that aren’t liver functions, so we’re left with our correct answer, answer choice D: Detoxify drugs.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Questions 30-33

30) Before we jump into the details, let’s quickly think about the note. Mendelian refers to the inheritance of traits through genes, and was originally proposed by Gregor Mendel. Mendel states that a cell holds a gene containing two alleles, with one allele inherited from each parent. This means that only two alleles, one dominant and one recessive, could exist for a given gene. In this question, red is dominant (we’ll denote R), while brown is recessive (we’ll denote r). First set of offspring show a roughly 3:1 ratio of red to brown offspring. Given that ratio, we can fairly confidently say the two initial, red beetles were heterozygotes. Why is that? We can draw out our Punnet square. We have a heterozygous father (Rr) and a heterozygous mother (Rr):

  R r
R RR Rr
r Rr rr

As I mentioned, the ratio of red beetles to brown is 3:1 given the above square. That’s the F1 generation. We’re interested in the F2 generation to answer this question. We have two red F1 beetles (meaning either RR or Rr from the above square) that are crossed, and we want to know the probability that both red and brown beetles will appear in the F2 generation. How would that be possible? If both parents had one recessive allele. A red beetle with an RR genotype cannot have a brown offspring. Only Rr beetles mating could yield a brown offspring, so we need to find the probability both the father and mother have Rr genotypes, given we’re looking at red beetles. Think back to our F1 generation and the Punnet square above. The red beetles were either RR or Rr, which I’ve highlighted below

  R r
R RR Rr
r Rr rr

Of the three highlighted beetles, only two of the three have the Rr genotype, while one has the RR genotype. So, the probability of both the red father and red mother having Rr genotypes is 2/3 x 2/3, or 4/9.

  1. 4/9. This matches the calculated value we came up with.
  2. 1/2. A student might incorrectly pick this answer choice by simply looking at the above Punnet square and noting ½ of the genotypes are Rr. Of course, we know rr corresponds to a brown beetle so that genotype doesn’t factor into our probability. We’ll stick with option A as our best answer. 
  3. 2/3. This answer choice corresponds to the probability a red beetle has an Rr genotype. Answer choice A is the best option still.
  4. 3/4. This answer choice corresponds to the probability of picking a red beetle in the F1 generation. Answer choice A still corresponds to our calculated value and is our best answer.

31) To solve this question, we have to make sure to pay attention to the numbers in the question stem and what they represent. Let’s break down the question stem a little bit at a time. An exon is the region of a transcribed gene present in the final functional mRNA molecule. Introns are portions of a gene that are included in pre-mRNA transcripts, but are removed during RNA processing and rapidly degraded. Alternative splicing allows for the production of various protein isoforms from one single gene coding. In our question, we have two isoforms that are 16 or 17 amino acid residues long. I mentioned we have to be careful with our numbers. We can think of these isoforms in terms of base pairs as well. Each amino acid is going to be 3 bases, so we multiply our number of amino acids by 3 to get 48 to 51 bases. The only way to get 48 or 51 base pairs is to include both exons 1 and 4, but to eliminate either exon 2 or exon 3. If we eliminate exon 2, we get a length of 48 base pairs, if we eliminate exon 3, we get a length of 51 base pairs. Given this information and this breakdown, let’s jump into the four options and see if we can identify the best technique.

  1. Synthesize cDNA from Protein X mRNA using primers overlapping exons 1 and 4, followed by gel electrophoresis and band visualization. Timing here is important as the complementary DNA is synthesized from mRNA. You’ll see some of the other answer choices focus on DNA instead of RNA. In this case the complementary DNA will not have the extra introns, but will contain exons. Exons 1 and 4 are expressed in both isoforms, which is why we have primers overlapping these exons. Use PCR to make DNA copies which should differ by 3 pairs and migrate differently along the gel. In theory you’d be able to see different bands, but 48 and 51 base pairs are very close together, so it’s a little tough to visualize. Regardless, this method would still work.
  2. Separate Protein X mRNA by gel electrophoresis and visualize band pattern using a DNA probe complementary to exon 3. While the technique here is promising, the DNA probe should be complementary to exons 1 and 4, like we mentioned in the breakdown of the question. This technique would only really show you the presence of one of the two isoforms we are looking for.
  3. Perform a PCR of Protein X genomic DNA using primers overlapping exons 1 and 3, followed by gel electrophoresis and band visualization. This method is not going to work because we’re focused on the expression of the two isoforms. PCR of genomic DNA is not going to give us the necessary information about expression. We can rule out this answer choice.
  4. Perform a restriction digest of Protein X genomic DNA using an endonuclease that cuts in the middle of exon 2, followed by gel electrophoresis and band visualization. Reasoning here is going to be the same as answer choice C. We are focused on mRNA or protein, and not DNA. We can eliminate answer choices C and D both for focusing on DNA. We can stick to our correct answer, answer choice A.

32) We’re shown properties of two enzymes and we’ll note both enzymes have the same substrate so we can directly compare the results given in the table. Let’s do a quick overview of the 4 columns we’re looking at:

  • Kcat measures the number of molecules of substrate turned over into product per unit time (usually seconds) and is proportional to efficiency. The higher your kcat, the more reactions happen per unit time. 
  • Kd is the dissociation constant. When we have a smaller Kd, that corresponds to a higher affinity of an enzyme for its substrate. A higher Kd means lesser affinity.
  • The Hill coefficient is a measure of cooperativity in a binding process. A Hill coefficient of 1 indicates independent binding, a value of greater than 1 shows positive cooperativity binding.
  • Isoelectric point, by definition, is the pH at which the net charge of a molecule is neutral.

We’ve got our definitions and our values handy, so we can jump into the answer choices.

  1. The maximal velocity of the reaction catalyzed by Enzyme X is higher than that of the reaction catalyzed by Enzyme Y. Vmax is the maximal velocity at saturating substrate concentrations, and it depends on Kcat. Looking at the values of Kcat, we actually see a higher value for enzyme Y than for enzyme Z. This answer choice contradicts what we’re told in the question stem. 
  2. Enzyme X has a higher molecular weight than Enzyme Y. This answer choice is out of scope. It has nothing to do with the values given to us in the question stem.
  3. Enzyme X exhibits cooperativity, whereas the activity of Enzyme Y does not. This answer choice has to do with the Hill coefficient. I mentioned a Hill coefficient of 1 (which corresponds to enzyme Y) indicates independent binding, but a value of greater than 1 (which corresponds to enzyme X) shows positive cooperativity binding. This answer choice is consistent with our breakdown.
  4. Enzyme X has a lower binding affinity for the substrate than Enzyme Y. Enzyme X has a lower Kd, but we mentioned a smaller Kd corresponds to a greater affinity of an enzyme for its substrate. We can eliminate this answer choice as well. Answer choice C remains our best option.

33) This is a standalone question that requires us to know about different Lineweaver-Burk plots. This should be general knowledge and something you have to commit to memory before test day. I’ve got the visual below that you should have memorized and the reasoning for each plot.

We can confidently say the plot in the question stem resembles the third plot on the right in our visual: noncompetitive inhibition. In noncompetitive, or allosteric inhibition, an inhibitor molecule binds to the enzyme at a location other than the active site (an allosteric site). The substrate can still bind to the enzyme, but the inhibitor changes the shape of the enzyme so it is no longer in an optimal position to catalyze the reaction. Let’s look for an answer choice that’s consistent with noncompetitive inhibition.

  1. irreversibly binds the enzyme. This is, like the name suggests, consistent with irreversible inhibition. The active site is made unavailable and increasing substrate would not have an effect. That’s not consistent with what we see in the question stem or in a Lineweaver Burk plot demonstrating noncompetitive inhibition.
  2. can only bind the enzyme after the enzyme has already bound substrate. The answer choice corresponds to uncompetitive inhibition. Uncompetitive inhibitors bind at a site other than the active site, and they do not bind to the enzyme until it has associated with the substrate to form the enzyme-substrate complex. Once the uncompetitive inhibitor has bound, the substrate remains associated with the enzyme. This is not consistent with our breakdown of the question and image in the question stem.
  3. binds the enzyme and the enzyme–substrate complex with the same affinity. This answer choice describes noncompetitive inhibition in which the inhibitor binds the enzyme allosterically. The inhibitor can bind both the enzyme and enzyme-substrate complex with the same affinity as it does not bind the active site. We like this answer choice better than answer choices A and B.
  4. prevents the binding of the enzyme to its substrate by occupying the enzyme’s active site. We just covered this in our previous answer choice, but noncompetitive inhibition does not involve occupying the enzyme’s active site. This describes competitive inhibition which yields a different Lineweaver Burk plot (shown in the visual above). The key to this question is really knowing and understanding the Lineweaver-Burk plots for inhibition. Make sure to have those down as you practice and before your exam!

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 5

34) We can pull up Figure 1 and go through each answer choice one-by-one. We’ll rely on some of the conclusions we can make about the data from the figure and reading the passage.

Intracellular bacteria levels go up in the first 2 hours. That’s expected because that’s when the host cells are incubated in the growth media containing VP. We also have increased levels of intracellular bacteria through the 3rd and 4th hour also. That’s because of the bacteria replicating inside the cell. That continues through hour 4. Hours 5-7 we see a decrease in intracellular bacteria. At this point, the bacteria are likely being exposed to that antibiotic and being killed off. That ultimately causes the subsequent decrease in intracellular bacteria in hours 5-7. 

  1. The number of host cells in the sample increased until carrying capacity was reached. Number of host cells in the sample wasn’t increasing. Another thing we want to note that might look confusing. We’re looking at changes in bacterial levels, not changes in host cells.
  2. Infection with VP is maximal at 5 hours of exposure. This answer choice is a direct contradiction. We can see in our figure that at 5 hours of exposure, we’ve already peaked in terms of intracellular bacteria levels. That level is actually going down. We don’t have additional infection with VP, because the growth media contains antibiotics, and won’t let any more extracellular VP grow.
  3. VP replicates inside the host cell before causing host cell lysis. This answer choice is consistent with our prediction. We initially have replication inside the host cell. This answer also explains why there’s a decrease in intracellular bacteria in hours 5-7. If we have host cell lysis, VP is exposed to antibiotics. That VP is either going to be washed away in the hourly rinse, or the antibiotics end up killing them off. We can keep this answer choice and eliminate answer choices A and B because both contradicted the experimental setup and results.
  4. Rate of VP entry into host cells increases then decreases over time. This answer choice contradicts the setup in the experiment and our breakdown. Additional VP isn’t entering host cells because extracellular VP is being killed by antibiotics. We’re not seeing an increase in VP entry past the initial 2 hours. We can eliminate answer choice D, so we’re left with our correct answer, answer choice C: VP replicates inside the host cell before causing host cell lysis.

35) To answer this question, we’ll see what the author mentions about the amino acid residue, then we’ll use our knowledge of amino acids to identify the specific amino acid.

If you’re shaky on amino acid structures, I’d pull up a list with their structures as you go through this question. Make sure to know those structures for your exam because they can be easy points. For this question we have to know the structures to know which amino acid has an amide functional group. Amides have carbonyl groups linked to nitrogen atoms. Let’s see which of the answers matches our criteria.

  1. glutamine. Right away, this answer choice matches our criteria. Glutamine is a good answer for the time being, but we want to be thorough and review the remaining answer choices.
  2. leucine. This contradicts our breakdown. In fact, there aren’t any nitrogen atoms in leucine’s side chain. We can eliminate answer choice B.
  3. arginine. Arginine does have a nitrogen in its side chain, but we don’t have that carbonyl group. We can eliminate answer choice C.
  4. tyrosine. Another answer choice that contradicts our breakdown. No amide functional group, so no deamidation. We can eliminate answer choice D and we’re left with the correct answer, answer choice A: Glutamine

36) To answer this question, we can briefly go back to the passage. 

We have Experiment 1 here, but we’re not going to read the whole thing. Make sure to really break down any experiments like this in your initial readthrough. If you ever go back to the passage, it should only be for specific details (often quantitative). X-axis is usually our independent variable. The time the samples were collected isn’t changing based on other variables, and it’s independent of the rest of the variables. Dependent variable is usually on the Y-axis. The levels of intracellular bacteria at any given time are dependent on when these samples are actually collected. We see the dependent variable vary as a result of how time varies. We’re looking for an answer that talks about: The time, in hours, that the researchers collected the samples of host cells.

  1. Length of time host cells were exposed to the bacteria. This answer choice is something that’s controlled in the experiment, but is limited to just the initial 2 hours. We’re still tracking the change in the dependent variable later in the experiment, and that’s being tracked in relation to total time, not just exposure time. 
  2. Amount of cells collected in each sample. This is another answer choice that’s controlled, but not necessarily affecting the dependent variable. The dependent variable is measured as colony forming units per milliliter. That means regardless of the amount of cells collected, we’re looking at the same units: per milliliter.
  3. Time at which the cell samples were collected. This answer choice matches our prediction. This is also the X-axis on Figure 1. The independent variable is the time, in hours, that the researchers collected the samples of host cells. As that independent variable varies, we track the change in our dependent variable. We can eliminate answer choices A and B because both of those variables were controlled, and neither is our independent variable. 
  4. Amount of intracellular bacteria measured at each time point. This answer choice is our dependent variable. We mentioned this is what’s actually being tracked as time passes. That means we can eliminate this answer choice as well because it contradicts our breakdown. We’re left with our best answer, answer choice C: Time at which the cell samples were collected.

37) This answer is going to come from the passage, but is also going to tie into question 35 where we talked about the amino acid residue at position 61 in Rac. Let’s revisit the passage, and we’ll see what we can look for to show that VopC modifies Rac at residue 61.

It says VopC contains a catalytic domain that irreversibly activates host cell GTPase Rac through the deamidation of the side chain of the residue at position 61. We touched on this when we went through Question 35. There’s deamidation-that means an amide functional group is removed in the side chain of the residue at position 61. Deamidation happens and we have the addition of a water group and removal of an ammonia group. We don’t have to worry about the alanine part of the question stem. If VopC modifies Rac, the samples should have the presence of ammonia because of the deamidation being successful.

  1. CO2. Carbon dioxide doesn’t match the removal of an ammonia group, or even the addition of the water. 
  2. NH3. This answer choice matches our breakdown. We did a quick breakdown of deamidation in amino acids. We established we’re dealing with an amide functional group, the addition of water, and the removal of an ammonia group. This answer choice is consistent with that breakdown, so we can eliminate answer choice A.
  3. O2 (molecular oxygen). Reasoning here is going to be the same as answer choice A. We broke down deamidation, and there’s a removal of an ammonia group. IF VopC modifies Rac, we’ll see ammonia in the sample, not molecular oxygen. We can elimnate answer choice C.
  4. H2S (hydrogen sulfide). Same as answer choices A and C. This also contradicts our breakdown of the question. IF VopC modifies Rac, that means we have deamidation of the side chain of the residue at position 61. Deamidation means removal of an ammonia group and addition of water. We can eliminate answer choice D, so we’re left with our correct answer, answer choice B: ammonia.

38) To answer this question, we have to think back to Figure 2 from the passage.

In Figure 2 we’re measuring proteins by Western blot analysis. From your initial readthrough, you should have noticed the aldolase row had bands that were all the same thickness. That shows there’s roughly the same amount of aldolase protein in each sample. Given that information, we looked at total Rac in the middle row. We don’t know if this Rac is activated or inactivated. We’re only seeing that it’s present. The bands are similar in thickness, meaning we have roughly the same amount of protein. However, if we had different levels of protein there, that would mean some strains affect total amount of Rac (not just activation or inactivation). By showing that we have the same levels, we can test for Rac activation without adjusting for any possible upregulation or general differences in total Rac.

  1. To confirm that the antibody used to detect GTP–Rac was working in each sample. This answer choice doesn’t answer the specific question being asked. Total Rac isn’t going to be detected by an antibody used to detect GTP-Rac. Even if we confirmed the antibody worked like that, this answer doesn’t explain why we needed to measure the total amount. 
  2. To determine if the same amount of each sample was loaded on the gel. This answer choice is reasonable, but we also used aldolase to do the same thing. Measuring total Rac helped confirm this, but that wasn’t the actual experimental reasoning behind it. Still, we like this answer choice better and answer choice A. 
  3. To act as a control for comparing rates of Rac activation by VopC. The total Rac isn’t telling us anything about the rate of activation. We mentioned in our breakdown that this is just measuring total Rac-nothing to do with activation, so we can eliminate answer choice C.
  4. To determine if VopC affects Rac expression levels. This answer choice is consistent with our prediction, and our thinking. We’re making sure we have the same amount of total Rac in each sample, so in a sense answer choice B is on the right track. But answer choice B doesn’t provide the correct reasoning. The reason we’re making sure the same amount of protein was found in each band is to see if Rac expression levels changed at all. If we had discrepancies, then we’d have multiple variables fluctuating at once, and we’d have to refine the experiment. We can eliminate answer choice B. It didn’t provide the correct reasoning for why the total amount of Rac was measured in experiment 2. We’re left with our correct answer, answer choice D: To determine if VopC affects Rac expression levels.

39) Based on what we learned from the passage, we want to identify the structure of GTP. The researchers confirmed in Experiment 2 that when Rac is bound to GTP, it’s activated. More specifically, how is Rac activated? Through deamidation by VopC, and binding to GTP

We know GTP and ATP are “double-ringed” why do we know that? Adenine and guanine are purines. We can eliminate our single-ringed answers: answers B and C. Those are CTP and UTP, and cytosine and uracil are pyrimidines.

We’re looking at structures in answer choices A and D. Structure D we recognize as ATP, that’s one of the most common molecules we’ll see in biochemistry. We can eliminate that answer choice. We have a guanine instead of that adenine in structure A. Structure A is GTP. We can stick with our best answer, answer choice A.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 6

40) To answer this question, we’ll go back to the passage, and specifically Experiment 2. Experiment 2 involved isoelectric focusing and autoradiography, so we’ll break those down using our general knowledge. 

Here we have Experiment 2. We’re told we have to incubate our wildtype protein with radioactive ATP, then with either, or both of the protein kinases. 2nd sentence is the important one. It says Isoelectric focusing and autoradiography were used to detect phosphorylation (Figure 2). 

The above image gives you a visual representation of isoelectric focusing. We’re running our samples through a gel with a pH gradient. Gels are set in a buffer, in a container with a cathode on one end, anode on the other. Proteins are added to the solution, and current is applied; compounds are separated by their net charge.

We’re also told we use autoradiography, which means we’re utilizing the radiolabeled ATP to detect phosphorylation. The bands we see in figure 2-we’re visualizing radiolabeled ATP. The samples interact with the very negative ATP, and bands show up at the positive end. We also have 2 bands of migrated proteins that have been phosphorylated. 

Ultimately, we need both aspects of our experiment set up properly, meaning we need to have our gel and container set up properly for isoelectric focusing. We also need to be able to detect radiolabeled ATP and any phosphorylated, radiolabeled protein.

  1. Amplify the samples by PCR before running on the gel. This answer choice doesn’t match our breakdown. We don’t need to amplify the samples before running the gel. We’re focused on charge, and the amount of phosphorylation. No need to amplify the samples-which is done traditionally to DNA to make multiple copies of DNA segments.
  2. Establish a stable pH gradient in the gel before adding samples. This answer choice is referencing the setup of the experiment. We talked about running our samples through a gel with a pH gradient. That pH gradient has to be established before experimentation so we can actually conduct the experiment. That pH gradient allows us to separate proteins in our experiment and get the results in Experiment 2. We can hold on to this answer choice, it’s superior to answer choice A.
  3. Denature the samples so the proteins unfold into a linear chain. In this case, we’re focused on the overall charge on the molecule, and movement across the gel. We have separation in Experiment 2, because of overall charge differences. The visualizing that follows is because of that separation, meaning proper setup of isoelectric focusing. As well as having a radiolabeled molecule we can detect using autoradiography. We need those 2 aspects of the experiment working. We can eliminate answer choice C because it’s not as direct of an answer as answer choice B. Answer choice B is our superior answer choice: it addresses Experiment 2, and how visualizing works specifically.  
  4. Treat the samples with a chemical that will add an overall negative charge to each protein. This is describing something that happens in SDS PAGE. In this case ATP is a negatively charged molecule, but is that necessary to conduct our experiment? We can still utilize isoelectric focusing and autoradiography, even without the negative charge. We need the isoelectric focusing set up properly, meaning the stable pH gradient. We also need some sort of radiolabeled molecule we can detect using autoradiography. We can eliminate answer choice D. We’re left with our best answer, answer choice B: Establish a stable pH gradient in the gel before adding samples

41) To answer this question, we can quickly review the 2 figures, and see what we can conclude about phosphorylation. 

We have figure 1 on the left, figure 2 on the right. We’re going to go through some broad conclusions we can come up with. These are likely observations you made during the readthrough, but we want to be thorough here.

Figure 1: When we have wild-type CREB327 (meaning lanes 2-4) we have phosphorylation in the presence of PKA alone, but we have even more phosphorylation in the presence of PKA plus GSK-3. That combination led to our thickest band. There are no radiolabeled protein bands in any other lanes.

Figure 2. We don’t have any migration in the GSK-3 only lane. We have migration in the other two, non-control lanes. We have more phosphorylated protein detected in lane 4, which is our wild type CREB327 incubated with PKA then GSK-3. In both figures we see phosphorylation when incubated with PKA only. We see more activity when incubated with PKA and GSK-3. And in Figure 2 specifically, that’s PKA then GSK-3. 

  1. GSK-3 also phosphorylates CREB327WT at residue 119. This answer choice contradicts our breakdown. We looked at Figure 1 and we saw no phosphorylated protein in the GSK-3 only lanes. Not even in the wildtype. We don’t see any phosphorylation by GSK-3 alone. 
  2. GSK-3 phosphorylates CREB327WT at a faster rate than PKA. This ties into our previous answer. GSK-3 doesn’t phosphorylate CREB327 wildtype at all, without the presence of PKA. We’re going to group this with answer choice A for now, but neither one is a great answer choice. We’re still looking for a better option.
  3. Phosphorylation of CREB327WT by PKA occurs at more than one residue. This answer choice doesn’t match what we see in the passage. When we have CREB327-119, we see no phosphorylation by PKA. That implies phosphorylation takes place at that single residue. All 3 answer choices contradict what we saw in the passage and what I said in the breakdown of the question. 
  4. Phosphorylation of CREB327WT by GSK-3 can only occur after phosphorylation with PKA. This conclusion is consistent with my breakdown. If we have phosphorylation with PKA, but we also have GSK-3 present, we see additional phosphorylation. We saw that in both of our figures. If we only have GSK-3 present, then we have no phosphorylation. That means we need the presence of PKA first, and we can also have phosphorylation of CREB327 wildtype by GSK-3. We can keep this answer choice, we can eliminate answer choices A-C, those all contradicted our passage. We’re left with our correct answer, answer choice D. Phosphorylation of CREB327WT by GSK-3 can only occur after phosphorylation with PKA

42) To answer this question, we’ll have to remember a few key points from the passage, but also use our general knowledge. We remember from the passage, researchers made variants of wild-type CREB327 including CREB327-119, which can’t be phosphorylated at serine 119. We want an amino acid that can be chosen as a substitute for residue 119. A few criteria we want to meet here: first and foremost, we don’t want an amino acid we associate with phosphorylation. That means threonine and tyrosine are out. Those are commonly tested on the MCAT as the other amino acids that are phosphorylated. And that’s in addition to serine. Next, we don’t want any charged amino acids, or any extremely large amino acids. Why is that? We don’t want to disrupt the amino acid chain. We just want to make a substitution. No charges, and no steric hindrance. So that means no tyrosine, no tryptophan, and no phenylalanine, if possible, and no electrically charged side groups.

  1. Tyrosine. When we see tyrosine, that’s two red flags. Tyrosine is a large amino acid, so we might have steric hindrance. Tyrosine is also phosphorylated. 
  2. Glutamate. Glutamate has a negatively charged side group. Not exactly ideal, but a better answer than answer choice A. Answer choice A presented an answer choice that could be phosphorylated, which was our main criteria.
  3. Alanine. Alanine is uncharged, and it has the second smallest amino acid side chain. No phosphorylation either, so alanine hits all of our checkpoints. We can keep answer choice C, we can eliminate answer choice B-that was an electrically charged amino acid, and contradicts our breakdown.
  4. Threonine. Threonine is the 2nd most commonly tested amino acid when it comes to phosphorylation on the MCAT. That was the main point of our breakdown: we want an amino acid that isn’t phosphorylated. We can eliminate answer choice D, and we’re left with our correct answer, answer choice C: alanine. 

43) To answer this question, we can quickly recap what we said about the two isoforms in the passage, and then use our general knowledge to explain different isoforms of a protein. In the passage, the author mentions CREB exists as two isoforms in cells, one being CREB327. That’s what we focused on during a lot of the passage. What do we know about isoforms? Essentially, we have two functionally similar proteins. Even though they’re functionally similar, amino acid sequence isn’t the same. That can mean a few different things. Same genes, but alternative splicing (so multiple isoforms can be made from one single gene coding). That’s the main thing we think when we think isoform. It could also mean possibly different transcription start sites, and overall genetic differences, because of a difference in amino acid sequence.

  1. Pre-mRNA is transcribed from one CREB allele versus the other. A few things we want to point out: different alleles aren’t going to cause different isoforms. Also, pre-mRNA becomes mRNA after processing. mRNA is actually translated, not pre-mRNA. 
  2. CREB mRNA transcripts with different combinations of exons are generated. This answer choice is the definition of splicing. Exons of a gene can be included, or excluded in the mRNA that’s going to be translated. That discrepancy allows for multiple isoforms to exist. We like answer choice B more than we liked answer choice A. 
  3. Different post-translational modifications of CREB occur in the endoplasmic reticulum. The different we brought up in our prediction is from transcriptional differences, not translational. We have spliced mRNAs, and the proteins that are translated will have different amino acid sequences. For the sake of the MCAT, post-translational modifications are more for chemical modifications, or additions of proteins. If we didn’t have answer choice B, this could be a viable answer. But we’re still looking for an answer choice that mentions alternative splicing, just because we’re dealing with isoforms. Answer choice B remains superior.
  4. Pre-mRNA of various lengths are transcribed based on different termination sequences in the CREB gene. Pre-mRNA becomes mRNA after processing. mRNA is actually transcribed, not pre-mRNA. We can also eliminate this answer choice, so we’re left with our correct answer, answer choice B: CREB mRNA transcripts with different combinations of exons are generated

44) Let’s look at the data in Table 1 and see what observations we can come up with. We want to keep in mind that we’re focusing on phosphorylation. 

Control group has the least percent conversion, meaning least transcriptional activity. 

Wild-type has the highest percent conversion, meaning the highest transcriptional activity. 

Now to the variants: CREB 327 115 means no phosphorylation at serine 115 where we normally have GSK-3 phosphorylation. 

119 means no phosphorylation at serine 119 where we normally have PKA phosphorylation. 

Neither one shows a percent conversion as high as the wild type.

But, there’s much higher percent conversion, and transcriptional activity in CREB327-115. That’s the variant where PKA can still phosphorylate serine 119. 

We have much lower percent conversion, and less transcriptional activity in CREB327-119. That’s the variant where GSK-3 can still phosphorylate serine 115. That percent is actually close to the control percent. 

Neither one gets up to that 27% like the wild type, but CREB327-115 has much higher transcriptional activity because of PKA phosphorylation.

  1. Phosphorylation by GSK-3 has no effect on CREB327WT activation levels. This answer choice contradicts table 1. When we theoretically only have PKA activity, which is the second row, CREB327115, we only have 8.5% conversion. When we have both PKA activity and GSK-3 activity, which the wild type row, we see 27%, which is a huge difference. We also saw from Experiments 1 and 2, the two protein kinases work together. This answer choice doesn’t support the data in Table 1.
  2. PKA has a higher specificity for CREB327WT than CREB327-115. This answer choice is out of scope. We didn’t make any mention to specificity in the passage, or the experiment in general. PKA binds at serine 119, so there shouldn’t be any differences in specificity for wild type vs. CREB327-115. Still, this isn’t a direct contradiction of Table 1. We’ll keep answer choice B, we can eliminate answer choice A.
  3. Phosphorylation by PKA can partially activate CREB327. This answer choice is supported by the data in Table 1. Look at CREB327-115. We see phosphorylation, but only to a certain degree. We don’t get that same 27% we see in the wildtype. We can keep this answer choice. It’s supported by the data in Table 1. We can eliminate answer choice B because we said that was out of scope.  
  4. The kcat for the phosphorylation of CREB327 by GSK-3 is higher than the kcat for phosphorylation by PKA. Kcat is the turnover number, which isn’t going to be supported by Table 1. Even if this were true, there’s no way to support much enzyme kinetics-related using Table 1. We can eliminate answer choice D. We’re left with our correct answer, answer choice C: Phosphorylation by PKA can partially activate CREB327.

45) In other words, how can the control group be set up to see if autophosphorylation happens?

In Experiment 2, researchers were concerned by the possibility of autophosphorylation. What does that mean? When we have the protein kinases in the presence of ATP, we have phosphorylation of the kinase, by itself. How can we see if autophosphorylation is happening, instead of phosphorylation of CREB327? We remove CREB327 from the equation. Have the protein kinases incubated with ATP, and see if autophosphorylation happens. If it does, then the setup has to be adjusted. If it doesn’t then we can conduct the experiment normally.

  1. CREB327WT with ATP but no PKA or GSK-3. This is the opposite of what we’d want. We can’t see if autophosphorylation happens if we don’t even have the protein kinases in the control group. This wouldn’t show us much, because there are no kinases that can catalyze the phosphorylation.
  2. PKA and GSK-3 with ATP but no CREB327WT. This answer choice matches our prediction. We put our protein kinases together with ATP, and make sure no autophosphorylation takes place. The only way we can do that, is by also making sure there’s no substrate to phosphorylate. That’s consistent with this answer choice so we can eliminate answer choice A for being a direct contradiction of our breakdown.
  3. PKA and GSK-3 with no ATP or CREB327WT. This answer choice is incomplete. We have the protein kinases together, but we also need the phosphate group on ATP. Without the ATP, we can’t have phosphorylation, and no autophosphorylation in this case. We can eliminate this answer choice because it contradicts our breakdown of the question.
  4. CREB327WT with no ATP, PKA, or GSK-3. This could be a control the researchers used, but it doesn’t answer the specific question being asked. We’re trying to account for autophosphorylation. We can only see if that happens if we have our protein kinases together, with ATP. This answer choice also contradicts our breakdown, so we can eliminate answer choice D. We’re left with our correct answer, answer choice B: PKA and GSK-3 with ATP but no CREB327WT

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 7

46) To answer this question, we’ll need some information from the passage initially. The author talks about FSHRs in the first paragraph. Then we’ll use our general knowledge to determine tissue types. 

We have our first paragraph here, and we’re focused on location of FSHRs. Very first part of the passage says Follicle stimulating hormone (FSH) is a peptide hormone known to activate FSH G protein-coupled receptors (FSHRs) on ovarian cells. However, FSHRs have recently been found on osteoclasts.

We’re given two locations that I’ve highlighted: ovarian cells, and osteoclasts. Ovarian cells are epithelial cells, so we’re going to predict epithelial tissue. Osteoclasts are found in connective tissue, think of bone. No other locations are mentioned in the passage, and the question explicitly mentioned we want an answer based on what we read in the passage.

Based on this information, only option 1 and 2 (connective and epithelial) are correct. If we look at our 4 answer choices, that corresponds to only one correct answer choice. Answer choices A, B, and D are all incorrect. We have our correct answer, answer choice C: I and II only

47) To answer this question, we’ll likely have to recall some information from the initial readthrough of the passage, but we can pull up Figure 2 first before getting into additional details.

We’re testing the binding of an antibody designed to bind the specific sequence of FSHpep, which is a part of FSH. Top graph shows absorbance and binding in FSHpep and we have increased binding and increased absorbance. 

In our bottom graph, we’re seeing the affinity of the antibody for FSH. As expected, decreased absorbance and binding. However, what this graph allows us to do is compare that to the affinity of the peptide alone. 

  1. The Kd for binding of FSHpep by FSH-Ab is lower than the Kd for binding of FSH by FSH-Ab. This answer choice matches our breakdown. Equilibrium dissociation constant (KD) gives us relative binding affinity. A smaller Kd value means a greater binding affinity, which is consistent with this conclusion and Figure 2. We have increased absorbance and increased binding of FSHpep by FSH-Ab.
  2. IgG has a higher affinity for FSH than for FSHpep at all concentrations tested. IgG was used as a control, so we have to assume the control is the same amount in both the top and bottom graphs. That’s standard experimental procedure. We can’t get tricked by the absorbance scales. Note the scale on the y-axis changes from top graph to bottom graph. We can eliminate answer choice B because it contradicts the experimental setup from the passage.
  3. FSH binding sites for FSH-Ab are saturated when FSH reaches a concentration of 10µg. This answer choice contradicts Figure 2. We’re not explicitly told if we have saturation at concentration of 5 micrograms, but we don’t see absorbance go up from that point. IF we were to pick a concentration, we wouldn’t pick 10 micrograms. There’s no change in absorbance from 5 micrograms to 10 micrograms, so we’d pick 5 micrograms. We can eliminate answer choice C.
  4. The affinity of FSH for FSH-Ab is greater than the affinity of FSHpep for FSH-Ab. This is the opposite of our breakdown and what we’re seeing in the passage. Increased absorbance is indicative of increased binding. Instead, the affinity of FSHpep for FSH-Ab is greater than the affinity of FSH for FSH-Ab. We can eliminate this answer choice for contradicting Figure 2. We’re left with our correct answer, answer choice A: The Kd for binding of FSHpep by FSH-Ab is lower than the Kd for binding of FSH by FSH-Ab.

48) To answer this question, we’ll explain the difference between FSH and FSHpep, and why absorbance levels differ. Increased absorbance is indicative of increased binding. The antibody is designed to bind the specific sequence of FSHpep. Literally, the antibody binds that small sequence from FSH, so it makes sense that isolating that sequence would cause increased binding and absorbance levels. In FSH, that sequence exists, but it’s within the entire peptide hormone. We don’t know exactly how large FSH is, but it’s easier to bind that specific sequence than find that specific sequence in the entire molecule. 

  1. The FSH-Ab binds non-specifically to more than one site on FSHpep. This answer choice directly contradicts our breakdown of the question. The author said FSH-Ab binds specifically to the sequence of FSHpep, which means right away we’re not liking this answer choice.
  2. IgG competitively inhibits the binding of FSH-Ab to FSH. IgG was our control, and shouldn’t have an effect on the experiment being conducted. If IgG competitively inhibited binding, it would likely inhibit the binding to FSHpep also. 
  3. The FSH-Ab cooperatively binds FSH. Going by the bottom graph in Figure 2, we see a linear increase in absorbance. Cooperative binding would mean absorbance and binding increase significantly after the initial binding. We’d see a sigmoidal graph, and a sharp increase instead of the linear trend.
  4. The tertiary structure of FSH limits FSH-Ab binding interactions. This answer choice would explain the difference in absorbance and binding levels. Let me reiterate, FSHpep is a short peptide sequence, and the antibody binds that sequence specifically. FSH is an entire protein, so binding doesn’t happen as readily. Think of it this way: if you have a two piece jigsaw puzzle, versus a 100 piece jigsaw puzzle, it’s a lot easier to find the piece that fits. We can eliminate answer choices A, B, and C. They all contradicted the passage, and specifically the data in Figure 2. 

49) To answer this question, we’ll review the 13-residue sequence, or the structure of FSHpep in the passage. We’ll use our amino acid knowledge to find which amino acid substitutions would have the greatest effect. There are countless substitutions that can take place, so after we break down Figure 1 we can start comparing our 4 answer choices.

Figure 1 shows the structure of FSHpep. We’re also told the underlined and bolded residues are found at the receptor-hormone interface. Those are likely going to be the most important. A few things we want to focus on: charge, size, polarity. The newly substituted amino acid has to have significantly different properties to disrupt the usual binding. Let’s go through our 4 options:

Answer choice A says substitute lysine for arginine

Answer choice B says substitute arginine for aspartate

Answer choice C says substitute valine for leucine

Answer choice D says substitute glutamine for asparagine.

First thing we notice. Only answer choices A and B show substitutions in the underlined and bolded residues. We’re going to focus on those two answers. Answer choice A: Lysine and arginine both have positively charged side chains. Not a huge difference there. Let’s compare to answer choice B. Answer choice B shows a substitution from an amino acid with a positively charged side chain, to an amino acid with a negatively charged side chain. That’s a big difference! That means answer choice B causes the greatest effect. 

50) Be careful with the verbiage here. We’re going to have 3 reasons why FSHpep was included in the ELISA experiment, then 1 reason that does not describe why FSHpep was included. We want to find that one reason because it’s our correct answer. 

The antibody FSH-Ab is designed to bind the specific sequence of FSHpep. That’s why in the experiment, we had essentially maximum binding between the antibody and FSHpep. What was the point of even including FSHpep in the experiment? It was to get a baseline, or a sense of the binding between the peptide and the antibody. We can compare that to the FSH numbers, and get more information about how the antibody binds FSH. The FSHpep is almost like a control.

  1. To act as a positive control to confirm that the assay was functional. This is a possible function of FSHpep. Just by FSHpep being present and the experiment showing absorbance, we can tell the assay was functional. The measurable fluorescent product forming showed us the experiment worked.
  2. To provide a baseline against which to evaluate the affinity of FSH-Ab for FSH. This is the main purpose I mentioned in the breakdown of the question. By designing an antibody for a specific sequence, then having that antibody bind that specific sequence, researchers have a baseline against which to evaluate the affinity of the antibody for the protein in question, FSH. This matches our reasoning more than answer choice A, we can eliminate answer choice B. Why are we eliminating answer choice B? Remember, we’re looking for reasons that do not describe the inclusion of FSHpep.
  3. To generate data to support that the FSH-Ab binds to the receptor-binding domain of FSH. This is also reason for including FSHpep. We want to ensure the antibody is binding to the specific region in question. Even though FSH-ab binds to FSH, we don’t know if the antibody binds the receptor-binding domain. FSHpep helped us confirm the antibody binds that domain. This is similar to answer choice A. Both are actually good reasons for using FSHpep, but not our primary reason like answer choice B. We can hold on to both answers for now. 
  4. To determine the amount of FSH-Ab that is most effective as a therapeutic treatment. This is a good answer to this specific question. Again, we’re not getting lost with the verbiage in the question stem. A therapeutic treatment would imply the antibody worked in the body. Testing on artificially produced FSHpep is not the way to test that. We have that FSHpep sequence in FSH, but it doesn’t exist alone in the body. We used FSHpep as a control or baseline in the body for the reasons presented in answer choices A-C. That means we can eliminate those answers and keep our best answer, answer choice D: To determine the amount of FSH-Ab that is most effective as a therapeutic treatment.

51) This is a broad question, so our answer could theoretically come from what the author mentioned in the passage, or this could be like a standalone question. We can do a quick recap of what was said about FSH in the passage. The author tells us FSH is a peptide hormone. It stimulates the development of ova, the female reproductive cell. The follicular phase of the menstrual cycle begins with an increase in FSH. We have a lot of different, important functions in the female body-mainly controlling the menstrual cycle and stimulating the growth of eggs in the ovaries. We also talked about FSH activating FSHRs, which can stimulate bone resorption. 

  1. FSH does not require transport proteins to remain soluble in the bloodstream. This answer choice ties into the peptide hormone part of our breakdown. Peptide hormones are hydrophilic and soluble in blood. That means they can travel freely in the bloodstream, without transport proteins. 
  2. FSH enters the bloodstream by diffusing across the plasma membrane of endocrine cells. FSH is a peptide hormone, not a lipid hormone. Lipid-derived hormones can diffuse across plasma membranes. Amino acid-derived and peptide hormones cannot. Peptide hormones like FSH are too large, and they have polar groups that don’t allow for this diffusion across the membrane. We can eliminate answer choice B.
  3. FSH is synthesized from cholesterol. Another answer choice that isn’t talking about peptide hormones. Steroid hormones are synthesized from cholesterol, not peptide hormones. We can also eliminate answer choice C because it contradicts what we know from our science content.
  4. FSH is derived from a single amino acid. This is talking specifically about amino acid-derived hormones like epinephrine, norepinephrine, or melatonin. This entire question is just asking us to distinguish between the three major types of hormones. We eliminate answer choices B, C, and D. Those are examples of either lipid or amino acid-derived hormones. We can stick with the only answer choice that describes peptide hormones, answer choice A:  FSH does not require transport proteins to remain soluble in the bloodstream

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Questions 52-56

52) We have two different analyses here: First one involves SDS-PAGE without the reducing agent, while the second one involves SDS-PAGE with the reducing agent. SDS will denature proteins, so SDS-PAGE gels are used in experiments to separate proteins by size. Proteins are further denatured by the addition of a reducing agent that breaks down the disulfide bonds between cysteine residues. To answer this question, it helps to visualize what’s happening in the second analysis:

As I mentioned previously, SDS will denature proteins and we can separate proteins by size. That’s exactly what happens in the first analysis, but any disulfide bonds would still be intact. In the second analysis, this situation changes. Because we see two bands, we can assume the reducing agent breaks down disulfide bonds; now there are two bands representing different masses. Let’s see which prediction listed in our answer choices is supported by the results in our question stem:

  1. is posttranslationally glycosylated. While this may or may not be true, this prediction is not explicitly supported by what we’re told in the question stem. We can’t eliminate this answer choice just yet, but we still likely want a better option.
  2. contains a high proportion of charged residues. This is similar to answer choice A. This may or may not be true, but we’re not entirely concerned with whether the statement is the truest. We want to know which answer choice is supported by the question stem. SDS-PAGE separates based on molecular weight, and charge is not going to factor into the results in our question stem. 
  3. contains no disulfide bonds. While options A and B were unable to be eliminated, this seems like the worst option we’ve seen so far. As we broke down in our breakthrough, the fact that there were two bands in the second analysis suggests the protein did contain disulfide bonds. We can eliminate this answer.
  4. is composed of multiple subunits. This answer choice seems likely for the same reason answer choice C was unlikely. Adding the reducing agent broke down the disulfide bond that was previously holding the multiple subunits together. After breaking down the subunits, we can see the different bands show up in the gel. Answer choice D is going to be the best answer choice within the context of this question.

53) This is a standalone question that deals exclusively with knowing your biology content. Be careful with the verbiage-we want an interaction that does NOT contribute to stabilization. Three of our answer choices will contribute, our correct answer will NOT. 

Tertiary structure of a protein is the overall the three-dimension folding driven largely by interactions between R groups.

Some of the key things you’re expected to know are: R group interactions that contribute to tertiary structure include hydrogen bonding, ionic bonding, dipole-dipole interactions, and London dispersion forces – basically, the whole gamut of non-covalent bonds. Disulfide bonds are covalent bonds that contribute to tertiary structure. Proline is more rigid than most amino acids and can’t act as a hydrogen bond donor; proline can disrupt alpha helix and beta sheet formation and form kinks. 

  1. Disulfide bond. This is something we explicitly mentioned in the breakdown of the question. Disulfide bonds are covalent bonds that contribute to tertiary structure. When we see disulfide bonds we think sulfhydryl groups and cysteine amino acids.
  2. Phosphodiester bond. The phosphodiester linkage forms the backbone of the double helix structure of DNA and the single strand structure of RNA. It’s between a ribose sugar and phosphate. This is a good answer choice for the time being because it’s dealing more with a bond we see in nucleotides, and not an interaction that’s contributing to the stabilization of the tertiary structure of a protein. 
  3. Hydrogen bond. Hydrogen bonding is one of the interactions I mentioned in the breakdown of the question. Hydrogen bonding will contribute to tertiary structure and, in general, when we see hydrogen bonding in MCAT material we’re thinking FON (fluorine, oxygen, nitrogen) atoms. Answer choice B is going to remain the best option here.
  4. Salt bridge. Salt bridges are not something we explicitly mentioned in our breakdown, but salt bridges do contribute to the stabilization of tertiary structure. These occur through ionic interactions between charged amino acids. Proteins will fold so positively and negatively side chains will be adjacent. We’re still going to stick with answer choice B as our best answer.

54) This question boils down to knowing how KM and Vmax are affected by adding an inhibitor. We don’t have to go in-depth about each type of inhibitor or the reasoning behind any changes in KM or Vmax, but we do have to know how the ratio changes. We can look at the Lineweaver-Burk plots for inhibition to help us answer this question:

Note from this image, the ratio of KM/Vmax is going to change in both competitive inhibition and noncompetitive inhibition. In uncompetitive inhibition, the slope is unaffected and the ratio remains unchanged. 

  1. Competitive. Reference the image above and we see KM changes while Vmax is unaffected, so the ratio will be altered.
  2. Uncompetitive. This matches what we said in the breakdown of the question. The ratio will be unchanged in uncompetitive inhibition; both KM and Vmax decrease.
  3. Noncompetitive. Reference the image above and we see KM is unaffected while Vmax is reduced, so the ratio will be altered.
  4. Mixed. In mixed inhibition we can see KM increase or be reduced, while Vmax decreases. The ratio can altered. We can stick with our correct answer: answer choice B.  

55) Uncompetitive inhibitors bind at a site other than the active site, and they do not bind to the enzyme until it has associated with the substrate to form the enzyme-substrate complex. Once the uncompetitive inhibitor has bound, the substrate remains associated with the enzyme. Any situation that would increase substrate-enzyme complex would make a good answer choice in this situation.

Option 1 mentions a decrease in substrate concentration. If we have less substrate, the substrate-enzyme complex is less abundant so we see less of an effect from the uncompetitive inhibitors. The awesome thing is, by eliminating option 1 we can automatically eliminate answer choices A and C.

Option 2 mentions an increase in substrate concentration. Reasoning here is going to be the opposite of option 1. If we have more substrate, the substrate-enzyme complex is more abundant, so we see more of an effect from the uncompetitive inhibitors. Given option 2 is correct here, we can look at our two remaining answers and see only answer choice D remains as a viable option. Let’s quickly go through option 3.

Option 3 involves increasing inhibitor concentration. Reasoning here will be similar to option 2. The effect of the inhibitor comes from binding substrate-enzyme complex. If we have more inhibitor, more enzyme can be bound and inhibited. This is also a good option, so we stick with the correct answer, answer choice D: II and III only. 

56) This question boils down to knowing the properties of transmembrane helices. Transmembrane helices contain a larger number of hydrophobic amino acids, so we’re looking for an answer choice that contains the most hydrophobic amino acids. We can go back to our amino acid visual:

We’re looking at the nonpolar amino acid classifications here. Those groups contain the more hydrophobic amino acids like alanine, valine, leucine, methionine, and isoleucine. We also have our nonpolar, aromatic R group amino acids: phenylalanine, tyrosine, and tryptophan. All we’re doing to answer this question is finding the answer choice with the most overlap with our list of hydrophobic amino acids.

  1. Ala–Ile–Phe–Val–Leu. This answer choice contains 5 amino acids that I explicitly mentioned in the breakdown of the question. Good answer choice to start. 
  2. Ala–Thr–Lys–Asn–Leu. Two of the amino acids listed here we mentioned in our breakdown, but three we did not. Answer choice A is going to remain superior here.
  3. Lys–Thr–Arg–Asn–His. This is similar to answer choice B. Answer choice A had more hydrophobic amino acids.
  4. Val–Thr–Pro–Tyr–Ser. Only two amino acids here that were listed in our breakdown. We’ll stick with our correct answer, answer choice A. A transmembrane helix of a protein will contain many hydrophobic amino acids. 

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 8

57) Said differently, which of the 4 answer choices correctly estimates relative tumor cell growth in each of the 3 groups discussed in the question stem? Some big takeaways from the passage: Induction of pRB correlates to increased apoptosis, except in the absence of Bax protein, but that part’s not relevant here. We also have increased apoptosis when the inducible copy of RB_SP is expressed. Again, not relevant here. So, we expect higher levels of pRB will mean increased apoptosis, and decreased tumor growth, or tumor volume on the Y-axis.

It’s actually easiest to compare all four answer choices at once here. We expect percent apoptosis to increase from left to right as we go from no pRB, to basal levels of pRB, to additional induced amounts of pRB. But we’re focused on tumor growth. We expect tumor volume to decrease from left to right. The two are inversely proportional, and for this question, we’re focused on tumor volume. That means the only answer that matches our breakdown is answer choice C. Answer choices A, B, and D are all incorrect trends, and they contradict the information we gathered from the passage.

58) To answer this question, we can pull up Figure 1. 

Some of the biggest takeaways from the passage: pRB induction means increased TNF alpha-stimulated apoptosis, but that’s only true for 2 of the 3 cell lines: Wild type and Bak deficient. That’s not true for Bax deficient, which we see all the way to the right. Let’s consider the 4 answer choices and see which is best supported by Figure 1:

  1. TNFα stimulates nuclear to mitochondrial translocation of pRB. This answer choice wasn’t mentioned in the passage, especially in regards to TNF alpha. We’re focused on an answer choice that can be supported by the data in Figure 1. What I don’t want you to do is pick an answer choice because you don’t remember it from the passage, so you assume it must be correct. AAMC throws these in as distractors, because they’re neither explicitly wrong, nor are they answering our question correctly. We’ll keep the answer for now for comparison sake. We still want to look at the other options.
  2. pRB exerts its effects on TNFα-stimulated apoptosis via the extrinsic apoptotic pathway. We have another answer similar to answer choice A. When we read this statement then look at Figure 1, can we say the statement is supported? Not only is this out of scope when considering Figure 1, but the extrinsic apoptotic pathway wasn’t mentioned in the passage. We covered the mitochondrial pathway, which is the intrinsic pathway. Neither A nor B stand out as good answers thus far.
  3. TNFα-stimulated apoptosis correlates negatively with induced pRB levels in Bak–/– cells. This answer choice is the opposite of our breakdown. Look at the Bak deficient cells-that’s the middle cell line in the Figure. We’re focused on the rightmost bar. We have TNF alpha stimulated apoptosis correlating positively with induced pRB levels. That bar shows a higher percent apoptosis than the others of the same cell line.
  4. pRB induction requires BAX, but not BAK, for TNFα-stimulated apoptosis. This answer choice is consistent with our breakdown. We said pRB induction means increased TNF alpha-stimulated apoptosis, but for just 2 of the 3 cell lines: wild type and Bak deficient. Howeber, when Bax is deficient, we don’t have increased TNF alpha-stimulated apoptosis. In fact, there’s no effect. That means pRB induction requires BAX. We can eliminate answer choices A-C because A and B were out of scope and C contradicted what we see in Figure 1. We’re left with our correct answer, answer choice D: pRB induction requires BAX, but not BAK, for TNFα-stimulated apoptosis.

59) Be careful with the verbiage because we want an approach that is least effective. This is a very broad question, but ultimately, it’s going to boil down to locating a specific protein within a cell. That can be done in a multitude of ways. We want an approach that’s least effective. 

  1. Fusion of a fluorescent tag to pRB. This actually sounds like a very good way to determine the localization of a protein within a cell. If we have fluorescence, we can literally see and identify the protein, and its location. 
  2. Use of a fluorescent probe that hybridizes to rb transcripts. This answer choice could work in theory if we were focused on mRNA, not the actual protein. mRNA eventually goes to ribosomes for translation, and that’s when we have proteins made and directed to their destination. This answer choice doesn’t address the specific question being asked, so this is the front-runner for least effective. 
  3. Cellular fractionation to isolate different organelles. Cellular fractionation involves separating cellular components into many parts. If we isolate different organelles, we can identify where our proteins of interest are found. This is also an effective method so we eliminate this answer choice.
  4. Pull down of pRB and identification of interacting proteins by mass spectrometry. Pull down assays can be used to detect the activation status of specific proteins. They can also give more information about specific protein interactions, which can provide a clue into location. We can also eliminate answer choice D so we’re left with our correct answer, answer choice B: Use of a fluorescent probe that hybridizes to rb transcripts.

60) Rhodamine 123 is a dye that binds to polarized membranes-such as the inner mitochondrial membrane. Apoptosis disrupts the mitochondrial membrane and that binding, so the variant where we see the highest percent apoptosis would result in the lowest level of staining. To answer this question, we can look at Figure 2. Remember, we want to find the variant where we have the highest percent apoptosis, and therefore the lowest level of staining.

We want to find the highest percentage of apoptosis, which we can do here using Figure 2. Percent apoptosis is highest when the inducible copy of RB_SP is expressed, so we want an answer similar consistent the small pocket variant.  

  1. pRB induced. This is the 2nd highest bar in our Figure 2. We like this answer choice for now
  2. RB_N. This answer choice is the opposite of what we want. We said the highest percent apoptosis would mean the lowest level of staining. We can eliminate answer choice B. 
  3. RB_SP. This answer choice matches our breakdown. It’s a similar height to answer choice A, but answer choice C shows slightly higher apoptosis. That means we can now eliminate answer choice A.
  4. RB_C. This is similar to answer choice B. It’s the opposite of what we want. We said the highest percent apoptosis would mean the lowest level of staining. We can also elimnate answer choice D. We’re left with our correct answer, answer choice C: RB_SP. 

61) To answer this question, we want to know the function of the caspase activator discussed in the passage, or the function of cytochrome C. Cytochrome C is an essential component of the electron transport chain. Cytochrome C is a mobile electron carrier than can only carry one electron at a time. It delivers electrons to the last complex of the electron transport chain, complex IV

  1. Proton pumping. This is tangentially related to the electron transport chain, which is likely why it’s included here. The complexes in the electron transport chain will pump protons to generate a proton gradient. That’s not the primary function of cytochrome C. 
  2. Heme biosynthesis. Cytochromes are a family of related proteins that have heme prosthetic groups containing iron ions. Cytochrome C is a hemeprotein, but primary function is not heme biosynthesis.
  3. Calcium signaling. Calcium levels and calcium release can affect cytochrome C release-but this function has more to do with apoptosis than electron transport. Cytochrome C is influenced by calcium levels, it’s not the one responsible for calcium signaling. 
  4. Electron transport. This answer choice matches our breakdown. We said cytochrome C is an essential component of the electron transport chain. This answer also matches the specific question being asked. We can eliminate answer choices A-C, none of those identified the primary function of cytochrome C. We’re left with our correct answer, answer choice D: Electron transport. 

62) To answer this question, we can break down what we know about apoptosis, and the presence of cytochrome C in different parts of the cell, given different circumstances. We want to find the option that shows the graphic corresponding to the Western blot analysis, given the situations described in the question stem. 

We have cells incubated with either 0 or 500 nanomolar pRB, in the presence or absence of BAX and BAK proteins. We expect apoptosis when we have pRB present, along with BAX and BAK. But what does that mean for the presence of cytochrome C?

In the supernatant we have lighter, less dense objects. Pellet is the bottom, where heavier, denser objects will end up. Cytochrome C in intact mitochondria will be found in the pellet. If we have apoptosis, cytochrome C leaves the mitochondria, and instead is found in the supernatant. That means in situations when we expect apoptosis, we’ll have cytochrome C levels in the supernatant. Again, that’s when we have pRB present, along with BAX and BAK. We want an answer choice that shows pRB can induce apoptosis when BAX is present. In all other situations, we should only see cytochrome C in the pellet.

Answer choices A and B correctly show cytochrome C proteins in the supernatant when we have pRB present, along with BAX and BAK. The only difference is we also see proteins present in other lanes in answer choice B. Answer choice A is the only answer choice that’s consistent with our breakdown and what we’re looking for in an answer choice.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 9

63) Note the verbiage here, the question stem emphasizes “NOT.” That means 3 answer choices will play a role, but 1 answer choice will not. To answer this question, we’ll go through the information we have about STAT3, and which protein domains will play a role in signaling. STAT3 is a nuclear factor that’s phosphorylated by JAK2. Phosphorylated STAT3 forms homodimers that activate transcription of target genes. These were the main points about STAT3 from the passage. Specifically, we want to focus on it being a nuclear factor, so we want a domain that isn’t used in the nucleus.

  1. Nuclear localization domain. We just mentioned in our readthrough, STAT3 is a nuclear factor, a nuclear protein. That means that nuclear localization domain will play a role in signaling. Specifically, for nuclear translocation. Remember, we want an answer choice that will not play a role in STAT3 signaling. 
  2. Signal sequence domain. This answer choice wasn’t mentioned in the passage like answer choice A. Signal sequences are found in proteins that are destined toward the secretory pathway. That means they’ll be secrete from the cell, or inserted into cell membranes. Not something we expect from STAT3, which actually makes this a good answer choice. That means we can eliminate answer choice A which contradicted what was said in the passage. 
  3. DNA binding domain. This is similar to answer choice A. DNA binding domain is required to bind to regulatory regions of target genes. Remember, phosphorylated STAT3 molecules activate transcription of target genes. We can eliminate this answer choice also for contradicting the information from the passage.
  4. Protein binding domain. This answer is going to tie into the formation of homodimers. Remember phosphorylated STAT3 molecules form homodimers that activate transcription of target genes. STAT3 is also recruited to the LEPRb/JAK2 complex. There’s likely a protein binding domain here. We can eliminate this answer choice for contradicting the passage as well. We’re left with our correct answer choice, answer choice B: signal sequence domain

64) To answer this question, we’re going to find an amino acid substitution that will mean upregulation of leptin signaling. What this likely means, is we want to find a negative regulator of leptin signaling, and provide a substitution that would inhibit its negative regulation. How would that happen? Likely by substituting an amino acid that is unlike the amino acid for which it’s substituting. For example, a positive amino acid for a negative amino acid. Or an amino acid with a smaller side chain, for an amino acid with a larger side chain.

Quick glance at our answer choices shows we’re substituting from Y, which is the abbreviation for tyrosine, to F, which is the abbreviation for phenylalanine. That means we’re going from a polar side chain with a hydroxyl group at the end, to a nonpolar side chain, without that hydroxyl group. That means we’ll amend our breakdown just a bit. We’re focused instead on the location of these substitutions more than the amino acids themselves.

  1. Y570F within JAK2. LEPRb binds to JAK2, resulting in JAK2 autophosphorylation at Y570 and activation. We’d be affecting this autophosphorylation of JAK2. Preventing JAK2 activation wouldn’t affect leptin signaling itself. Remember, we’re focused exclusively on leptin signaling and its regulation. 
  2. Y705F within STAT3. STAT3 is recruited to the LEPRb/JAK2 complex, and JAK2 phosphorylates STAT3 at Y705. We’d be affecting that phosphorylation. That would affect the number of STAT3 molecules that form homodimers, but wouldn’t directly affect leptin signaling. 
  3. Y985F within LEPRb. SOCS3 binds to Y985, blocking recruitment of STAT3 to the LEPRb/JAK2 complex. We’d be preventing that recruitment of STAT3 to the complex. We talked about this during the passage. This step is essential to negative feedback. If we don’t have that binding and negative regulation, leptin signaling isn’t downregulated as normal. We can keep this answer choice for now and now eliminate answer choices A and B. Those affected other players from the passage, but answer choice C is our best answer choice so far.
  4. Y1138F within LEPRb. JAK2 phosphorylates Y1138 within a YXXQ amino acid motif. A binding site for STAT3. Again, we’re preventing this binding, but not directly affecting leptin signaling. That was the key in all of these answer choices. We can eliminate answer choice D as well, we’re left with our correct answer, answer choice C. Y985F within LEPRb

65) To answer this question, we’re picking a property or mechanism that’s exclusive to adipocytes. That means we’ll first decide what the author mentions about adipocytes in relation to leptin. Then we’ll use our general knowledge about gene expression and how leptin works. The author mentions in the passage: Leptin is encoded by the ob gene and is primarily expressed in adipocytes in response to feeding. Every cell in the body will have the ob gene, but why do we have expression in adipocytes one? Transcription factors function to regulate genes to make sure they’re expressed in the right cell, at the right time, and in the right amount.

  1. the ob gene. This answer choice implies only adipocytes have the ob gene. In reality, the basic genetic code is the same for all cells in the human body. We expect all cells would have the ob gene, not just adipocytes. We need to find an answer that talks about making sure genes are expressed in the right cell.
  2. a promoter for the expression of the ob gene. A promoter is an actual DNA sequence-that would be found in every cell. This isn’t specific to adipocytes, so both answer choices A and B are found in all cells. We need an answer that talks about making sure genes are actually expressed in the right cell, at the right time and amount.
  3. enhancers for the expression of the ob gene. An enhancer is also an actual DNA sequence-again found in every cell. This is going to be similar to answer choice B. We need an answer that talks about making sure genes are actually expressed in the right cell, at the right time and amount. 
  4. nuclear factors for the expression of the ob gene. Nuclear factors are closely related transcription factors. We mentioned transcription factors function to regulate genes to make sure they’re expressed in the right cell, at the right time, and in the right amount. That sounds exactly like the answer we’re looking for. We can eliminate answer choices A-C. Answer choice D is our correct answer.

66) To answer this question, we’re going to find amino acids that are most similar, and therefore causing the least negative effects following a substitution. Like we’ve done previously, we want to find amino acids that are the most similar. That means if we have a positive amino acid, we’re substituting another positive amino acid to minimize interference. If we have a smaller side chain, we replace it with another smaller side chain. Note all 4 answer choices: we’re substituting glutamine for another amino acid. Glutamine is a polar amino acid with a neutral side chain, no net positive or negative charge. We want a substitution by a similar amino acid.

  1. glutamine to glutamic acid. Glutamic acid is an acidic amino acid, it has a negatively charged side chain. There’s a significant difference between the two amino acids.
  2. glutamine to glycine. Glycine is a nonpolar amino acid, and has the simplest side chain of the amino acids. We said glutamine is polar, which is a direct contrast. 
  3. glutamine to asparagine. Both glutamine and asparagine are polar amino acids. They have similar side chains, so this is our best answer so far. We can eliminate answer choices A and B.
  4. glutamine to alanine. This is similar to answer choice B. Alanine is a nonpolar amino acid with a simple side chain. Glutamine is polar, which is a direct contrast. We’re left with our correct answer, answer choice C: Glutamine to asparagine.

67) To answer this question, we’re going to focus on dimerization and the ideal situations and environment in which we find dimerization take place. Dimerization occurs in the cytosol of the cell, which is a hydrophilic region. Dimerization typically occurs due to hydrophobic interactions. Why is that? The amino acids are not interested in interacting with water molecules in the cytosol. Hydrophilic, polar amino acids would be less likely to dimerize because of these interactions with water.  

  1. Polar amino acids. Polar amino acids are more likely to interact with water molecules, rather than other polar amino acids. We won’t form strong bonds between polar amino acids, because both amino acids would be willing to break that bond and interact with water. 
  2. Hydrophobic amino acids. This answer choice matches our prediction. We said dimerization typically occurs due to hydrophobic interactions. Neither amino acid is going to interact with water-instead these amino acids are free to bond to one another, and we have dimerization. We can keep answer choice B and eliminate answer choice A because it contradicts our breakdown.
  3. Positively charged amino acids. This is also similar to answer choice A. Charged amino acids, like positively charged amino acids, would also interact with water molecules. Both amino acids are willing to interact with water, so a strong bond and dimerization is not likely. We can eliminate this answer choice.
  4. Negatively charged amino acids. This is also similar to answer choice A. Negatively charged amino acids would interact with water molecules. Both amino acids are willing to interact with water, so a strong bond and dimerization is not likely. We can also eliminate answer choice D. We’re left with our correct answer, answer choice B: hydrophobic amino acids.  

68) To answer this question, let’s consider what we know about different isoforms. Isoforms are functionally similar, but their amino acid sequence isn’t the same. That can mean same genes, but alternative splicing. What is alternative splicing?  Alternative splicing is a process that occurs during gene expression. It allows for the production of multiple isoforms from a single gene coding. Alternative splicing can occur due to the different ways in which an exon can be excluded, or included in mRNA. It can also occur if portions on an exon are excluded or included, or if there is an inclusion of introns.

  1. exons of the LEPR gene. This answer choice is consistent with our breakdown. Different isoforms exist because of different combinations of exons from alternative splicing. We like this answer choice for now, so we’ll hold on to it and look at our additional answer choices.
  2. isoforms of the LEPR gene. The isoforms are produced via alternative splicing, not by using other isoforms of the LEPR gene. The isoforms are all going to have different amino acid sequences. This answer choice contradicts our breakdown.
  3. promoters of the LEPR gene. We’re not focused on promoters of the LEPR gene, instead, the exons. Promoters are sequences that proteins bind to initiate transcription. We’re not going to have different promoters here, those will remain the same. We can also eliminate answer choice C because it contradicts our breakdown.
  4. protein cleavage sites of LEPRb. This would actually involve breaking the peptide bonds between amino acids in proteins. Instead, we’re focused on including and excluding exons in mRNA. This answer choice contradicts our prediction, and the definition of alternative splicing, so we’re left with our best answer, answer choice A: exons of the LEPR gene. 

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Questions 69-73

69) Make sure to be careful with the verbiage here. We want a conclusion that is NOT valid. At first glance, Protein X is broken into subunits on the left. We’re likely looking at SDS gel electrophoresis under non-reducing condition. Gel B is more likely the native gel and Protein X is not denatured into its subunits.

Looking closer at Gel A, the WT lane shows bands at 50 and 20 kDa. For the Variant, we see a single band at 70 kDa. We can’t say with 100% certainty, but it’s likely the 50 kDa and 20 kDa subunits are covalently linked and that’s why we have the 70 kDa band in the Variant lane. If we’re considering Protein X, it likely has two 50 kDa subunits and two 20 kDa subunits that make up the 140 kDa protein. Let’s use this information and dive into our answer choices. Remember, we want a conclusion that is not valid!

  1. Gel B is a native gel. We covered this in the breakdown of the question. Protein X isn’t denatured into its subunits, so Gel B is more likely the native gel. This is a valid conclusion, which makes it an incorrect answer.
  2. Protein X is a tetramer composed of 2 identical heterodimers. This is something we also covered in our breakdown of the question. We have a 140 kDa protein that’s made of two 50 kDa subunits and two 20 kDa subunits. The 50 kDa and 20 kDa subunits are covalently linked and we see them run together in the Variant lane in Gel A. Answer choice B is also a valid conclusion, which makes it an incorrect answer.
  3. The variant form contains covalently linked subunits. This is something we already touched on in answer choice C and our breakdown. We mentioned the 50 kDa and 20 kDa subunits are covalently linked. Answer choice C is also a valid conclusion, which makes it an incorrect answer.
  4. The largest subunit of Protein X is approximately composed of 120 amino acids. Largest subunit of Protein X is the 50 kDa subunit. The average molecular weight of an amino acid is 110 Da. Dividing 50 kDa (50,000 Da) by the average weight of an amino acid (110 Da) yields approximately 455 amino acids. Even given the fact we did some approximation and the fact amino acid sizes can vary, there are likely many more AAs than proposed in this answer choice. Because answer choice D is not a valid conclusion, we can pick this as our best answer.

70) This question just comes down to knowing the four different methods listed. We’ll be smart about eliminating answer choices as we make a verdict about one of the four methods. For example, if we find option I is valid, we’ll narrow our options down to A and C only.

Which methods separate proteins based on their charge?

  • SDS-PAGE: SDS-PAGE gels are used in experiments to denature and separate proteins by size. This method does not separate based on charge. That means right away we can eliminate options A and C. That means we’ve only looked through ¼ of our options, but have a 50% chance of getting the question correct, just by guessing. Luckily we can improve that chance by going through our additional questions!
  • Isoelectric focusing: Isoelectric focusing separates proteins based on pI, meaning this is a valid option. Option II is found in all four answer choices, so it’s no surprise this is a valid option!
  • Ion-exchange chromatography: Both of our remaining answer choices (B and D) contain option III, so good chance this is a valid option. In fact, ion-exchange chromatography can separate proteins according to their charge.  This method is based on the attraction between oppositely charged ions, so a cationic stationary phase is used to separate anions in cation-exchange chromatography, and an anionic stationary phase is used to separate cations in anion exchange chromatography.
  • Affinity chromatography: Affinity purification separates a specific protein based on an affinity tag designed for that chromatographic process. This is not based on charge, so only options II and III are valid methods.

We broke down all four of the methods listed and found only options II and III are valid methods by which to separate proteins by charge. That corresponds to our correct answer: Answer choice B.

71) To answer this question, we can rely on our external knowledge. 

Throughout the citric acid cycle, the starting molecule oxaloacetate (which has 4 carbons) is progressively transformed into several different molecules (as carbon atoms are added to and removed from it), but at the end of the cycle it always turns back into oxaloacetate to be used again. This specific question is asking about the citric acid cycle, but from a specific point: from α-ketoglutarate to oxaloacetate. Why is that important? Because there is an NADH produced in a prior citric acid cycle step that we’re not counting here. If we remove that NADH from our count, we’ll be left with 2 NADH and 1 FADH2. The answer that corresponds to these three molecules is answer choice C. So many pathways and topics are connected, and whenever you have the opportunity you should connect them and review them on a macro and micro level:

72) This is similar to the previous question because it comes down to knowing pathways and knowing specific details about this content. Review the pathways below and answering the question just comes down to identifying the enzyme both have in common.

  1. Phosphoglucomutase: Glycogenolysis & glucogenesis. Not what we’re looking for.
  2. Glucose 6-phosphatase: This is found in both pathways we’re looking at here. This is going to be our correct answer. 
  3. Hexokinase. Hexokinase is present in glycolysis, but not in gluconeogenesis.
  4. Glucokinase. Reasoning here is going to be similar to answer choice C. This is only present in glycolysis. We can stick with answer choice B as the correct answer. 

73) Catalytic efficiency is given by Kcat/Km and measures how well an enzyme is able to “grab” its substrate and turn it into product. This means you want the greatest number of reactions happening, for the least amount of input of time/substrate. That’s basically what kcat/km says. Also note the test-maker says “assuming the same isoenzyme concentration.” Vmax is equal to the product of Kcat and the concentration of the enzyme. That means for the purposes of this question, we can consider that as Vmax goes, Kcat will also go.

Kcat, the catalytic constant, measures the number of molecules of substrate turned over into product per unit time (usually seconds). Km is our Michaelis constant. It’s used to determine the affinity between an enzyme and substrate. 

The higher your kcat is, the more reactions happen per unit time (i.e. Vmax) 

All of this is taking place when conditions are optimal, and you have enough substrate. So, Kcat is positively proportional to efficiency; 

The higher km is, the more concentration of substrate you need before you reach optimal conditions for kcat, so km is inversely proportional. 

We have all of our parameters organized in the table, so we can find the highest ratio of Vmax to Km to see which isozyme has the highest catalytic efficiency. Divide and we get the following:

  1. 2×107
  2. 5×106
  3. 1×105
  4. 6×104

Answer choice A is the largest value and has the highest ratio.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 10

74) To answer this question, we can go through the dissociation constants of prorenin for 4 variants in Table 1.

We have Table 1 shown here, and we want to know which amino acid residues of PRR are involved in binding to prorenin. From the passage, we should recall that the only Kd value we see that didn’t change much from the wildtype is the 2nd one in our table: residue 109. We had valine substituted for glutamic acid, which is a change from a nonpolar amino acid to an acidic, negatively charged amino acid. Despite that substitution, we don’t see much change in dissociation constant from the wild type. That likely means that residue is not essential to binding prorenin. All of the other Kd values changed much more. Those substitutions changed the binding affinity, meaning those other residues are involved in binding to prorenin. Only residue we’re not going to pick is residue 109. 

  1. Residue 109
  2. Residue 140
  3. Residue 201
  4. Residue 269

That means options 2, 3, and 4 are correct. We can find the only answer choice that matches those options: Answer choice D: II, III, and IV only. 

75) C17XA is a competitive inhibitor. 

It competes with the normal substrate for the active site of activated prorenin. In competitive inhibition, Km increases, Vmax is unaffected. Our graph is going to show the same y-intercept with and without the inhibitor. Again, that’s because Vmax is unaffected. A Km increase means the slope of the line with the inhibitor will be greater, we’ll see a steeper slope, meaning more vertical movement. 

  1. Answer choice A. Slope of the lines here is consistent with our breakdown. Note how the lines are labeled. Either with or without inhibitor. However, we don’t have the same y-intercept, which we said is the case in competitive inhibition. 
  2. Answer choice B. This is consistent with our breakdown. Steeper line with inhibitor? Check. Same Y-intercept? Check. Consistent with increased Km and unaffected Vmax? Check. We can keep this answer choice, we can eliminate answer choice A because that incorrectly showed inconsistent Y-intercepts. 
  3. Answer choice C. This graph is showing uncompetitive inhibition. That means we have reduced Km and Vmax. That’s inconsistent with our breakdown and our general knowledge. 
  4. Answer choice D shows noncompetitive inhibition. That means Km is unaffected, so we have the same X-intercept, but Vmax is reduced. We can also eliminate this answer choice because it’s inconsistent with our prediction. We’re left with our correct answer, answer choice B.

76) To answer this question, we want to know how blood pressure changes in relation to overexpression of PRR. We’re looking at both angiotensin II dependent and independent pathways. We can look at Figure 1, and keep in mind we have overexpression of PRR.

We’re considering overexpression of PRR. The graph is measuring relative levels of ROS, which is an intermediary molecule responsible for increased blood pressure.

Let’s look at the control first. Overexpression of PRR is the gray bar, and we have increased relative levels of ROS, or increased blood pressure.

Adding Losartan, the antagonist, changes nothing compared to what we saw in the control group. 

Adding prorenin causes increased ROS in both species: transfected with empty vector and transfected with PRR. We don’t know if this is angiotensin II dependent or independent.

Adding the combination of prorenin and the antagonist raised ROS for the empty vector compared to the control, but not for the overexpressing specimen. That means Losartan was working. Losartan is an angiotensin II receptor antagonist, which means overexpression of PRR increases blood pressure in part through an angiotensin II-dependent pathway. Losartan is able to influence that effect, so that shows it’s an angiotensin II-dependent pathway.

We have one more thing to consider. We also had an increase in ROS in the empty vector when treated with prorenin and the combination. Losartan didn’t decrease the effect in the 4th treatment group. The antagonist failed to decrease this effect, meaning there’s another way ROS is increasing-an independent pathway.

That means we have two potential answers here: overexpression of PRR increases blood pressure in part through an angiotensin II dependent AND independent pathway. The key here was the increases in ROS and Losartan influence. 

  1. increases blood pressure in part through an angiotensin II-dependent pathway.
  2. increases blood pressure in part through an angiotensin II-independent pathway.
  3. decreases blood pressure in part through an angiotensin II-dependent pathway.

That means we’re keeping options I and II. Only one answer choice matches those two options: Answer choice D: I and II only.

77) To answer this question, we’re finding amino acids that are likely to interact with one another. We’re going to identify the residue at position 201 of PRR. Then we’ll use our general knowledge to find an amino acid that would most likely interact with that residue.

We have Table 1 here, so let’s identify the amino acid at position 201. Amino acid we had before that substitution shown in Table 1 is aspartate (D). What do we know about aspartate? It’s acidic, and electrically charged: we have a negative charge on that side chain. We want an amino acid that would interact with aspartate. That would likely be a basic amino acid, and also electrically charged. Ideally, we’d have an amino acid that has a positive charge on its side chain: either histidine, arginine, or lysine.

  1. Alanine. Alanine doesn’t match our breakdown, it’s a neutral, nonpolar amino acid. We’ll keep it for now, but look for a better answer.
  2. Glutamate. Glutamate is also negatively charged, just like aspartate. We don’t want two negative amino acids interacting. We can eliminate answer choice B. Answer choice A is still our best choice.
  3. Arginine. Arginine was one of the 3 amino acids we mentioned in our prediction. It’s basic, and it has that positive charge. We like this option, so we’re going to keep this as our best answer choice and eliminate answer choice A. 
  4. Glutamine. Glutamine is a polar, uncharged amino acid. Similar to answer choice A, we don’t want a neutral amino acid. At least not for our best answer choice. That means we can eliminate answer choice D and we’re left with our best answer, answer choice C: Arginine.

78) To answer this question, we’re going to identify the relationship between cDNA and enzymes. cDNA is complementary DNA. It’s DNA that’s been reverse transcribed from RNA. We’re going to find an answer choice that’s not used in cloning of wildtype-PRR complementary DNA. We can eliminate any answers that would be used.

  1. DNA polymerase. DNA polymerase is an enzyme that adds nucleotides during replication of the DNA. It’s necessary for DNA replication and amplification. This enzyme was used in the cloning process.

  1. RNA polymerase. RNA polymerase is used to synthesize mRNA using a DNA template. But in this case, that’s not necessary. We explicitly need DNA polymerase from answer choice A, but not RNA polymerase. We can eliminate answer choice A. 
  2. DNA ligase. DNA ligase can connect two strands of DNA together. We’re told we have cloning. Ligase is going to be necessary in the ligation of complementary DNA to DNA vector-that’s necessary in the experiment. We can eliminate answer choice C. 
  3. Reverse transcriptase. Reverse transcriptase converts single-stranded RNA to double-stranded DNA. We need reverse transcriptase to transcribe cDNA from RNA; that’s how we get cDNA vectors for the experiment to work. We can eliminate this answer choice because reverse transcriptase was used in cloning. We’re left with our correct answer, answer choice B: RNA polymerase.

79) To answer this question, we’ll go to our passage which will tell us whether we have increased or decreased affinity following the substitution. We’ll focus on the amino acids themselves, and what the substitution represents in terms of the properties of each amino acid.

Here we have Table 1. We’re focused on the 3rd variant down, W140L. That’s a substitution of tryptophan with leucine. Comparing Kd in the wildtype with this variant, we notice an increase. A larger Kd value, means less binding affinity. Smaller Kd value, means a greater binding affinity.

Larger Kd value is less binding affinity, meaning decreased stability. And we’re substituting tryptophan (a nonpolar, aromatic, uncharged amino acid) with leucine (another nonpolar and uncharged, but non-aromatic amino acid). 

     a.  Elimination of a π-stacking interaction decreases stability of the complex. Pi-stackind involves interactions between interactive rings which contain pi bonds which is consistent with our breakdown. We lost aromatic tryptophan and substituted with leucine. Second part of the answer choice is consistent with our prediction because of the larger Kd value. 
     b.  Elimination of an ionic interaction decreases stability of the complex. We just mentioned tryptophan is a neutral amino acid-it likely won’t form an ionic interaction. We see charged amino acids made ionic interactions. Even though the decreased stability is consistent with our breakdown, first part of this answer choice actually contradicts our breakdown, so we can eliminate answer choice B.
  c/d.  We can look at both of these answers at the same time: 

Addition of a hydrophobic interaction increases stability of the complex AND Addition of a hydrophilic interactions increases stability of the complex. We already mentioned these would be eliminated because the Kd value in the table increased, meaning decreased binding affinity. Decreased binding affinity means decreased stability, not increased stability. We can eliminate both of these answer choices for contradicting Table 1 and our breakdown. We’re left with our correct answer, answer choice A: Elimination of a π-stacking interaction decreases stability of the complex.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 11

80) To answer this question, we’ll reference the passage, but ultimately our answer is going to come from knowing content. We need to know which metabolic pathways correspond to different situations, so let’s go back to the passage.

The passage mentions After the depletion of the blood glucose pool, the initial source for energy replacement is from hepatic glycogen, which is sufficient to support basic metabolism for the first 12 hours. A few sentences later, we’re told hepatic glycogen stores are soon exhausted, and the glucose pool must be replenished from non-carbohydrate sources including fatty acids and glucogenic amino acids.

We have an initial breakdown of glycogen. That’s known as glycogenolysis. But following that initial breakdown, we have the glucose pool replenished from non-carbohydrate sources. Glucogenic amino acids can be converted into glucose through gluconeogenesis. We’re looking for an answer choice involving glycogenolysis and gluconeogenesis.

  1. Glycogenesis followed by glycogenolysis. Let’s break this answer choice down one part at a time. Glycogenesis is the formation of glycogen from sugar. We mentioned glycogenolysis is the breakdown of glycogen. The first part of our answer choice contradicts what we were told in the passage. We said glycogen is broken down, and that’s followed by gluconeogenesis. 
  2. Glycogenolysis followed by gluconeogenesis. This sounds familiar, this is actually the exact same answer as our breakdown, which is usually a good sign. We mentioned a breakdown of glycogen, and then replenishing the glucose pool from non-carbohydrate sources. I’m liking this option for now. 
  3. Gluconeogenesis followed by glycogenolysis. The components of this answer choice match our breakdown, but they’re in the opposite order. We expect glycogenolysis to occur first, so we can eliminate answer choice C for contradicting the passage.
  4. Gluconeogenesis followed by glycogenesis. Glycogenesis is not a part of our answer choice. We mentioned glycogenesis is the formation of glycogen from sugar, which is not taking place here. That means we can eliminate answer choice D as well. We’re left with our correct answer, answer choice B: Glycogenolysis followed by gluconeogenesis.

81) To answer this question, we’ll rely on our content, particularly about gluconeogenesis. This is almost like a standalone question that’s tangentially related to the passage and something we touched on in question 80. Infants exhaust hepatic glycogen stores, and the glucose pool has to be replenished from non-carbohydrate sources including fatty acids and glucogenic amino acids. The body needs to utilize gluconeogenesis to replenish the glucose pool. So essentially, we want to know which of the compounds listed can enter gluconeogenesis. Or, can any of the 4 options be converted into a compound that can enter gluconeogenesis? Either way that gets us to sustain blood glucose levels. Another thing we want to note is the passage explicitly says glucogenic amino acids. 

  1. Acetyl-CoA isn’t used for gluconeogenic purposes, but for ketogenic purposes. The pyruvate dehydrogenase reaction is irreversible. For each turn of the citric acid cycle, 2 carbons enter as acetyl-CoA and 2 leave as carbon dioxide. That acetyl CoA doesn’t meet our criteria.
  2. Lactate is a major non-carbohydrate precursor along with fatty acids and glucogenic amino acids. Lactate is also converted to glucose in the Cori cycle; we know it’s gluconeogenic. 
  3. 3 & 4 say oxaloacetate and alpha ketoglutarate. Alpha-keto glutarate can be turned into oxaloacetate which enters gluconeogenesis. So we’re keeping both of these answer choices as well. We’re left with our ideal options: choices 2, 3, and 4. 

Looking at our answers, we can pick the correct answer. Answer choice D says II, III, and IV only.

82) We’ll keep in mind the question is asking about which event is not a likely outcome. Always be careful with verbiage, and never let that wording be a reason you get a question wrong.

We only had one mention of glucagon in the passage. The author mentions there’s an increase in the level of glucagon and activation of its 66 kDa G protein-coupled receptors in the liver. And this happens following the mobilization of hepatic glycogen. What we want to do in this question is use our general knowledge to explain what happens after this binding.

We have activation of the GPCRs which is a fairly broad topic, so let’s touch on some key points and we’ll dive deeper when we go through our questions, as necessary. GPCRs have alpha, beta, and gamma elements. Once a ligand binds the receptor, there is a conformational change and the receptor activates the G-protein. This releases GDP from the α subunit and picks up GTP. The subunits split into alpha and beta/gamma subunits. These fragments can activate other proteins like ion channels or enzymes. Adenylate cyclase is the classic example of an enzyme activated by GPCRs. The adenylate cyclase produces the second messenger cyclic adenosine monophosphate which is AKA cyclic AMP.

  1. GDP binding to Gα subunit of the G protein. This is the opposite effect of what we expect upon activation. This activation promotes exchange of GDP for GTP on the G alpha subunit. This answer choice would be more fitting when it is time to inactivate the subunit, which given the wording of our question, makes this a good answer. Remember, we’re looking for an event that’s an unlikely outcome. 
  2. adenylate cyclase activity. Upon activation, there’s activation of downstream effectors, including adenylate cyclase. In this case, answer choice B is a likely outcome of glucagon binding its receptor. This contradicts our breakdown and what we’re looking for in the question stem.
  3. protein kinase A activity. We expect protein kinase A to be activated and activity to increase following increase in cAMP, which we expect in this situation.
  4. cAMP generation. We touched on this earlier, we have activation of downstream effectors, and these effectors will control concentration of secondary messengers like cAMP. Cyclic AMP is synthesized from ATP by adenylate cyclase. We can eliminate answer choice D. We’re left with our correct answer, answer choice A. GDP binding to Gα subunit of the G protein is NOT a likely outcome.

83) This question is referencing the infants the author mentions that are born to mothers with poorly controlled diabetes. These infants received an oversupply of glucose during fetal development, and the body is used to this oversupply. And later the neonate will have higher than normal serum insulin concentration. 

How do we counteract the elevated levels of circulating insulin to prevent brain injury? There is elevated insulin because the neonate is used to having elevated glucose supply. Insulin allows the cells to absorb glucose from the blood. Glucagon triggers a release of stored glucose from the liver. We’re expecting an answer that has something to do with glucagon, or possibly alpha cells-peptide hormones that counteract insulin.

  1. Insulin. This is the opposite of our breakdown. We have a surplus of insulin already, so this answer choice is unlikely. 
  2. Epinephrine. Epinephrine can stimulate the secretion of glucagon, which is ultimately what we’re looking for. This is a better answer than answer choice A, so we can hold on to answer choice B for now.
  3. Glucagon. This answer matches our breakdown exactly. We said this is the exact answer that would counteract the excess insulin. The author actually mentions in the passage glucagon can mobilize blood glucose to prevent brain injury. We can hold on to answer choice C because it’s a superior answer choice to answer choice B.
  4. Cortisol. Cortisol is usually released in response to low blood-glucose concentration, but it’s a steroid, not a peptide hormone. We’re left with our correct answer, answer choice C.

84) To answer this question, the only thing we have to remember from the passage is that the first 12 hours is when hepatic glycogen supports the basic metabolism of the newborn. We will use our knowledge of glycogen and enzymes to answer the question. 

We need an enzyme that allows a newborn infant to utilize hepatic glycogen in the first 12 hours. Essentially an enzyme involved in glycogenolysis. We’re breaking down glycogen into glucose. The enzyme we use to do this is glycogen phosphorylase. Glycogen phosphorylase breaks down glycogen into glucose.

  1. Glycogen phosphatase. This is a nonsense answer that just sounds similar to our correct answer. We don’t have glycogen phosphatases. Phosphatases catalyze the transfer of phosphate groups from phosphoproteins to water. Not what we’re looking for here.
  2. Answer choice B says Glycogen phosphorylase. This matches our prediction exactly, and this comes directly from AAMC’s content outline. Content category 1D lists glycolysis, substrates, and products. Glycogen enters the glycolytic pathway through the action of glycogen phosphorylase. Glycogen phosphorylase breaks down glycogen into glucose, so this is a superior answer to answer choice A. We still want to go through all of our answer choices, but often we have an answer that we feel is objectively correct like in this case. We don’t have to spend a ton of time on the additional answers here. We’re just looking for a possible superior answer.
  3. Glycogen hydrolase. These enzymes are more prevalent in different organisms, and not exactly what we’re looking for to break down hepatic glycogen. Hepatic glycogen is broken down by glycogen phosphorylase. 
  4. Glycogen kinase. These kinases are related to the addition of phosphate onto amino acid residues. Not so much what we’re looking for in step 2 and 3. We can eliminate answer choice D and we’re left with our correct answer, answer choice B-Glycogen phosphorylase.

85) This question is asking about the glycosidic bonds between glucose molecules in glycogen. How do we know it’s asking about glycogen? Because the passage mentions the newborn uses hepatic glycogen as an energy source. Quick glance at our answer choices to see the format we need to answer the question, and it looks like we want to explain the linear linkages and the linkages at branch points. Most of the glucose residues in glycogen are linked by alpha 1,4 linkages. Branches are created by alpha 1,6 glycosidic bonds. How can we distinguish between the two? We look at the confirmation of the compound, meaning we focus on the hydroxyl group on carbon 1. When the OH is above the glucose ring, we have alpha linkages. When the OH is below, we have beta linkages. 

  1. α(1→4) linkage linearly and β(1→6) linkage at branch points. This contradicts our prediction. The linkage numbers are correct, but we want alpha 1->6 linkages for branches, not beta.
  2. β(1→6) linkage linearly and α(1→4) linkage at branch points. Another answer than contradicts our breakdown. We want alpha linkages only. 
  3. α(1→4) linkage linearly and α(1→6) linkage at branch points. This answer choice is consistent with our breakdown. This is now our superior answer choice for this question.
  4. α(1→6) linkage linearly and α(1→4) linkage at branch points. The linkages here are the opposite of what we expect, so we can eliminate answer choice D as well. We’re left with our correct answer, answer choice C.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Questions 86-90

86) To answer this question, we’ll recall what we know about gluconeogenesis. We’re finding the answer choice that is NOT a precursor or substrate. This is purely a content question and we can start by looking at gluconeogenesis.

  1. Lactate. Lactate and glycerol are both starting material for gluconeogenesis and highlighted in red above. Note how they fit into gluconeogenesis and meet our criteria.
  2. Glycerol. See reasoning for answer choice A. Lactate and glycerol are both starting material for gluconeogenesis and highlighted in red above. Note how they fit into gluconeogenesis and meet our criteria.
  3. Oxaloacetate. We can see oxaloacetate as a substrate in gluconeogenesis.
  4. Phosphogluconate. This is the one answer choice that doesn’t fit into gluconeogenesis. Make sure to not get confused by PEP which sounds similar! We can stick with answer choice D as our best answer here.

87) This is a standalone question and we’re going to rely on our external knowledge. This question is twofold. We’ll need to determine the groups added to proteins by kinases. Then we’ll determine which type of enzyme removes this group. Kinases work to catalyze the transfer of phosphate groups from high-energy molecules like ATP to specific substrates. Phosphatases function to remove these phosphate groups. We often hear that a kinase phosphorylates a molecule while a phosphatase dephosphorylates a molecule.

  1. Phosphorylase. Phosphorylase catalyzes the addition of a phosphate group from an inorganic phosphate to an acceptor.
  2. Cyclase. Cyclase function sto catalyze reactions to form cyclic compounds.
  3. Phosphatase. This answer choice is consistent with our breakdown; a kinase phosphorylates a molecule while a phosphatase dephosphorylates a molecule. We can stick with answer choice C as our best answer so far. 
  4. Acetylase. Acetylase catalyzes the transfer of an acetyl group, not a phosphate group. We can stick with answer choice C as our best option.

88) To answer this question, we can reference the reaction catalyzed by succinyl-CoA synthetase. We expect increased levels of any products of this reaction:

GDP + phosphate + succinyl-CoA    GTP + succinate + CoA

Of the three options listed, only answer choices B and C list products of this reaction. The only thing AAMC is testing here is whether you know your content. The only answer choice consistent with the reaction is answer choice D. 

89) This is a standalone question that relies on knowing your content. We can pull up the pentose phosphate pathway here and not the results:

We have our pathway above and note NADPH and pentoses are generated. Let’s look through our answer choices.

  1. NADPH, which is used as a reductive agent in cellular respiratory processes. This answer choice is consistent with our breakdown and we can clearly see this is the case in our visual. The visual actually shows three functions of NADPH which is consistent with the reasoning here as well. 
  2. NADPH, which is used as an oxidative agent in cellular respiratory processes. While we like the NADPH listed in this answer choice, the reasoning here is not consistent with what we’re looking for. We keep answer choice A as the superior answer.
  3. NADH, which is used as a reductive agent in cellular respiratory processes. We have to be careful here to not mix NADPH and NADH. While NADPH is used as a reductive agent, this answer choice is wrong because it lists NADH.
  4. NADH, which is used as an oxidative agent in cellular respiratory processes. Similar reasoning as answer choice C. We want to stick with the answer consistent with the pentose phosphate pathway. Answer choice A remains our best answer choice.

90) To answer this question, we can pull up Lineweaver Burk plots for inhibition and note the effect on Vmax.

This question can be answered within a matter of seconds, just by knowing the above content and information. Vmax is altered by options II, III, and IV (noncompetitive, uncompetitive, and mixed inhibition). Answer choice D will be the best answer.

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Passage 12

91) To answer this question, we can reference Experiment 1 and Figure 1.

In Graph A, C75 will decrease cell viability. At a higher concentration, and the longer the exposure, there’s less cell viability.

In Graph B, we see relative MMP decrease at all concentrations of C75. As we go further to the right and longer exposure, we see that relative MMP decrease more and more.

These results could mean these two graphs are related. In fact, it’s likely. Note in Graph B, mitochondrial membrane potential decreases after 2 hours, but cell viability hasn’t decreased yet in Graph A. It’s possible that decrease in mitochondrial membrane potential is what directly results in the decrease in cell viability. 

  1. Changes in the MMP likely affect cell viability. This answer choice is consistent with our breakdown. We said mitochondrial membrane potential decreased after 2 hours of exposure with C75, but cell viability hasn’t decreased yet in Graph A. That meant it’s possible the decrease in mitochondrial membrane potential likely affect cell viability. 
  2. Changes in cell viability likely affect the MMP. This is the opposite of our breakdown. We saw MMP decrease before decrease in cell viability. We already went through the reasoning in answer choice A. We can eliminate answer choice B.
  3. MMP is unrelated to cell viability. This answer choice also contradicts our prediction and what we saw in experiment 1. Both MMP and cell viability decreased after treatment with C75. The change in MMP likely affects cell viability. We can also eliminate answer choice C. 
  4. Cell viability can be restored by decreasing MMP. This answer choice also contradicts what we see in figure 1. When we have decreased MMP, that corresponds to a decrease in cell viability. Cell viability isn’t restored, rather decrease in mitochondrial function adversely affects cell viability. We’ll stick with our correct answer here, answer choice A: Changes in the MMP likely affect cell viability

92) To answer this question, think back to the passage, and specifically Experiment 1. We learned from Experiment 1, cell viability and relative MMP decrease in the presence of C75-including when treated with 50 micromolar of C75. Why are we adding a free radical scavenger? Free radicals are unstable atoms that can damage cells and cause illness and aging.

By adding NAC, the researchers are likely seeing if free radicals play a role in the decline in cell viability and relative MMP. If we don’t see that same decline following the addition of NAC, that could indicate the decline depends on free radicals.

  1. mitochondrial respiration plays a role in mediating the effects of C75. This answer choice is out of scope. The experiment is set up in a way where the researchers see the effect of the presence of free radicals, and then in the absence of possible free radicals. Researchers introduced this new variable to specifically test for its effect. Let’s see if we can find a more direct answer.
  2. reactive oxygen species play a role in mediating the effects of C75. This answer choice matches our breakdown. Reactive oxygen species are radicals that can cause cell damage and negative effects. Like we said in our breakdown, these researchers are just seeing if free radicals play a role in the decline in cell viability and relative MMP. The way that can happen is through introducing NAC. I’m liking this answer choice, but let’s keep comparing.
  3. transport proteins play a role in mediating the effects of C75. This is similar to answer choice A. Why did the researchers add this specific variable to the experiment? It was to test if ROS were involved in the results in Experiment 1. We did a thorough breakdown of Experiment 1, and we know this new variable isn’t added to study transport proteins. We can eliminate answer choice C.
  4. G protein signaling plays a role in mediating the effects of C75. The reasoning here is going to be the same as answer choice C. We have to recognize the importance of adding NAC, and that was to see if free radicals play a role in the decline in cell viability and relative MMP. The focus isn’t G protein signaling. We can eliminate answer choice D as well, it’s out of scope. It’s not addressing the experimental setup given in the question stem. We’re left with our correct answer, answer choice B: reactive oxygen species play a role in mediating the effects of C75

93) To answer this question we’ll decide which options are necessary to validate the results in Figure 2. We’ll pull up Experiment 2 and Figure 2 while looking at these 3 options. 

Left side we have Experiment 2 and Figure 2. Right side we have the three options from which we choose. Quick recap, siRNA is small interfering RNA. It essentially interferes with expression of genes, and prevents translation. That’s why we can knock down the various proteins. 

Option 1 says Evaluating the protein levels of FASN and mtKAS following siRNA treatment. Would this be necessary? Yes, it would. We have to see whether the siRNAs work or not. If protein levels don’t even change and the siRNAs don’t prevent new protein from being made, then siRNAs didn’t affect anything in the experimental setup. Researchers have to include this control experiment.

Option 2 says Evaluating the siRNAs specificity for FASN and mtKAS by assessing the cellular levels of unrelated proteins. This option is also necessary. We have to make sure the siRNAs aren’t affecting anything outside of the specific enzymes we’re focused on. If other proteins are being affected, it’s possible the results of the experiment are unrelated to the silencing of mtKAS or FASN.

Option 3 says Evaluating the siRNAs specificity by showing that a non-specific siRNA has no effects on the protein levels of FASN and mtKAS. This option is also necessary. We have to make sure the specific siRNAs we’re using are causing the effects on protein levels, and it’s not just any siRNA. It’s possible protein expression could be reduced because of any siRNA. We have to perform option 3 to make sure non-specific siRNA isn’t reducing the expression of our two enzymes. 

That means our best guess is all 3 options. We can compare all of our 4 answer choices at once. We’re left with our correct answer, answer choice D: I, II and III.

94) To answer this question, we want to know which reactions would be affected by C75 treatment. C75 is a substrate analog. We know it affects FASN, and also inhibits mtKAS. We actually get results showing the effects of mtKAS in Experiment 1. mtKAS is a mitochondrial enzyme and involved in the synthesis of lipoic acid, which is a cofactor for several enzyme systems. Knowing reaction cofactors isn’t exactly high-yield material, but that’s what this question is boiling down to: either knowing the locations of each answer choice, or by knowing the cofactors. 

  1. fructose 6-phosphate to fructose 1,6-bisphosphate. This answer choice we can associate with glycolysis, which a metabolic pathway in the cytosol. The reaction is catalyzed by phosphofructokinase, but there’s no need for the lipoic acid cofactor. 
  2. fumarate to malate. We can associate this answer choice with the citric acid cycle. The reaction’s catalyzed by fumarase and we associate it with the mitochondrial matrix. No lipoic acid cofactor here either, so we can keep comparing. 
  3. pyruvate to acetyl-CoA. We can associate this answer choice with acetyl-CoA formation, and this is the first reaction where lipoic acid is also a cofactor. It’s a cofactor for pyruvate dehydrogenase, meaning C75 would ultimately prevent this reaction from proceeding at its normal rate. We can eliminate answer choices A and B, those contradicted what the author mentions in the passage about lipoic acid.
  4. oxaloacetate to phosphoenolpyruvate (or PEP). We can associate this answer choice with gluconeogenesis. Oxaloacetate is decarboxylated and then phosphorylated to form phosphoenolpyruvate using the enzyme PEPCK. Another reaction where we don’t have lipoic acid as a cofactor. We can eliminate answer choice D. We’re left with our correct answer, answer choice C: pyruvate to acetyl-CoA.

95) In the passage we read about the deletion of the gene encoding for mtKAS. Let’s revisit that excerpt and jump into our choices.

First sentence here says: The deletion of the gene encoding for mtKAS results in dissipation of the electrochemical gradient in mitochondria due to an increase in membrane permeability. In other words, deletion of the gene encoding for mtKAS increased membrane permeability. Increased permeability means the proton gradient won’t exist in the same capacity.

In the electron transport chain, electrons are passed from one molecule to another as a series of redox reactions. The energetically “downhill” movement of electrons through the chain causes pumping of protons into the intermembrane space by the first, third, and fourth complexes. If we have increased permeability, protons can move back into the matrix. Again, meaning the gradient dissipates.

  1. protons in the mitochondrial matrix. This is the opposite of our breakdown. We expect protons in the mitochondrial matrix will increase, not decrease. We have increased permeability, which allows for protons to freely move back into the matrix.
  2. protons in the mitochondrial intermembrane space. This is consistent with what we’re looking for in an answer choice. Normally protons would be pumped into the mitochondrial intermembrane space. They wouldn’t be able to freely move back into the matrix down their gradient. The increased permeability allows for protons to leave the intermembrane space, meaning we have a decrease in protons. We can eliminate answer choice A because it contradicted information in the passage.
  3. electrons in the mitochondrial matrix. This answer is out of scope. The movement of electrons isn’t related to the increased permeability or the proton gradient we just talked about. Answer choice B remains superior.
  4. electrons in the mitochondrial intermembrane space. Similar to answer choice C here. Movement of electrons isn’t related to that increased permeability, or the proton gradient. We can also eliminate answer choice D for being out of scope. We’re left with our correct answer, answer choice B: decreased level of protons in the mitochondrial intermembrane space

96) To answer this question, we’re going to have to explain whether there’s some upregulation or positive feedback here. Possibly some negative feedback if there’s too much lipoic acid formed. We’ll study the graphs in our question stem, then use our knowledge of feedback to classify the relationship. 

Top graph shows mRNA level, bottom shows protein level. After increasing concentration of lipoic acid, we have an increase in relative mtKAS mRNA levels. That’s indicative of positive regulation, or positive feedback. 

  1. negative feedback at the level of transcription. This answer choice would only be correct if we had a decrease in relative mtKAS mRNA levels. We’re also asked about transcription. Even though that wasn’t part of our breakdown, we’re not going to panic. A decrease at the mRNA level would indicate negative feedback at the level of transcription. 
  2. negative feedback at the level of translation. This answer choice would be correct if we had a decrease in relative mtKAS protein levels, but no concurrent change in transcription. We don’t have a decrease in protein levels, so this answer choice contradicts the graph in the question stem.
  3. positive feedback at the level of transcription. This is consistent with our breakdown-we said we see positive feedback. The key to the second part of this answer choice is the mRNA graph. We have an increase in mRNA level in the top graph. That coincides with an increase in protein level in the bottom graph. Why is that? Because we have positive feedback at the level of transcription. There’s more RNA being transcribed, and also more translation happening. Ultimately, we know the feedback is at the level of transcription, because of the increase in mRNA. We can eliminate answer choices A and B. Both incorrectly stated there’s negative feedback.
  4. positive feedback at the level of translation. This answer choice would be correct if we had the same protein level graph, but no increase in mRNA levels. That would mean we have positive feedback at the level of translation. The way this is set up currently, we have feedback at the level of transcription. More mRNA being transcribed, and subsequently more translation and more protein. We can eliminate answer choice D for contradicting the question stem. We’re left with our correct answer, answer choice C: positive feedback at the level of transcription

 

Section Bank: Biological and Biochemical Foundations of Living Systems: Questions 97-100

97) To answer this standalone question, we have to rely on external knowledge. The author tells us we’re looking at reactions in the citric acid cycle. On the left we have succinate which is easy to identify because it a symmetric molecule. On the right side we’re looking at malate. In the citric acid cycle, succinate undergoes an oxidation reaction and uses FAD as a prothetic group. Enzyme is succinate dehydrogenase. Product of this reaction is intermediate X, or fumarate.

Fumarate undergoes a hydration reaction, enzyme is fumarase, and product is malate. We want an answer choice consistent with malate being intermediate X. Looking at our answer choices, only answer choice C matches our breakdown.

98) This is going to be similar to Question 97. We’re relying on external knowledge of the pentose phosphate pathway.

The enzyme that catalyzes the production of 6-phosphogluconolactone can be seen in the image above: Glucose 6-phosphate dehydrogenase. The only answer choice that matches this answer choice is answer choice A. Make sure to review the image above if you are not confident in knowing the pathway thoroughly.

99) For questions like these, we always want to be careful with which strand we’re looking at and which strand we’re looking for. It’s easy to mix up coding strands, primary strands, or complementary strands, but it’s imperative not to. The template strand is the strand from which the sequence of DNA is copied during the synthesis of mRNA. The coding strand is the opposite strand, so it is identical to the base sequence of the RNA transcript produced, but with uracil. Whenever we’re looking at a gene, one strand is the coding strand, while the other strand is the template strand. For this question, we’re asked about the mRNA.

  1. 5′–GACAUGGACUCGCUA–3′  This answer choice matches the breakdown of our question. The coding strand and mRNA are identical, but we replace any T with U. We like this answer choice for the time being.
  2. 5′–CUGUACCUGAGCGUA–3′ This answer choice incorrectly lists the complementary strand to our correct answer backwards. If we look at this strand backwards (from 3’ to 5’) then we get answer choice C.
  3. 5′–UAGCGAGUCCAUGUC–3′ This answer choice incorrectly lists the complementary strand to our correct answer.
  4. 5′–AUCGCUCAGGUACAG–3′ If this answer choice were written from 3’ to 5’ it could be correct. However, this is written in the wrong direction. Answer choice A remains the best option.

100) Gene expression is the process by which information from a gene is used to be converted into a functional product, such as a protein, and this can be analyzed by using a variety of methods. We want to determine which of the four answer choices cannot be used to analyze gene expression. We can think about what we’re looking for specifically by thinking back to the central dogma of molecular biology: DNA is transcribed into RNA, which is translated into protein. We’re looking for if a gene has been expressed, so we can look at RNA and/or protein.

  1. Western blotting. Western blots use protein-specific antibodies to recognize proteins of interest. Analyzing the protein allows researchers to detect a gene is expressed. We want an answer choice that lists a technique that cannot be used to analyze gene expression. Not a strong answer.
  2. Northern blotting. Northern blotting allows researchers to detect RNA sequences in a sample. Once again, this allows you to analyze gene expression, so not a good answer to this specific question.
  3. Southern blotting. Southern blotting allows for detection of a specific DNA sequence in a sample. Analyzing DNA may show the presence of a gene, but not expression which we’re looking for. This is our best answer choice so far.
  4. Reverse transcription PCR. This is helpful when using mRNA to synthesize a complementary DNA strand. PCR is used to amplify that synthesized DNA. In this case we’re using RNA to make this DNA, so by analyzing this specific DNA, you can analyze gene expression. Answer choice C remains our best answer because Southern blotting cannot be used to analyze gene expression.


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