Exam 2 B/B Solutions: Passage 1
1) To answer this question, we’ll go back to the passage and look at Figure 1 that shows us CRY1 and XPA levels. One thing I want you to note is the test-maker says “information in the passage suggests.” For a question like this, that doesn’t automatically mean go back to the passage. Ideally you get to the point where you don’t have to flip back to the passage for every question, unless you’re looking for a specific detail or quantitative value. In this case, we’ll do the former.
We have Figure 1 above which shows relative levels of XPA and CRY1 proteins. We can see the two are inversely related. As CRY1 levels increase, we have a decrease in XPA levels. What this likely means is CRY1 represses the expression of the XPA-encoding gene.
- activating XPA protein activity. This is the opposite of what we see in Figure 1. When we have increased CRY1 levels, we actually see a decrease in XPA levels.
- activating translation of XPA-encoding transcripts. This is going to be similar to answer choice A. We see a decrease in XPA levels as CRY1 levels increase.
- repressing replication of the XPA-encoding gene. This is a better answer than answer choices A and B, however, replication has more to do with synthesis of DNA and not the protein expression we’re seeing.
- repressing transcription of the XPA-encoding gene. This is going to be our best answer. When we have increased CRY1 levels, we have decreased XPA levels. CRY1 is actually repressing the expression of this protein, which suggests we’re repressing transcription of the XPA-encoding gene. Answer choice D is going to be our best answer here.
2) The test-maker references Figure 3 from the passage, but we should be able to answer this question just using our external knowledge. Even though this is a passage-related question, we can almost treat it like a standalone question. We’re going to decide which of our four answer choices are highly proliferative.
- Adipocytes. Adipocytes are essential in energy storage, but they are not highly proliferative. The cells will fluctuate in size, but not unusually in number.
- Cardiac muscle cells. Cardiac muscle cells are going to be similar to answer choice A. We don’t expect proliferation here past the embryonic stage.
- Gastrointestinal epithelial cells. We can think about the environment in which GI epithelial cells have to work. The pH levels are rough and there are different pathogens coming in and out. These cells have to proliferate to replace cells. This is going to be our best answer at this point.
- Neurons. Neurons are not going to divide, so this is factually incorrect. Answer choice C is going to be our best answer.
3) Before we answer this question, let’s do some quick background. The structure of DNA is called a double helix, which looks like a twisted staircase.
The sugar and phosphate make up the backbone of the nucleotides and are bonded together by phosphodiester bonds. In this case, we have a lesion that causes a break in the DNA strand, or that backbone we talked about. Polymerase is used to resynthesize the strand (which we’re told has already occurred in the question stem), while the backbone is corrected by ligase and bonded together by phosphodiester bonds. The only answer choice here that matches our breakdown is going to be answer choice D: phosphodiester bond. You want to be careful here to not pick hydrogen bond. While there is hydrogen bonding earlier in the process of correcting this lesion, we’re concerned with the process after the strand has been resynthesized. Answer choice D is our best option.
4) The passage starts out talking about CPDs and now we’re asked to identify an example of a CPD. A good place to start would be to look at some purines and pyrimidines with which we’re already familiar.
We have our pyrimidines in the bottom of our figure. They’re all single-ringed which is a big identifying factor.
Left side of answer choice A shows a pyrimidine, but we said we’re looking for single-ringed structures.
Neither side of this structure shows a pyrimidine. We’re looking for a dimer that contains two single-ringed pyrimidines.
This answer choice matches our breakdown. We said pyrimidines are single-ringed. This dimer contains pyrimidine monomers.
This is similar to answer choice A. Left side of answer choice D shows a pyrimidine, but we said we’re looking for single-ringed structures. We can stick with answer choice C as our best option.
Exam 2 B/B Solutions: Passage 2
5) To answer this question, we’ll have to consider Reaction 1 from the passage and how we can increase the production of products.
If we increase reactants (AIP and H+) then we should see an increase in phosphine production. In this case, our question is asking in terms of different pHs. What do we know about pH? Acidic solutions have a pH less than 7, with lower pH values corresponding to increasing acidity. The strength of an acid refers to how readily an acid will lose or donate a proton. A strong base is the converse of a strong acid; whereas an acid is considered strong if it can readily donate protons, a base is considered strong if it can readily deprotonate from other compounds. That means our best option is going to be an acidic environment so we have more protons (one of our reactants in Reaction 1).
- pH < 4 Strong acidic answer to start. Like I mentioned, the best option is going to be an acidic environment so we have more protons (one of our reactants in Reaction 1).
- 4 < pH < 7 This is also a viable answer, but it’s not as good as answer choice A. We want the largest amount of phosphine production, which means we want the most acidic environment in this case.
- 7 < pH < 10 This is a basic pH which is the opposite effect of what we’re looking for.
- pH > 10 This is a strongly basic pH which is certainly not the environment we are looking for in our answer. We can stick with answer choice A as our best option.
6) To answer this question, we have to use information we learned in the passage, but we’ll also have to know basic information about amino acids. This is one of the highest-yield topics that every student should know. Make sure to know the 1 and 3-letter abbreviations of every amino acid, the basic properties, and anything that makes an amino acid unique (structure, bonding, charge, etc). What do we know about how phosphine reacts? We can pull up an excerpt from the passage.
I’ve highlighted a key point here. It is thought that phosphine reacts with sulfhydryl groups. That should immediately make you think of one amino acid: cysteine. Cysteine has that thiol group and we’re expecting phosphine can react with this group.
We have all of our amino acids listed in the visual above. The only amino acid that matches our breakdown and our criteria is answer choice B: Cysteine.
7) There are a few things we’ll have to consider to answer this question. We’re going to need to reference the passage and what we were told about ATP levels. Researchers “found that AlP exposure resulted in a 65% decrease in ATP levels and a 48% decrease in the rate of ATP synthesis.” We have quite a bit of decrease in ATP levels, but we’re also told total cellular concentration of ATP in the AlP-exposed rat liver cells was equal to the control value. How is that possible? Clearly there is some compensating going on for the decreased ATP levels. We can also glance at Table 1 from the passage:
Table 1 Electron Transport Chain Complex Activity in Response to AlP
|ETC component||Control activity (U/mL)||Activity after AlP exposure (U/mL)|
Despite the ETC activity decreasing after AlP exposure, we still have ATP concentration equal to the control. That means a different pathway is upping its ATP production to account for the decrease in ETC activity.
- Glycolytic flux is increased after AlP treatment. This answer choice sounds good at first glance. ETC activity decreases, so we compensate. The flux through glycolysis increases and supplies the necessary ATP.
- Glycolytic flux is decreased after AlP treatment. This is the opposite of what we expect. If we had significantly decreased ATP levels, this may have been a viable option, but answer choice A is still our best option here.
- Citric acid cycle flux is increased after AlP treatment. The citric acid cycle works in conjunction with the ETC, so if we have decreased activity of the ETC, the citric acid cycle cannot properly compensate and supply the necessary ATP.
- Citric acid cycle flux is decreased after AlP treatment. This ties into answer choices B and C. We saw ATP production didn’t actually decrease, and we’re expecting glycolytic flux is increased, not citric acid cycle flux. We can stick with answer choice A as our best answer.
8) This question is asking about the experimental setup and the results that were gathered and communicated in Table 1.
Table 1 Electron Transport Chain Complex Activity in Response to AlP
|ETC component||Control activity (U/mL)||Activity after AlP exposure (U/mL)|
We have control activity and activity after AlP exposure, but we have activity at three different complexes. We’re told in the passage, “The activities of these complexes were determined independently of each other.” This just sounds like great experimental procedure. Most of you have either researched in a lab or are at least familiar with laboratory techniques. Everything you do is done so methodically and deliberately so you know you have results that are relevant to what you’re trying to do.
In this specific case, we know complexes I, II, and IV are all related. If we have inhibition or if we affect complex I, we’ll affect our results in complexes II and IV as well. We want to test each complex independently for this purpose. We want our results to give us as much information without having to make guesses or assumptions.
- Complex stability is lost if the complexes are able to interact structurally. This is not something we expect to happen. We know the complexes all work in conjunction with one another. Stability is not going to be lost if they are able to interact.
- The complexes have different cellular locations, and it is not feasible to isolate them together. This is incorrect. We can see the locations are in close quarters in the image above.
- The complexes all use the same substrates, so their use must be monitored separately. This is not entirely true. The complexes are all related, so there is some overlap to their function. However, they do not all use the same substrate. Rather they are contingent on the complexes that come before them and they all work together.
- The reactions catalyzed by the complexes are coupled to one another. This is consistent with our breakdown. If we affect complex I and II, for example, we also affect complex IV. If we do that, we don’t have a clear picture of how the experiment affects complex IV because we can’t separate the effect from complexes I and II, versus the effect of AlP alone. We can stick with answer choice D as our best option.
Exam 2 B/B Solutions: Questions 9-12
9) This is a standalone question that involves knowing the function of different organelles and cells, and phagocytosis.
Phagocytosis is the process by which large particles, such as cells, large particles or viral particles, are taken in by a cell. The membrane from the body of the cell and surrounds the particle, eventually enclosing it creating a vesicle. Once the vesicle containing the particle is enclosed within the cell, the vesicle merges with a lysosome for the breakdown of the material in the newly formed compartment (endosome).
- Nucleus. The nucleus is the control center of the cell. The nucleus of living cells contains the genetic material that determines the entire structure and function of that cell
- Golgi apparatus sorts and packages materials before they leave the cell to ensure they arrive at the proper destination.
- Lysosome is a membrane-bound cell organelle that contains digestive enzymes for breaking down parts of the cell, as well as material that has been taken into the cell by phagocytosis. This is what we’re looking for to answer this question.
- Endoplasmic reticulum is an organelle that is responsible for the synthesis of lipids and the modification of proteins. Answer choice C remains the best option.
10) This question is asking us to utilize different parts of our external knowledge. We’ll have to know about phosphofructokinase and its function. Phosphofructokinase is the main enzyme controlled in glycolysis. High levels of ATP, citrate, or a more acidic pH decrease the enzyme’s activity. Specifically, ATP binds an allosteric site on the enzyme to inhibit its activity. We know we’ll have option II in our correct answer which means we can eliminate answer choices A and B right away. We have to decide whether options I or III are correct.
Feedback inhibition involves the use of a reaction product to regulate its own further production. That’s exactly what we see in this situation. Option I is also correct, so we’ll stick with answer choice C as our best answer. A big thing you want to remember is glycolysis control begins with hexokinase, which catalyzes the phosphorylation of glucose; its product is glucose-6- phosphate, which accumulates when phosphofructokinase is inhibited.
11) We can look at the concentration of chlorine in the pond water versus in the cytoplasm. The concentration of chlorine in the cytoplasm is much higher than in the pond water. That means moving chlorine ions into the cells from low concentration to high concentration is going to require active transport.
- Osmosis Osmosis is the diffusion of water across the membrane. It is important for cells as the movement of water can change the cell’s volume. We’re focused on the movement of chlorine ions.
- Diffusion Diffusion involves moving from an area of high concentration to an area of low concentration, called a concentration gradient, until the concentration is equal. We’re looking for the opposite process.
- Active transport Active transport requires energy to move substances against a concentration gradient, from an area of low concentration to high concentration. That’s what needs to happen to move the chlorine ions. Answer choice C is going to be our best option.
- Facilitated diffusion facilitated diffusion is another example of passive transport. A concentration gradient exists that would allow these materials to diffuse into the cell without expending cellular energy. However, these materials are ions or polar molecules that are repelled by the hydrophobic parts of the cell membrane.
12) We’re going to focus on the blocking of voltage-gated potassium channels and how that would affect the action potential.
If we block voltage-gated potassium channels, we can still get to the depolarization phase above, but the repolarization would not be possible. Note in the image above, repolarization occurs as voltage-gated potassium channels are opened. If we don’t have that happening, we would prevent repolarization and have a prolonged action potential.
- It would inhibit the initiation of an action potential. We can see in our visual above, the action potential would still be initiated, but it would be prolonged if we block voltage-gated potassium channels.
- It would shorten the refractory period. The refractory period corresponds to hyperpolarization, not repolarization. If we have potassium channels blocked, we’re preventing repolarization, not hyperpolarization. We can eliminate this answer choice.
- It would prolong the action potential. As we mentioned in the breakdown of the question, repolarization occurs as voltage-gated potassium channels are opened. If we don’t have that happening, we have a prolonged action potential and prevent repolarization.
- It would prevent depolarization. This is going to be similar to answer choice A. Depolarization is still going to take place because the sodium channels are not blocked. We can eliminate this answer as well. Answer choice C is the best answer.
Exam 2 B/B Solutions: Passage 3
13) To answer this question, we’ll have to go back to the passage and reference where the author talks about properties of Wnt proteins.
What do we know about isoelectric point? The pH at which the net charge is neutral is called the isoelectric point (pI). We can go through some basic background as well. Amino acids all contain the same backbone, which has both an acidic (-COOH) and a basic (-NH2) group. Each amino acid also has a functional group attached to the backbone which can be positive, negative, neutral, or polar in nature. The backbone and functional groups give a protein its overall charge. At a pH below the protein’s pI, a protein will carry a net positive charge; above its pI, it will carry a net negative charge. In this case, an isoelectric point of 9 means the protein is positively charged at physiological pH.
- They are composed of multiple subunits. The passage indicates Wnt proteins are secretory proteins and gives us isoelectric points. Make sure to not confuse the information given in the passage about Frizzled with the information about Wnt proteins.
- They have a positive charge. We went over this in our breakdown of the question. Wnt proteins have a positive charge based on what we know from the passage.
- They are synthesized in the smooth endoplasmic reticulum. This answer choice is another tricky one. If it mentioned rough ER instead of smooth ER, this could be a potentially correct answer.
- They fold into their tertiary structure in the cytoplasm. This is going to be similar to answer choice C. The location is incorrect here; we expect folding to occur in the rough ER.
14) To answer this question we can reference the information about β-catenin from the passage.
- multiple subunits. We’re not given many clues into β-catenin’s structure in the question stem or the passage. Let’s look for a more appropriate and fitting answer.
- very few disulfide bonds. Reasoning here is going to be similar to answer choice A. We’re not given many clues into β-catenin’s structure.
- a nuclear localization sequence. Even though this talks about an amino acid sequence on β-catenin, this ties into the function of β-catenin rather than its actual structure. A nuclearl localization sequence tages a protein to be imported into the cell nucleus. Why is that? So β-catenin can actually activate expression of Wnt target genes. We’re told this is a function of β-catenin in the passage. This answer choice answers the specific question better than answer choices A and B.
- a high proportion of surface-exposed nonpolar residues. Reasoning here is going to be similar to answer choices A and B. We’re not given specific information about the structure of β-catenin, but we also know it’s not likely to have surface-exposed nonpolar residues if β-catenin is going from the cytoplasm to the nucleus. We can stick with answer choice C as our best answer.
15) To answer this question, we can reference the passage, and specifically Figure 1 from the passage. The author did a great job of illustrating and breaking down the Wnt signaling pathway in the absence and presence of Frizzled activation. We’re focused on what happens with β-catenin in addition to covalent modification.
- bound by a proteasome to initiate degradation into short peptides. We can compare the left and right sides here in Figure 1. Left side is in the absence of Frizzled activation where we’re focused. In ubiquitination, the protein ubiquitin is appended to another protein of interest, marking it for degradation by the proteasome. This is a good answer choice for the time being.
- translocated into the Golgi body for secretion through exocytosis. There is no evidence or reasoning in the passage that implies this is true. Biggest clues we’re given are going to be in Figure 1; answer choice A remains the best option.
- engulfed by a lysosome where it is hydrolyzed by proteases. A lysosome is a membrane-bound cell organelle that contains digestive enzymes for breaking down parts of the cell, as well as material that has been taken into the cell by phagocytosis. The author does not mention lysosomes have to engulf β-catenin.
- stored in vesicles until the signaling pathway is activated. Nothing in the passage suggests β-catenin is stored in vesicles until the pathway is activated. This question boiled down to using the scope of the question and what we learned in the passage. Answer choice A is going to be our best option.
16) This question is going to involved knowing the properties of different amino acids. We’re told “Wnt proteins…bind and activate the G protein-coupled receptor Frizzled, whose structure includes seven transmembrane α-helical domains.” Key word here is going to be transmembrane. Transmembrane proteins span the entirety of the cell membrane and cross the phospholipid bilayer. What does that mean for the proportion of amino acid residues? There are going to be more hydrophobic amino acid residues.
This is one of the highest-yield visuals for the exam. Make sure you know your amino acids, their abbreviations, and basic structures and properties.
- Nonpolar this answer choice is consistent with our breakdown of the question. We expect we’ll have hydrophobic amino acid residues. “transmembrane” was a key word here.
- Polar uncharged Polar is the opposite of what we expect from our breakdown. Charge is not going to be a deciding factor here. Answer choice B remains the best option.
- Positively charged like we went over in answer choice B, charge is not going to be the biggest deciding factor; we’re expecting a high proportion of nonpolar amino acid residues. Answer choice A remains the best answer.
- Negatively charged like we went over in answer choice B, charge is not going to be the biggest deciding factor; we’re expecting a high proportion of nonpolar amino acid residues. Answer choice A remains the best answer.
Exam 2 B/B Solutions: Passage 4
17) To answer this question, we’ll have to go back to the passage and get details about Protein X. We can revisit Figure 1.
We have three lanes here: native condition, denaturing condition, and denaturing/reducing conditions. The denaturing condition breaks down Protein X to an extent, but it’s broken down further under reducing conditions. This should immediately make you think sulfur linkages.
A covalent disulfide bond (sulfur linkage) can form between the sulfur-containing R groups of two cysteine molecules (called sulfhydryl groups). Disulfide bonds between cysteine residues can affect protein folding and stability. Disulfide bonds form between cysteine residues under oxidizing (high pH) conditions. Disulfide bonds can be broken under reducing (low pH) conditions. That’s exactly what we’re seeing in this case. We have disulfide bonds broken and the different sizes show up in lane 3.
- thiol groups of methionine residues. While thiol groups is a good start, we have sulfur-containing R-groups of cysteine molecules form disulfide bonds, not methionine.
- thiol groups of cysteine residues. This matches our breakdown exactly. This is likely going to be our correct answer; let’s keep comparing.
- hydroxyl groups of serine residues. When we have reducing conditions, we think thiol groups and cysteine residues. Answer choice B remains the best option.
- hydroxyl groups of threonine residues. Reasoning here is going to be the same as answer choice C. Knowing key points about different amino acids is very important to doing well on the exam. Answer choice B remains the best option.
18) To answer this question, we can pull up Figure 2
What’s the biggest takeaway from Figure 2? When we have ANS alone and ANS+Protein X, we see similar fluorescence. However, in the presence of DPC, we see a substantial increase in fluorescence. We also know ANS exhibits “increased fluorescence upon binding to hydrophobic surface residues of proteins.” We see a correlation between the presence of DPC and this increased fluorescence from binding to hydrophobic surface residues of proteins.
- DPC phosphorylates these amino acids. There is no evidence in the passage or in the question stem that this is true. We know in the presence of DPC, there is increased fluorescence from binding to hydrophobic surface residues of proteins.
- DPC hydrolyzes these amino acids. Reasoning here is identical to answer choice A. We have little evidence to support this claim.
- DPC exposes these amino acids. This answer choice is consistent with what we see in the passage. I mentioned we see a correlation between the presence of DPC and this increased fluorescence from binding to hydrophobic surface residues of proteins. We have additional exposure of hydrophobic residues on the surface, and that is why we see the increased fluorescence in Figure 2.
- DPC suppresses these amino acids. This is the opposite of what we see in the passage and Figure 2. The hydrophobic amino acids are exposed and that’s why we see the increase in fluorescence in Figure 2. We can stick with answer choice C as our best option.
19) This question is going to tie into what you just answered in Question 18.
When we look at Figure 2, we see a correlation between the presence of DPC and this increased fluorescence from binding to hydrophobic surface residues of proteins. When we have ANS alone and ANS+Protein X, we see similar, lower fluorescence. We know ANS exhibits “increased fluorescence upon binding to hydrophobic surface residues of proteins.” In the presence of DPC we have hydrophobic surface residues and higher fluorescence in Figure 2, while in the absence of DPC we have hydrophilic surface residues and much lower fluorescence levels in Figure 2. Our reasoning here is going to be consistent with options I and III. Option II is the opposite of what we know to be true. Surface amino acids of Protein X are mostly hydrophobic in the presence of DPC, which is why we see that increased fluorescence.
20) What event will most likely occur if Protein X is inserted into the inner membrane of mitochondria? To answer this question, we can revisit the passage, and specifically the effect of Protein X on the dye-loaded liposomes. Liposomes have phospholipid bilayers, just like we see in the inner membrane of mitochondria, so we will likely see a similar effect in the inner membrane.
Protein X caused the release of the fluorescent dye from the liposomes, meaning the phospholipid-bilayer is not permeable enough to release some of the dye. What would happen if we made the inner membrane of mitochondria permeable?
If that inner membrane is permeable, we no longer have the same effect of the electrochemical/proton gradient which is essential to make ATP.
- The citric acid cycle will cease to function. The citric acid cycle takes place in the matrix of the mitochondria. It captures the energy stored in the chemical bonds of acetyl-CoA from the products of glycolysis, trapping it in the form of high-energy intermediate molecules. The trapped energy from the citric acid cycle is then passed on to oxidative phosphorylation, where it is converted to ATP, a usable form of cellular energy.
- The electron transport chain will cease to function. This seems more direct than answer choice A as the ETC may be affected more than the citric acid cycle. We would still likely have electrons passed down the electron transport chain and releasing energy. Some of that energy is used to pump hydrogen ions, moving them out of the mitochondrial matrix and into the intermembrane space. Normally that pumping establishes an electrochemical gradient, but in this case that gradient will be affected.
- The proton gradient across the inner membrane will dissipate. This is exactly what we expect to happen. We already saw the effect on the dye-loaded liposomes and we expect something similar to happen with the inner membrane. As the membrane becomes more permeable, we have more free-movement of protons and no electrochemical gradient.
- The pH of the intermembrane space will decrease. The opposite is likely true as protons will be free to move out of the intermembrane space. pH would only decrease if there were additional protons. We can stick with answer choice C as our best answer.
Exam 2 B/B Solutions: Passage 5
21) This question comes down to reading the passage carefully, then applying our external knowledge. We can reference the enzyme encoded by the tdh2 gene in the passage.
At this point, we can restate the question as: “GAPDH catalyzes the reversible conversion of:” and we can focus on using our external knowledge. We’ve essentially turned this into a standalone, content question. We can pull up the steps of glycolysis and narrow down our answer choice.
Glyceraldehyde-3-phosphate dehydrogenase is the enzyme in reaction 6 in our visual above as we go from glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate. This matches answer choice B.
22) To answer this question, we can reference Figure 1 and note the effects of the deletion of hst3 or rtt genes on the lifespan of yeast.
Figure 1: Effects of the deletion of hst3 or rtt alone or simultaneously on the lifespan of yeast
We have our wildtype which is the clear box data points in Figure 1. We note when we have deletion of the rtt genes, we have minimal change in lifespan. When we have deletion of hst3, however, we have a decrease in lifespan; that remains true even when we delete rtt genes on top of deletion of hst3.
- The deletion of the hst3 gene compensates for the ∆rtt-induced decrease in lifespan. This goes against what we see in the passage. There is no real ∆rtt-induced decrease in lifespan. The decrease comes from the deletion of the hst3 gene.
- The deletion of the hst3 gene has no effect on the ∆rtt-induced increase in lifespan. The deletion of the hst3 gene decreases lifespan. There is no ∆rtt-induced increase in lifespan. We see very little effect.
- The deletion of the rtt gene compensates for the ∆hst3-induced decrease in lifespan. The deletion of the rtt gene cannot compensate for the ∆hst3-induced decrease in lifespan. Not the upside down triangle in Figure 1 and the decreased lifespan.
- The deletion of the rtt gene has no effect on the ∆hst3-induced decrease in lifespan. This answer choice is consistent with our breakdown of the question, and what we see in Figure 1. We have the ∆hst3-induced decrease in lifespan, but the deletion of the rtt gene has minimal effect there. Answer choice D is our best answer.
23) This question corresponds to Passage 5, but it is almost like a standalone question. The test-maker is trying to get us to determine the experimental method used to analyze posttranslational modifications. Post-translational modifications are the chemical modifications a polypeptide chain receives after it is translated that convert it to the mature protein. In histones specifically, modifications such as acetylation or methylation can alter how tightly DNA is wrapped around them, while methylation of DNA changes how the DNA interacts with proteins, including the histone proteins that control access to the region.
Western blots use protein-specific antibodies to recognize proteins of interest. Gel electrophoresis is used to separate proteins onto a gel, and then the proteins are transferred onto a solid support matrix. This solid matrix is then exposed to antibodies specific to the protein of interest, which can be visualized via fluorescence or an enzymatic reaction. Antibodies can detect specific proteins, and also proteins with specific post-translational modifications like acetylation in this case.
- Western blot this matches our breakdown of the question.
- Southern blot southern blotting is used to analyze DNA
- Northern blot northern blotting is used to analyze RNA
- RT-PCR RT-PCR combines reverse transcription of RNA into DNA and the usual amplification of DNA targets that we see in PCR
24) Reactive oxygen species (ROS) are highly reactive oxygen molecules with unstable electrons. That includes the superoxide anion, hydroxide, and hydrogen peroxide. Antioxidants work to counteract excessively produced ROS. Essentially, the researchers can adjust the amount of antioxidant enzymes relative to the base-level (by decreasing and overexpressing the enzymes) to analyze the effect of ROS. Additionally, we mentioned hydrogen peroxide as an example of ROS. All three of the options listed here are consistent with what we’re looking for to answer this question. We can stick with answer choice D as our best answer. Note, if you only know about hydrogen peroxide here, you could eliminate answer choices A and B right away. From there, you can note that options I and II have the opposite effect and can help analyze the effect of ROS, so only answer choice D is a viable answer.
Exam 2 B/B Solutions: Questions 25-28
25) First thing we’re going to do is break down the function of vasopressin, or antidiuretic hormone (ADH).
The intake of fluids is stimulated by specialized osmoreceptors in the brain that detect dehydration. These receptors communicate with the pituitary gland to stimulate the release of vasopressin, which in turn communicates with the kidneys to reduce urine production by reclaiming water. Vasopressin increases the water permeability of renal collecting duct cells. It allows water to be reabsorbed from the collecting duct urine to blood. This leads to the excretion of more concentrated urine. How exactly does this happen? Through increased transcription and insertion of aquaporins into the apical membrane of collecting tubule and collecting duct epithelial cells. That’s exactly what we’re looking for in this question. Aquaporins allow water to move down their osmotic gradient and out of the nephron. This ultimately increases the amount of water re-absorbed from the filtrate back into the bloodstream.
- Collecting duct This answer choice matches our breakdown. Vasopressin increases the water permeability of renal collecting duct cells. It allows water to be reabsorbed from the collecting duct urine to blood. This is a good answer for the time being.
- Proximal tubule is involved in the resorption of sugar, sodium and chloride ions, and water from the glomerular filtrate. Answer choice A remains the best option for this specific question.
- Bowman’s capsule is a cup-like sack at the beginning of the tubular component of a nephron that performs the first step in the filtration of blood to form urine.
- Ascending loop of Henle reabsorbs ions from the filtrate into the interstitial fluid, but does not match what we’re looking for specifically to answer this question. We can stick with answer choice A as our best answer.
26) This is a standalone question that relies on your external knowledge. There’s no associated passage, so we’ll have to know about the peripheral nervous system and myelin-forming cells. Oligodendrocytes and Schwann cells are types of neuroglia cells that produce myelin sheath, which acts as insulation along the axon; myelin increases the speed of conduction in the axon. Schwann cells are neuroglia cells that support neuronal function by increasing the speed of impulse propagation. Oligodendrocytes in the central nervous system and Schwann cells in the peripheral nervous system are types of neuroglia cells that produce myelin. We have to be careful here to focus on the peripheral nervous system, so we’re going to pick Schwann cells over oligodendrocytes.
- Microglia Microglial cells monitor and maintain the health of neurons by detecting injuries to the neuron.
- Astrocytes Astrocytes support and brace the neurons and anchor them to their nutrient supply lines. They also play an important role in making exchanges between capillaries and neurons.
- Schwann cells this matches our breakdown exactly. We said Schwann cells in the peripheral nervous system are types of neuroglia cell that produce myelin. This is exactly what we’re looking for in an answer.
- Oligodendrocytes Oligodendrocytes in the central nervous system produce myelin. We want an answer choice that talks about peripheral nervous system. Answer choice C is our best answer.
27) Even though this is the more biology-centric section of the exam, we always have to be ready to deal with some chemistry. We’re asked to solve for the concentration of salt in sea water and given the equation for osmotic pressure above. We will have to rearrange the equation to isolate for the concentration of solute, M.
Π = iMRT
We can divide both sides of the equation by iRT to isolate M:
Plug in the numbers we’re given. Π is 24 atm, i is 2, R is 0.08 L•atm/mol•K, and T is 25°C or 300 K.
M = (24 atm) / (2 x 0.08 L•atm/mol•K x 300 K) = 0.5 M
This was a math problem where we did minimal rounding or approximation. We solved for an exact concentration that corresponds to our correct answer, answer choice B: 0.5 M.
28) This is a basic standalone question that just asks you to define a term. A prion is a self-propogating, misfolded protein and it’s a pathogen that’s smaller than a virus. They are proteins with a defective structure that also trigger other proteins to adopt this faulty structure. They have no genetic material at all, and are responsible for a variety of different neural diseases. The only answer choice that matches this definition is answer choice C: Protein.
If you need to review subviral particles, make sure to go to our website and note the information we have found is the most important: jackwestin.com/resources/mcat-content/viral-life-cycle/prions-and-viroids-subviral-particles
Exam 2 B/B Solutions: Passage 6
29) This question relies on external knowledge. We want to know where blood goes after it leaves the small intestine. We should know from our digestive system content that blood comes directly from the small intestine to the liver where it is detoxified. The liver detoxifies blood by removing ammonia, alcohol and drugs or toxins by metabolizing them.
There’s not much additional information here we need to dive into, and this question was not too tricky. If you need to review liver content, make sure to go to our website and note the information we have found is the most important jackwestin.com/resources/mcat-content/digestive-system/liver
Correct answer here is D: Liver. Blood from the small intestine will be transported to the liver.
30) We learned in the passage “the causative agent of typhoid fever is the bacterium Salmonella typhi. S. typhi enters the body via contaminated food or water and then crosses the epithelial cells of the small intestine.” We want to know why an antibiotic may become less effective over time in treating a bacterium. What do we know about bacteria and prokaryotes in general? The short generation time, random mutations, and mechanism of genetic recombination allow some prokaryotes to have a high degree of genetic adaptability and acquisition of antibiotic resistance. That’s likely what’s happening here, and that’s something we see fairly commonly. As bacteria like S. typhi are exposed to antibiotics, they can become resistant to it. As that happens, the antibiotic, like chloramphenicol, will become less effective in combatting the bacteria.
- against chloramphenicol in ΔF508 heterozygotes. While this would be a possible reason for chloramphenicol to be less effective in this sample, we have to think about what this question is asking specifically. We’re not asked about heterozygotes or homozygotes specifically. We’re asked about the antibiotic in general becoming less effective. We want a more general answer.
- against chloramphenicol in wild-type homozygotes. Reasoning here is going to be the same as answer choice A. We’re not asked about heterozygotes or homozygotes specifically. We’re asked about the antibiotic in general becoming less effective. We want a more general answer.
- for chloramphenicol resistance in populations of S. typhi. This sounds exactly like what we’re looking for here. If we have chloramphenicol resistance in the causative agent of typhoid fever, then chloramphenicol is not as effective of a treatment. Answer choice C sounds like a great option.
- for ΔF508 CFTR, which cannot bind chloramphenicol. This is going to be similar to answer choices A and B. We’re asked a very general question here. While the question is related to the passage, this is almost like a standalone question where the test-maker is asking you to dig into the passage more than is needed. Answer choice C is the best answer choice.
31) I always advocate for reading the question carefully to ensure you know exactly what’s being asked. It’s almost impossible to get a question correct unless you know what the test-maker is asking. In this case, we’re asked the probability of a child homozygous for the G542X allele. We know the mother is heterozygous for G542X, but we’re given no such information about the father. We have to be extra careful. We’re told the father is a male heterozygous for the ΔF508 allele and NOT for G542X. In order for the offspring to be homozygous, the father would have to be a carrier as well. Because there’s only one G542X between the two parents, there is no way the child can be homozygous for the G542X allele. There is a 0% chance the child would be homozygous for the G542X allele.
32) This is going to involve going back to the passage and find necessary information. We’re told information about individuals homozygous for the ΔF508 allele.
Individuals that are heterozygous would not exhibit CF. Furthermore, we’re given information about heterozygous superiority.
By combining these two bits of information from the passage, we can form a predicted answer.
- More susceptible to typhoid fever than wild-type homozygotes and has CF. This is the opposite of what we said would be true. The individual would be more resistant to typhoid fever than wild-type homozygotes, and would not have CF.
- More susceptible to typhoid fever than ΔF508 homozygotes and does not have CF. Reasoning in the first part of this answer is incorrect. The individual is less susceptible to typhoid fever.
- More resistant to typhoid fever than ΔF508 homozygotes and has CF. While the first part of this answer choice is correct, we know from the passage the individual does not have CF.
- More resistant to typhoid fever than wild-type homozygotes and does not have CF. This answer choice matches the passage and our consolidated, predicted answer exactly. Answer choice D is going to be our best option.
Exam 2 B/B Solutions: Passage 7
33) This is something I always emphasize any time I read a passage. We take for granted that we deal with different experiments and studies as we navigate the MCAT, but there’s always a reason behind these studies. Why is the scientist or researcher putting in so much time and effort to conduct this study? The author actually makes our lives a bit easier by mentioning the purpose later in the passage.
We’re told pituitary effects of NPY are examined, and LH levels were measured in different treatments. We know from the passage that GnRH and neuromodulators may regulate LH secretion. In the study, as different treatment conditions are used (independent variables), blood LH levels are measured (dependent variable). The conditions include a control (saline) and a condition where we already know LH secretion is modulated (GnRH). With the additional treatments, we can see the effect of NPY on LH secretion in different situations.
- to determine whether LH can modulate GnRH or NPY secretion. We’re not testing whether LH can modulate GnRH or NPY secretion. LH is our dependent variable.
- to determine whether GnRH can modulate NPY secretion. We’re testing GnRH and NPY treatments separately and together to see the effect on LH secretion, not the effect on NPY secretion.
- to determine whether NPY can modulate LH secretion. This answer choice matches the reasoning in our breakdown. Answer choice C is going to be our best answer.
- to determine whether NPY can modulate GnRH secretion. This si going to be similar to answer choice B. We are focused on LH secretion, not NPY and/or GnRH secretion. We can stick with answer choice C as the best option.
34) Releasing factors are secreted and promote the release of specific hormones (like LH). We want to know why NPY is not classified as a releasing factor. We can bring up Table 1 from the passage, and something the author says in the passage that might be a clue.
Neuromodulators do not stimulate LH release on their own, but they modulate the responses of secretory cells. How does this relate to what we see in the experimental results?
When we compare the control (saline) treatment and NPY alone treatments, we have similar LH levels. It’s not until we add GnRH that NPY modulates LH secretion.
- no effect on LH secretion. This is not true because we can see GnRH + NPY shows an increase in blood LH levels.
- no effect on LH secretion without GnRH. This is consistent with our breakdown and what we see in Table 1. Once we have GnRH, that’s when we have NPY modulate LH secretion.
- no effect on LH secretion in combination with GnRH. This contradicts what we see in Table 1 and what we said in our breakdown of the question. Answer choice B remains superior.
- the same effect on LH secretion as GnRH. The author mentions in the passage that neuromodulators modulate responses of the secretory cells of the anterior pituitary to GnRH. We don’t have the same effect on LH secretion.
35) LH plays a role in preparing the body for pregnancy. It plays a role in the development of ova, as well as in the induction of ovulation and stimulation of estradiol and progesterone production by the ovaries. GnRH from the hypothalamus leads to LH production by the pituitary gland. However, if we have a lack or deficiency in GnRH or LH, then we might have infertility in some cases. Luckily for us, we learned in this passage that levels of LH can be influenced by GnRH and neuromodulators.
We saw that GnRH + NPY treatment led to significantly greater blood LH levels. If we identify an individual with a deficiency in the GnRH receptor system, NPY treatment with GnRH can potentially increase blood LH levels to normal. Answer choice A is going to be the best answer here.
36) This is a content question in a format that AAMC loves to use. We have a cause and effect. NPY is part of the cause, the effect is the generation of the preovulatory LH surge. How do we know there’s a cause and effect relationship? Take away the cause. By removing NPY from the equation, we can see if there is still generation of the preovulatory LH surge and subsequent ovulation. If the LH surge is still present, then NPY may not be necessary. If it is not present, then NPY is necessary.
- When the actions of NPY are blocked in female rats, the LH surge and ovulation do not occur. This is exactly what we talked about in the breakdown of the question. Remove NPY from the equation and see if the LH surge and ovulation still occur. This is a good answer choice.
- NPY release from the hypothalamus increases just prior to the preovulatory LH surge. We’re dealing with a cause and effect situation, but this answer is not telling us if NPY is necessary for the LH surge or ovulation. Answer choice A remains superior.
- NPY can enhance GnRH-stimulated LH secretion in female rats during the preovulatory period. This is a true statement, but does it answer the specific question being asked? We don’t know if this is a direct cause in this case. Answer choice A is still the best.
- NPY has no effect on GnRH-stimulated LH secretion in male rats. This doesn’t help answer our question because it’s an observation seen in male rats. We’re looking for a cause/effect for the generation of the preovulatory LH surge and ovulation. This is going to be in female rats. Answer choice A remains our best option.
37) This is a simple math problem that requires picking the correct numbers from Table 1.
To answer this question, we can compare blood LH levels in the GnRH+NPY group (7.03 ng/mL) with the levels in the GnRH alone group (4.74 ng/mL):
(7.03 ng/mL) / (4.74 ng/mL) = 1.48, or an increase of 48%. Note, the MCAT doesn’t give you complicated numbers. We can increase see 4.74 ng/mL is increased by roughly 1/2 to get to 7.03 ng/mL. You’re not going to be asked to pick between 47%, 48%, and 49% on a question like this because the test-maker knows you won’t have a calculator. We can stick with answer choice B as our best answer.
38) This question requires us to analyze the experiment and data given to us in the passage, but also apply external knowledge to explain the effects on blood estrogen levels. We want to know the effects of treatment with NPY alone. We can reference Table 1.
The NPY alone group actually saw a slight decrease or minimal change in blood LH levels. We typically see a positive feedback loop with estrogen and LH where additional LH will cause additional production of estrogen, estrogen will trigger the anterior pituitary to release more LH. However, in this case we won’t see any changes. NPY won’t influence LH levels, so we’re expecting blood estrogen levels will remain unchanged as well. Answer choice A best represents this:
Exam 2 B/B Solutions: Passage 8
39) To answer this question, we can pick out a few key points from the passage. CatB and CatL are mammalian proteases. We’re also given some additional details about the fusion in the second paragraph.
We want to note that the entry of an EboV nucleocapsid into the host cell cytoplasm starts with the fusion of the viral membrane with the host cell membrane. Additionally, we’re told inhibitors of endosomal acidification block EboV infection, which is another big clue. An endosome is a membrane-bound compartment inside a eukaryotic cell.
Endocytosis is a type of active transport that moves particles, such as large molecules, parts of cells, and even whole cells, into a cell. The plasma membrane of the cell forms a pocket around the target particle. The pocket pinches off, resulting in the particle being contained in a newly-created intracellular vesicle formed from the plasma membrane. Once the vesicle containing the particle is enclosed within the cell, the vesicle merges with a lysosome for the breakdown of the material in the newly formed compartment (endosome). The cells will take in the virus through endocytosis, and that is done through the use of endosomes. Answer choice C is going to be our best answer for this question.
40) To answer this question, we can pull up the data presented in Figure 1.
Figure 1 shows us the effect of CatB or CatL (or both) on VSV-EGP infectivity. We can see CatB, CatL, and CatB + CatL along the x-axis along with a few additional columns. Lane 1 is the wildtype, Lanes 2-4 involve expression of CatL, and Lanes 5-8 involve a cell with a lack of both CatB and CatL.
- Yes, because VSV-EGP infects cells expressing CatL better than it infects cells not expressing CatL. We can look at lanes 2 and 5 here to compare bacterial cells with and without CatL. There’s no significant difference in % of cells infected.
- Yes, because VSV-EGP infects cells expressing CatB better than it infects cells not expressing CatB. While this may or may not be true, we’re focused on CatL expression and not CatB expression.
- No, because VSV-EGP does not infect CatB–/– cells expressing CatL better than it infects CatB–/– cells not expressing CatL. This is consistent with what we see in Figure 1. We can look at lanes 2 and 5 again to compare bacterial cells with and without CatL. We don’t see any significant difference between the two. This is our best answer choice so far.
- No, because VSV-EGP infects cells expressing both CatB and CatL better than it infects cells expressing CatB but not CatL. This is a true statement, but it’s out of scope and similar to answer B. There’s no clear evidence here that CatL expression is sufficient for VSV-EGP infection. We have a combination of both CatB and CatL and we don’t know the effect of each.
41) This question is going to be similar to question 40 in that we’re asked to read the Figure presented in the passage.
Optimal infection in this case is going to be the greatest percent of total cells infected. We have the highest percentage in lane 3 when we have CatL but no CatB initially, then CatB gene is introduced. We also have a higher level when we introduce CatB + CatL genes. Optimal VSV-EGP infection therefore requires both CatB and CatL: Answer choice C.
42) We can break down what this question is asking at its core by breaking down two main components. We know from the passage that CatB and CatL are mammalian proteases. Protease enzymes break down proteins into smaller fragments, often via hydrolysis. We want to know their effect on Ebola virus glycoprotein (EGP). Glycoporteins, like the name suggests, are proteins with glycans attached to amino acid side-chains. If we have proteases introduced to proteins, we expect the proteases enzymes will break down the proteins into smaller segments. Correct answer here will be answer choice D.
43) We’re told EGP is similar to sGP, but we have the insertion of a nucleotide to the open reading frame. That’s also known as a frameshift mutation and can change the reading frame. We also know this is not a nonsense mutation that would act as a stop codon. Proteins are translated from the N-terminus to C-terminus, so if we add an extra nucleotide to the open reading frame, we expect the C-terminus to change, but the N-terminus will not. The N-terminus should just contain methionine, the start codon.
- amino-terminal sequence as sGP, but a different primary carboxy-terminal sequence. This matches our breakdown exactly. We expect the C-terminus to change, but the N-terminus to have the same sequence.
- carboxy-terminal sequence as sGP, but a different primary amino-terminal sequence. This is the opposite of what we expect to happen. In fact, if we changed the primary amino-terminal sequence, we might actually see a change in carboxy-terminal sequence as well. Answer choice A remains the best option.
- sequence as sGP except that EGP has one additional amino acid. We have to note the difference between amino acid and nucleotide. Only an additional nucleotide is being added, not necessarily an entire amino acid (three nucleotides).
- sequence as sGP except that EGP has one less amino acid. Reasoning here is going to be the same as answer choice C. We do not have three nucleotides removed in this situation, but rather we have the addition of a single nucleotide. Answer choice A remains the best option.
Exam 2 B/B Solutions: Questions 44-47
44) This is a content question that comes down to knowing the citric acid cycle.
We have the formation of L-malate, succinate, and alpha-ketoglutarate. Phosphoenolpyruvate is involved in glycolysis and gluconeogenesis. That leaves answer choice D as our correct answer here. Make sure to review your content if you are not comfortable with this!
45) First thing we need to do here is review glycolysis and the number of moles of ATP produced by the consumption of a single mole of glucose.
For every mole of glucose, there is a net production of 2 moles of ATP. That means 5 moles glucose generates 10 moles of ATP. We also have to determine the total number of molecules. We can use Avogadro’s number. For every mole of ATP, there are 6 x 1023 molecules/mole. Multiply (10 moles) x (6 x 1023 molecules/mole) = 6 x 1024 molecules.
This was a math problem where we did minimal approximation and no rounding. That means we can stick with our correct answer, answer choice B: 6 x 1024 molecules.
46) This is a standalone question that, like most, is testing our knowledge of content. We know both eggs and sperm play an essential role in the reproduction process, but we have to know how they differ and how they’re similar. We’re focused on how they’re similar.
- cell size. An egg is many times larger than a sperm cell. Although both contain a haploid number of chromosomes, the egg is much wider and there is a lot more cytoplasm. The cell size is not similar.
- genome size. This is something we touched on in answer choice A. Both the egg and sperm contain a haploid number of chromosomes. This is going to be a good answer choice for the time being. Let’s see if the additional answers are differences or similarities.
- the time required for development. Egg cell production begins while the female fetus is in the uterus and these eggs drop into the uterus monthly, later in life. Sperm cells are constantly produced and replenished. Answer choice B is still the best similarity.
- the numbers produced by a single individual. The number of sperm cells produced vastly outnumber the eggs produced by individuals. This is a stark difference, so we’ll stick with the biggest similarity listed: Answer choice B, genome size.
47) This is a content question that involves knowing which organs are derived from which layer. We’re focused on the mesoderm.
- Brain we can check our visual and note the nervous system is derived from the ectoderm. Right away this is a strong answer choice. Note we’re looking for an organ that is NOT derived from mesoderm.
- Heart the mesoderm gives rise to heart and the circulatory/lymphatic systems. Answer choice A is going to remain the best option.
- Kidney checking our visual, the kidneys are derived from the mesoderm as well.
- Skeletal muscle the skeletal system is derived from the mesoderm, so answer choice A is going to be the best option. The brain is the only organ listed that is not derived from mesoderm.
Exam 2 B/B Solutions: Passage 9
48) To answer this question, we’ll go back to the passage for some key information. We’re told CTLs use secretory lysosomes called lytic granules to secrete cytolytic proteins. That’s how “killing by CTLs” is done.
In Table 1, we see how inactivating different proteins affects the function of the CTLs. If we inactivate Rab27a, we do not have the normal killing by CTLs. However, the question asks “secretion of lytic granules does NOT require.” Rab27a looks like it IS required, because with Rab27a inactivation we have no killing by CTLs. On the other hand, myosin Va is inactivated by mutation, but killing by CTLs is not affected. That means the secretion of lytic granules does not require myosin Va. Best answer choice here is going to be the one that matches our breakdown exactly, answer choice B: myosin Va.
49) This is going to be similar to Question 48 in that the answer is mostly going to come from reading the passage carefully and picking out the necessary details. We’re told myosin Va is a motor protein on microfilaments. Motor proteins are non-enzymatic proteins that perform mechanical movement in cells or muscles.
Myosin is a fibrous protein that forms (together with actin) the contractile filaments of muscle cells and is also involved in motion in other types of cells. Motor proteins like myosin molbe along microfilaments by interacting with actin, so our correct answer here will be answer choice B: Actin.
50) While this is a passage-related question, we will likely be able to use external knowledge to answer this question. We’re told secretory lysosomes have functional components in common with conventional lysosomes. We’ll lay out what we know about conventional lysosomes and then find an answer choice that is consistent with our breakdown.
Lysosomes are a cell’s “garbage disposal.” Hydrolytic enzymes within the lysosomes aid the breakdown of proteins, polysaccharides, lipids, nucleic acids, and worn-out organelles. These enzymes are active at a much lower pH than that of the cytoplasm. Therefore, the pH within lysosomes is more acidic than the pH of the cytoplasm. In general, we can say lysosomes contain digestive enzymes for breaking down parts of the cell, as well as material that has been taken into the cell by phagocytosis.
- ribosomes. Ribosomes were not one of the key components we brought up in our breakdown.
- Krebs cycle enzymes. Krebs cycle enzymes were not one of the key components we brought up in our breakdown
- RNA and DNA polymerases. This is similar to answer choices A and B. We’re not expecting polymerases which form polymers.
- degradative enzymes that function at low pH. This is going to be the ideal answer. We said lysosomes contain digestive enzymes for breaking down parts of the cell; the pH within lysosomes is more acidic than the pH of the cytoplasm. Answer choice D is going to be the best answer choice.
51) We know from our passage melanosomes are a type of secretory lysosome that transport melanin pigments. We’re also told melanosomes are transported along microtubules to the cell periphery where the melanosomes transfer to microfilaments prior to the release of melanin from the cell. Do we have to go much further here? It’s unlikely. We’re asked about where microtubules originate in and radiate from. We can answer this question using our general knowledge.
A centriole is a cylinder of nine triplets of microtubules, held together by supporting proteins. Centrioles are best known for their role in centrosomes, structures that act as microtubule organizing centers in animal cells. A centrosome consists of two centrioles oriented at right angles to each other, surrounded by a mass of pericentriolar material, which provides anchoring sites for microtubules.
The centrosome is duplicated before a cell divides, and the paired centrosomes seem to play a role in organizing the microtubules that separate chromosomes during cell division. Our best answer here is going to be that microtubules originate from centrosomes. Answer choice A is our best answer.
52) We know from the passage, CTLs use secretory lysosomes called lytic granules to secrete cytolytic proteins. We can break down what we know about how T-cells work and antigen presentation.
Antigen presentation broadly consists of pathogen recognition, phagocytosis of the pathogen or its molecular components, processing of the antigen, and then the presentation of the antigen to naive (mature but not yet activated) T cells. The ability of the adaptive immune system to fight off pathogens and end an infection depends on antigen presentation. T cells must be presented with antigens in order to perform immune system functions. The T cell receptor is restricted to recognizing antigenic peptides only when bound to appropriate molecules of the MHC complexes on APCs. Each type of T cell is specially equipped to deal with different pathogens. Helper T cells receive antigens from MHC II on an APC, while cytotoxic T cells receive antigens from MHC I. The only answer choice here that matches our breakdown is going to be answer choice A.
Exam 2 B/B Solutions: Passage 10
53) To answer this question, we can pick out some key information from the passage. The author mentions “Foxp3 binds specific sequences in the ErbB2 promoter, and deletion of these Foxp3-binding sites increases expression from the ErbB2 promoter in Foxp3-expressing cells.” Binding to the promoter insinuates we’re affecting synthesis at the transcriptional, or mRNA, level. We know from earlier in the passage that cancerous mammary epithelium has many times more mRNA than noncancerous mammary epithelium. That likely means Foxp3 is inhibiting synthesis of ErbB2 mRNA, and that is why there is 8- to 12-fold less mRNA in noncancerous mammary epithelium.
- inhibiting synthesis of ErbB2 mRNA. This answer choice matches our breakdown exactly. A key component was mRNA instead of protein, and inhibiting synthesis of ErbB2 in noncancerous epithelium. This is a good answer choice.
- stimulating synthesis of ErbB2 mRNA. This answer choice incorrectly mentions stimulating synthesis of ErbB2 mRNA instead of inhibiting synthesis.
- inhibiting synthesis of ErbB2 protein. We mentioned in the breakdown we’re dealing with transcription, not translation. That’s why our correct answer lists mRNA, not protein.
- stimulating synthesis of ErbB2 protein. We mentioned in the breakdown we’re dealing with transcription, not translation. That’s why our correct answer lists mRNA, not protein. Answer choice A is going to remain the best option.
54) This question is going to be similar to Question 53 because we will have to pick our key information from the passage to answer this question. We know from the very beginning of the passage that scurfin (sf) allele is a likely culprit in developing mammary tumors, while Foxp3 inhibits synthesis of ErbB2 mRNA. Cancerous mammary epithelium has more much mRNA for ErbB2 than does noncancerous mammary epithelium. The question relies on analyzing each answer choice individually rather than coming up with a predicted answer first, so we can jump right into our answers.
- The Foxp3 genes deleted This would be the opposite effect that we’re looking for. We mentioned Foxp3 inhibits synthesis of ErbB2 mRNA. If Foxp3 genes are deleted, we expect the mice are less likely to survive the longest.
- One of the ErbB2 genes amplified This is similar to answer choice A. If we have additional ErbB2, that corresponds to cancerous mammary epithelial cells. This would also decrease survival.
- A Foxp3-expressing plasmid introduced This matches our breakdown of the question. Foxp3 inhibits synthesis of ErbB2 mRNA. Cancerous mammary epithelium has more much mRNA for ErbB2 than does noncancerous mammary epithelium. Answer choice C is going to be our best option so far.
- The Foxp3 binding sites deleted in the promoter of one of the ErbB2 genes Reasoning here is the same as answer choices A and B. We said Foxp3 inhibits synthesis of ErbB2 which we’d likely see in the mice that survive the longest. Answer choice C remains our best answer choice.
55) This question is as simple as going back to the first part of the passage and noting how scurfin allele affects the likelihood that mice develop mammary tumors.
The author explicitly tells us that a majority of female Foxp3sf/+ mice (60%) that carry one inactivating scurfin (sf) allele and one wild-type (+) allele of the Foxp3 gene, spontaneously develop mammary tumors by 2 years of age. This question was as simple as reading carefully the first time you went through the passage, or knowing exactly where to find this necessary information quickly. The best answer here is going to be answer choice A: 1.
56) This question is tangentially-related to the passage because we’re dealing with the human homolog of ErbB2, which was discussed in the passage. Ultimately, we’re looking at the structure of an antibody and asked which regions are highly specific for the extracellular domain of HER2. We can pull up a labeled antibody here:
The above figure is labeled with all of the important parts we’ll need to answer out question. Specifically, the area where the antigen is recognized on the antibody is known as the variable domain or variable region. That’s labeled in the figure above, and it corresponds to Regions 1 and 3 in the question stem. We’re going to pick the answer choice that corresponds to the variable regions, answer choice B: Regions 1 and 3 only.
Exam 2 B/B Solutions: Questions 57-59
57) This is a standalone question that relies on our knowledge of the human male reproductive system. Most of the male reproductive system is located outside of the body (external structures) and includes penis, scrotum, epididymis, and testes. The internal organs or the accessory organs include the vas deferens, seminal vesicles, prostate gland, and bulbourethral glands.
Let’s go through our four answer choices here. We can define each structure and its function.
- Testis this is the location for testosterone production. The coiled collection of tubes within the testes is the seminiferous tubules. Within these tubules, spermatogenesis takes place.
- Urethra is the duct by which urine is conveyed out of the body from the bladder. The vas deferens transports mature sperm to the urethra in preparation for ejaculation. Male urethra conveys sperm, in addition to urine.
- Epididymis is located at the back of the testis and connects it to the vas deferens. This is where the sperm fully matures and becomes motile, and it functions to store and carry sperm. This is going to be our best answer because this is where sperm become motile and capable of fertilization.
- Prostate gland this is a walnut-sized structure located below the urinary bladder in front of the rectum. It also contributes to semen production. Our best answer here is going to be answer choice C: Epididymis.
58) We can refer to our general knowledge to answer this question. The mechanism of contraction is the binding of myosin to actin, forming cross-bridges that generate filament movement.
We know muscle contraction happens when myosin binds to actin. ATP is used to detach the myosin from actin, but if the body is no longer making ATP and there’s no ATP available, then we can’t have the release of the myosin head from the actin filament. A lot of you have probably heard of rigor mortis. Essentially, the muscles stiffen after death and stay stiff for quite some time. The lack of ATP means myosin adheres to actin, but the myosin head cannot be released from the actin filament. Only answer choice D is consistent with our visual and our breakdown.
59) This question boils down to knowing different experimental setups you see on the MCAT. Proteins can be chemically denatured before running them on the gel and are separated exclusively by size. The chemical SDS denatures proteins, and so denaturing SDS-PAGE gels are used for these experiments.
Proteins can be further denatured by the addition of a reducing agent that breaks down the disulfide bonds between cysteine residues (these are not fully broken by SDS alone). When we have a protein that does not break down fully, oftentimes we need a reducing agent to break down the disulfide bonds.
- Increasing the gel running time this would not affect whether the disulfide bonds in the protein are broken down or not.
- Adding a reducing agent this is exactly what we said in our breakdown of the question. The reducing agent breaks down the disulfide bond in the homodimeric protein.
- Using a higher voltage this does not solve our disulfide bond issue either. Answer choice B remains superior.
- Removing the SDS removing SDS would mean the protein doesn’t break down at all in the first place. We actually want something that would increase the chances of observing the mobility expected for the 22.5-kDa monomer. The way we do that is answer choice B: Adding a reducing agent.