Exam 1 B/B Solutions: Passage 1

1) For passages like this one, it’s always imperative to understand what is going on in the experiments, but not worry about memorizing any specific details. For example, note any trends or outliers in the experimental results, but you don’t have to memorize tumor volumes at every time interval. Along those same lines, we want to look at Figure 1 to start jumping into some observations about HSP110ΔE9.

We can see the effect of HSP110ΔE9 on tumorigenesis by comparing with HSP110WT. What biggest trend do we see? Relative to the wild type, HSP110ΔE9 slows tumor growth. The tumor volume shown along the Y-axis as time passes is increasing much more slowly for the cells expressing HSP110ΔE9. We know cancer as a disease in which the cells of a tissue undergo uncontrolled proliferation, but the HSP110ΔE9 allele appears to be cancer-suppressing.

1. Cancer-promoting and dominant to HSP110WT. Right away, this does not come across as a good answer because it says the HSP110ΔE9 allele promotes cancer. If this were the case, we would see tumor volume increase relative to HSP110WT in Figure 1. This is not a good answer choice. However, it does mention the HSP110ΔE9 allele being dominant to HSP110WT which appears correct. There is a clear difference in tumor volume in the HSP110ΔE9 versus HSP110WT mice because the HSP110ΔE9 allele is dominant to HSP110WT.
2. Cancer-promoting and recessive to HSP110WT. This answer choice is incorrect on two accounts. We said the HSP110ΔE9 allele is dominant to HSP110WT. Additionally, the allele is cancer-suppressing.
3. Cancer-suppressing and dominant to HSP110WT. This answer choice matches our breakdown of the question. The HSP110ΔE9 allele appears to be cancer-suppressing, and there is a difference in tumor volume in the HSP110ΔE9 versus HSP110WT mice because the HSP110ΔE9 allele is dominant to HSP110WT.
4. Cancer-suppressing and recessive to HSP110WT. First part of this answer choice is consistent with our breakdown, but the second half is incorrect. We said the HSP110ΔE9 allele is dominant to HSP110WT. Answer choice C is going to be our best answer here. 

2) This is going to be similar to Question 1 from this question set where we’ll need to go back and find the necessary information from the research results to answer our question. This question asks us to relate the size of the deletion in the HSP110 T17 microsatellite to our four answer choices and pick the choice that inversely related. How should we attack this question? We can start with Table 1 which shows us the Effect of T17-deletion size on relative expression of HSP110ΔE9 in MSI CRC tumor cells.

First thing we want to note is the independent variable in this case is the T₁₇-deletion size (bp). We’re increasing by one bp and seeing the effect on HSP110ΔE9 mRNA expressed. We can see a clear trend as deletion size increases. HSP110ΔE9 mRNA expressed as a % of total HSP110 mRNA expressed is also increasing. As deletion size increases, the amount of HSP110ΔE9 mRNA expressed relative to the total amount of HSP110 mRNA expressed in MSI CRC primary tumor cells increases. We want an inverse correlation for this question, so we can say as deletion size increases, the amount of wild-type HSP110 (HSP110WT) expressed relative to the total amount of HSP110 mRNA expressed in MSI CRC primary tumor cells decreases.

1. the amount of HSP110ΔE9 protein expressed. This is an incorrect correlation. As the size of the deletion in the HSP110 T₁₇ microsatellite increases, we see the amount of HSP110ΔE9 mRNA expressed relative to the total amount of HSP110 mRNA also increases. This is not an inverse correlation.
2. the number of mature HSP110WT transcripts synthesized. This is consistent with our breakdown. We said as deletion size increases, the amount of wild-type HSP110 (HSP110WT) expressed relative to the total amount of HSP110 mRNA expressed in MSI CRC primary tumor cells decreases. Answer choice B correctly shows an inverse relationship.
3. the frequency of the omission of HSP110 Exon 9 during splicing. We are told in the passage “The larger HSP110 T17 deletions cause Exon 9 to be omitted from the final sequence during pre-mRNA processing.” That means as deletion size increases, more Exon 9 is omitted. This is not an inverse correlation.
4. the extent of premature translation termination in Exon 10. We are told in the passage, “The larger HSP110 T17 deletions cause Exon 9 to be omitted from the final sequence during pre-mRNA processing. This generates a premature stop codon in Exon 10.” That means as deletion size increases, more Exon 9 is omitted, and that generates a premature stop codon/translation termination in Exon 10. This is not an inverse correlation, so answer choice B is going to be our best answer. 

3) When it comes to science questions, we always make sure to pick up on the big picture of every passage, but we have to be aware the test-maker can test on smaller details from the passage as well. That’s going to be the case in this question. The author points out colorectal cancers are caused by defects in the mismatch repair system. Additionally, in some cases, there can also be microsatellite instability. There’s a distinction here between defects in the MMR and MSI which is important. The author also provides a detail in Paragraph 2: “People with MSI CRC have HSP110ΔE9 transcripts in cancerous tissue only.” We’re not focused on defects in the MMR here (which may be genetic), we’re focused on the CRC mutation that results in the synthesis of HSP110ΔE9 (which is present in cancerous tissue only, or somatic cells). The tumors themselves cannot be passed on from the man to the child. This means that based on what we read in the passage, that corresponds to answer choice A: 0%.

4) We are told in the passage the mismatch repair system corrects certain DNA replication errors. Think of the MMR like someone that scrutinizes your every move. If there is a mistake, the MMR system is there to call it out! We want to go through the pairings in the question stem and find any mismatches.

According to our base pairings, dAMP and dTMP would be a proper pairing. However, Option I. dTMP and dCMP and Option II. dGMP and dAMP are mismatches. That leaves only one correct pair (Option III). We can stick with Options I and II only as our correct answer here for the pairings that would be recognized by the MMR system during DNA replication. Answer choice C is correct. 

5) To answer this question, we can go back to Paragraph 2 from the passage where the author talks about Intron 8 of HSP110.  

First part of the paragraph mentions deletions of 3 to 8 base pairs in the 17-thymine nucleotide microsatellite (T₁₇) located in Intron 8. The larger HSP110 T₁₇ deletions cause Exon 9 to be omitted from the final sequence during pre-mRNA processing. This generates a stop codon in Exon 10 and the mutant protein HSP110ΔE9. How might Exon 9 be omitted? That sounds like classic alternative splicing. Alternative splicing is a process that occurs during gene expression and allows for the production of multiple proteins (protein isoforms) from a single gene coding. Alternative splicing can occur due to the different ways in which an exon can be excluded from or included in the messenger RNA. It can also occur if portions on an exon are excluded/included or if there is an inclusion of introns. Introns are non-coding and will be removed during rNA processing, but intron 8 most likely has a splice acceptor site that influences the splicing (or lack thereof) taking place.

1. Stop codon. Intron 8 itself does not have a stop codon, but rather it generates a stop codon in Exon 8. Introns themselves are non-coding, and the passage is not implying there is a stop codon in Intron 8.
2. Splice acceptor site. This matches our breakdown exactly. Alternative splicing allows for the production of multiple proteins based on whether exons (like Exon 9) are included or omitted.
3. HSP110 gene promoter. This is going to tie into what we said in answer choice A. Introns are non-coding and removed during RNA processing; the passage is not implying there is a gene promoter in Intron 8.
4. Partial coding sequence of HSP110. Introns are non-coding, and the passage is not implying there is a coding sequence of HSP110 in Intron 8. We can stick with answer choice B as our most likely conclusion. 

6) We’re asked about heat shock protein 110 and we specifically want to identify the type of protein. Heat shock proteins are produced in response to heat shock and stress and they function to stabilize and refold proteins. We can define each of the protein types listed as answer choices and see which one is most applicable to heat shock proteins and HSP110 specifically.

1. Adhesion. Adhesion proteins are glycoproteins that mediate cell-cell connections. They’re found on the membranes of cells and interact in the space between the cells, holding the membranes together.
2. Chaperone. Chaperone proteins assist in folding parts of a protein which is consistent with the function of heat shock proteins. Heat shock proteins function to stabilize and refold proteins which are functions of chaperone proteins.
3. Clathrin. Clathrin works to form vesicles to assist in transport within cells. Not something consistent with the function of HSPs. 
4. Enzyme. Enzymes catalyze chemical reactions by lowering activation energy barriers and converting substrate molecules to products. Once again, not the function of HSPs. We can stick with answer choice B as our best answer choice.

 

Exam 1 B/B Solutions: Passage 2

7) We’re starting the question set with a pseudo-standalone question here. The passage did talk about tryptophan, but this question is focusing on the properties of tryptophan (and amino acids) which come from our external knowledge. I always urge you to take the time to go through all of your amino acids and review their structures, properties, and abbreviations. This is high-yield material!

We can utilize the visual above that shows all the amino acids we want to know. Tryptophan is classified as nonpolar and has the aromatic side chain. We know from the passage, AT1 is a neutral amino acid transporter. That means if we want to predict an answer, the amino acid that is most likely to also be transported by AT1 will be either phenylalanine or tyrosine based on our amino acid classifications. If we don’t have an aromatic option, we will still make sure to pick a neutral amino acid.

1. Phenylalanine. This option matches our breakdown. We mentioned phenylalanine, like tryptophan, is one of our neutral, aromatic amino acids. Phenylalanine would be recognized by the same transport mechanism. This is a strong starting point, but we still want to compare with the other answer choices.
2. Lysine. Lysine has the positively charged side group and is not one of the aromatic amino acids. Answer choice A remains superior.
3. Arginine. Arginine, like lysine, has the positively charged side group and is not one of the aromatic amino acids. 
4. Glutamate. Glutamate is a negatively charged amino acid and also not aromatic. We can stick with answer choice A as our best option.

8) We can start this question by focusing on Paragraph 1 where the author introduces AT1 and its function.

AT1 functions as a transmembrane protein and simultaneously transports tryptophan against its concentration gradient and a sodium ion along its concentration gradient into the cytoplasm of intestinal epithelial cells. We can go through and define our four answer choices within the context of transmembrane proteins and AT1. The mRNA is most likely to contain a sequence coding for one the four genetic factors listed.

A. Signal sequence. Transmembrane proteins require specific modifications that occur in the rough endoplasmic reticulum. The mRNAs for these proteins contain a specific signal sequence that directs them to dock onto the ribosomes on the RER. Ribosomes transfer the proteins into the lumen of the RER where they undergo structural modifications. These modified proteins will be incorporated into membranes or secreted from the cell. That signal sequence is a necessary, specific sequence found in the mRNA of proteins that allows for docking onto the RER. This is a good answer choice. 

B. Introns. Introns are intervening sequences within a pre-mRNA molecule that do not code for proteins and are removed during RNA processing by a spliceosome. Mature AT1 mRNA will not likely contain introns.

C. Promoter. To begin transcribing a gene, RNA polymerase binds to the DNA of the gene at a region called the promoter. We don’t have these promoter sequences in mRNA

D. Nuclear localization signal. Proteins with a nuclear localization signal are recognized by nuclear pores for entry into the nucleus. Our transmembrane protein is not likely to be entering the nucleus, so we can eliminate this answer choice. We’re left with answer choice A as our best answer.

9) For passages like this one, it’s always imperative to understand what is going on in the experiments, but not worry about memorizing any specific details. For example, note any trends or outliers in the experimental results, but you don’t have to memorize the exact percent weight loss in Table 1 for every situation. We can look at the data in Table 1 which shows the effects of Ace2 genotype and diet on colitis (through monitoring weight loss) and circulating tryptophan. Quick glance at our answer choices shows we’re focused on changes in weight. That means we’ll pull up Table 1 here:

Ace2 genotype	Diet	Percent weight loss 4 days after start of DSS treatment (%)	Tryptophan blood levels (relative to control)
+/y (WT)	SD*	2	1.0
PFD	22	not determined
SD + G–T	2	1.4
–/y	SD*	10	0.3
PFD	20	not determined
SD + G–T	2	1.0

We are focused on the protein-free diet and its effect on weight. We see both the +/y and the -/y mice lost weight. The +/y and the -/y mice lost 22 and 20 percent respectively after start of DSS treatment.

1. Both the +/y and the –/y mice gained weight. This answer choice is the opposite of what we see in the passage. We see both the +/y and the -/y mice actually lost weight, not gained weight.
2. The +/y mice lost weight, and the –/y mice gained weight. This answer choice is only half right. Both groups of mice lost weight. 
3. The +/y mice gained weight, and the –/y mice lost weight. This answer choice is only half right. Both groups of mice lost weight.
4. Both +/y and the –/y mice lost weight. This answer choice matches our prediction exactly. In Table 1, we see both the +/y and the -/y mice lost weight. The +/y and the -/y mice lost 22 and 20 percent respectively after start of DSS treatment. We can stick with answer choice D as our correct answer.

10) While this question is related to the passage we just read, the answer is going to come primarily from our external knowledge. NAD is a dinucleotide that contains an adenine nucleobase and nicotinamide. Essentially, we have a deficiency in the latter nucleotide. NAD exists in an oxidized form, NAD⁺, and a reduced form, NADH. We specifically want to know when nicotinamide nucleotides are neither oxidized nor reduced during cellular respiration. If you need to review any pathways prior to the exam, I highly recommend using the content outline on our website. I’ve added some visuals that help demonstrate an example of where we see oxidation/reduction in the incorrect answers. Quick glance at our answer choices shows we’ll be touching on a few major pathways and we want to decide if nicotinamide nucleotides are oxidized or reduced.

A. Glycolysis

B. Chemiosmosis. Chemiosmosis is used to generate 90 percent of the ATP made during aerobic glucose catabolism. It is also the method used in the light reactions of photosynthesis to harness the energy of sunlight in the process of photophosphorylation. During chemiosmosis, the free energy from the series of reactions that make up the electron transport chain is used to pump hydrogen ions across the membrane, establishing an electrochemical gradient. Nicotinamide nucleotides are neither oxidized nor reduced in this step. Answer choice B is going to be our correct answer.

C. Citric acid cycle

D. Electron transport chain

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