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MCAT Content / AAMC MCAT Practice Exam 4 Bb Solutions

AAMC FL4 BB [Web]

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.

 

Exam 4 B/B Solutions: Passage 3

13) While this question is tangentially related to the information in the passage (the entire second half of the passage focuses on ghrelin to an extent), this answer is ultimately going to come from our general knowledge. The author reiterates something we should know from our content review: ghrelin is a hormone secreted in the stomach that acts on the hypothalamus and promotes feelings of hunger and also regulates body weight. The way I learned it, is “ghrelin” sounds a little like “growl,” which happens when we’re hungry, so I will never forget the connection. It might sound silly, but find the best way to get these little things to stick for you as well! While there may be a difference in ghrelin sensitivity in individuals that are obese and those that are lean, we expect higher ghrelin secretion in lean individuals (less energy store, more “growling”), and lower ghrelin secretion in obese individuals. This is also consistent with what we see in Figure 1 where the lean individual has higher levels of circulating ghrelin than their obese counterpart with no underlying condition.

  1. The hypothalamus secretes less ghrelin in individuals who are lean because their metabolic needs are greater than those of individuals who are obese. Recall from our breakdown of ghrelin, ghrelin is a hormone secreted in the stomach that acts on the hypothalamus. In fact, the author actually mentions this in the passage as well in Paragraph 4. We saw in Figure 1, lean individuals secrete more ghrelin, and they will have lower metabolic needs. This is a factually incorrect answer choice.
  2. Ghrelin secretion is suppressed in individuals who are obese because their bodies have greater energy stores than do the bodies of individuals who are lean. This answer choice is consistent with what we said in our breakdown of the question and what we saw in Figure 1. We have reduced ghrelin levels in obese individuals compared to their lean counterparts because the obese individuals have greater energy stores and less need to trigger the hunger response (Note this question is focused on ghrelin levels like we see in Figure 1, and not sensitivity which could change our reasoning). This is our best answer choice.
  3. The hypothalamus secretes more ghrelin in individuals who are lean because their metabolic needs are less than those of individuals who are obese. The reasoning for this answer choice is factually correct, but we have to note a key detail that we touched on in our breakdown of the question and our explanation for answer choice A: ghrelin is a hormone secreted in the stomach that acts on the hypothalamus. This answer choice is factually incorrect. Answer choice B remains our best option.
  4. Ghrelin secretion is suppressed in individuals who are obese because their bodies have fewer energy stores than do the bodies of individuals who are lean. First part of this answer choice starts out great, however, individuals who are obese have greater energy stores than do individuals who are lean. This answer choice is factually incorrect. We can stick with answer choice B as our best answer choice.

14) This question is similar to the previous question in this set in that we’re ultimately going to rely on our general knowledge to answer, but we still need to pick out some key information from the passage to make sure we’re answering this specific question correctly. Note, the test-maker says “Based on the passage” which doesn’t always mean we rely exclusively on the passage, but we likely pick out some key details.

First thing we want to note is the author talks about the makeup of ghrelin in Paragraph 4. We’re told ghrelin is a 28-amino-acid peptide hormone that’s actually cleaved from a longer protein. However, we’re focused on mRNA, which means we’re talking nucleotides. Each amino acid of the 28 is coded by 3 nucleotides. However, we can’t forget one major detail: the STOP codon! That means at minimum we’re looking at:

(28 amino acids x 3 nucleotides/amino acid) + (3 nucleotides STOP codon) = 87 nucleotides.

Why do I say minimum? Because the author explicitly tells us ghrelin is cleaved from a longer peptide chain. 87 is going to be the very least number of nucleotides, when in reality that number is going to be higher. Let’s glance at our answer choice:

  1. more than 28 amino acids. The test-maker mentions we’re talking about the composition of mRNA so we’re dealing in nucleotides, not full amino acids.
  2. exactly 84 nucleotides. This answer choice incorrectly forgets to include the 3 nucleotides from the STOP codon, and also disregards the fact the ghrelin is cleaved from a longer peptide chain.
  3. exactly 87 nucleotides. At first glance this may sound like a viable answer, but we remember that key detail from the passage: ghrelin is cleaved from a longer peptide chain. We expect the actual number to be higher than 87, not exactly 87.
  4. more than 87 nucleotides. This answer choice is consistent with our breakdown. We know the number of nucleotides has to include the STOP codon, so we’re looking at 87 nucleotides just from the cleaved portion, but ghrelin is cleaved from a longer peptide chain meaning we have more than 87 nucleotides. Answer choice D is our best answer choice.

15) At first glance, this question is in line with what we’ve had to do so far in the question set. We’ll likely go back to the passage for specific information, but ultimately, we will rely on our general knowledge to get to our final answer. The author tells us a few key characteristics of Prader-Willi syndrome including GH deficiency, cognitive disabilities, and (once they reach the hyperphagic stage) excessive eating. Hormone replacement therapy should, therefore, address the hormone deficiency the author mentions, which is GH deficiency. Growth hormone stimulates growth, cell reproduction and regeneration. Hormone replacement therapy would be used to make sure this still happens in the children who have Prader-Willi syndrome.

  1. giantism. This is the opposite of what we would expect. Hormone replacement (additional GH) would be used to encourage growth, cell reproduction, and cell regeneration. Hormone replacement therapy would encourage growth, not prevent giantism.
  2. weight loss. While this could be a possible issue when younger, we know these children have reached the hyperphagic stage. The author mentions that corresponds to excessive eating. This is not a likely answer choice.
  3. short stature. This answer choice is consistent with our breakdown. Growth hormone stimulates growth, cell reproduction and regeneration. By using hormone replacement therapy, we can ensure the child is still able to grow and we avoid short stature. This is going to be our best answer choice so far.
  4. muscle rigidity. This is going to be similar to our reasoning for answer choice B. this could be a potential issue in younger children, but we’re asked about children who have Prader-Willi syndrome and have reached the hyperphagic stage. At this stage we have excessive eating and the lack or degeneration of muscle is not a concern. Answer choice C is our best answer choice.

16) Similar to the questions we’ve come across in this set, we’re going to use a hybrid of information from the passage and our general knowledge. The passage focused on ghrelin and the author explains that ghrelin acts on the hypothalamus to increase appetite and regulate body weight. Now we want to jump a bit more into the mechanism by which this happens and the conditions that would suppress ghrelin secretion. Ghrelin secretion isn’t simply controlled by the presence of food in the stomach or distension. Let’s consider what we know about hunger and regulating metabolic processes. Hormones primarily involved in body mass regulation that decrease food intake and increase energy expenditure are insulin and leptin. Insulin stimulates the production of leptin by adipose tissue that decreases appetite and gives the feeling of satiety (opposite feeling of ghrelin). Leptin then decreases the secretion of insulin and enhances tissue sensitivity to insulin, leading to glucose uptake for energy utilization or storage. When and why do we expect suppressed ghrelin secretion? Increased insulin (feeling of satiety) or decreased glucagon.

  1. Increased levels of circulating glucagon. This is the opposite of what we expect to suppress ghrelin secretion following a meal. Following a meal, we have decreased levels of circulating glucagon and increased levels of circulating insulin.
  2. Decreased osmolarity in the ileum. This is similar to answer choice A because we have to consider what happens following a meal. The ileum is the final section of the small intestine and functions mainly to absorb vitamin B12, bile salts, and any products of digestion that were not absorbed by the jejunum. We expect increased osmolarity following a meal.
  3. Increased levels of circulating insulin. This answer choice is consistent with our breakdown of the question and our general knowledge. Following a meal, we expect increased levels of circulating insulin. That means a feeling of satiety and suppression of ghrelin (which brings feelings of hunger). This is going to be our best answer choice.
  4. Decreased levels of circulating insulin. This is essentially the same thing as answer choice A. We expect the opposite of answer choices A and D to be true. Following a meal, we will have decreased levels of circulating glucagon and increased levels of circulating insulin. Answer choice C remains our best answer choice.

17) Just like the other questions in our question set, this answer choice relies a bit on what we read in the passage, but we can ultimately use our general knowledge to pick the correct answer. We have an individual that initially gains weight, then loses a significant amount of weight. We can think about the fasting ghrelin levels while considering what we know from Figure 1 in the passage. Figure 1 showed that lean individuals that are fasting have higher levels of circulating ghrelin than do obese individuals that are fasting. We expect the individual in the question stem (when they first gain weight) will have lower levels of circulating ghrelin. However, as they lose the significant amount of weight, we expect the individual will have higher levels of circulating ghrelin.

  1. increase as the individual gained weight, then decrease as the individual lost weight. This is the opposite of what we expect. We can reference Figure 1 again to think about what we actually expect in the situation described in the question stem. Figure 1 showed that lean individuals that are fasting have higher levels of circulating ghrelin than do obese individuals that are fasting. That means both parts of this answer choice are incorrect.
  2. decrease as the individual gained weight, then increase as the individual lost weight. This is exactly what we said in our breakdown of the question and what we saw in Figure 1. Figure 1 showed that lean individuals that are fasting have higher levels of circulating ghrelin than do obese individuals that are fasting. We expect the individual in the question stem (when they first gain weight) will have lower levels of circulating ghrelin. However, as they lose the significant amount of weight, we expect the individual will have higher levels of circulating ghrelin. This is our best answer choice.
  3. increase as the individual gained weight and remain high as the individual lost weight. This goes against what we said in our breakdown and what we saw in Figure 1. We see different levels of circulating ghrelin in lean vs obese individuals. We don’t expect ghrelin levels to remain similar.
  4. decrease as the individual gained weight and remain low as the individual lost weight. Reasoning here is similar to answer choice C. This goes against what we said in our breakdown and what we saw in Figure 1. We see different levels of circulating ghrelin in lean vs obese individuals. We don’t expect ghrelin levels to remain similar. We can stick with answer choice B as our best option.

 

Exam 4 B/B Solutions: Passage 4

18) This is a passage-based question and the test-maker explicitly asks “based on the passage.” A few things we want to notice right away is we’re looking for a subunit that is LEAST suitable for generation of a vaccine. Three of our answer choices will be suitable, while our correct answer is not suitable for generation of a vaccine. Additionally, we grouped the subunits into an A component and B component. Three of our answer choices are associated with the B component, and only one answer choice is associated with the A component. While this isn’t an automatic reason to pick answer choice A, we do want to note that is the one answer choice that is not part of the B component. What we want to note is which of the subunits would produce an immune response (without an immune response we don’t have an effective vaccine), but also be safe for use in humans. We’re told the S1 subunit is actually inserted into the cytoplasm, while the B component simply binds to cell surfaces. The ADP ribosyl transferase (S1 subunit) is toxic, while the B component (contains S2, S3, S4, and S5 subunits) subunits would be safe for use. This confirms our initial instincts that we’ll end up picking the option that is not grouped in with the others. We can pick the only answer choice that is not part of the B component, answer choice A: S1.

19) This is a passage-based question, but we’ll also use our general knowledge and what we know about ribosomes. Ribosomes are macromolecular structures composed of rRNA and polypeptide chains. They are formed of two subunits (in bacteria and archaea, 30S and 50S; in eukaryotes, 40S and 60S), that bring together mRNA and tRNAs to catalyze protein synthesis. The author tells us in the passage that treatment of pertussis includes erythromycin which we’re focused on in this question. Erythromycin inhibits bacterial protein synthesis by binding to the 23S rRNa component of the large subunit of the bacterial ribosome. The large subunit is the 50S subunit in bacteria.

  1. 50S. This answer choice matches our breakdown of the question and what we see in our visual. The large subunit is the 50S subunit in bacteria.
  2. 60S. The 60S subunit is the large subunit in eukaryotes and not relevant to answering this question.
  3. 70S. Bacteria have 70S ribosomes that are made of the 50S and 30S subunit. The author tells us in the passage erythromycin binds to the 23S rRNA component of the large subunit (50S) of the bacterial ribosome.
  4. 80S. Eukaryotic ribosomes are 80S ribosomes. This is similar to answer choice B as it is not relevant to the question being asked. We can stick with answer choice A as our best option.

20) This is a classic AAMC test question that relies on our knowledge of amino acids, their structures, and their properties. The author tells us amino acid D is at position 54.

We essentially want to know which amino acid would least prevent expression of the PTx operon, or disrupt the existing structure the least. Aspartate has a negatively charged R-group which is the same as glutamate. Theoretically, substituting another amino acid with a negatively charged R-group would least likely prevent the expression of the PTx operon.

  1. Glycine. Glycine has the simplest R-group with a single hydrogen atom. Going from a negative charge to a neutral charge can have an effect.
  2. Asparagine. Asparagine has a positive charged R group, so this is actually the opposite of what we had initially at positive 54. Answer choice A is going to be the best answer choice so far.
  3. Glutamate. This answer choice matches what we said in our breakdown of the question. Note in our amino acid visual: aspartate has a negatively charged R-group which is the same as glutamate. This is going to be our best answer choice.
  4. Alanine. Reasoning here is going to be the same as answer choice A. Going from a negative charge to a neutral charge can have an effect more significant than what we’d get going from aspartate to glutamate. Answer choice C is going to be our correct answer.

21) When assembled for translation, ribosomes have three binding sites that accommodate tRNAs: The A site, the P site, and the E site. Take a look at the diagram below to see how these are arranged relative to each other:

Incoming aminoacyl-tRNAs (a tRNA with an amino acid covalently attached) enter the ribosome at the A site. The peptidyl-tRNA (a tRNA carrying the growing polypeptide chain) is held in the P site. The E site, located right next to the P site, holds empty tRNAs just before they exit the ribosome.

We’re told in the passage that erythromycin inhibits bacterial protein synthesis by binding to the 23S rRNA component of the large subunit of the bacterial ribosome. What does this do? It interferes with aminoacyl transferase function, preventing tRNA transfer between the tRNA sites of the ribosomes. Aminoacyl transferase transfers tRNA bound at the A site to the P site and eventually the E site. The only answer choice that lists this order correctly is answer choice D: the A site to the P site of the ribosome.

 

Exam 4 B/B Solutions: Passage 5

22) This is a passage-related question that relies on us going back to the passage for specific details. Quick glance at our answer choices confirms that we’ll be looking at 4 different compounds either discussed in the passage, or related to a compound discussed in the passage. Keep in mind we’re looking to mimic the effect of exercise in undifferentiated muscle cells in vivo. The author starts the passage by telling us that muscle cells are equipped with inherent capacity to perceive and respond to biochemical stimuli and convert them into molecular events. We’re told later in the passage that researchers simulated an in vitro biomechanical stimulus environment and investigated the effect of CTS on differentiation by measuring MYOD1 in the presence and absence of TNF-α. Essentially, the researchers use CTS to simulate exercise in this case. Muscle cells respond to this biochemical stimulus.

We also know MYOD1 plays a role in regulating muscle differentiation (Paragraph 2) by binding to consensus sequences of the regulatory region of muscle-specific genes like Cdkn1a. Induction of Cdkn1a results in commitment to differentiation of previously undifferentiated cells. Some big takeaways from Figure 1 before we jump into our answer choices: CTS upregulates MYOD1 (which is a positive regulator of muscle cell differentiation), while TNF-α downregulates MYOD1 mRNA levels.

  1. An MYOD1-specific microRNA. miRNA functions to control gene expression through RNA silencing and post-transcriptional regulation. MicroRNA will bind mRNA to mark the mRNA to either be destroyed or recycled. This downregulation of MYOD1 would have the opposite effect of what we’re looking for. Recall we said CTS is our exercise simulator in the passage and upregulates MYOD1 mRNA levels.
  2. An activator of NOS expression. We have to make a few connections from the passage here. The author tells us TNF-α induces the expression of NOS production. However, TNF-α downregulates MYOD1 mRNA levels. Because TNF-α induces the expression of NOS production, this answer choice is going to be the opposite of what we’re looking for.
  3. A repressor of Cdkn1a expression. This is something I touched on in the breakdown of the question. MYOD1 plays a role in regulating muscle differentiation by binding to consensus sequences of the regulatory region of muscle-specific genes like Cdkn1a. Induction of Cdkn1a results in commitment to differentiation of previously undifferentiated cells. A repressor of Cdkn1a would reduce differentiation.
  4. An antagonist of TNF-α signaling. This answer choice is consistent with our breakdown of the question. From Figure 1: CTS upregulates MYOD1 (which is a positive regulator of muscle cell differentiation), while TNF-α downregulates MYOD1 mRNA levels. If we have an antagonist of TNF-α signaling, this means upregulation of MYOD1 (the same as we would see in CTS-the way researchers simulated exercise in the passage).

23) This is a passage-based question that ultimately is going to rely on our external knowledge to answer the question. The biggest thing we need to recall from the passage is the author says NG-nitro-arginine methyl ester (L-NAME) is a competitive inhibitor of NOS activity. At this point this becomes a purely content-based question:

We can find the only answer choice that matches the Lineweaver-Burk plot for competitive inhibition above, and that’s answer choice B.

24) This is a passage-based question, but relies almost exclusively on our general knowledge to answer. The central dogma in biology describes the flow of genetic information from DNA to RNA to proteins. DNA is transcribed into RNA and RNA is translated into a polypeptide chain (protein). For this question, we’re focused on translation, which is the process where RNA is used to create a new protein.

Ideally, because we’re looking into transcriptional regulation, we find an answer that deals with measuring mRNA levels.

  1. Assessing Cdkn1a mRNA levels by RT-PCR. RT-PCR (reverse transcription PCR) functions to measure the amount of a specific RNA in a sample. This is consistent with exactly what we’re looking for.
  2. Assessing Cdkn1a mRNA levels by Southern blotting. Southern blotting is a laboratory technique used to detect a specific DNA sequence in a blood or tissue sample. Answer choice A remains superior.
  3. Assessing Cdkn1a protein levels by quantitative PCR. Quantitative PCR functions to quantify the concentration of DNA species in a reaction or sample. Answer choice A remains superior.
  4. Assessing Cdkn1a protein levels by Western blotting. Western blots use antibodies to recognize proteins that have been separated by gel electrophoresis and transferred to a membrane. We can stick with answer choice A as our best answer choice.

25) This is another passage-based question, but we’ll ultimately rely on our general knowledge to pick the correct answer. We’re focused on the nuclear localization sequence from Paragraph 2 in the passage: NDAFEITKRC. We can find the net charge of the sequence by noting any charges on the amino acids in the sequence.

The sequence has three positively charged amino acids and three negatively charged amino acids giving us a net charge of 0. Only answer choice B is consistent with this answer choice.

26) This is a question that AAMC tends to ask fairly often, and it’s also something I like to answer as I read through the passage, whenever possible. Why is that? Because the conclusion is one of the most important parts of any research. The researchers put in time and effort to answer a question or dive into a problem, and the conclusion is essentially what they are looking to take away from the time and effort put into the experiment. Additionally, we have visual data shown in Figure 1 which means we need to be able to read a graph properly and get to that conclusion. We already touched on this a bit in Question 22, so we can recap what we said there.

The author starts the passage in Paragraph 1 by telling us that muscle cells are equipped with inherent capacity to perceive and respond to biochemical stimuli and convert them into molecular events. We’re told later in the passage that researchers simulated an in vitro biomechanical stimulus environment and investigated the effect of CTS on differentiation by measuring MYOD1 in the presence and absence of TNF-α. Essentially, the researchers use CTS to simulate exercise in this case. Muscle cells respond to this biochemical stimulus.

MYOD1 plays a role in regulating muscle differentiation (Paragraph 2) by binding to consensus sequences of the regulatory region of muscle-specific genes like Cdkn1a. Induction of Cdkn1a results in commitment to differentiation of previously undifferentiated cells. Biggest takeaways from Figure 1: CTS upregulates MYOD1 (which is a positive regulator of muscle cell differentiation), while TNF-α downregulates MYOD1 mRNA levels. TNF-α is a proinflammatory cytokine that induces expression of NOS and NO. We’ll keep these general observations in mind and go through the four answer choices.

  1. a biomechanical stimulus reverses the effects of inflammatory cytokines. Recall researchers used CTS to simulate exercise. CTS upregulates MYOD1 while TNF-α downregulates MYOD1 mRNA levels. TNF-α is a proinflammatory cytokine, so a biochemical stimulus reverses the effects of inflammatory cytokines. This is an accurate answer choice based on the data in Figure 1.
  2. NOS activity reverses the effects of inflammatory cytokines. TNF-α is a proinflammatory cytokine that induces expression of NOS and NO. We can see from columns 1 and 3 in Figure 1 TNF-α downregulates MYOD1 mRNA levels. However, when we have L-NAME as well (column 5) we see MYOD1 mRNA levels increase once again. That means inhibiting NOS activity (via a competitive inhibitor) reverses the effect of TNF-α.
  3. inflammatory cytokines reverse the effects of L-NAME. While we said the opposite is true, we cannot test this conclusion based on Figure 1. To do so, we would need another lane with the presence of L-NAME and no TNF-α to establish a baseline. Answer choice A remains the most viable option.
  4. NOS activity is essential for MYOD1 expression. This is similar to answer choice C. If we were to test this, we would need another lane with the presence of L-NAME and no TNF-α or CTS to establish a baseline. The research and data would have to be expanded. Only answer choice A is a conclusion that can be drawn from the data presented in Figure 1.

27) This question is very similar to Question 25. This is another passage-based question, but we’ll ultimately rely on our general knowledge to pick the correct answer. We’re focused on the nuclear localization sequence from Paragraph 2 in the passage: NDAFEITKRC. Ultimately, we have to know the identities and properties of amino acids to answer this question.

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. The nuclear localization contains a single threonine (T), but no serine (S) or tyrosine (Y). That means there is one phosphorylation site; answer choice B is our correct answer choice.

 

Exam 4 B/B Solutions: Questions 28-31

28) This is a standalone question that relies on our knowledge of different biochemical techniques. The best way to answer this question is to define each of the four answer choices and identify the answer choice that separates proteins independently of their charge.

  1. Native PAGE. Native PAGE is a technique for separating proteins based on both size and conformation. The mobility of proteins in Native PAGE is reliant on both size and intrinsic charge.
  2. Gel filtration chromatography. In gel filtration chromatography, proteins as separated based on size. As the solution travels down the column some smaller particles get stuck in the pores. The larger molecules simply pass by the pores because they are too large to enter the pores. This is exactly what we’re looking for to answer this question.
  3. Ion exchange chromatography. Ion exchange chromatography is a technique used to separate proteins according to their charge. This is the opposite of what we’re looking for in this question. 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.
  4. Isoelectric focusing. Isoelectric focusing is a laboratory technique in which a protein migrates along a membrane until it reaches its pI. a pH gradient is created on top of a membrane. One end of the membrane is connected to a negatively charged cathode, while the other end of the membrane is connected to a positively charged anode. When proteins are dolloped onto the membrane, their baseline charge will determine whether they will be attracted to the cathode or to the anode. We can stick with answer choice B as the technique that separates proteins independently of their charge.

29) This is a standalone question that forces us to visualize and use some basic math. We’re told three different tripeptides were isolated, but none of the tripeptides share a common amino acid. We don’t need to get into specifics here (in fact, we don’t know the specifics of the tripeptides), but we can refer to the tripeptides as A, B, and C. We want to know how many different combinations we can have with A, B, and C without any repeats. If each tripeptide is in a specific position, it can combine with the other two tripeptides in two different ways. For example:

A in the first position means ABC or ACB. B in the first position means BAC or BCA. C in the first position means CAB or CBA. Those are all the possible combinations we can come up with. The total number of possible structures possible for the full-length peptide is 3 x 2 = 6 possible ways. Answer choice C is our best option.

30) This is a standalone question that focuses on our knowledge of mendelian genetics. The key to this question is something we’re told in the question stem: we have an asymptomatic man who is heterozygous for the cystic fibrosis allele. What does that tell us about the cystic fibrosis allele? It’s recessive-the recessive allele is able to be covered up by a dominant trait. Despite having the allele, the man does not have cystic fibrosis. Alternatively, the woman has cystic fibrosis. Considering cystic fibrosis is due to a recessive allele, the woman must have homozygous recessive allele. How do we know that? Because that is the only way she can have cystic fibrosis. We can draw out our Punnet square here. We have a heterozygous father (Cc) and an affected mother (cc):

C c
c Cc cc
c Cc cc

The probability of having a child with cystic fibrosis given a heterozygous father (Cc) and an affected mother (cc) is 50%. The only answer choice that matches this probability is answer choice B: 50%.

31) This is a standalone question that has to do with our knowledge of heterochromatin and X-chromosome inactivation. Heterochromatin is the part of the chromosome in which the DNA does not have coding genes. An example of heterochromatin is the X chromosome inactivation in mammalian females where the X chromosome is packed into transcriptionally inactive heterochromatin. This is because female mammals have two X chromosomes and to prevent duplicates, one of them is inactivated. Because of this, it is one of the last chromosomes to undergo replication in mitosis.

  1. It does not replicate. While one of the last chromosomes to undergo replication, inactive X-chromosomes will still replicate.
  2. Its chromatin structure is less condensed than that of an active X chromosome. This is the opposite of what is true. The heterochromatin region of the chromosome is highly condensed.
  3. It is one of the last chromosomes to replicate. This answer choice is consistent with our breakdown of the question. We said it is one of the last chromosomes to replicate because female mammals have two X chromosomes and to prevent duplicates, one of them is inactivated. This is our best answer choice.
  4. It is highly transcribed. This is the opposite of what is true as well. We said in our breakdown the inactive X chromosome is packed into transcriptionally inactive heterochromatin. Answer choice C best applies to the inactive X chromosome in mammalian females.

 

Exam 4 B/B Solutions: Passage 6

32) This is a passage-based question that relies on us going back to the passage for specific details. We can focus on Reaction 1 from the passage:

We’re asked about increasing the initial concentration of pyruvate which causes an increase in the equilibrium concentration of acetylphosphate. In other words, we’re increasing the initial concentration of a reactant, and this is causing an increase in the concentration of a product. This is a classic example of Le Châtelier’s principle. Le Châtelier’s principle is an observation about chemical equilibria of reactions. It states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state. According to Le Châtelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. That’s exactly what is happening in this situation. By the same logic, reducing the concentration of any product will also shift the equilibrium to the right. The converse is also true. If we add an additional product to a system, the equilibrium will shift to the left, in order to produce more reactants. Or, if we remove reactants from the system, equilibrium will also be shifted to the left.

  1. Reaction spontaneity. While a reaction may be spontaneous or not spontaneous, the change in concentration of the reactants shifting the equilibrium toward the side of the products is an example of Le Châtelier’s principle.
  2. Le Châtelier’s principle. This answer choice matches our breakdown of the question. Le Châtelier’s principle says adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products, just like we see in our question stem.
  3. Heat released by the reaction. This is similar to answer choice A as it is out of scope here. The amount of heat released by the reaction does not explain how increasing concentration of pyruvate causes an increase in the equilibrium concentration of acetylphosphate.
  4. The increase in free energy. We have a spontaneous reaction here which means free energy actually decreases. This is the opposite of what we see in Reaction 1. We can stick with answer choice B as our best option.

33) This is a passage-based question, but relies on our general knowledge. The equilibrium constant is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. If we have an equation aA + bB ⇌ cC + dD, we can write out the equilibrium constant expression:

However, before we write anything for this specific question, what we need to pay attention to is the state of each reactant and product. Pure solids and liquids are not included when writing out this equilibrium constant.

In this case, water is not included when writing out the equilibrium constant because it is a liquid. The equilibrium constant given the format presented above is going to be:

[Acetylphosphate][CO2][H2O2] / [Pyruvate][Pi][O2]

The only answer choice that matches this constant is answer choice A.

34) While this is another passage-based question, the key to answering this question is understanding the role of a negative control. A negative control is included in an experiment, but is not expected to change because of the experiment itself or influence the results of the experiment. It functions like a baseline and is not subject to the treatment of the experiment. By using a negative control, researchers can look for confounding variables and get more clarity about positive results. For example, if we’re experimenting with a treatment for patients with a specific illness, the negative control would be used to see what happens if no treatment is administered. Once we have that baseline, we can note any changes that occur because of the positive control.

In the case of the experiment in the passage, we can focus on AECs which are incubated in media and necessary to detect 53BP1 and γH2AX, regardless of strain. Note in Figure 1 we have the negative control where we still note the expression of 53BP1 and γH2AX in the AECs. That gives us a baseline and shows what happens if cell media is not inoculated with one of the three virulent strains-or in unconditioned media. We still want to compare the cell types using AECs, but we do not want any of the three strains present, so our negative control was unconditioned media with AECs.

  1. unconditioned media without AECs. No AECs essentially means no way to test expression of 53BP1 and γH2AX.
  2. unconditioned media with AECs. This answer choice is consistent with our breakdown. We still want to compare the cell types using AECs, but we do not want any of the three strains present, so our negative control was unconditioned media with AECs.
  3. conditioned media without AECs. Conditioned media means we are no longer dealing with a negative control.
  4. conditioned media with AECs. Conditioned media means we are no longer dealing with a negative control. The only answer choice that is consistent with our answer choice is answer choice B.

35) This is a passage-based question and the test-maker is making sure we really understand the data given to us in the passage. The author tells us that researchers hypothesized that secreted bacterial factors like hydrogen peroxide are the primary causes of DSBs in host cell DNA. For this question, not only do we have to understand the results of experiments 1 and 2, but we’re also asked the best way to link the results of the two to show that S. pneumoniae can cause apoptosis of host cells by contributing to DSBs. Ideally, when we read the passage we can pick out key conclusions, trends, and outliers that come from the data. Here we’ll be making sure we understand the results in both experiments and are able to relate the two.

In Paragraph 1, the author mentions that DSBs activate a kinase pathway that phosphorylates histone H2AX to form γH2AX. That γH2AX recruits DNA repair proteins, such as 53BP1, to the damaged area. Therefore, in Experiment 1, we’re looking at expression of 53BP1 and/or γH2AX, and generation of DSBs. In Experiment 2, we’re testing apototic cell % in the presence and absence of catalase, which decomposes H2O2 to water.

Think back to the initial purpose of the experiment. The author tells us that researchers hypothesized that secreted bacterial factors like hydrogen peroxide are the primary causes of DSBs in host cell DNA. To link experiments 1 and 2, we can measure the formation of H2O2 in each strain. The presence of 53BP1 and/or γH2AX indicates DNA damage, and exposure to H2O2 induces apoptosis.

  1. Measure levels of H2O2 produced by each strain. This is the best method by which to link the results of experiments 1 and 2. The presence of 53BP1 and/or γH2AX indicates DNA damage, and exposure to H2O2 induces apoptosis. Answer choice A will be our best answer choice.
  2. Determine if other factors are secreted by each strain. While this might be beneficial information, this does not link the results of experiments 1 and 2.
  3. Determine if H2O2 is produced by other S. pneumoniae strains. Similar to answer choice B, while this might be beneficial information, this does not link the results of experiments 1 and 2.
  4. Measure the number of DSBs that occur in AECs cultured with each strain. Once again, we’re focused on linking the results of experiments 1 and 2. Only answer choice A does that effectively. We can stick with answer choice A as the best answer.

 

Exam 4 B/B Solutions: Passage 7

36) This is a passage-based question that focuses on something we’re introduced to in Paragraph 3. The first two paragraphs involved the author getting into the functions and details of CCK, but then we get into a hypothesis relating abnormalities in CCK secretion in patients with bulimia. In Paragraph 2, we’re told CCK triggers satiety centers within the brain upon ingestion of food and thereby inducing a sensation indicating enough food has been eaten. When we get to Paragraph 3, we’re told bulimic individuals consume abnormally large amounts of food in one sitting. That gets into the connection between CCK and bulimia. Without CCK triggering satiety centers within the brain, it would theoretically be easier to consume abnormally large amounts of food in one sitting like we see in bulimic individuals.

  1. Fewer CCK-sensitive vagal afferent nerves. In Paragraph 3 the author tells us scientists have hypothesized there are abnormalities in CCK secretion, not in vagal afferent nerves.
  2. Excess bile release in response to CCK. Reasoning here is going to be similar to answer choice A. We want an answer choice consistent with abnormalities in CCK secretion. Only once CCK is secreted does it release bile from the gallbladder into the duodenum.
  3. Decreased release of CCK in response to food. This is consistent with the scientists’ hypothesis and what we said in our breakdown of the question. Without CCK triggering satiety centers within the brain, it would theoretically be easier to consume abnormally large amounts of food in one sitting like we see in bulimic individuals. This is going to be our best answer choice.
  4. Hypersensitivity to the presence of CCK. Once again, this ties into what we said for answer choices A and B. We’re focused on a defect in CCK secretion, not hypersensitivity to CCK. We can stick with answer choice C as our best option.

37) This is a passage-based question that relies on picking out some key information. The first two paragraphs went into the functions and mechanism of CCK. We’re told CCK facilitates digestion by releasing into the duodenum bile from the gallbladder and pancreatic juices rich in enzymes. We’re also told CCK is known to trigger satiety centers within the brain upon ingestion of food and thereby induce a sensation indicating that enough food has been eaten. This involves simulation of vagal afferent fibers. Quick glance at our answer choices shows we’re focused on the gallbladder and hepatopancreatic sphincter, however, so we’ll focus on CCK facilitating digestion and the release of bile. This is the point where we get into our external knowledge. Bile that is stored in the gallbladder is secreted into the small intestine where it aids in digestion. The hepatopancreatic sphincter controls the flow of digestive juices, including bile, into small intestine.

  1. Relaxation of muscle in wall of gallbladder and relaxation of hepatopancreatic sphincter. While the second part of this answer choice is consistent with what we’d expect following the release of CCK, we expect contraction of muscle in the wall of the gallbladder. We have contraction as bile exits the gallbladder. We expect relaxation of the sphincter so digestive juices can flow into the small intestine.
  2. Contraction of muscle in wall of gallbladder and relaxation of hepatopancreatic sphincter. This answer choice is consistent with our breakdown of the question and what we know from our general knowledge. Following the release of CCK (which facilitates digestion by releasing bile from the gallbladder, we expect contraction of muscle in the wall of the gallbladder. We have contraction as bile exits the gallbladder. We expect relaxation of the sphincter so digestive juices can flow into the small intestine. This is going to be our best answer choice.
  3. Relaxation of muscle in wall of gallbladder and contraction of hepatopancreatic sphincter. This is exactly the opposite of what we would expect. We expect contraction of muscle in the wall of the gallbladder and relaxation of the hepatopancreatic sphincter.
  4. Contraction of muscle in wall of gallbladder and contraction of hepatopancreatic sphincter. While we expect contraction of muscle in the wall of the gallbladder, we actually expect relaxation of the hepatopancreatic sphincter. Answer choice B remains our best option.

38) This question is going to be answered using only information from the passage. The author tells us in the passage, CCK triggers satiety centers within the brain upon ingestion of food and thereby inducing a sensation indicating enough food has been eaten. We expect CCK treatment would initiate this sensation more quickly, and food consumption would decrease.

  1. This answer choice is the opposite of what we would expect. CCK triggers satiety centers within the brain upon ingestion of food and thereby inducing a sensation indicating enough food has been eaten. Therefore, we expect the CCK treated rats to be full much faster.
  2. This answer choice shows no difference in the size of the meal between the control group and the CCK treated group. We expect the CCK treated rats to be full much faster. Additionally, if we compare answer choices B and D, we can see they’re saying the same thing. We’re told the Y-axis is in grams, but not given any quantitative values. B and D are essentially identical answer choices and we know every MCAT question only has one correct answer.
  3. This is exactly what we’re looking for an answer choice. The Control group will eat a certain amount of food, but CCK triggers satiety centers within the brain upon ingestion of food. The CCK treated rats will feel like they have eaten enough food much more quickly. We can stick with answer choice C as our best option.
  4. We mentioned in the explanation for answer choice B that answer choices B and D are saying the same thing. We’re told the Y-axis is in grams, but not given any quantitative values. Both are incorrect as the CCK treated rats will feel like they have eaten enough food much more quickly and consume less food.

39) While this is a passage-based question, we can approach this like a pseudo-standalone question. The author is essentially just asking, “what do peptidases digest?”. Peptidases are enzymes that function to break down peptides into amino acids. We want an answer choice that mentions the digestion of proteins into smaller fragments.

  1. fats. The question stem mentions secretion of peptidases which involves the digestion of proteins. This would be the correct answer if the question had instead said lipases.
  2. protein. Peptidases are enzymes that function to break down peptides into amino acids. This is going to be our best answer choice.
  3. nucleotides. The question stem mentions secretion of peptidases which involves the digestion of proteins. This would be the correct answer if the question had instead said nucleases.
  4. carbohydrates. The question stem mentions secretion of peptidases which involves the digestion of proteins. This would be the correct answer if the question had instead said carbohydrases. We can stick with answer choice B as our best option.

 

Exam 4 B/B Solutions: Passage 8

40) This question is tangentially-related to the passage, but we should be able to answer this question using our general knowledge of osmoregulation. Osmoregulation is the active regulation of the osmotic pressure of bodily fluids to maintain the homeostasis of the body’s water content. Although the kidneys are the major osmoregulatory organ, the skin and lungs also play a role in the process. That means that while a large amount of water is excreted through urine, we also will excrete water through the skin and lungs.

  1. the extra water was stored as blood. This is not what we expect from the water not excreted as urine. We can actually see in Figure 1, that blood volume does not fluctuate much with water consumption. Extra water being stored as blood would mean blood volume increases and blood gets diluted.
  2. water was consumed with the food that was eaten. This answer choice has more to do with the timing of the water intake. We still do not have a significant change in blood volume or osmolarity according to our graphs. Instead, we expect water was excreted through the skin and lungs like we mentioned in our breakdown of the question.
  3. the extra water was excreted by the intestine. Typically, water is absorbed in the intestines and not excreted by the intestine. Excretion by the intestine would mean diarrhea or watery stool, which the author did not mention.
  4. water was excreted via the skin and the lungs. This answer choice is consistent with osmoregulation. Although the kidneys are the major osmoregulatory organ, the skin and lungs also play a role in the process. We expect the difference between water intake and urine excretion is due to water being excreted via the skin and lungs.

41) This is another passage-based question, similar to Question 40. Osmoregulation is the process by which the osmotic pressure in the urine is regulated by the concentration of water and ions. In our question stem, the osmolarity of urine is much higher than average blood osmolarity. That means a lot less water is being excreted relative to what we see in the blood. More ions are being removed from the body relative to the water being removed. We want an explanation for this high osmolarity of urine.

  1. the kidneys are secreting very little Na+. This is the opposite of what we see in our question stem. If the kidneys are secreting very little sodium ion, we expect osmolarity of urine to be much lower. In the question stem we’re told osmolarity is much higher compared to blood osmolarity. Answer choice A is unlikely.
  2. the kidneys are acting to conserve water. This is a viable option. Our bodies function to maintain homeostasis through the regulation of the concentration of our urine. If we have a much higher concentration, that means we are retaining much more water (there is less water in the urine compared to ions). This is a likely scenario.
  3. the subjects are on a low-protein diet. This is the opposite of what we expect. Typically, high levels of protein will result in higher urine osmolarity, not the other way around. Answer choice B remains superior.
  4. the subjects are dehydrated. This is a tricky answer choice because dehydration would mean higher osmolarity in urine. However, we would also have higher blood osmolarity as well. The author explicitly mentions in Paragraph 1 that dehydration induces thirst sensations, a decrease in body fluid volume, and an increase in blood osmolarity. We have to make sure the option we pick answers the specific question being asked. Only answer choice B correctly gives a reason for the higher osmolarity of urine while maintaining normal blood osmolarity.

42) This is a passage-based question and something the author touched on in Paragraph 1. There are three types of receptors that detect changes in body water: osmoreceptors that are sensitive to changes in osmolarity, volume receptors that detect changes in the volume of blood, and oral receptors that mediate the sense of thirst (dry mouth).

This is an entirely new experiment now, and we want to know what happens following infusion of a salt solution that induced thirst and a desire to drink. We can think about the three types of receptors. We expect increased blood osmolarity following intravenous infusion of salt which would mean the body needs a way to lower osmolarity (that can happen by drinking water). Blood volume would increase after the infusion meaning the volunteers are less likely to drink. Lastly, an intravenous infusion would not cause oral receptors to trigger the desire to drink.

  1. blood volume receptors, as a result of increased blood volume after the infusion. Increased blood volume after the infusion would mean the volunteers are less likely to drink. Only if blood volume drops would this be a viable solution.
  2. osmoreceptors, as a result of increased blood osmolarity after the infusion. This answer choice is consistent with what we’re told in the question stem and our breakdown of the question. We expect increased blood osmolarity following intravenous infusion of salt which would mean the body needs a way to lower osmolarity. That can happen through drinking water. This is our best option so far.
  3. a sensation of dryness in the mouth by oral receptors. We mentioned this in our breakdown of the question; an intravenous infusion would not cause oral receptors to trigger the desire to drink. Answer choice B remains our best option.
  4. the NaCl itself, because salt makes a person thirsty regardless of osmolarity. This answer choice is factually incorrect. If the osmolarity were low enough, or the same osmolarity as the blood, then we would not expect increased thirst. We can stick with answer choice B as our most likely reason for thirst stimulation.

43) This is another passage-related question that requires us to also use external knowledge. In the passage, the author tells us dehydration induces thirst sensations and a decrease in body fluid volume simultaneously with an increase in blood osmolarity. Cardiac output is the amount of blood the heart pumps through the circulatory system in a minute and is given in liters per minute. This output depends on stroke volume (volume of blood ejected from each ventricle due to contraction of the heart muscle) and heart rate.

  1. A decrease in stroke volume. We mentioned in our breakdown of the question that cardiac output depends on stroke volume and heart rate. To maintain cardiac output under dehydration conditions, we would need an increase in stroke volume.
  2. A decrease in heart rate. This answer choice is along the same lines as answer choice A. Cardiac output depends on stroke volume and heart rate. To maintain cardiac output under dehydration conditions, we would need an increase in heart rate.
  3. An increase in heart rate. This answer choice is consistent with our breakdown of the question and the reasoning we used to eliminate answer choices A and B. Cardiac output is the amount of blood the heart pumps through the circulatory system in a minute and is given in liters per minute. If we need to maintain this cardiac output under dehydration conditions we would need an increase in stroke volume or heart rate. Answer choice C is our best answer choice.
  4. An increase in blood viscosity. Increasing blood viscosity is not going to adjust for the low blood volume during dehydration. Answer choice C is the superior, more direct answer choice.

44) Similar to the other questions in this question set, this is a passage-based question that will ultimately rely on our external knowledge to get to the correct answer. We have to think about exercise and active muscles. Think of going to the gym and lifting weights. The first set of an exercise might be easier, but muscle use can quickly overwhelm the ability of the body to deliver oxygen. When oxygen supplies are running low, many prokaryotes and eukaryotes use lactic acid fermentation to regenerate NAD+. Lactate dehydrogenase is the enzyme that catalyzes this process converting pyruvate to lactate. That additional lactate increases osmolarity of the venous blood.

Muscle fibers must switch to anaerobic metabolism and produce lactic acid, at which point the muscle begins to fatigue. Anaerobic respiration causes decreased pH and reduced affinity of hemoglobin for oxygen. We want an answer choice consistent with the formation of lactic acid and increase in osmolarity of the venous blood.

  1. lactate concentration in plasma. This answer choice is consistent with our breakdown. Muscle fibers switch to anaerobic metabolism and produce lactic acid. We have an increase in osmolarity of venous blood which is consistent with what we’re looking for in this question.
  2. Oconcentration in plasma. We expect the opposite to be true and there is not an increase in oxygen concentration in the plasma. O2 is reduced in venous blood like we mentioned in the breakdown of the question.
  3. oxyhemoglobin concentration in red blood cells. 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 A remains the best option.
  4. blood pressure in systemic arteries. An increase in blood pressure in systemic arteries is not going to influence osmolarity of the venous blood. If this answer choice instead talked about BP in veins we would consider this as a viable option. However, answer choice A is going to be the best option in this case.

 

Exam 4 B/B Solutions: Questions 45-48

45) This is a standalone question that relies on our knowledge of a specific enzyme. Retroviruses like HIV are a unique class of viruses. The genetic material of retroviruses is single-stranded RNA with two copies per viral particle. The RNA must be converted into double-stranded DNA by an enzyme called reverse transcriptase. It’s aptly named, as it reverses the normal flow of DNA to RNA to protein in cells; instead, it converts single-stranded viral RNA to double-stranded DNA.

This is a basic content question that relied on our knowledge of reverse transcriptase. We can pick the only answer choice consistent with our breakdown of the question, answer choice A: RNA to DNA.

46) This is another standalone question that relies on our general content knowledge. We’re told the solid line in each of the four answer choices represents an uncatalyzed chemical reaction. The same chemical reaction is catalyzed by an enzyme and shown as a dotted line in our correct answer. We want to find the answer choice that correctly shows this dotted line.

Enzymes are catalysts that are able to lower activation energy without being used up or being destroyed in the reaction. When an enzyme binds its substrate, it forms an enzyme-substrate complex. After binding between the enzyme and substrate takes place, one or more mechanisms of catalysis lowers the energy of the reaction’s transition state by providing an alternative chemical pathway for the reaction. Ultimately, we want an answer choice consistent with the idea that catalysts work to increase the rate of a reaction by lowering activation energy.

  1. This answer choice starts out well as activation energy is lower in the dotted line. However, enzymes do not change the energy of the reactants or products themselves. Answer choice A incorrectly shows the energy of the final products as higher in the catalyzed reaction.
  2. This answer choice is consistent with our breakdown of the question. Activation energy decreases but the energy of the reactants and products themselves does not change. This is going to be our best answer choice so far.
  3. This answer choice incorrectly shows an increased activation energy. We expect the opposite to be true. Answer choice B remains our best option.
  4. This answer choice is similar to answer choice A. The energy of the reactants and products themselves doesn’t change, rather the activation energy decreases. Only answer choice B is consistent with our breakdown of the question and definition of an enzyme.

47) This is a standalone question that is based on our knowledge of the sodium-potassium pump, or Na+/K+ ATPase Recall that sodium-potassium pumps bring two K+ ions into the cell while removing three Na+ ions per ATP consumed. What is the importance of the last bit of this sentence? The consumption of ATP implies active transport (movement of ions into a region of higher concentration, against their concentration gradient).

Given our definition and breakdown, options I and IV are the only correct options. Options II and III incorrectly mention ions moving with their concentration gradient. Answer choice B is going to be our correct answer: I and VI only.

48) This is a standalone question that involves being able to read the graph in our question stem. We’re told cell division slows as the embryo enters gastrulation; the slowing appears to be around 4 or 5 on our y-axis. However, what does that 4 and 5 represent? We have to note that the Y-axis is given in Log10, so at a value of 4, the number of cells is 104 or 10,000. At a value of 5, the number of cells is 105 or 100,000. The only answer choice that is consistent with this range is answer choice D: 10,000 and 100,000.

 

Exam 4 B/B Solutions: Passage 9

49) This is a passage-based question that involves going back to the passage to pick out some key information. In Paragraph 3, the author explains that PptB is a phosphotransferase and adds phosphoglycerols onto pilin subunits. Phosphotransferases function to catalyze phosphorylation reactions. We want an answer choice that explains a change in a property of pilin after adding this phosphoglycerol onto pilin.

  1. Increase in number of disulfide bonds. This is usually only a viable option when we have a passage that deals with cysteine residues. In this case, it is out of scope. There is no mention of cysteine or disulfide bonds, nor is there any way to assume we will get an increase in the number of disulfide bonds.
  2. Decrease in molecular weight. The opposite of this is actually true. If we add a phosphoglycerol to pilin, we should see an increase in molecular weight, not a decrease.
  3. Increase in enzymatic activity. Like we mentioned in the breakdown of the question, we’re focusing on a change in a property of pilin. The addition of the phosphoglycerol is not going to increase enzymatic activity in pilin. PptB functions as the enzyme (phosphotransferase) in the reaction.
  4. Decrease in isoelectric point. This is going to be a viable option. The transfer of a negatively charged phosphate group is going to decrease isoelectric point. Think about negatively charged amino acids. The presence of additional acidic groups is going to lower the isoelectric point. Answer choice D is our best option.

50) This is almost like a pseudo-standalone question once we get a key piece of information from the passage. The author tells us that PptB is a phosphotransferase and adds phosphoglycerols onto pilin subunits. 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.

The only answer choice listed that can be phosphorylated is answer choice C: Serine.

51) This is a passage-based question that involves knowing some details about the initial immune response to N. meningitidis. We’re told bacteria will enter the blood, migrate to the brain, and cross the blood-brain barrier. The subsequent immune response is what causes meningitis. We want to go through our four answer choices, define each one, and decide which option would mount the initial immune response to N. meningitidis that results in meningitis. We’re likely focused on glial cells in the CNS because of the location of the immune response (bacteria migrate to the brain), and we’re also thinking innate immunity instead of adaptive. The cells of the adaptive immune system are a type of leukocyte called a lymphocyte. B cells and T cells are the major types of lymphocytes involved in adaptive immunity.

  1. B cells. B cells are lymphocytes that produce antibodies and are responsible for immune response, but not in the brain.
  2. Cytotoxic T cells. A cytotoxic T cell is a lymphocyte that kills cells that are damaged. Similar to answer choice A, the location for cytotoxic T cells is not the brain.
  3. Helper T cells. Helper T cells activate B cells, which proliferate and produce antibodies specific to the antigen. Once again, we’re looking for an answer choice synonymous with the brain.
  4. Microglia. Microglial cells are macrophage cells that monitor and maintain the health of neurons by detecting injuries to the neuron. Most importantly for this question, microglia are distributed throughout the CNS. This is going to be our best answer choice.

52) This is a pseudo-standalone question that is tangentially related to what we covered in the passage. We’re focused on different types of intercellular junctions and connections:

Best way to answer this question is to go through the four different options listed as answer choices and define each one. We want to pick the best answer that represents the intercellular connections we find in the blood-brain barrier. The blood-brain barrier is highly selective and molecules are not able to cross into the CNS. While vital nutrients are able to cross, toxins and pathogens are usually not able. We want an answer choice that is consistent with this.

  1. Desmosomes. Animal cells may contain junctions called desmosomes which act like spot welds between adjacent epithelial cells. A desmosome involves a complex of proteins and some of these proteins extend across the membrane, while others anchor the junction within the cell.
  2. Gap junctions. Gap junctions are channels between neighboring cells that allow for the transport of ions, water, and other substances between cells. This is not what we’re looking for in our answer choice.
  3. Intercalated discs. Intercalated discs are synonymous with cardiac muscle, not with the blood-brain barrier. This answer choice is out of scope.
  4. Tight junctions. Tight junctions create a watertight seal between adjacent cells. At the site of a tight junction, cells are held tightly against each other by many individual groups of tight junction proteins called claudins. These claudins interact with a partner group on the opposite cell membrane. The groups are arranged into strands that form a branching network, with larger numbers of strands making for a tighter seal. That’s what we have at the blood-brain barrier. The purpose of these tight junctions in the CNS is to keep liquid from escaping between cells and preventing toxins and pathogens from entering. This allows a layer of cells to act as an impermeable barrier. Answer choice D is our best option.

 

Exam 4 B/B Solutions: Passage 10

53) This is a passage-based question so we will be picking out some key details from the passage. Ultimately, we’re going to be using our general knowledge to identify the correct answer, however. The author tells us that GABA binds protein receptors that function as chloride channels in neurons. After GABA binding, these receptors allow chloride ions to enter cells. We know chloride ions are negatively charged, so the cytoplasm of the oocytes will become more negative. We can pull up a graphic helping us visualize this additional negative charge:

Chloride ions entering the cells will cause membrane potential to be more negative. We want an answer choice consistent with this.

  1. Oocytes become hyperpolarized as chloride ions enter them. This answer choice is consistent with what the author tells us in the passage. Hyperpolarization involves the oocyte’s membrane potential to become more negative. This happens through the influx of the chloride anion.
  2. Oocytes become depolarized and produce an action potential from sodium ion entry. This answer choice is out of scope. We’re focused on the movement of chloride ions and hyperpolarization, not sodium ions and depolarization. Answer choice A remains superior.
  3. Oocytes become hyperpolarized as chloride ions exit them. First part of this answer choice starts out promising, but we know chloride ions enter oocytes, not exit them. We can stick with answer choice A as our superior option.
  4. No change occurs in potential; chloride ions enter and sodium ions exit the oocytes. This answer choice is also out of scope. Reasoning here is going to be similar to answer choice A. We’re focused on the movement of chloride ions, not sodium ions. We can stick with answer choice A as our best option.

54) This is a pseudo-standalone question that asks us to identify the effects of introducing an amber codon. A stop codon is one of three codons (UAA, UAG, and UGA) that signals the end of translation. If there is a mutation which occurs and causes an early stop codon in an mRNA sequence this is called an amber codon. The example in our question stem says the third codon in the coding region is replaced with TAG. mRNA is going to be the same, but we replace T with U. We know UAG is one of three codons that signals the end of translation, and full-length receptors are not made. The only answer choice consistent with this is answer choice C which says functional full-length receptors will not be synthesized.

55) This is another pseudo-standalone question. We’re told we have newly synthesized GABA-receptor subunits move to the cell surface. Glancing at our answer choices, we want an answer choice in terms of the rough ER, endosomes, lysosomes, vesicles, or Golgi complex in some combination of three. We recall GABA receptors (proteins) are located in the cell membrane and are synthesized in the rough ER; proteins can also be modified here. Vesicles will transport proteins into the lumen of the Golgi apparatus. As the proteins and lipids travel through the Golgi, they undergo further modifications that allow them to be sorted. These newly-modified proteins and lipids are then tagged with phosphate groups or other small molecules so that they can be routed to their proper destinations. The modified and tagged proteins are packaged into secretory vesicles that bud from the trans face of the Golgi. While some of these vesicles deposit their contents into other parts of the cell where they will be used, other secretory vesicles fuse with the plasma membrane and release their contents outside the cell.

  1. Rough endoplasmic reticulum, endosome, and Golgi complex. An endosome is a membrane-bound compartment inside a eukaryotic cell that functions to bring materials from outside the cell into the cell. We’re focused on the movement of receptor subunits to the cell surface, not on material coming into the cell.
  2. Rough endoplasmic reticulum, Golgi complex, and secretory transport vesicle. This answer choice is consistent with the breakdown we came up with. This is a superior answer to option A.
  3. Golgi complex, endosome, and lysosome. Reasoning here is similar to answer choice A. An endosome is a membrane-bound compartment inside a eukaryotic cell that functions to bring materials from outside the cell into the cell. We’re focused on the movement of receptor subunits to the cell surface, not on material coming into the cell. Answer choice B remains superior.
  4. Golgi complex, rough endoplasmic reticulum, and secretory transport vesicle. While the different organelles here are correct, they are out of order. The receptor subunits are synthesized in the RER before making their way to the Golgi complex. Answer choice B is going to remain our correct answer.

56) While this is another passage-based question, the key to answering this question is understanding the role of a negative control. A negative control is included in an experiment, but is not expected to change because of the experiment itself or influence the results of the experiment. It functions like a baseline and is not subject to the treatment of the experiment. By using a negative control, researchers can look for confounding variables and get more clarity about positive results. For example, if we’re experimenting with a treatment for patients with a specific illness, the negative control would be used to see what happens if no treatment is administered. Once we have that baseline, we can note any changes that occur because of the positive control, or any independent variables.

  1. Addition of GABA to mock-transfected frog oocytes. A negative control functions to not generate a response, but rather give us a baseline. This method simply shows that transfection process itself does not affect any results and we have a good sense of the baseline if we have the addition of GABA. This is a strong answer choice.
  2. Transfection of wild-type GABA receptor into insect oocytes. This also does not help us as much because we want to focus on frog oocytes, not adding additional receptors into insect oocytes. A negative control provides us with a baseline of sorts so transfection into insect oocytes doesn’t help us here.
  3. Addition of excess GABA to Rdl-transfected oocytes. The author tells us in the passage Rdl is the mutant form of the GABA receptor gene. This answer choice involves adding GABA to these Rdl-transfected oocytes but Rdl is not responsibe to GABA. Answer choice A remains superior.
  4. Transfection of wild-type GABA receptor into Rdl-expressing oocytes. This would essentially introduce the wild-type GABA receptor into an organism that has a mutant form of the GABA receptor gene. Once again, this is not a negative control. Answer choice A is our best option.

 

Exam 4 B/B Solutions: Questions 57-59

57) This is a standalone question that asks us to relate base pair length and codons. We should know from our general knowledge that each codon is formed of three nucleotides. If we have a mutant allele that is the result of the deletion of two codons, the length is:

(2 deleted codons) x (3 nucleotides / 1 codon) = 6 nucleotides

The mutant allele is the result of the deletion of two entire codons. The base pair length of the mutant allele is six fewer nucleotides compart to that of the wild-type allele.

Answer choice D is our best answer choice.

58) Unless we’re told otherwise, we can usually assume complete dominance for these types of genetics questions. Complete dominance is a condition where the dominant allele completely masks the effect of the recessive allele. That means both alleles must be recessive for the recessive allele to be expressed. In this specific case, the red-eyed, dark-bodied female exhibits only dominant traits. We know she has at least one dominant allele. However, we don’t know if she is homozygous or heterozygous for either trait. The only way we can do so is by mating the female to a male that exhibits recessive phenotypes for both traits. If the two mate and their offspring don’t produce any recessive phenotypes, we know the female is homozygous for the trait. If the two mate and their offspring produce half of each phenotype, we know the female is heterozygous for the trait.

  1. Red eyes, dark body. These are both dominant traits. The only way we would see any recessive phenotypes is if both parents are heterozygous. Unless that is the case, we wouldn’t be able to tell if the mother or father are heterozygous or homozygous.
  2. Red eyes, yellow body. While this would give us information about the body genotype, we would run into a problem with the eyes. Red eyes are dominant, so we can’t easily determine if the female is heterozygous for eye color.
  3. White eyes, dark body. This is similar to answer choice B. While this would give us information about the eyes genotype, we would run into a problem with the body color. Dark bodies are dominant, so we can’t easily determine if the female is heterozygous for body color.
  4. White eyes, yellow body. This is going to be our best option. We know the male exhibits recessive phenotypes for both traits. If the two mate and their offspring don’t produce any recessive phenotypes, we know the female is homozygous for the trait. If the two mate and their offspring produce roughly half of each phenotype, we know the female is heterozygous for the trait. Answer choice D is our correct answer choice.

59) This is a standalone question that relies on our general knowledge of genetics. Our correct answer choice is going to explain the likely reason for the failure of a gamete to receive a copy of a particular chromosome. Best way to answer this question is to define all four of our answer choices first. We can find the one that best answers the specific question being asked.

  1. Recombination. Recombination involves the formation of genetic combinations in offspring that are not present in the parents. That happens from independent assortment and crossing over and introduces genetic diversity into gametes during meiosis. This does not explain the failure of a gamete to receive a copy of a particular chromosome.
  2. Linkage. Linkage is the property of genes of being inherited together. The segregation of alleles into gametes can be influenced by linkage, but is not responsible for the failure of a gamete to receive a copy of a particular chromosome.
  3. X inactivation. X inactivation has to do with inactivation of a copy of an X chromosome. The inactive X chromosome is packaged into heterochromatin. This inactivation does not explain the failure of a gamete to receive a copy of a particular chromosome.
  4. Non-disjunction. Nondisjunction occurs when chromosomes fail to segregate during meiosis; when this happens, gametes with an abnormal number of chromosomes are produced. This can occur in Anaphase 1 and Anaphase 2 and can lead to gametes with too many or too few chromosomes. This is the exact answer we’re looking for and is most likely responsible for the failure of a gamete to receive a copy of a particular chromosome.


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