Exam 4 B/B Solutions: Passage 1

1) This is a passage-related question that functions almost like a pseudo-discrete. The author introduces cytochrome P450 enzymes in Paragraph 3 and we get some background information:

Cytochrome p450 enzymes appear a few times in AAMC’s practice material but it’s not explicitly listed on the content outline. They are membrane-bound proteins that enable oxidation reactions. We want an answer choice that deals with redox reactions. When the body detects a new drug/toxin or increased concentrations of a drug, the body increases the concentration of cytochrome P450 to metabolize the drug effectively. Cytochrome P450 alters the activity of drugs by acting as an oxidizing agent to allow drugs and toxins to be metabolized.

1. phosphorylating them. We mentioned in our breakdown that we’re looking for an answer choice that deals with redox reactions. Typically, kinases will be the ones phosphorylating proteins in the body.
2. dephosphorylating them. We mentioned in our breakdown that we’re looking for an answer choice that deals with redox reactions. This answer choice is describing phosphatases which function to dephosphorylate.
3. oxidizing them. This answer choice matches our breakdown of the question and the cytochrome P450 enzyme background information we went over. Cytochrome P450 enzymes function as monooxygenases. When the body detects a new drug/toxin or increased concentrations of a drug, the body increases the concentration of cytochrome P450 to metabolize the drug effectively. Cytochrome P450 alters the activity of drugs by acting as an oxidizing agent to allow drugs and toxins to be metabolized. This is our best answer choice.
4. reducing them. While cytochrome P450 itself is reduced, the enzymes function to alter the activity of drugs by oxidizing them. This is going to be the opposite function of what we’re looking for. Answer choice C is our best answer choice.

2) This is a passage-based question and we can focus specifically on Paragraph 2 where the author talks about phosphorylation of STAT5b proteins:

The author specifically tells us JAK2 is a tyrosine kinase. A kinase is an enzyme that catalyzes the transfer of phosphate groups, and the tyrosine kinase receptor specifically is an example of a receptor enzyme. The tyrosine kinase receptor transfers phosphate groups from ATP to tyrosine molecules. Our specific question here wants to know what atoms on tyrosine are exchanged for this phosphate group on ATP. We can recall the structure of tyrosine:

Right away, we can eliminate answer choices B and C because tyrosine has a nucleophilic hydroxyl group we see highlighted above in blue. That hydroxyl group attacks the gamma phosphate group:

The hydrogen atom from that hydroxyl group ultimately is exchanged for the phosphate group on ATP. The only answer choice that matches this is answer choice A.

3) This is a pseudo-standalone question, but touches on something we covered in a previous question: P450 enzymes. When the body detects a new drug/toxin or increased concentrations of a drug, the body increases the concentration of cytochrome P450 to metabolize the drug effectively. In this case, we’re dealing with an antibiotic that is metabolized by estrogen-sensitive P450 enzymes within the liver. Where do we expect estrogen to be more prevalent: in males or females? Females have higher levels of estrogen. Estrogen is a hormone responsible for the appearance of secondary sex characteristics of human females at puberty, and the maturation and maintenance of the reproductive organs in their functional state. Erythromycin being metabolized by estrogen-sensitive P450 enzymes within the liver would therefore have a shorter half-life in adult females.

1. This visual correctly shows the shorter half-life in adult females relative to the half-life in males. Erythromycin is metabolized by estrogen-sensitive P450 enzymes in the liver, so females (higher estrogen levels) will metabolize erythromycin more quickly.
2. This answer choice incorrectly shows both graphs having equal relative half-lives. We expect the half-life of erythromycin in adult females to be shorter than the half-life of erythromycin in males. Answer choice A remains the best option.
3. Reasoning here is going to be similar to answer choice A. This answer choice incorrectly shows both graphs having equal relative half-lives. The only different between this answer choice and answer choice B is the size of the bars in the graph, but with no axis on the graph, these are essentially the same answer. We can eliminate these two answers for saying the same thing. We expect the half-life of erythromycin in adult females to be shorter than the half-life of erythromycin in males. Answer choice A remains the best option
4. This answer choice incorrectly shows relative half-life as higher in females although we expect the half-life of erythromycin in adult females to be shorter than the half-life of erythromycin in males. This is the opposite of what we’re looking for in an answer choice. Answer choice A is going to be our best option.

4) While this question is tangentially related to a topic we talked about in the passage, we’re ultimately using our general knowledge to answer this question. We’re asked about transcription factors, so we can go over some background information before jumping into our answer choices. A transcription factor, like STAT5b, is a protein that controls the rate of transcription of genetic information from DNA. Transcription factors bind to the promoter region and then help recruit the appropriate polymerase to begin transcription.

The completed assembly of transcription factors and RNA polymerase bind to the promoter, forming a transcription pre-initiation complex. Transcriptional repressors bind to promoter or enhancer regions and block transcription whereas the transcriptional activators promote transcription. Enhancer regions are binding sequences, or sites, for transcription factors. When a DNA-bending protein binds to the enhancer, the shape of the DNA changes, which allows interactions between the activators and transcription factors to occur.

1. can dimerize. While this is possible in transcription factors, we’re looking for a defining characteristic. We’re looking for something we mentioned in our breakdown of the question about transcription factors. Ideally something consistent with how transcription factors control the rate of transcription of genetic information from DNA.
2. can phosphorylate other proteins. This answer choice is thrown in as a distractor because we talked about kinases in the passage. This is not a defining characteristic of transcription factors.
3. contain a DNA binding domain. This answer choice is consistent with our breakdown of the question. Transcription factors contain a DNA binding domain and through this binding, they can control the rate of transcription of genetic information from DNA. This is going to be our best answer choice.
4. are present within the nucleus of the cell. This is another answer choice that does not answer the specific question being asked. We’re asked for a defining characteristic, but this is something that can be applicable to more than just transcription factors. We can stick with answer choice C as our best answer choice.

 

Exam 4 B/B Solutions: Passage 2

5) This is a passage-based question initially, but we will ultimately use our general knowledge to pick the correct answer. We can go back to Paragraph 1 from the passage to note where the author introduces ACC2 and provides basic background information:

ACC2 specifically regulates fatty acid oxidation. β-oxidation is the catabolic breakdown of fatty acids to produce energy and takes place in the matrix of the mitochondria. The only answer choice that matches this location is answer choice A. ACC2 is most likely compartmentalized to the mitochondria where β-oxidation takes place.

6) This question has a similar setup to our previous question because we will be referencing the passage for some key information, but we’ll ultimately rely on our general knowledge to come up with our final answer.

We can start with Acc2 -/-. We’re told in the passage ACC2 regulates fatty acid oxidation through the conversion of acetyl CoA to malonyl-CoA (shown in Reaction 1), so the absence of ACC2 can adversely affect fatty acid oxidation.

Next, we move on to Cpt1 -/-. Carnitine palmitoyltransferase 1 is an integral part of the carnitine shuttle which functions to transport fatty acids that impermeable to the mitochondrial membranes into the mitochondrial matrix. Once in the matrix, β-oxidation takes place. However, the absence of Cpt1 means there are no fatty acids in the mitochondria, so oxidation is not happening.

1. increased insulin secretion. This answer choice is tangentially related to what we would see, but we likely wouldn’t see increased insulin secretion. We do anticipate seeing increased glucagon as a result of the decreased β-oxidation, but not necessarily insulin. When blood glucose levels decrease, glucagon is released to promote beta-oxidation, not insulin.
2. decreased fatty acid oxidation. This answer choice is consistent with our breakdown of the question. We can focus our attention on CPT1 which we mentioned is an integral part of the carnitine shuttle that functions to transport fatty acids into the matrix. When we don’t have this shuttle, we don’t have the necessary fatty acids in the mitochondria and oxidation is not happening.
3. decreased triglyceride synthesis. This gets into the effect of the absence of ACC2. The author presents Reaction 1 in the passage which we can pull up here:
   Absence of ACC2 means more Acetyl-CoA available for fatty acid synthesis (catalyzed by ACC1). Increased fatty acid synthesis would mean the opposite of this answer choice. We expect increased triglyceride synthesis.
4. increased malonyl-CoA production. This ties into our reasoning for answer choice B. The absence of CPT1 means no carnitine shuttle transporting fatty acids into the mitochondria and no β-oxidation or formation of acetyl CoA. We will likely see decreased malonyl-CoA production.

7) This is a pseudo-standalone question that is tangentially related to the passage. Note, the author gives us Reaction 1 after Paragraph 1 in the passage.

This is a reaction we should know from reviewing fatty acid synthesis.

Specifically, we can focus on Step 1, where acetyl-CoA carboxylase adds a carboxyl group to acetyl CoA in the formation of malonyl-CoA

1. This answer choice incorrectly shows the addition of a phosphate group We want an answer choice that shows the addition of a carboxyl group.
2. This answer choice incorrectly shows the addition of a hydroxyl group. We want an answer choice that shows the addition of a carboxyl group.
3. This answer choice is consistent with our breakdown of the question; acetyl-CoA carboxylase adds a carboxyl group to acetyl CoA in the formation of malonyl-CoA. As we saw in our visuals, this is going to be our correct answer choice.
4. This answer choice incorrectly shows the addition of a methyl group. We want an answer choice that shows the addition of a carboxyl group. Answer choice C is our best option.

8) This is a passage-based question that relies on revisiting the passage for key information. I don’t always advocate going back to the passage, but for specific details like we need in this question, it’s better to flip back for a few seconds than try and memorize every detail in your first readthrough. The author tells us in Paragraph 3 of the passage, “AMPK-mediated phosphorylation of mouse ACC1 (at residue 79) and ACC2 (at residue 212) inactivates these enzymes.” Phosphorylation inactivates ACC2 at residue 212, but we want a strain that expresses a constantly active variant. Key thing we want to point out is we have phosphorylation. Phosphorylation is a form of protein modification and regulation. We want to substitute with an amino acid that cannot be phosphorylated so we have a constantly active variant. Which amino acids are phosphorylated? For the sake of the MCAT: serine and threonine. Sometimes we include tyrosine.

1. We mentioned in our breakdown of the question, tyrosine can be phosphorylated.
2. We mentioned in our breakdown of the question, serine can be phosphorylated
3. This is going to be our best answer choice. Phosphorylation inactivates ACC2 at residue 212, but we want a strain that expresses a constantly active variant. We want to substitute with an amino acid that cannot be phosphorylated so we have a constantly active variant. We can substitute with alanine. This is going to be our best answer choice.
4. We mentioned in our breakdown of the question, threonine can be phosphorylated. Answer choice C remains our best option.

 

Exam 4 B/B Solutions: Questions 9-12

9) This is a standalone question that is testing our knowledge of the makeup of different molecules. This is a fairly broad and open-ended question so we can quickly glance at our answer choices. Our four options show we’re going to be distinguishing between DNA and proteins, and deciding which contain phosphorous and which contain sulfur. This sounds like a variation of the Hershey-Chase experiment. The Hershey-Chase experiment proved DNA is genetic material by using the isotopic marker ³²P, which labels DNA. They infected a bacterium where it replicated the ³²P as genetic material demonstrating DNA is genetic material.

Why do we get these specific results? Because DNA helices contain sugar-phosphates (and not sulfur) and enters the cell:

Alternatively, proteins can form disulfide bonds, but do not enter the cell (only outside the host):

The only answer choice consistent with our breakdown is answer choice C: only protein, not DNA, can enter the host cell.

10) This is a standalone question that relies on using our general knowledge. While it’s always a good idea to review and know biochemistry pathways, it’s more common that we see big picture questions like this than questions about specific details of these pathways. That being said, knowing the entire pathway obviously provides us with the necessary information to accurately answer this question. We want to split up the number of ATP into ATP consumption and production in the preparatory and payoff phases of glycolysis:

During the preparatory phase (phosphorylation of glycose and conversion to glyceraldehyde 3-phosphate) We note 2 ATP molecules are consumed: 1 molecule each in Step 1 and Step 3. No ATP molecules are produced.

During the payoff phase, we note 4 ATP molecules are produced: 2 molecules ATP each in Step 7 and Step 10. The only answer choice that matches our breakdown and these numbers is answer choice A: Two ATP molecules consumed, four ATP molecules produced.

11) This is a standalone question that asks about actively contracting skeletal muscle tissue. We can think about what this contraction entails. Skeletal muscle mainly attaches to the skeletal system via tendons to maintain posture and control movement. For example, contraction of the biceps muscle, attached to the scapula and radius, will raise the forearm. Think of going to the gym and lifting weights-that involves actively contracting skeletal muscle tissue. The first set of an exercise might be easier, but muscle use can quickly overwhelm the ability of the body to deliver oxygen. Muscle fibers must switch to anaerobic metabolism and produce lactic acid, at which point the muscle begins to fatigue. Anaerobic respiration therefore causes decreased pH and reduced affinity of hemoglobin for oxygen. Additionally, that increased temperature, such as from increased activity of the skeletal muscle, causes the affinity of hemoglobin for oxygen to be reduced. We want an answer choice consistent with this breakdown of actively contracting skeletal muscle tissue.

1. increases as a result of an increase in plasma temperature. While it is true we expect increased temperature, we said we expect a decrease in the affinity of hemoglobin for oxygen in the muscle tissue.
2. increases as a result of an increase in plasma PO₂. This answer choice is the opposite of our breakdown. We expect affinity to decrease, and we actually have to use anaerobic respiration. This is going to be our worst answer choice thus far.
3. decreases as a result of a decrease in plasma pH. This answer choice is consistent with our breakdown, and is factually correct, unlike answer choices A and B. Note, answer choice A mentioned that increase in temperature, but incorrectly stated that affinity would increase. We always want to pick the best answer, and an answer that is factually correct. Answer choice C is going to be our best answer choice.
4. decreases as a result of a decrease in plasma PCO₂. First part of this answer choice is consistent with our breakdown of the question. We do expect affinity will decrease, but not because of a decrease in plasma PCO₂. This effect would be seen in the opposite carbon dioxide environment. As the level of carbon dioxide in the blood increases, more H+ is produced, and the pH decreases. The increase in carbon dioxide and subsequent decrease in pH reduce the affinity of hemoglobin for oxygen. Answer choice C remains our best option.

12) This is a standalone question that asks about our knowledge of Henry’s Law. We can do a quick breakdown here before jumping into the answer choices.

Henry’s law states that the concentration of gas in a liquid is directly proportional to the solubility and partial pressure of that gas.

C= kH x P where C is the concentration of our dissolved gas, kH is the constant (which we’re interested in finding here), and P is the partial pressure of the gas. The higher the partial pressure of the gas, the higher the number of gas molecules that will dissolve in the liquid. The concentration of the gas in a fluid is also dependent on the solubility of the gas in the liquid. In this specific case, we already know [CO₂], but we do not know P, the partial pressure of our gas.

1. Atmospheric pressure. We have to be careful here and not immediately pick this answer because it mentions pressure. While we are looking for an answer that deals with pressure, we’re focused on the partial pressure of a gas, not atmospheric pressure.
2. Volume of the solvent. If we didn’t know the concentration of CO₂, this might be a viable answer, but this volume is not necessary in this case to calculate the Henry’s Law constant.
3. Partial pressure of the gas. This answer choice is consistent with our breakdown. We mentioned we already know concentration and we’re trying to find the Henry’s Law constant. The only missing piece is this partial pressure. This is going to be our best answer choice.
4. Vapor pressure of water. This answer choice is similar to answer choice A. We’re not going to get tricked and pick this answer choice simply because it mentions pressure. Instead, we can focus on our best answer, answer choice C: Partial pressure of the gas.

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