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MCAT Content / AAMC MCAT Practice Exam 1 Bb Solutions

AAMC FL1 BB [Web]

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

11) First thing we want to do as we go through this question is gather necessary details about Ace2.

Let’s think about X-linked (or sex-linked) genes, which are referring to genes on the X chromosome in humans. Because males only have one X chromosome, they only inherit one allele of sex-linked genes, making males more susceptible to diseases like hemophilia caused by sex-linked genetic mutations. As with any question like this, we can draw out our Punnet square according to our question stem. We have a heterozygous mother (X’X) and an affected father (X’Y):

  X’ X
X’ X’X’ X’X
Y X’Y XY

If a man with a mutant copy of Ace2 has a child with a woman that is heterozygous for the mutant Ace2 allele, there is a 25% chance the child will be a female that is homozygous for the mutant Ace2 allele.

 

Exam 1 B/B Solutions: Questions 12-15

12) This is a standalone question that relies exclusively on our external knowledge. To answer, we will need to review muscle contraction.

Note step 4 in the visual: myosin binds actin after Ca2+ binds troponin, exposing myosin-binding sites. We can stick with our correct answer, answer choice D: Ca2+.

13) This is a standalone question that relies exclusively on our external knowledge. We can pull up the steroid structure here and count the total number of fused rings:

The total number of fused rings present in a steroid is answer choice C: 4

14) This is a fairly lengthy standalone question, so we have to make sure we don’t miss any key information. Epilepsy can result in motor seizures because of a specific cause (massive synchronous firing of neurons). The epileptic focus is the site of the brain where the actual seizure originates-this synchronous firing is in a small area of the cerebral cortex. However, excitation spreads from the focus, so we want to pick a change in the epileptic focus that would treat epilepsy. To treat the epilepsy, we’ll want to make sure we focus on the cause: the massive synchronous firing of neurons. For reference as we go through our answer choices, we can pull up a visual detailing action potential here. It’s a good idea to reference the visual as you go through each answer choice so you can get a sense of the effect of each proposed change in the epileptic focus.

  1. An increase in the neuron-firing threshold. We said in our breakdown, the cause we’re trying to reduce or minimize is the massive synchronous firing of neurons in that small area of the cerebral cortex. One possible way to do that would be to increase the neuron-firing threshold. We can reference our visual as well. Note the dotted threshold line going horizontally across our image. By increasing that threshold that needs to be surpassed, we can effectively reduce the massive synchronous firing of neurons and the excitation that spreads from the focus. This is a strong answer choice.
  2. An increase in extracellular Na+ concentration. An increase in extracellular sodium concentration leads to additional sodium influx as well. The ions will move to adjust for the charge difference, and the threshold is reached more quickly. Answer choice A remains a more solid answer. 
  3. A decrease in axon–membrane permeability to negative ions. This ties into what we saw in answer choice B. By not allowing negative ions into the cell, we have more positive potential and the threshold is reached more quickly.
  4. A decrease in the length of the depolarization stage. This could lead to the opposite effect of what we’re looking for from the drug. A decrease in the length of the depolarization stage can mean the neuron is able to fire again that much more quickly. We want to stick with our best option that effectively reduces the massive synchronous firing of neurons: Answer choice A.

15) This is a standalone question that relies on our knowledge of the kidneys and filtration.

It’s imperative you know what each structure in this visual does and how concentration is affected within each structure. Glomerular filtrate is at its highest concentration near the bottom of the loop of Henle before the ascending loop, and again before it’s excreted as urine. We can go through the four structures given as answer choices and determine which one is consistent with our breakdown.

  1. Proximal convoluted tubule. The proximal convoluted tubule has a relatively low concentration. We can compare with our additional answer choices to see if we can get any better answers. Specifically, we want something that mentions the bottom of the loop of Henle or the collecting duct right prior to excretion.
  2. Distal convoluted tubule. The distal convoluted tubule has a similar concentration to the proximal convoluted tubule. Answer choice A remains superior. 
  3. Cortical portion of the collecting duct. The collecting duct system can be divided into cortical and medullary ducts. Cortical ducts will receive filtrate and filtrate will descend into medullary collecting ducts. 
  4. Medullary portion of the collecting duct. The medullary portion of the collecting duct is the latter portion and the answer choice where glomerular filtrate reaches its highest concentration. Of the four options listed, answer choice D is our best answer. 

 

Exam 1 B/B Solutions: Passage 3

16) This is a passage-based question and asks about Experiment 1 specifically. We want to know the part of the experiment that does NOT address whether membrane composition has an effect on Na+K+ ATPase activity. Be careful with the verbiage here. Three of our answer choices will address membrane composition’s effect on Na+K+ ATPase activity. Our correct answer will not address a possible effect.

  1. showed less temperature dependence in the 14:1 liposome than the 14:0 liposome. We can see in our two graphs that Na+K+ ATPase activity is less affected as temperature increases in the monounsaturated PC (14:1) than in the saturated PC (14:0). That means different membrane composition does have an effect.
  2. was highest in the 14:0 liposomes at all temperatures. This is going to be similar to answer choice A because we’re seeing a difference in in Na+K+ ATPase activity between the 14:0 and 14:1 liposomes at all temperatures. This means different membrane composition does have an effect.
  3. increased with temperature in both the 14:1 liposome and the 14:0 liposome. We want to be careful here because this is not comparing the different membrane compositions. Answer choices A and B correctly compared the different membrane compositions, while answer choice C makes a general observation about both sets of liposomes. Increased activity in both sets of liposomes does not address whether membrane composition has an effect on Na+K+ ATPase activity. The key here is whether the presence/lack of cholesterol and the saturation of the PCs is compared. In this case we’re not looking at any differences. That makes answer choice C our best answer so far.
  4. was greater at all temperatures when cholesterol was present. This is going to be similar to answer choices A and B where we’re actually seeing the effect of cholesterol versus no cholesterol on Na+K+ ATPase activity. Answer choice C is the only answer choice that does not address whether membrane composition has an effect on Na+K+ ATPase activity.

17) The passage focused primarily on Na+K+ ATPase, and activity of Na+K+ ATPase. To answer this question, we may need key information from the passage, but we also have to know about enzyme activity in general. Three of our answer choices will measure the activity of Na+K+ ATPase, but our correct answer does NOT measure this activity. We’ll keep in mind Na+K+ ATPase is an enzyme that functions to bring two K+ ions into the cell while removing three Na+ ions per ATP consumed.

  1. Measuring the rate of ATP hydrolysis. We mentioned in our breakdown that Na+K+ ATPase is an enzyme that functions to bring two K+ ions into the cell while removing three Na+ ions per ATP consumed. If we have more ATP hydrolyzed, there’s more Na+K+ ATPase activity. We can eliminate this answer choice because it does measure Na+K+ ATPase activity.
  2. Measuring the free energy of the ion transport. To measure the activity of the enzyme, we want to focus on the kinetics of the reaction. Reactions can usually be influenced by two factors: the relative stability of the products (i.e. thermodynamic factors) and the rate of product formation (i.e. kinetic factors). Measuring the free energy of ion transport is a thermodynamic property, while enzyme activity is a kinetic property. This is going to be our best answer choice so far.
  3. Measuring the rate of ADP production. This is going to be similar to answer choice A. If we have a higher rate of ADP production, it is because we have increased Na+K+ ATPase activity. More ATP is being used and more ADP is being produced. We can eliminate this answer choice because it does measure Na+K+ ATPase activity.
  4. Measuring the change in ion concentration within the liposome. Na+K+ ATPase is an enzyme that functions to bring two K+ ions into the cell while removing three Na+ ions per ATP consumed. By measuring changes in ion concentration, we can effectively measure Na+K+ ATPase activity. Answer choice B is going to remain our best option for this question. 

18) This question is part of a passage-based question set, but we can consider this like a pseudo-standalone question. We can pull up a figure that helps us visualize action potentials and note the function of the Na+K+ ATPase. 

We’re focused near the bottom right of the figure. The K+ channels close, and the Na+/K+ transporter restores the resting potential.

  1. Stimulation of the action potential. A stimulus from a sensory cell or another neuron causes the target cell to depolarize toward the threshold potential. If the threshold of excitation is reached, voltage-gated Na+ channels open, and the membrane depolarizes. This is not a function of the Na+K+ ATPase.
  2. Depolarization of the membrane. This ties into our previous answer. If the threshold of excitation is reached, voltage-gated Na+ channels open, and the membrane depolarizes. This is not a function of the Na+K+ ATPase.
  3. Hyperpolarization of the membrane. At the peak action potential, K+ channels open and K+ begins to leave the cell. At the same time, Na+ channels close. The membrane becomes hyperpolarized as K+ ions continue to leave the cell. The hyperpolarized membrane is in a refractory period and cannot fire.
  4. Restoration of the resting potential. K+ channels close, and the Na+/K+ transporter restores the resting potential. This is going to be our best answer choice. Answer choices A-C did not list functions of the Na+K+ ATPase during a neuronal action potential. 

19) This question is part of a passage-based question set, but we can consider this like a pseudo-standalone question. Na+K+ ATPase is an enzyme that functions to bring two K+ ions into the cell while removing three Na+ ions per ATP consumed:

The only answer choice consistent with our breakdown and our visual is going to be answer choice A: Na+ is transported out of the cell; K+ is transported into the cell.

 

Exam 1 B/B Solutions: Passage 4

20) To answer this question, we can go back to the passage where the author introduces p65 and cRel

We’re told these proteins are transcription factors. A transcription factor is a protein that, like the name suggests, controls the rate of transcription of genetic information from DNA. This happens through binding a promoter region. Let’s go through some background. 

DNA binding proteins help to regulate protein production, cell growth and division, and storing DNA inside the nucleus. Histones, repressors, and activators are examples of proteins that bind to DNA. Eukaryotes also require several other proteins, called transcription factors, to first bind to the promoter region, and then help recruit the appropriate polymerase. The completed assembly of transcription factors and RNA polymerase bind to the promoter, forming a transcription pre-initiation complex. Upon activation, p65 and cRel control the level of IL-6 mRNA by binding DNA. Only answer choice B matches our breakdown. 

21) The test-maker explicitly tells us we’ll need to use the structure of STN to answer this question. We can pull up the structure here, and we’ll use external knowledge to explain how the structure ties into the mechanism for entry into the cell.

Let’s make some general observations. We have a simple, planar, non-polar, hydrophobic molecule. We want to know how this structure will enter into the cell. We can go through our four options while keeping the structure in mind. Remember, we’re not losing sight of the specific question being asked. We want to know the most likely mechanism by which it enters into the cell. 

  1. Active transport. Active transport requires energy to move substances against a concentration gradient, from an area of low concentration to high concentration. The structure of STN is not going to cause a need for active transport. Instead, this would only be the correct answer if STN was going from low to high concentration in the cell.
  2. Receptor mediated endocytosis. Receptor-mediated endocytosis employs receptor proteins in the plasma membrane that has a specific binding affinity for certain substances and works similarly to phagocytosis. This does help carry large particles across the cell membrane. We do have a larger molecule, but it has a simple, planar structure that is usually not indicative of RME.
  3. Diffusion directly through the membrane. Generally, only molecules that are small and hydrophobic can diffuse through the cell membrane. However, molecules that are planar are also more likely to diffuse through the membrane. Our initial observations about STN were simple, planar, nonpolar, and hydrophobic. These properties are all consistent with being able to diffuse directly through the membrane. This is going to be our best answer choice.
  4. Passage through an ion channel. This answer choice would only be tempting if we were dealing with an ion (STN is not an ion). We can eliminate this answer choice as well. We’re left with answer choice C as our best answer.

22) This is a passage-based question, so we can pull up the relevant portion of the passage here to get more clues into the event that directly activates CARD11. 

Looking at Figure 1, I’ve highlighted the activation of CARD11 by PKC. We’re told in the passage PKC is protein kinase C. A kinase is an enzyme that catalyzes the transfer of phosphate groups; kinases phosphorylate molecules. The only answer choice listed that is consistent with PKC activating CARD11 is answer choice D: Phosphorylation. 

23) We’re told in Paragraph 1, “The NF-κB proteins p65 and cRel are transcription factors that can be continuously activated when mutations occur to proteins upstream in the signaling pathway.” This is something that ties into Question 20 from this question set as well. A transcription factor is a protein that, like the name suggests, controls the rate of transcription of genetic information from DNA. This happens through binding a promoter region. Upon activation, p65 and cRel control the level of IL-6 & IL-10 mRNA by binding DNA.

  1. They cause disruption of the mitochondria. We actually know from the passage, activation of NF-κB proteins p65 and cRel leads to increased cell proliferation and suppression of apoptosis (a form of programmed cell death). The mitochondria plays a role in apoptosis, so this answer choice is unlikely.
  2. They contain a p65/cRel binding site in their promoter region. This answer choice is consistent with our breakdown. A transcription factor is a protein that, like the name suggests, controls the rate of transcription of genetic information from DNA. This happens through binding a promoter region. This is a strong answer choice. 
  3. They have accumulated mutations that alter function. We are told that NF-κB proteins p65 and cRel are transcription factors that can be continuously activated when mutations occur to proteins upstream in the signaling pathway. Their activation leads to increased cell proliferation and suppression of apoptosis. The genes themselves do not have mutations. Answer choice B remains a superior answer.
  4. They bind to STN in the cytoplasm. This answer choice is out of scope. The author does not bring up binding to STN, nor the location of any binding. We want an answer choice that is correct according to the passage. Answer choice B is our best answer.

 

Exam 1 B/B Solutions: Passage 5

24) To answer this question, we can focus on the background information the author provides us about P-gp. We’re focused on the mechanism by which P-gp facilitates drug resistance.

P-gp is an ABC transporter and prevents drugs from accumulating to cytotoxic levels. P-gp also exhibits ATPase activity, which makes sense. The ATP-binding cassette (ABC) transporter is a classic example of active transport. ABC transporters use ATP to transport a variety of different substrates across cell membranes, most commonly out of the cell. In this case, P-gp uses ATP to actively transport drugs out of the cell. 

  1. P-gp binds to antitumor drugs in the presence of ATP and degrades the drugs. This is describing a ligand which is a chemical molecule that binds to a receptor. We know we’re dealing with a transporter which functions to transport substrates across membranes, not degrade them. The author does not mention the degradation of the drugs.
  2. P-gp serves as a pump and uses active transport to move antitumor drugs outside the cell. This answer choice matches our breakdown. The ABC transporter is a classic example of active transport. ABC transporters use ATP to transport substrates, often out of the cell. ABC transporters are examples of primary active transport which uses a chemical energy source like ATP to move solutes against their concentration gradient. This is a strong answer choice.
  3. P-gp prevents the entry of anti-tumor drugs into the cell. This goes back to our definition in our breakdown of the question. ABC transporters will use ATP to transport drugs out of the cell, they do not function to prevent the entry of drugs into the cell. The author does not mention the lack of entry into the cell by anti-tumor drugs. Answer choice B remains superior.
  4. P-gp causes increased membrane permeability, which causes antitumor drugs to exit the cell. Increasing membrane permeability is not a one-way ordeal. The author didn’t explicitly mention there was a higher concentration of drugs inside or outside of the cell, but increasing permeability would increase the ability of the drug to both go into or out of the cell. P-gp instead uses ATP to move antitumor drugs to exit the cell. We can stick with answer choice B as our best answer.

25) This question is asking about a specific detail from the passage. We’re looking at MBCD-treated MDR cells after exposure to 100 µM cholesterol, which means we can focus on Figure 2 where we see the cholesterol-dependent ATPase activity of native MDR and MDR cells pre-treated with MBCD for 1 hour at 37°C. 

We see that as concentrations of cholesterol are increased to 100 µM cholesterol, ATPase activity increases in MBCD-treated MDR cells. ATPases function to catalyze the decomposition of ATP into ADP + Pi. More ATPase activity means more ATP is hydrolyzed, and more ADP and inorganic phosphate are produced. The only answer choice that matches our breakdown is answer choice B: ADP + Pi

26) The best way to approach this question is to revisit the part of the passage that focuses on P-gp. To answer our question, we’re focused on the location of P-gp within the plasma membrane.

The passage tells us P-gp is localized in cholesterol-rich domains within the membrane. That’s our biggest clue into the location of P-gp within the plasma membrane. Given AAMC’s format, it’s unlikely our final answer will be as simple as “cholesterol-rich domains,” but rather we’ll rely on our external information to connect our answer choices to cholesterol-rich domains within the membrane. There are areas of high cholesterol concentration in the membrane where the composition of the proteins, carbohydrates and different lipids differ from the rest of the membrane – these are called lipid rafts and they are thought to be specialized microdomains in the membrane.

  1. Associated with lipids on the cytoplasmic side only. First part of our answer choice starts our promising. We do expect P-gp is associated with lipids, however, we’re not told a specific side on which we would find P-gp. That means at first glance, this might be an extreme answer choice because we’re limiting the location, but it is still a decent answer choice.
  2. Associated with lipids on the extracellular side only. This is similar to answer choice A as we do expect P-gp is associated with lipids, however, we’re not told a specific side on which we would find P-gp. This is also an extreme answer choice. We can dive further into answer choices A and B if we do not find a superior answer choice.
  3. Peripheral to the plasma membrane. We are not expecting P-gp is a peripheral protein. Peripheral proteins and carbohydrates form specialized sites on the cell surface that allow cells to recognize each other. That is not consistent with the cholesterol-rich domain or function of P-gp from the passage.
  4. Within a lipid raft. A lipid raft is an area in the membrane where there is a high concentration of cholesterol, and a different composition of carbohydrates, proteins and other lipids compared to the rest of the membrane. Because we know P-gp is localized in cholesterol-rich domains, we expect them to be localized within a lipid raft. Answer choice D is our best answer.

27) Like the other questions in this set, this will come down to revisiting key information from the passage. The initial readthrough of the passage might not focus on smaller details, but at the very least, we should be able to recall where in the passage we’ll find the necessary information. As the author gets into Experiment 1, we’re told MBCD is a cyclic oligosaccharide. 

At this point, this becomes a standalone question. We can rely on external knowledge to answer. An oligosaccharide is a carbohydrate chain formed by several monosaccharides connected by glycosidic linkages. The only answer choice consistent with this description is answer choice D: carbohydrates.

 

Exam 1 B/B Solutions: Questions 28-31

28) This is a standalone question that relies on our knowledge of microtubules. If we prevent the formation of microtubules, one of the four mitotic processes in our answer choices would not occur. Three of the answer choices would still happen, while our correct answer would no longer occur after exposure to the drug. Recall mitosis steps consist of prophase, metaphase, anaphase, and telophase. 

Microtubules are hollow tubes that are composed of tubulin proteins and help the cell transport materials within itself and resist shape changes. Microtubules are the largest element of the cytoskeleton and commonly used by eukaryotic cells. The walls of the microtubule are made of polymerized dimers of α-tubulin and β-tubulin, two globular proteins. They help the cell resist compression, provide a track along which vesicles move through the cell, and pull replicated chromosomes to opposite ends of a dividing cell. In mitosis specifically, microtubules form the mitotic spindle, which is responsible for the capture of chromosomes and align them at the center during prometaphase. In the first part of anaphase, the kinetochore microtubules shorten and draw the chromosomes toward the spindle poles. In the second part, the astral microtubules, which anchor to the cell membrane, pull the poles further apart.

  1. Formation of new cell plate. Cytokinesis is the actual splitting of the cell membrane; animal cells pinch apart, while plant cells form a cell plate that becomes the new cell wall. Mitosis steps consist of prophase, metaphase, anaphase, and telophase.
  2. Replication of DNA during interphase. Mitosis steps consist of prophase, metaphase, anaphase, and telophase. During interphase, the cell grows, and DNA replicates. 
  3. Breakdown of the nucleolus during prophase. Mitosis begins in prophase, where the chromosome condenses into chromatids. A centromere connects each chromatid to its copy, making the linked pairs look like X’s. The nuclear envelope starts to disintegrate, mitotic spindle begins to assemble, and centriole pairs move toward opposite poles of the cell. The breakdown of nucleolus does not involve microtubules.
  4. Movement of the chromosomes toward opposite poles of the cell during anaphase. This answer choice matches our breakdown. During anaphase, sister chromatids physically separate at the centromere and pull towards opposite poles of the cell by the mitotic spindle. Microtubules are essential in anaphase. We can stick with answer choice D as our correct answer.

29) The first step of the formation of the urine is glomerular filtration in the glomerulus. The glomerulus is the site in the nephron where fluid and solutes are filtered out of the blood to form a glomerular filtrate. The process of glomerular filtration filters out most of the solutes, particularly large solutes like proteins, due to the high blood pressure and specialized membranes in the afferent arteriole. 

The glomerulus forces solutes out of the blood by pressure.

  1. passive flow due to a pressure difference. The glomerulus forces solutes out of the blood by pressure. Fluid and solutes are filtered out of the blood and into the space made by Bowman’s capsule which surrounds the glomerulus. 
  2. passive flow resulting from a countercurrent exchange system. This is not happening in the glomerulus, but rather in different parts of the nephron like we see in our visual. For example, we can see in our visual that we have movement of water and electrolytes in the loop of Henle, but we have passive flow due to the pressure difference in the glomerulus.
  3. active transport of water, followed by movement of electrolytes along a resulting concentration gradient. There is no active transport or ATP involved in the movement of water. Different electrolytes can be moved actively, but not in the glomerulus. 
  4. active transport of electrolytes, followed by passive flow of water along the resulting osmolarity gradient. Once again this is not correct in the glomerulus. There is movement of water in the loop of Henle and active transport of electrolytes, but we’re focused on the initial filtration step in the glomerulus of the mammalian kidney. Answer choice A is our best answer.

30) To answer this question, we will look at the figure provided and determine the half-life of urea. We’re told in the question stem itself, “The half-life of urea is the time it takes for its concentration to fall by one-half,” so we can analyze our graph. At a time of 0 hours, we have an initial plasma concentration of urea of 200 ng/mL. The time it takes to get to half of that concentration (100 ng/mL) is 0.5 hours along the x-axis. If we keep going, the time it takes to get to half of 100 ng/mL is another 0.5 hours along the x-axis. This tells us the half-life of urea is 0.5 hours. We want to know the length of time equal to five half-lives for urea. We can multiply:

5 half-lives x (0.5 hours / 1 half-life) = 2.5 hours. This is a math problem that involved no rounding or approximation. We can find the answer choice that matches our calculated value: Answer choice B: 2.5 hours.

31) Enzymes catalyze chemical reactions by lowering activation energy barriers and converting substrate molecules to products. Enzymes bind with these substrates, which are chemical reactants. A specific chemical substrate matches this site like a jigsaw puzzle piece and makes the enzyme specific to its substrate. 

When an enzyme binds its substrate, it forms an enzyme-substrate complex. This complex lowers the activation energy of the reaction and promotes its rapid progression by providing certain ions or chemical groups that actually form covalent bonds with molecules as a necessary step of the reaction process. Enzymes also promote chemical reactions by bringing substrates together in an optimal orientation, lining up the atoms and bonds of one molecule with the atoms and bonds of the other molecule. This can contort the substrate molecules and facilitate bond-breaking. The active site of an enzyme also creates an ideal environment, such as a slightly acidic or non-polar environment, for the reaction to occur. The enzyme will always return to its original state at the completion of the reaction. One of the important properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its products (substrates).

  1. co-localizing substrates. This is consistent with our breakdown. We said enzymes promote chemical reactions by bringing substrates together in an optimal orientation, lining up the atoms and bonds of one molecule with the atoms and bonds of the other molecule. 
  2. altering local pH. This is consistent with our breakdown. We said the active site of an enzyme creates an ideal environment, such as a slightly acidic or non-polar environment, for the reaction to occur.
  3. altering substrate shape. This is consistent with our breakdown. We said enzymes can contort the substrate molecules and facilitate bond-breaking.
  4. altering substrate primary structure. Enzymes are not responsible for changing the primary structure of a protein. Whenever we do have the change in structure of a protein, it’s not the enzyme itself that is changing the structure. Furthermore, this is not one of the methods by which enzymes alter the rate of chemical reactions. That makes answer choice D our best answer choice listed.

 

Exam 1 B/B Solutions: Passage 6

32) To answer this question, we will have to go back to our passage and pick out some key information. The author tells us in the passage, “Ornithine is an amino acid that is found in cells, but not incorporated into proteins.” We want a reason for why ornithine isn’t likely found in proteins synthesized in vivo. 

  1. There is no codon for it in the standard genetic code. This is consistent with what we learned in the passage. We know ornithine is found in cells, but is not incorporated into proteins. Genetic code is a set of nucleotides that code for specific amino acids during protein synthesis. These are in the form of triplets that are known as codons. No codons code for ornithine in the genetic code which is why ornithine isn’t found in proteins synthesized in vivo. This is a strong answer choice, let’s keep comparing.
  2. It cannot form a peptide bond. This answer is out of scope. While it may or may not be true that ornithine cannot form a peptide bond, this does not mean ornithine is not found in proteins synthesized in vivo. That being said, the structure presented in the passage (R = –CH2CH2CH2NH2) suggests that ornithine can form a peptide bond. Answer choice A is superior because it provides a better, more direct answer to the specific question being asked. 
  3. It is not available in the diet. Not all amino acids in our bodies come from our diet. 10 of the amino acids are considered essential amino acids for humans since the human body cannot produce them. Nonessential amino acids can be synthesized in the body and can also be formed from precursors. There is ornithine in the cells, so presumably ornithine is either coming from the diet, or it is possible it is being formed from its precursors. This is not the fundamental reason ornithine is unlikely to be found in proteins synthesized in vivo. 
  4. It has a net positive charge in aqueous solution. Given there are positively charged amino acids that are found in proteins synthesized in vivo, this is an unlikely answer choice. We can stick with answer choice A as our best, most likely answer choice.

33) First thing we can do to answer this question is pull up Equation 1 so we can look into the role of ornithine decarboxylase. 

We’re focused on the enzyme ornithine decarboxylase. Enzymes catalyze chemical reactions by lowering activation energy barriers and converting substrate molecules to products. Decarboxylase enzymes specifically catalyze the decarboxylation of amino acids like ornithine.

  1. Catalyst. We mentioned in our breakdown ornithine decarboxylase is an enzyme that catalyzes the reaction shown in Equation 1. A catalyst is a substance that increases the rate of a chemical reaction by lowering the activation energy without being used up in the reaction.
  2. Cofactor. Cofactors are inorganic ions that assist an enzyme in its catalytic activity. We said in our initial breakdown of the question, ornithine decarboxylase is the enzyme itself, not a cofactor. Answer choice A remains superior. 
  3. Substrate. A reactant in a chemical reaction is called a substrate when acted upon by an enzyme. We can reference our breakdown and the visual in our breakdown to see the relationship and difference between a substrate and enzyme
  4. Activator. Enzyme activators can bind to enzymes to alter activity, but that does not describe ornithine decarboxylase. We mentioned ornithine decarboxylase is an enzyme and catalyst. Answer choice A is our best answer. 

34) This is tangentially related to Question 33 in this question set as it’s asking about the ornithine decarboxylase reaction. We can pull up the specifics of the reaction here and go through our four answer choices to find the reason for the extensive research.

We’re told ornithine is found in cells, but not incorporated into proteins. Earlier in the question set we mentioned this likely means there is no codon for ornithine in the standard genetic code. We’re told in mammals and many other organisms, this reaction is an early event in cell division.

  1. a good source of ornithine. The reaction is not a good source of ornithine, but rather ornithine is converted to diaminobutane and bicarbonate via the reaction. 
  2. an early event in cell division. This is consistent with what we’re told in the passage. Cell division is a very important process, so it stands to reason this reaction would be studied extensively by researchers. This is going to be superior to answer choice A which incorrectly said the reaction is a good source of ornithine. 
  3. unique to mammals. The passage mentions the presence of ornithine in mammals and many other organisms. This answer choice is too extreme. The ornithing reaction is not unique to mammals. 
  4. one in which a chiral reactant is converted to an achiral product. While this might be something that certain researchers would want to look into, note we’re focused on biomedical researchers and why they would want to study the reaction extensively. Cell division is the more relevant process and more interesting to biomedical researchers. The most likely reason for the interest of biomedical researchers is because the ornithine decarboxylase reaction is an early event in cell division. Answer choice B is going to be our best answer. 

35) We’re jumping into another question in this set that focuses on the ornithine decarboxylase reaction. 

Enzyme assays are used to measure enzymatic activity, and we have to explain enzyme specificity to answer this question. An enzyme is uniquely suited to bind to a particular substrate to help catalyze a biochemical reaction. The assay takes advantage of this specificity.

The enzyme’s active site binds to the substrate. Since enzymes are proteins, this site is composed of a unique combination of amino acid residues. The positions, sequences, structures, and properties of these residues create a very specific chemical environment within the active site. A specific chemical substrate matches this site like a jigsaw puzzle piece and makes the enzyme specific to its substrate. 

  1. requires radioactive ornithine of high specific activity. Specific activity is the activity in a quantity of a radionuclide. There is no indication in the passage that the assay requires radioactive ornithine of high specific activity. Instead, we’re focused on the specificity of the enzyme. The assay measures enzymatic activity, and our focus is on enzyme specificity. 
  2. generates diaminobutane of high specific activity. This is going to be similar to answer choice A. Specific activity is the activity in a quantity of a radionuclide. There is no indication in the passage that the reaction generates diaminobutane of high specific activity. The specificity in this case refers to the specificity of the enzyme. 
  3. can distinguish ornithine decarboxylase activity from the many other enzymatic reactions in a cell. This is consistent with our breakdown. An enzyme is uniquely suited to bind to a particular substrate to help catalyze a biochemical reaction. The assay has to be able to distinguish the activity of the ornithine decarboxylase from that of other enzymatic reactions. The reason it can do that is because of enzyme specificity. This is going to be our best answer choice here.
  4. can measure the small amount of ornithine present in a cell. The assay being highly specific has more to do with the unique properties of the enzyme, not with the assay being so precise that it measures the small amount of ornithine present in a cell. We can stick with answer choice C as our best answer. 

 

Exam 1 B/B Solutions: Passage 7

36) To answer this question, we can refer to the key information in the passage that references how alcohol is related to cP-450.

We’re told that cP-450 usually metabolizes barbiturates, but alcohol acts as a competitive inhibitor of barbiturate metabolism. 

cP-450 usually metabolizes barbiturates; alcohol acts as a competitive inhibitor of cP-450 in the mechanism shown in our visual above, and cP-450 can’t metabolize barbiturate. We’re told in the case of the alcoholic, an autopsy revealed that if alcohol had not been present, the barbiturates would have been metabolized rapidly enough to be eliminated by the body, and death would not have occurred. Because alcohol inhibited the cP-450, barbiturates were not metabolized rapidly enough. The only answer choice that matches our breakdown and is consistent with what the author tells us in the passage is answer choice B: inhibited the cP-450.

37) This is a pseudo-standalone question. Even though it’s part of the question-set related to this passage, we are going to focus on our external knowledge, and specifically our knowledge of acetyl CoA. Why do I say that? Because we’re told how acetyl coenzyme A is formed using components from the passage, but ultimately this question is about the acetyl CoA itself. 

A. the Krebs (citric acid) cycle. Acetyl-CoA is the reactant needed in the citric acid cycle. This is going to be our correct answer choice. Typically, we see acetyl CoA produced by the oxidation of pyruvate, which is the end product of glycolysis. That acetyl CoA is involved in the first step of the Krebs cycle (visual below)

B. glycolysis. Acetyl CoA is produced by the oxidation of pyruvate, which is the end product of glycolysis. However, acetyl CoA is not needed to participate in glycolysis.

C. electron transport. Acetyl CoA itself is not used directly in the electron transport chain. Answer choice A remains superior.

D. oxidative phosphorylation. Oxidative phosphorylation is the final stage of cellular respiration where the combined action of the electron transport chain and chemiosmotic coupling result in ATP production. Once again, acetyl CoA is not fed into oxidative phosphorylation. Only answer choice A lists a correct answer. Acetyl coenzyme A that is formed can participate in the citric acid cycle. 

38) To answer this question, we can consider the key information we learned about cP-450 to answer this question in the

First of all, why do expect greatly increased cP-450? Because there are additional toxins (from smoking) that must be metabolized. We know that process is often catalyzed by cP-450. However, is there always cP-450 ready to go? We’re actually told cP-450 is inducible. When the body detects a new toxin or increased concentrations of a toxin, the body increases the concentration of cP-450 to metabolize the toxin more effectively. How does the body increase concentration of this protein? Transcription of cP-450 increases and, to an extent, the amount of protein directly increases. 

  1. DNA sequences that code for cP-450. We don’t expect increased DNA itself. What we expect to change would be DNA that is expressed or not. This is an incorrect answer choice.
  2. mRNA sequences that code for cP-450. Messenger RNA or mRNA is a type of RNA that is copied from a DNA sequence during transcription and is used to create a protein during translation. When we have increased concentrations of cP-450, these correlate to increased concentrations of mRNA sequences that code for cP-450. In order to have higher levels of protein, we also expect higher levels of mRNA. This is a correct answer choice. 
  3. rRNA that process cP-450. While mRNA is a unique form of RNA that carries information about the sequence of a protein by encoding amino acids, rRNA is ribosomal RNA and does not encode amino acids. rRNA instead serves as the building blocks of ribosomes. We can stick with answer choice B as the best answer. mRNA levels and protein levels are most closely related of our answer choices. 
  4. tRNA that are specific for cysteine. We’re told in the passage that many isoenzymes of cP-450 have been isolated, all of which contain iron with 4 coordinate bonds to nitrogen atoms in the heme group, and 1 coordinate bond to a cysteine residue in the protein portion of the enzyme. tRNA is a type of RNA that carries amino acids to a growing polypeptide chain by recognizing the codons in an mRNA sequence. There is 1 coordinate bond to a cysteine residue in the protein portion of the enzyme. This does not imply that when there are increased concentrations of cP-450 we have increased tRNA specific for cysteine. Answer choice B is our best answer choice.

39) First thing we’ll do to answer this question is to note the part of the passage that talked about the case in which the chronic alcoholic died of a barbiturate overdose. 

The author says in the beginning of the paragraph, “While sober, the alcoholic was having trouble falling asleep and, therefore, took the recommended dosage of sleeping pills (barbiturates).” Despite taking the sleeping pills, the alcoholic was still having trouble sleeping. This question comes down to explaining the reasoning behind the trouble sleeping despite taking the sleeping pills. The best explanation here is going to be an increased concentration of cP-450 in the alcoholic’s body prior to taking the sleeping pill. We’re told earlier in the passage, “cP-450 is inducible. That is, when an organism is challenged by a new toxin or by increased concentrations of a toxin, the organism can increase its concentration of cP-450 and thus metabolize and excrete the toxin more effectively.” The alcoholic often drinks, and therefore has increased concentrations of toxins in the blood. That means cP-450 concentration is already high and will metabolize the sleeping pills quickly, at least relative to someone that takes the same dosage and is not an alcoholic. That means the reason no drowsiness was initially felt by the alcoholic is because the previous abuse of alcohol had induced the cP-450. That elevated cP-450 concentration metabolized the sleeping pills quickly; no drowsiness was felt.

  1. denatured the cP-450. This is the opposite of what we expected to have happened. The previous abuse of alcohol actually meant there were increased cP-450 levels. 
  2. inhibited the cP-450. Similar to answer choice A, this is the opposite of what we expected to have happened. The previous abuse of alcohol actually meant there were increased cP-450 levels. cP-450 wasn’t inhibited until alcohol was consumed.
  3. reduced the cP-450. Similar to answer choice A once again; this is the opposite of what we expected to have happened. The previous abuse of alcohol actually meant there were increased cP-450 levels.
  4. induced the cP-450. This is consistent with our breakdown. The reason no drowsiness was initially felt by the alcoholic is because the previous abuse of alcohol had induced the cP-450. That elevated cP-450 concentration metabolized the sleeping pills quickly.

 

Exam 1 B/B Solutions: Passage 8

40) To answer this question, we can go back to our passage and find the key information the author tells us about the pericytes used in the experiments. We specifically want to know which phase of the cell cycle in which we’d find the pericytes.

We can pinpoint what the author says about the pericytes used in the experiment. They were growth-arrested, meaning they were treated so they wouldn’t divide, but other metabolic processes would function normally.

Growth arrest, or G0, is also viewed an extended G1 phase where the cell stops growing. Not all cells undergo mitotic phase. Cells in the G0 phase are not actively preparing to divide. The cell is in a quiescent stage that occurs when cells exit the cell cycle, and that’s what we see in the pericytes. 

Some cells enter G0 temporarily until an external signal triggers the onset of G1. No more DNA replication or cell division happens at this phase. The cells that never or rarely divide include mature cardiac muscle and nerve cells, and they remain in G0 permanently. 

In this case, we can say the pericytes were either in the G0 phase or interphase.

  1. Telophase. Telophase is a phase of mitosis and occurs during cell division. During telophase, chromosomes arrive at opposite poles and unwind into thin strands of DNA, the spindle fibers disappear, and the nuclear membrane reappears.
  2. Metaphase. Metaphase is a phase of mitosis and occurs during cell division. During metaphase, sister chromatids align along the middle of the cell by attaching their centromeres to the spindle fibers.
  3. Anaphase. Anaphase is a phase of mitosis and occurs during cell division. During anaphase, sister chromatids physically separate at the centromere and pull towards opposite poles of the cell by the mitotic spindle.
  4. Interphase. This matches our initial breakdown of the question. We can reference our visual as interphase is the phase of the cell cycle between divisions. Cells enter the G0 inactive phase after they exit the cell cycle when they are not actively preparing to divide. The cell can still conduct normal functions, but it is not dividing. We can stick with answer choice D as our best answer.

41) This is a pseudo-standalone question that is only tangentially related to our passage. We can do a brief description about erythrocytes and see what we think about the student’s hypothesis. Red blood cells, or erythrocytes, are specialized cells that circulate through the body delivering oxygen to other cells. Erythrocytes are formed from stem cells in the bone marrow. In mammals, these red blood cells are small, biconcave cells that, at maturity, do not contain a nucleus or mitochondria. EC growth would therefore not be affected by the DNA from circulating erythrocytes because these erythrocytes do not have nuclei or DNA.

  1. No; the DNA in circulating erythrocytes is needed to help transport O2 through the capillaries. Erythrocytes do not contain a nucleus at maturity. Instead, erythrocytes contain hemoglobin, a protein that uses iron to carry oxygen.
  2. No; circulating erythrocytes do not contain DNA. This answer choice matches our breakdown. Mature erythrocytes do not contain a nucleus or mitochondria. EC growth would therefore not be affected by the DNA from circulating erythrocytes because these erythrocytes do not have nuclei or DNA.
  3. Yes; DNA is responsible for cell division in most cells. Right away we can eliminate this answer choice because we said the student’s hypothesis is unreasonable. Mature erythrocytes do not contain a nucleus or DNA that would affect EC growth. Answer choice B remains superior.
  4. Yes; circulating erythrocytes carry DNA nutrients through the capillaries. This is an incorrect statement. Mature erythrocytes do not contain a nucleus or DNA that would affect EC growth. The only answer choice that matches our breakdown is answer choice B.

42) To answer this question we can go back to the passage and provide a little background for this question. Then we can jump into the reasoning behind the pericytes being growth-arrested in Experiment 1. 

We are told pericytes were growth-arrested, meaning they were treated so they wouldn’t divide, but other metabolic processes would function normally. This question is asking the reasoning behind this. To get to the reasoning, we can look at the results of the experiment.

We have growth-arrested cells (pericytes, smooth-muscle cells, and fibroblasts) mixed directly with ECs in three different containers. As time passes, the number of cells in each container is counted and the results are shown in the Figure. Our dependent variable here is the number of cells in each container. If the pericytes are allowed to divide themselves, the number of cells in the containers would not be controlled. We would not know if the change in number of cells was due to the change in pericytes or the ECs. By treating the pericytes so they would not divide, the researchers can attribute the change in cell numbers to the growth of the ECs and not have to consider and account for the change in the number of pericytes. 

  1. pericyte growth would not interfere with the measurement of EC growth. This answer choice matches our breakdown exactly. By treating the pericytes so they would not divide, the researchers can attribute the change in cell numbers to the growth of the ECs and not have to consider and account for the change in the number of pericytes. We can focus on our dependent variable (EC growth) without needing to account for pericyte growth.
  2. pericytes would not inhibit EC growth. This is what the researchers are testing in the experiment. They want to measure growth/inhibition of growth of ECs in different environments, including with pericytes. Answer choice A remains superior. 
  3. metabolic wastes of the pericytes would not interfere with the measurement of EC growth. While metabolic waste might be something the researchers would consider, when cells are growth-arrested their metabolic processes still function normally. We would still have the usual metabolic wastes.
  4. ECs would not cause the pericytes to grow. This is out of scope. The pericytes are growth-arrested, and the researchers are not focused on stopping the pericytes from growing due to being mixed with ECs. The researchers care about the growth of ECs and simply controlled for the growth of pericytes to conduct their experiment. Answer choice A remains the best answer choice.

43) To answer this question, we can go back to Figure 1 from the passage. We’re going to note the cells that are most important in the exchange of O2 between the blood and the surrounding tissues which is going to require us using external knowledge. 

Endothelial cells are specialized cells that allow the permeability of selective materials through the walls of the blood vessels. We’re also told in the passage they define the lumen of capillaries and are associated with pericytes, which are contractile cells that line capillaries and venules. ECs form barriers between blood vessels and the surrounding tissues. They also help in the movement of chemicals that facilitate in vasoconstriction and vasodilation, such as epinephrine. From Figure 1, we can see endothelial cells are going to be most important in exchanging O2 (and selective materials) between the blood and surrounding tissues. 

  1. Pericytes. Pericytes are elongated contractile cells, but are not responsible for the exchange of O2 between the blood and surrounding tissues.
  2. Endothelial cells. This answer choice is our best answer. ECs form barriers between blood vessels and the surrounding tissues.
  3. Smooth-muscle cells. Smooth muscle is responsible for involuntary movement and is found on the walls of major organs. This is out of scope.
  4. Fibroblasts. Fibroblasts are cells that generate any connective tissue that the body needs, as they can move throughout the body and can undergo mitosis to create new tissues. This answer choice is also out of scope. Answer choice B is our best answer. 

 

Exam 1 B/B Solutions: Questions 44-47

44) This is a standalone question that is asking us to compare and contrast bacteria and animal viruses. A virus is about ten times smaller than a typical bacteria cell, and at least 100 times smaller than a typical eukaryotic cell. Viruses cannot replicate by themselves and must infect a host cell to hijack their reproductive machinery to replicate; they cannot replicate themselves. Bacteria can make their own food, move, and reproduce. Viral infections are systemic, while bacterial infections are usually localized.

  1. are obligate parasites. This is consistent with our breakdown. Viruses cannot replicate by themselves and must infect a host cell to hijack their reproductive machinery to replicate.
  2. lack DNA. Certain viruses contain DNA, so this answer choice is factually incorrect.
  3. assimilate carbon. Viruses depend on their hosts to survive and cannot assimilate carbon. This answer choice is factually incorrect.
  4. require essential vitamin supplements for growth. Viruses must infect a host cell to hijack their reproductive machinery to replicate. They do not require specific nutrients, but rather entire hosts. We can stick with answer choice A as the best answer here.

45) This is a standalone question that relies on our knowledge of telomeres and bacterial cells. Telomeres are the capping ends present at the ends of the chromosomes that protect DNA during replication. 

Telomeres help protect DNA during replication and prevents its degradation. The biggest thing we want to notes is that telomeres are capping ends present at the end of chromosomes. Bacterial DNA is circular, so there is not the same need for telomeres to act as capping ends. Telomeres are therefore not as important to bacterial cells.

  1. do not replicate. This is factually incorrect as bacterial cells and chromosomes do replicate.
  2. are circular. This answer choice is consistent with our breakdown. Telomeres help protect DNA during replication and prevents its degradation. Bacterial DNA is circular, so there is not the same need for telomeres to act as capping ends. This is going to be our best answer so far.
  3. replicate quickly and efficiently. This is out of scope. The speed and efficiency with which bacteria replicate does not explain the need for telomeres. We can stick with answer choice B as a superior answer choice here.
  4. are composed of single-stranded DNA. Bacteria will use double-stranded DNA, but still do not have the same need for telomeres to act as capping ends. We can stick with answer choice B as our best answer choice. 

46) This is a standalone question and we’re given the structures of thymine and adenine. We know from our external knowledge that the nitrogen bases A and T (or U in RNA) always go together (and C and G always go together), forming the 5′-3′ phosphodiester linkage found in nucleic acid molecules. 

A phosphodiester bond is a linkage that occurs when two hydroxyl groups in a phosphate molecule react with hydroxyl groups on ribose sugars to form two ester bonds. The phosphate groups form bonds between the 3rd carbon (3’) of the ribose sugar of one base and the 5th carbon (5’) of the ribose sugars of the other base, forming the phosphate backbone of a DNA strand. The only answer choice that is consistent with our breakdown is answer choice D: A bond between the phosphate of the adenine (5’) and the sugar of the thymine (3’).

47) This is a standalone question that relies on our knowledge of action potentials. We can pull up a visual that goes through the necessary details here and determine what happens during the movement of sodium ions into a neuron.

We have voltage-gated sodium channels open and sodium rushes into the neuron. This movement of sodium ions causes the membrane to depolarize. We can see from our visual the other answer choices are all incorrect. We’re left with answer choice D as our best answer.

 

Exam 1 B/B Solutions: Passage 9

48) This is a question that is tangentially related to the passage, but is ultimately a pseudo-standalone question. We touched on Addison’s disease during the passage, but the effect of aldosterone deficiency on serum ion levels should also be general knowledge.

We’re told in Paragraph 2, “Aldosterone, the primary mineralocorticoid, maintains ionic balance by causing conservation of Na+ and excretion of K+.” A lack of aldosterone should, therefore, have the opposite effect on these particular ions. A lack of aldosterone means less conservation of sodium ions and less excretion of potassium ions. 

  1. Na+ ions. A lack of aldosterone means less conservation of sodium ions. We should see a decrease in the serum level of sodium ions.
  2. Cl ions. Similar to answer choice A, when we have a lack of aldosterone, we have less conservation of chloride ions. We should also see a decrease in the serum level of chloride ions. 
  3. K+ ions. This answer choice is consistent with our breakdown and what we learned in the passage. A lack of aldosterone means less conservation of sodium ions and less excretion of potassium ions. We also know from our general knowledge, a lack of aldosterone means less excretion of potassium ions, so no decrease in serum levels of potassium ions.
  4. HCO3 ions. This is going to be similar to answer choices A and B. Aldosterone increases absorption of bicarbonate ions. An aldosterone deficiency will cause a decrease in the serum levels of bicarbonate ions.

49) This is going to be another passage-based question that relies on picking out key information from the passage. However, we will likely rely on our external knowledge of hormones and the role of the adrenal cortex to answer this question. We know from our general knowledge that ACTH is one of the seven main hormones produced by the anterior pituitary. A hypothalamic factor stimulates the release of ACTH. We want to know what happens to the secretion of the hypothalamic factor in a patient with Addison’s disease. Let’s quickly revisit essential information from the passage. The author tells us in Paragraph 2, “Addison’s disease occurs when cells of the adrenal cortex are destroyed, leaving the gland unable to secrete either glucocorticoids or mineralocorticoids.” The adrenal cortex is the largest part of the adrenal gland, which is an endocrine gland. Most endocrine glands are under negative feedback control that acts to maintain homeostasis. This negative feedback prevents deviation from an ideal value. In this case, however, the cells of the adrenal cortex are destroyed, so we likely don’t get that same negative feedback that’s necessary to maintain that ideal value. Normally, negative feedback would prevent the additional secretion of hypothalamic factor when levels are ideal. In this case, that negative feedback isn’t happening so we’d have higher than normal secretion of the hypothalamic factor.

  1. be lower than normal. As we mentioned in the breakdown of the question, negative feedback prevents deviation from an ideal value. In this case, the cells of the adrenal cortex are destroyed, so we likely don’t get that same negative feedback that’s necessary to maintain that ideal value. We expect the secretion of the hypothalamic factor will be lower than normal.
  2. be higher than normal. This answer choice is consistent with our breakdown. Normally, negative feedback would prevent the additional secretion of hypothalamic factor when levels are ideal. In this case, that negative feedback isn’t happening so we’d have higher than normal secretion of the hypothalamic factor. Answer choice B is our best answer choice so far.
  3. be unchanged. This is unlikely as we’ve lost the negative feedback that typically prevents deviation from the ideal values. We expect the secretion of the hypothalamic factor will be higher than normal.
  4. increase before disease onset and decrease thereafter. This is the opposite of our breakdown. In a patient with Addison’s disease, the secretion of hypothalamic factor will be higher than normal. We expect an increase after disease onset. Answer choice B remains our superior answer choice.

50) This question is related to something we read in the passage. The author mentions, “Addison’s disease occurs when cells of the adrenal cortex are destroyed, leaving the gland unable to secrete either glucocorticoids or mineralocorticoids.” To counteract this effect, a patient may need a replacement dose of glucocorticoids. However, if there is too high a replacement dose, we could see some adverse effects. Let’s break down the function of glucocorticoids. Glucocorticoids are a set of steroid hormones that are synthesized in the adrenal cortex and regulate glucose metabolism. The author actually tells us cortisol, the body’s primary glucocorticoid, functions to stimulate gluconeogenesis in the liver by activating DNA transcription to produce liver enzymes, and by mobilizing amino acids from muscle tissue. If a patient is given too high a replacement dose of glucocorticoids, we expect the effect of these functions would be increased. 

  1. muscle mass. Cortisol functions to mobilize amino acids from muscle tissue. We would see the opposite effect if a patient is given too high a replacement dose of glucocorticoids. The patient will instead see decreased muscle mass.
  2. muscle weakness. This answer choice is consistent with our breakdown. Cortisol functions to mobilize amino acids from muscle tissue. If we have too high a replacement dose, we see increased mobilization of amino acids from muscle tissue and ultimately muscle weakness.
  3. red blood cell count. High levels of cortisol can decrease the activity of white blood cells. Glucocorticoids do not act on cardiac muscle which might be why the test-maker included this answer choice. Cortisol would mobilize amino acids from skeletal muscle tissue instead. Answer choice B remains the best option.
  4. heart rate. This is a shorter-term effect of epinephrine, not a longer-term effect of cortisol which functions to regulate blood sure levels for the body. Answer choice B is our best option.

51) This is essentially a standalone question that relies on knowing about gluconeogenesis. Gluconeogenesis is the synthesis of new glucose molecules from pyruvate, lactate, glycerol, or the amino acids alanine or glutamine. This process takes place primarily in the liver during periods of low glucose, that is, under conditions of fasting, starvation, and low carbohydrate diets. Gluconeogenesis is considered as the reverse process of glycolysis, but with different enzymes.

  1. blood glucose levels are high. The opposite of this answer choice is true. Gluconeogenesis is the synthesis of new glucose molecules. If blood glucose levels are already high, we expect a low rate of gluconeogenesis.
  2. cortisol release is inhibited. The author actually tells us cortisol, the body’s primary glucocorticoid, functions to stimulate gluconeogenesis in the liver by activating DNA transcription to produce liver enzymes. If cortisol release is inhibited, we don’t have that same gluconeogenesis stimulation in the liver.
  3. the body’s stores of carbohydrates are low. This answer choice is consistent with our breakdown of the question. Gluconeogenesis is the synthesis of new glucose molecules from pyruvate, lactate, glycerol, or the amino acids alanine or glutamine. This process takes place primarily in the liver during periods of low glucose. Answer choice C is our best option so far.
  4. the body’s stores of proteins are low. Gluconeogenesis is the synthesis of new glucose molecules from pyruvate, lactate, glycerol, or sometimes amino acids. If the body’s stores of proteins are low, there aren’t additional amino acids with which to synthesize new glucose molecules. We expect the opposite to be true. Answer choice C is our best option.

 

Exam 1 B/B Solutions: Passage 10

52) This question begins a set of passage-related questions by asking about insulin, which is something we’re very familiar with from our studies. In the question stem, the test-maker tells us insulin stimulates the first step in the glycolytic pathway; we’re asked the effect of insulin on glucose uptake in liver cells. Glycolysis is a metabolic pathway that occurs in the cytosol of cells and breaks down glucose into two three-carbon compounds and generates energy. Insulin stimulates glycolysis in hepatocytes which leads to decreased cellular concentration of glucose. To compensate, hepatocytes need to increase glucose uptake.

  1. hinders glucose uptake by increasing the cellular concentration of glucose. This answer choice says the opposite of our breakdown. Insulin stimulates glycolysis in hepatocytes which leads to decreased cellular concentration of glucose. To compensate, hepatocytes need to increase glucose uptake.
  2. aids glucose uptake by decreasing the cellular concentration of glucose. This answer choice is consistent with our breakdown. Insulin stimulates glycolysis in hepatocytes which leads to decreased cellular concentration of glucose. To compensate, hepatocytes need to increase glucose uptake. This is going to be our best answer choice. 
  3. hinders glucose uptake by using the ATP needed by the glucose transporter proteins. First part of this answer choice contradicts our breakdown, so answer choice B remains superior.
  4. aids glucose uptake by providing the ATP needed by the glucose transporter proteins. Insulin does not provide or make ATP. Glucose transport happens across the hepatocyte membrane and down its concentration gradient. There is no ATP needed by glucose transporter proteins. We can stick with answer choice B as our best answer.

53) The key takeaway from our question stem is that the test-maker tells us exercise promotes the insulin-independent uptake of glucose. We’re told in the passage, “In diabetes, the ability of many types of cells to take up glucose is compromised, leading to elevated blood glucose levels.” Despite having diabetes, exercise can promote insulin-independent uptake of glucose in all patients with diabetes. We expect regular exercise will reduce blood glucose levels in patients with both Type 1 and Type 2 diabetes. The only answer choice that matches our breakdown is going to be answer choice C.

54) To answer this question, we’ll have to rely on our external knowledge of different organelles. We’re talking about modifications to a translated polypeptide prior to being secreted from the cell. These modifications typically take place in the ER and/or Golgi. We can go through our answer choices and define the main functions of each and see which is consistent with the post-translational cleavage being discussed.

  1. Nucleus. The nucleus of living cells contains the genetic material that determines the entire structure and function of that cell. While the nucleus is obviously important, it is not the site for the modification and secretion of proteins. 
  2. Ribosomes. Ribosomes are macromolecules made up of rRNA and proteins that act as the sites of protein synthesis. We’re dealing with cleaving a translated polypeptide in our question stem. Ribosomes are not going to be the best option here.
  3. Endomembrane system. The endomembrane system is the set of membrane-bound organelles that are involved mainly in the modification and transportation of proteins. This includes the nuclear envelope, Golgi apparatus, endoplasmic reticulum, vesicles and lysosomes. This answer choice is consistent with our breakdown and what we’re looking for in our question stem. The question asked about modifications to a translated polypeptide prior to being secreted from the cell. The cleavage takes place in the endomembrane system.
  4. Cytoplasm. The modification described in the question stem takes place in organelles, not the cytoplasm surrounding the endomembrane system. We can eliminate this answer choice. Answer choice C is our best option. 

55) Glucose is the main fuel used by the brain. More than any other organelle, the brain needs glucose as its fuel source. In diabetes, the ability of many types of cells to take up glucose is compromised. Many tissues then metabolize fatty acids as an alternative energy source, and diabetes can also cause protein degradation. Despite these effects, the brain, moreso than any other organ, still receives adequate glucose. We want to explain the most likely cause of this adequate nourishment, despite the effects of diabetes. 

  1. less glucose than do other body tissues. The brain often metabolizes glucose more than any other body tissue. More than any other organelle, the brain needs glucose as its fuel source.
  2. insulin-independent transporters for the uptake of glucose. This answer choice would be consistent with what we covered in our breakdown. Normally, insulin promotes facilitated diffusion of glucose into cells from the bloodstream. In diabetes, the ability of many types of cells to take up glucose is compromised. By using insulin-independent transporters for the uptake of glucose, the brain is not subject to the same effects that are felt by other tissues in diabetes patients. This is going to be our best option so far.
  3. fatty acids for energy instead of glucose. The brain uses glucose as its fuel source and gets priority over other organs. The brain can use insulin-independent transporters for the uptake of glucose; it does not rely on ketone bodies.
  4. insulin-dependent transporters for the uptake of glucose. Normally, insulin promotes facilitated diffusion of glucose into cells from the bloodstream. In diabetes, the ability of many types of cells to take up glucose is compromised. Type 1 diabetics cannot produce insulin, while Type 2 diabetics have a compromised ability to respond to insulin. The brain must therefore rely on insulin-independent transporters. 

56) The passage covered diabetes extensively, along with some related symptoms. We were told a constant thirst and frequent need to urinate were symptoms an attending physician associated with diabetes. We’re given additional effects like compromised ability to take up glucose, metabolizing fatty acids as an alternative energy source, or protein degradation. We want to look at the four symptoms presented as answer choices. Three of the options will be likely symptoms, while our correct answer is the least likely to be a symptom. 

  1. Loss of appetite. Increased blood sugar and a compromised ability to take up glucose can actually increase appetite. This is not something that would be consistent with the symptoms described in our breakdown of the question.
  2. Sweet-tasting urine. In diabetes, the ability of many types of cells to take up glucose is compromised, leading to elevated blood glucose levels. The body will then excrete some of that excess glucose into the urine which leads to sweet-tasting urine (supposedly)!
  3. Unexplained weight loss. In diabetes, many tissues metabolize fatty acids as an alternative energy source and there is also protein degradation. This would be consistent with weight loss.
  4. Feelings of fatigue. This ties into our previous answer choice. In diabetes, many tissues metabolize fatty acids as an alternative energy source and there is also protein degradation. Protein degradation means feelings of fatigue, and the body is not as accustomed to utilizing fatty acids and ketone bodies as an energy source so there may be feelings of fatigue. Answer choice A is the least likely to be a symptom of diabetes mellitus. 

 

Exam 1 B/B Solutions: Questions 57-59

57) This is a fairly lengthy standalone question, so we want to make sure we pick out the most important pieces that are necessary to answer our question. We’re asked about a certain bacterium completing cell division and given specific genome masses depending on the media in which it is grown. We can use our external knowledge to first break down what we know about cell division and DNA replication.

DNA replication is semiconservative, meaning that each strand in the DNA double helix acts as a template for the synthesis of a new, complementary strand. During DNA replication, the double-stranded DNA helix unwinds with the help of the enzyme helicase, and the strands are separated. These strands act as templates for the formation of complementary “daughter” strands. After replication, two new double-stranded DNA molecules are formed, each containing one new daughter strand and one old “parental” strand. Thus, the process is semiconservative, as half of the resulting DNA is old and half is new. 

Prior to division, we have a genome mass in each bacterium of 5.5 fg (15N exclusively). 

After one round of division, each bacterium will have one strand 15N and one strand 14N. The genome mass will now be 5.45 fg. 

After another round of division, half of the bacterium will have one strand 15N and one strand 14N, while the other half have 14N exclusively. The half of the bacterium that have one strand 15N and one strand 14N have mass genome of 5.45 fg, while the half of the bacterium that have 14N exclusively have a mass of 5.4 fg. This was a math problem with no rounding or approximation. We can pick the only answer choice consistent with our breakdown: Answer choice B.

58) This is a standalone question that relies on our knowledge of genetics. There are two ways to go about answering this question. We know M is dominant and will cause deafness regardless of the other genes present. If we’re crossing Mm and mm, we know half the offspring will be Mm, meaning at least ½ of the offspring will be deaf. Right away we can eliminate answer choices A and B. Kk and Kk give you a 25% chance of getting the kk genotype (deaf), so the fraction of the offspring that is most likely deaf should be greater than ½. Only answer choice D is greater than ½. 

Alternatively, we can draw out the punnet square and get an exact fraction:

  KM Km kM km
Km KKMm KKmm KkMm Kkmm
Km KKMm KKmm KkMm Kkmm
km KkMm Kkmm kkMm kkmm
km KkMm Kkmm kkMm kkmm

10 of the 16 listed genotypes correspond to deaf offspring. That simplifies to answer choice D: 5/8

59) This is a standalone question that relies on us relating the number of amino acids to the number of bases. Every amino acid is encoded by three bases. In our question stem we’re given the total number of bases, so we simply divide:

1500 bases / (3 bases/amino acid) = 500 amino acids maximum 

This is a math problem that involved no rounding or approximation. We can find the answer choice that matches our calculated value: Answer choice A: 500 amino acids.



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