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.

1. 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.
2. 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. 
3. 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.
4. 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.

1. Adipocytes. Adipocytes are essential in energy storage, but they are not highly proliferative. The cells will fluctuate in size, but not unusually in number. 
2. 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. 
3. 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. 
4. 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.

A.

Left side of answer choice A shows a pyrimidine, but we said we’re looking for single-ringed structures.

B.

Neither side of this structure shows a pyrimidine. We’re looking for a dimer that contains two single-ringed pyrimidines.

C.

This answer choice matches our breakdown. We said pyrimidines are single-ringed. This dimer contains pyrimidine monomers.

D.

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).

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).
2. 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.
3. 7 < pH < 10 This is a basic pH which is the opposite effect of what we’re looking for. 
4. 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)
Complex I	26	13
Complex II	449	340
Complex IV	2.2	0.7

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.

1. 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. 
2. 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.
3. 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.

4. 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)
Complex I	26	13
Complex II	449	340
Complex IV	2.2	0.7

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.

1. 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.
2. 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. 
3. 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.
4. 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). 

1. 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
2. Golgi apparatus sorts and packages materials before they leave the cell to ensure they arrive at the proper destination.
3. 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. 
4. 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. 

1. 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.
2. 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.
3. 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. 
4. 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. 

1. 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.
2. 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. 
3. 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. 
4. 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.

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