What best explains why the nucleotide substitution in the mouse does not change its phenotype?

Darwin, C. On the Origin of Species by Means of Natural Selection, or the Preservation of Favoured Races in the Struggle for Life (London, John Murray, 1859)

Duret, L., & Arndt, P. F. The impact of recombination on nucleotide substitutions in the human genome. PLoS Genetics 4, e1000071 (2008) (link to article)

Eyre-Walker, A., & Keightley, P. D. The distribution of fitness effects of new mutations. Nature Reviews Genetics 8, 610–618 (2007) (link to article)

Galtier, N., & Duret, L. Adaptation or biased gene conversion? Extending the null hypothesis of molecular evolution. Trends in Genetics 23, 273–277 (2007)

Graur, D., & Li, W. Fundamentals of Molecular Evolution (Sunderland, MA, Sinauer Associates, 2000)

Green, P., et al. Transcription-associated mutational asymmetry in mammalian evolution. Nature Genetics 33, 514–517 (2003) doi:10.1038/ng1103 (link to article)

Kimura, M. Evolutionary rate at the molecular level. Nature 217, 624–626 (1968) doi:10.1038/217624a0 (link to article)

———. The neutral theory of molecular evolution: A review of recent evidence. Japanese Journal of Genetics 66, 367–386 (1991)

Lobry, J. R. Asymmetric substitution patterns in the two DNA strands of bacteria. Molecular Biology and Evolution 13, 660–665 (1996)

Lynch, M. The origins of eukaryotic gene structure. Molecular Biology and Evolution 23, 450–468 (2006)

———. The Origins of Genome Architecture (Sunderland, MA, Sinauer Associates, 2007)

Makalowski, W., & Boguski, M. S. Evolutionary parameters of the transcribed mammalian genome: An analysis of 2,820 orthologous rodent and human sequences. Proceedings of the National Academy of Sciences 95, 9407–9412 (1998)

Marais, G. Biased gene conversion: Implications for genome and sex evolution. Trends in Genetics 19, 330–338 (2003)

Meunier, J., et al. Homology-dependent methylation in primate repetitive DNA. Proceedings of the National Academy of Sciences 102, 5471–5476 (2005)

Sample Test Biology/Biochemistry Section Passage 1

1) To start us off on the bio/biochem section of the Sample test, we’re being asked to consider the effect of GTP hydrolysis to GDP when bound to Arf1. This passage was quite dense, but we can use the following information from the end of paragraph 2 to answer the question. “The Arf1 GTPase activating protein (GAP) catalyzes the conversion of Arf1-bound GTP to GDP and inorganic phosphate, thereby converting the protein to the inactive form.”

1. denaturation. – Denaturation would imply that Arf1 unfolds when GTP is hydrolyzed to GDP. We are simply told that it is converted to the inactive form, not that it unfolds.
2. activation. – This is the opposite of what the author says in the passage. The protein is inactivated, not activated.
3. inactivation. – We’ve come across the correct answer. The conversion of GTP to GDP when bound to Arf1 promotes its inactivation.
4. membrane association. – If you were tempted by this answer, you were probably thinking about the sentence that says “Upon GTP exchange, Arf1 undergoes a conformational change that releases the myristoylated N-terminus of the polypeptide chain from a structural groove in the protein and initiates localization to phospholipid bilayers.” However, the GTP exchange the sentence is referring to is the exchange of GDP for GTP which is the opposite of what is asked in the question stem would cause the protein to become activated. Answer choice C is the correct answer.

2) This question is a pseudodiscrete. Take a look at the structure of AMP in preparation for the answer choices.

1. Nucleotides – AMP, seen above, is a nucleotide. GTP is analogous to AMP but with two additional phosphates and a guanine instead of an adenine.
2. Amino acids – You should have the structure of each of the amino acids memorized by test day. None of them resemble GTP.
3. Lipids – Lipids are fats with plenty of hydrocarbons, not seeing much of that in GTP.
4. Carbohydrate – Some of the most notable examples of carbohydrates on the MCAT are glucose, lactose and sucrose, none of which resemble an entire GTP molecule. GTP is most closely related to nucleotides.

3) The passage states the following: “GTPase activating protein (GAP) catalyzes the conversion of Arf1-bound GTP to GDP and inorganic phosphate, thereby converting the protein to the inactive form.”

1. Transferase – A transferase transfers a functional group from one molecule to the next. GAP catalyzes the removal and release of a phosphate, not its transfer.
2. Phosphatase – A phosphatase cleaves a phosphate bond to remove a phosphate from its target. GAP removes an inorganic phosphate and is indeed a phosphatase.
3. Kinase – Kinases phosphorylate their target; GAP does the opposite and removes the phosphate.
4. Isomerase – An isomerase alters the structural arrangement of its substrate. GAP removes an inorganic phosphate which is the function of a phosphatase.

4) The second paragraph begins with “ADP-ribosylation factor I (Arf1) plays an essential role in vesicle formation and is responsible for the recruitment of cytosolic coat protein complexes (COPs) and subsequent retrograde transport from the Golgi apparatus.”

1. endoplasmic reticulum. – Arf1 recruits COP and when bound, facilitates retrograde transport. The target location for retrograde transport is the endoplasmic reticulum. Looking good.
2. cellular membrane. – While the prior paragraph notes that Arf1 initiates localization to phospholipid bilayers, this is before Arf binds COP.
3. nucleus. – Retrograde transport reverses the usual endoplasmic reticulum –> Golgi processing –> protein target location pathway. It will not take the protein to the nucleus.
4. cytosol. – The cytosol is the medium through which retrograde transport occurs, but is not its target. 

5) To answer this question, let’s go back to the passage. In the third paragraph, the author states “Brefeldin A (BFA), a lactone compound isolated from fungi, has been shown to inhibit Arf1-driven vesicle formation, resulting in reversible disruption of the Golgi apparatus.”

1. eukarya. – BFA reversibly disrupts the functioning of the Golgi apparatus. Eukaryotes have Golgi so in the absence of contradicting evidence, we like what we see because it would make for an effective drug.
2. viruses. – Viruses do not contain organelles, so they do not contain Golgi. This means that BFA would not be effective against viruses.
3. bacteria. – Like viruses, bacteria do not contain membrane-bound organelles, and the Golgi is a membrane bound organelle. 
4. archaea. – Archaea do not contain the Golgi apparatus. Of the answer choices provided, only eukaryotes contain Golgi, the target of BFA; BFA would only be effective against eukaryotes. 

Sample Test Biology/Biochemistry Section Passage 2

6) 

1. Performing the ischemia/reperfusion experiment using animals whose B (antibody-producing) cells are depleted and examining whether the degree of tissue damage is reduced – We’ve seen the MCAT test writers write tricky answer choices and here’s another example. If we’re not careful, this answer choice would tempt us because the passage presents an experiment showing the effect of antibodies (antibody B) for the beta subunit in neutrophil adhesion receptors. However, this answer choice refers to B cells, not the B antibody. This is an entirely different cell type and will not tell us if the neutrophils are the cause of the injury.
2. Performing the ischemia/reperfusion experiment using neutrophil-depleted animals and examining whether the degree of tissue damage is reduced –The goal of the new experiment is to provide support that neutrophils are causing the reperfusion injury. If the neutrophils in the animals are depleted, there are no neutrophils to cause the injury. If injury is minimized in the absence of neutrophils, it would be strong support of them being the causative agent. However, if reperfusion injury remained high, it would provide evidence against neutrophils causing the injury. This is a solid experiment to determine the impact of the presence or absence of neutrophils on reperfusion injury. We’ll read the other answer choices just in case. 
3. Repeating the experiment with another antibody directed against the entire alpha/beta heterodimer, and examining whether the degree of tissue damage is reduced – Recall that the heterodimer is for the neutrophil adhesion receptors and not the neutrophils themselves.  If the receptor has some mechanism by which to facilitate the reperfusion injury that is independent of the neutrophils, we would not be able to differentiate between the two mechanisms with this experiment. We’re looking for an answer choice that directly evaluates the effect of neutrophils on reperfusion injury.
4. Repeating the experiment with another antibody directed against the beta subunit, and examining whether the degree of tissue damage is reduced – Like answer choice C, this experiment would evaluate the role of the neutrophil adhesion receptor and not the neutrophils themselves. Answer choice B is the only answer to directly assess the impact that the presence or absence of neutrophils has on reperfusion injury.

7) 

1. adhering neutrophils to the endothelium. – In the second paragraph, the author states “The adherence of neutrophils was facilitated by an adhesion receptor on the neutrophil membrane. During reperfusion, adherent neutrophils released toxic products including oxygen-derived free radicals, proteases, and prostaglandin products.” The adhesion receptor facilitates adherence of neutrophils. Then, once they are adherent they can release their toxic products. 
2. transferring proteases from endothelium to neutrophils. – The proteases are released after they are adherent, and not before. Their adherence is the key to their mechanism of damage and as seen above, is where the beta subunit comes into play.
3. hydrogen bonding with the alpha subunit. – There is no evidence that hydrogen bonding between the two subunits is what promotes neutrophil-mediated reperfusion injury.
4. the generation of antibody against the subunit. – The presence of the beta subunit antibody decreased the degree of reperfusion injury. If the generation of antibodies were the function of the beta subunit, it would decrease the degree of injury, not promote it. Answer A best describes the role of the beta subunit in mediating reperfusion injury. 

8)

1. Antibody B is a high-affinity antibody; therefore, it will not be rejected by the patient. – Just because an antibody is high affinity doesn’t mean that it won’t be rejected by the patient. Affinity refers to binding propensity, whereas rejection is mitigated by a myriad of signaling and recognition molecules that will either believe the antibody is “self” or “foreign.”
2. Antibody B can block the initiation of events that result in the release of harmful, biologically active molecules. – Stopping a damaging cascade from even starting is a great way to prevent damage. The passage explains that reperfusion injury occurs roughly as follows: activation of neutrophils –> binding of neutrophils to adhesion receptor –> once adhered, release of toxic products –> damage. Preventing binding instead of waiting for the toxic products to be released means less damage before treatment and prevention kick in.
3. Antibody B is a very specific antibody; therefore, it will not recognize anything other than the beta subunit. – Even if this were true, it would not explain why the antibody is a better preventive method than the inhibitors. The correct answer needs to address the “why” based on the reperfusion injury cascade. Answer B is so far the best answer.
4. Antibody B exhibits a high half-life and can be used at any dosage at any time. – Again, this could very well be true, but it does not address how antibody B fits into the reperfusion damage cascade. B is the best answer choice.

9) 

1. Because the antibody was generated in the mouse, it can never be used in humans. – Never is a very strong word. Why would a mouse antibody be a bad fit for clinical use? What would cause it to not be used in humans? Hopefully there’s an answer choice that’s less extreme and more specific.
2. Because the antibody was generated in the mouse, repeated usage in the same patient would elicit the production of human anti-mouse antibodies. – The mouse antibody will not be recognized as “self” by the immune system, it’ll be recognized as “foreign.” When the immune system recognizes the mouse antibody as foreign, it will mount an immune response that will lead to the production of anti-mouse antibodies. Once this happens, the mouse antibody must be discontinued or risk allergic reactions. If the mouse antibody is known to cause an immune response and allergic reaction, it will not be good for clinical use.
3. Because the antibody was generated in the mouse, it will not recognize human antigens. – The antigens that an antibody recognizes are not limited to the same organism that created it. For example, we produce antibodies that recognize non-human antigens all the time.
4. Because the antibody was generated in the mouse, it can only be used in vitro. – We have no reason to believe that this antibody is limited to in vitro use only. Answer choice B is the only one to provide a clear and specific reason as to why the mouse antibody would not be used clinically.

10) 

1. The cell cannot release toxic products such as prostaglandins. – According to the question stem, the neutrophil is still able to migrate through endothelium, implying that it can still bind the endothelium. The release of toxic products occurs after the binding of the neutrophil to the endothelium via the adhesion receptor so the cell should still be able to release its toxic products.
2. The cell has only functional beta subunits. – Keep in mind that barring genes on the sex chromosomes and chromosomal abnormalities, we have two copies of each gene, one inherited from each parent. This means that it is possible for the point mutations mentioned in the question stem to be from one parent or the other or acquired somatically after birth. In either case, it’s possible to produce some functional and some nonfunctional proteins, and still be able to execute the function of interest. Clearly the neutrophil adhesion receptor has some functional subunits because the neutrophil is able to bind the endothelium, but that doesn’t mean it only has functional beta units and lacks any nonfunctional units.
3. The cell can bind to endothelium. – This is true, if the cell can migrate through endothelium, it is able to bind the endothelium and perform its function.
4. The cell has a defective cell membrane. –The focus of the question is not on the status of the cell membrane, but rather the functionality of the adhesion receptors. Also, based on the passage, we expect the injury or damage to come after the neutrophil has adhered and released its toxic products. Answer choice C directly addresses the function and role of the neutrophil adhesion receptor following the point mutations.

Sample Test Biology/Biochemistry Section Passage 3

11) For all the future doctors reading this solutions set, this is a great passage to reread if you have the time because you’ll actually learn a lot more about this topic once you’re in medical school and studying hematology. The key to answering this question is the location of the Factor VIII gene given in the passage: “The Factor VIII gene lies on the X chromosome” and is “recessive.” As noted in the passage, “men have only one X chromosome,” so men will make up the majority of the affected individuals. Because the gene is inherited on the X chromosome, it also means that males will need to inherit the mutated gene from their mother because their father will provide the Y chromosome.

                  I.         Normal sons born to hemophiliac fathers

                II.         Normal sons born to hemophiliac mothers

              III.         Hemophiliac sons born to normal fathers

              IV.         Hemophiliac daughters born to normal fathers

1. I and II only – Let’s look at RN I first. If a normal son is born to a father with hemophilia, this would not contradict the X chromosome or sex-linked recessive pattern of inheritance because the father will only provide the Y chromosome to his son. If the mother is affected or a carrier for the hemophilia, she can pass the affected/mutated gene on to her son. The question stem is asking us to find the pair that will fail to support or contradict the hypothesis of sex-linked inheritance; because RN I is consistent with a recessive X-linked pattern of inheritance, this answer choice is incorrect.
2. I and III only – As mentioned above, RN I is consistent with a recessive X-linked pattern of inheritance, so this answer choice is also incorrect. Recall that the question stem is asking us to find a pair that will contradict a recessive X-linked pattern of inheritance.
3. II and IV only – Now let’s assess RN II. If a normal son is born to a mother with hemophilia, it means that they inherited a wild type gene. However, if the mother has hemophilia, she has to have two copies of the mutated gene. RN II is not consistent with a recessive X–linked pattern of inheritance. This answer choice is looking good so far. Now on to RN IV; if a daughter with hemophilia is born to a normal father, there’s a problem. The father must pass his X chromosome on to his daughter for her to have two X chromosomes. However, if he is normal, he has a wild type gene on his X-chromosome. If his daughter has hemophilia, it means that she needed to have inherited two mutated genes which is not possible if her father only has a wild type gene to give her. RN IV is also inconsistent with a recessive X–linked pattern of inheritance making this the correct answer choice. See below for a Punnett square explanation.
4. III and IV only – We know from our assessment above that RN IV is inconsistent with a recessive X-linked pattern of inheritance. We’ll double check RN III just in case. RN III states that a hemophiliac son is born to a normal father; remember that the father only provides the Y chromosome to his sons. This means that for the son to have hemophilia, his mother is the one who needs to provide the mutated gene on the X chromosome. RN II is consistent with a recessive X-linked pattern of inheritance, so answer choice C is indeed the correct answer.

12) 

1. a soluble blood protein. – Whether the protein is functioning or not, it could still be a soluble blood protein. 
2. produced by a gene on the X chromosome. – We know that the Factor VIII gene is located on the X chromosome. Whether the gene is mutated and contains the allele for hemophilia or not, it will still be located on the X chromosome, so this answer choice is also incorrect.
3. able to relieve hemophilia symptoms. – If hemophilia is caused by a functionally deficient protein, replacing the mutated Factor VIII proteins produced by the body should provide beneficial results and relieve at least some symptoms. As an analogy, if you’re trying to filter water and the filter is not functional or not working, replacing the filter with one that is functional should improve water filtration.
4. encoded by a gene that contains introns. – Many genes contain introns which are spliced out prior to protein translation, so this wouldn’t really help us determine whether a functional deficiency is causing Factor VIII hemophilia. We’ll stick with answer choice C.

13) 

1. DNA → mRNA – This answer choice represents the process of transcription and post-transcriptional modifications—DNA → hnRNA or pre-mRNA → mRNA. Intron splicing occurs after transcription and prior to translation so this is looking good. See below for a review of eukaryotic transcription.
2. DNA → tRNA – tRNA carries the amino acids needed for translation but is not the transcript to be translated. Whoops, next!
3. mRNA → tRNA – mRNA is the transcript that has undergone splicing, but that mRNA is used in translation to produce the Factor VIII protein, not tRNA.
4. tRNA → protein – Again, it is the mRNA that will be used to produce the Factor VIII protein; tRNA is the carrier for amino acids used during translation to convert the mRNA to a polypeptide chain. Answer A is the only one that correctly associates splicing and transcripts.

14) To answer this question, we’ll need to consider the key differences between eukaryotic and prokaryotic transcription and translation. Unlike eukaryotes, prokaryotes couple transcription and translation such that there is no need for post-transcriptional modifications. As noted in prior questions, post transcriptional modifications are important for intron splicing. Let’s take a look at the answer choices.

1. are too small to incorporate the Factor VIII gene. – We have no reason to believe that bacteria are too small to incorporate the Factor VIII gene. Let’s see what the other answer choices are just in case.
2. possess no equivalent to the human X chromosome. – While it is true that bacteria do not contain an X chromosome, the question stem makes it clear that there is a precedent for inserting human genes into bacteria and successfully producing the corresponding proteins. This answer choice isn’t looking good either.
3. lack a membrane-bound nucleus. – Again, it’s true that bacteria don’t possess a membrane-bound nucleus. However, they are able to produce proteins just fine without one. Hopefully answer choice D is correct.
4. lack a mechanism for splicing out introns. – And in fact, answer choice D is correct! Prokaryotes do not have a nucleus so both their transcription and translation occur in the cytoplasm; these two are coupled and no post-transcriptional modification occurs. Intron splicing is a post-transcriptional modification that allows for the processing of hnRNA to mRNA. If there are no post-transcriptional modifications, the ribosome will not be able to produce the protein because there are non-coding regions (introns) still present.

Discrete Questions

15)

1. In blood, the concentration of H+(aq) is maintained at low levels by other equilibria. – According to Le Chatelier’s principle, decreasing the concentration of a product will cause a rightward shift in the reaction in order to restore equilibrium. If the concentration of protons is maintained low in the blood, then the reaction will shift forward and promote the dissociation of the organic acid. This answer choice is looking great.
2. In blood, the reaction is coupled to ATP hydrolysis to make it more favorable. – It would be incredibly unfavorable to have to pair ATP hydrolysis with every dissociation of an organic acid; that would be a huge source of energy consumption! There has to be a better answer choice.
3. In blood, the ionic strength of the solvent medium is much higher than pure water. – While increasing the ionic strength of a solvent can increase solubility, this is only true when there is no common ion present and protons are fairly common in reactions, so this answer choice is not a good one.
4. In blood, enzymes are used to catalyze the dissociation reaction. – Enzymes speed up a reaction, but they do not make the reaction more favorable. This is an important limitation to keep track of for test day. Answer choice A is then the best answer choice.

16) 

1. a 1-electron carrier. –  Cytochrome c moves electrons by going back and forth between Fe2+ (ferrous ion) and Fe3+ (ferric ion). The charge can only change by one at a time, meaning cytochrome c can only transfer one electron at a time. This is the correct answer.
2. a 2-electron carrier. – In order to be a 2-electron carrier, the iron ions would need to change charge by +/- 2.
3. a 3-electron carrier. – Like answer B above, in order to carry 3 electrons, cytochrome c would need to produce greater changes in oxidation state.
4. a 4-electron carrier. – This is the most extreme of the answer choices. Cytochrome c only causes a +/- 1 change in oxidation state, so it is a 1-electron carrier.

For a review of the electron transport chain, head over to https://jackwestin.com/resources/mcat-content/oxidative-phosphorylation/electron-transfer-in-mitochondria

17) A good PCR primer will have a high GC content at either end. Pause here and try to think about why this is the case. Welcome back! Higher GC content means stronger bonding (3 hydrogen bonds in a GC pairing compared to 2 H-bonds in an AT pairing), higher melting temperature and a more stable setup for PCR. Ideally, the GC content for a primer should be between 40-60%.

1. 5–ATTACGTTAACATGAAG–3 – Those 5’ and 3’ ends are looking pretty G and C deficient, what do you think?
2. 5–ATATCGTTAACAAATTG–3 – Like answer A, there’s a G thrown in at that 3’ end but not much else…
3. 5–GCTATAAAGATTGCAAA–3 – Getting warmer; the 5’ end has a G and C.
4. 5–GCATAGAAGCATTCCGC–3 – Now this makes for a “hot” product (the puns will never end…). The 5’ end has a G and C, and the 3’ end has double the GC content. These 5’ and 3’ ends will make for more stable bonds.

18) Nondisjunction occurs when the chromosomes do not correctly segregate during meiosis. Feel free to follow along on the figure below as we look through the answer choices.

1. Anaphase I – The chromosomes are pulled apart to opposite poles of the cell during anaphase. If the chromosomes will fail to segregate, this is when it will happen. 
2. Metaphase II – Metaphase II is when the sister chromatids line up in the “middle” so this is not the segregation step.
3. Prophase I – Crossing over occurs during prophase I; this is not the step we’re looking for.
4. Telophase II – Telophase occurs after the chromosomes and sister chromatids (I and II respectively) arrive at the opposite poles. This is too late; the best answer is A for anaphase I.

For a refresher on the differences between mitosis and meiosis, check out https://jackwestin.com/resources/mcat-content/meiosis-and-other-factors-affecting-genetic-variability/important-differences-between-meiosis-and-mitosis and for more on gene segregation (more directly applicable to this question) hop over to https://jackwestin.com/resources/mcat-content/meiosis-and-other-factors-affecting-genetic-variability/segregation-of-genes

Sample Test Biology/Biochemistry Section Passage 4

19) To answer this question, we can take a look at Fig. 1, the structure of ms2t6A, shown below with annotations. This cannot be overstated: you need to be incredibly comfortable with amino acids on test day.

1. Tyrosine – Tyrosine’s side chain includes a phenol group. There are no benzene rings in the amino acid outlined above.
2. Threonine – Threonine has two carbons coming off the backbone with an alcohol bound to the second-to-last carbon. The amino acid circled in yellow has been redrawn in a more familiar orientation in blue; both show two carbons attached to the backbone with an alcohol attached to the carbon before last.
3. Lysine – Lysine has four carbons with a terminal, positively charged amine group as its side chain.
4. Tryptophan – Tryptophan has not one, but two fused rings as part of its R group. The amino acid above has none. Threonine it is!

20) Before jumping back to Fig. 2, it’s important to recall that “Because Lys529 in firefly luciferase is essential for enzymatic activity, inaccurate translation of this codon results in a loss of luciferase activity.” This means that firefly luciferase activity, or FL activity in the figure, is a proxy for translation. More translation –> higher FL activity and less translation –> lower FL activity.

1. AAA codon in the wild-type background – The AAA codon in the WT background, like the AAG codon in the WT background, has the highest level of FL activity, so the highest level of successful translation and lowest level of mistranslation. This is the opposite of what we’re looking for.
2. AAA codon in the ΔyqeV background – While the AAA codon in the ΔyqeV bacteria or background has a lower FL activity than in the wild-type, it still has greater activity than the AAG codon in the ΔyqeV background.
3. AAG codon in the wild-type background – Like answer choice A, the AAG codon in the WT background has the highest FL activity to RL activity ratio meaning it has the lowest level of mistranslation.
4. AAG codon in the ΔyqeV background – As shown in the annotated figure above, the AAG codon in the ΔyqeV bacteria or background produces the lowest FL to RL activity ratio, meaning it has the least successful translation and the most mistranslation. The question asks for the highest level of mistranslation so this is the right answer.

21) The passage begins with “Variations in the human gene CDKAL1, which encodes CDKAL1, are associated with impaired insulin secretion and increased risk of type 2 diabetes (T2D).” A deficiency will likely manifest in a similar fashion as the variants, so the answer should reflect something impairing or compromising insulin secretion.

1. Cleavage of proinsulin – Cleavage of proinsulin, the precursor to insulin, would increase insulin levels. There’s no reason to think that this would impair insulin secretion and increase the risk of T2D. 
2. Synthesis of proinsulin – Like cleaving the precursor to insulin, producing insulin’s precursor would not impair secretion or increase the risk of T2D, but would instead promote normal insulin metabolism.
3. Misfolding of proinsulin – Ah, this looks much better. If proinsulin were misfolded, it would be more difficult to remove the C-peptide as shown in equation 1. To tie in specific passage evidence, we can consider the following: 
   Cdkal1–/– mice do not show the ms2t6A modification (second paragraph) which is required for the accurate translation of tRNALys (paragraph 1) –> no ms2t6A modification so mistranslation of lysine codons and there are two in the insulin gene (fourth paragraph) –> misfolded protein –> can’t correctly process proinsulin to insulin –> impaired insulin secretion. 
   The walkthrough above shows that we can support our answer choice with more specific information from the passage, but it can be a bit time consuming, right? Neither approach is inherently better than the other, but the first takes advantage of the fact that none of the other answer choices would produce the effect implied in the question stem (impaired insulin secretion due to CDKAL1 deficiency).
4. Ratio of wild-type to variant proinsulin – A CDKAL1 deficiency is unlikely to increase the ratio of wild-type proinsulin to variant insulin. Even if it did, more WT proinsulin would mean better insulin secretion, not impaired secretion. Only answer choice C produces the effect of interest.

22) If we read this question on test day and are unsure of where to start, we can take note of the condition(s) specified in the question stem: is the mice are nonfasting, we are interested in their response to glucose, and if we’re comparing wild-type to the knockout mice, then we want to know how the mice respond to changes in glucose when they have CDKAL1 deficiency.

Take a look at figures 3 and 4 to help us assess the upcoming answer choices.

1. Blood glucose levels – Will blood glucose levels rise or fall in the presence of insulin? They’ll fall right? Insulin will promote the incorporation of glucose transporters into the plasma membrane on insulin-sensitive cells and encourage the cellular uptake of glucose. Wild-type mice produce more insulin than Cdkal1–/– mice, so they will have higher glucose uptake and lower blood glucose levels than Cdkal1–/– mice. This is looking good.
2. Cellular glucose uptake – The question is asking what will be lower in WT mice when not fasting. In the previous explanation we established that the higher production of insulin in WT will increase cellular glucose uptake, not decrease it. 
3. Liver glycogen synthesis – Glycogen synthesis is neither the focus of the passage nor discussed in the passage. The question stem specifies “based on the passage” and is not a pseudodiscrete; answer choice A directly and correctly answers the question making it the best answer.  
4. Cellular protein synthesis – Again, the passage does not describe the effect of proinsulin or insulin on cellular protein synthesis. Some folks might be tempted to say that a fed state promotes cell function and growth, but answer choice A is the answer most directly related to the passage. The more “jumps” you need to make between the passage and the answer choice to answer the question, the less likely the answer choice is to be correct.

Sample Test Biology/Biochemistry Section Passage 5

23)

1. High-affinity transport (Kt) of L-alanine – Affinity refers to how readily the transporter will bind to the L-alanine. Under low-food conditions, we want the transporter to bind the L-alanine, so high affinity is preferable. 
2. Low-affinity transport (Kt) of L-alanine – Under low-food conditions, energy sources are scarce. The larvae should be readily transporting the L-alanine. A low affinity transporter would not bind the L-alanine as readily making it undesirable under low-food conditions.
3. High transport capacity (Jmax) of L-alanine – The question stem specifies low-food conditions. Under low-food conditions, capacity or how much L-alanine the transporter can move doesn’t matter. 
4. Low transport capacity (Jmax) of L-alanine – As in answer choice C, the maximum amount of L-alanine the transporter can move doesn’t matter if there isn’t much to move.

24) Transport affinity is an indicator of how readily the transport protein will bind the substrate. Analogy time! Imagine that my affinity for strawberries is how badly I want to eat a strawberry when I see one. If there were suddenly 12 of me, would my individual baseline desire to eat a strawberry when I see one change? No, my affinity for strawberries is a property that is dependent on who I am, not how many of me there are. On to the answer choices.

1. level off rapidly. – From our analogy above, we know that the transporter’s affinity for L-alanine should not change; it’s a property that depends on the identity of the transporter and not how many are present.
2. decrease. – Again, the affinity shouldn’t change as the concentration of the transporter increases.
3. not change. – Aha, this is the answer choice we’ve been looking for! Affinity is specific to the identity of the transporter and not its concentration.
4. quickly reach the maximum value. – Nope, we’ve established that affinity does not change with transporter concentration.

25) Our first instinct after reading this question may be to go back to Fig. 1—the question stem did mention it… However, if we’re looking to be as efficient as possible on test day, we need to recognize that Kt isn’t explicitly mentioned in the passage and that this question is a pseudodiscrete. Here, Kt is being used to represent Km, also known as the Michaelis constant

1. two times the maximal transport capacity. – The maximal transport capacity is better described by Vmax, not by Km.
2. the substrate concentration at one-half the maximal transport capacity. – Kt and therefore Km represent the substrate concentration required to reach half of Vmax, or the maximal transport capacity. This must be the correct answer.
3. the transport capacity at one-half the substrate concentration. – Km and therefore Kt is a concentration, not a capacity or Vmax.
4. the substrate concentration at one-third the overall transport rate. – While Km and Kt are concentrations, they represent the concentration required to reach half of Vmax, not one-third, making answer choice B the best answer.

26) Enzymes, ribosomes and proteins care about the conformation of their substrates and are stereospecific. Changing the substrate from an L conformation to D-alanine means the ribosomes used to produce proteins will be unable to incorporate alanine.

1. increase two-fold. – Something else would have to change to increase protein synthesis in the absence of the correct substrate. 
2. not change. – Would it be possible to make the same number of proteins if an entire amino acid pool is missing? Unfortunately not. 
3. decrease by one-half. – Oooh, this is tempting! We just said that we wouldn’t be able to keep up with protein synthesis if one of the amino acids is missing. However, a decrease by one-half implies that some of the original protein is being made, but if what is available is the wrong substrate, the original protein won’t be made! This answer choice moves in the right direction but isn’t quite right.
4. be inhibited. – This is a more complete answer than answer choice C in that it gets to the root of the problem: the ribosomes, which use L-alanine, will not be able to synthesize proteins with D-alanine. Saying there is a decrease is not enough, there will be inhibition of ribosomal protein synthesis.

27) Fig. 2 shows the protein and lipid content of the larval Linckia and Odontaster as a percent of dry weight, shown below in case we need to reference it.

1. Ambient water temperature – The ambient water temperature does not directly give us information about how long the starfish can withstand long-term nutrient deprivation. We would need information about how much energy is used at the different temperatures to use this information.
2. Average mass of an individual – Similar to answer choice A, mass does not directly tell us how long the starfish can survive nutrient deprivation. Something about the energy consumption would be required.
3. Average metabolic rate – Average metabolic rate would allow us to determine how much energy the starfish uses to stay alive. If we know how much energy is being used, we can determine how quickly the stored energy will run out. This is a great answer; as the average metabolic rate goes up in the face of a constant low food supply, the less time the organism can stay alive.
4. Duration of daily light exposure – Like answer choices A and B, this answer choice doesn’t give any indication as to how much energy is used, stored and produced. This means that it will not help us predict the starfishes’ ability to survive during prolonged nutrient deprivation.

Discrete Questions

28)

1. Anaerobic, because the final product (alcohol) would contain more energy than the final product of respiration (H2O) – The best measure of energy production during aerobic and anaerobic metabolism is the number of ATP and electron carriers produced. According to this measure, glucose metabolism under anaerobic conditions—glycolysis—only produces 2 ATP and electron carriers are not relevant because their conversion to ATP via the electron transport chain requires oxygen. This is much less energy produced than the 38 ATP during aerobic metabolism. The lower the energy production (anaerobic conditions), the slower the growth rate.
2. Anaerobic, because the cells would not have to produce the enzymes needed for the citric acid cycle – While this may be true, the energy produced under anaerobic conditions pales in comparison to the amount of energy produced during aerobic metabolism leading to slower growth during anaerobic metabolism. This decrease in energy production will play a greater role than the production of fewer enzymes.
3. Aerobic, because a much greater amount of ATP would be produced to provide energy for reproduction – This is true. Aerobic metabolism of glucose produces 19x more ATP (2 vs 38) than glucose metabolism under anaerobic metabolism. The much larger energy production allows for more reproduction and growth.
4. Aerobic, because the CO2 produced in fermentation would be toxic to the culture – If this were true, the carbon dioxide produced during aerobic metabolism would decrease growth rate, not increase it. We’ll stick with answer choice C.

For a brief review of net energy production when glucose is broken down, head over to https://jackwestin.com/resources/mcat-content/glycolysis-gluconeogenesis-and-the-pentose-phosphate-pathway/net-molecular-and-energetic-results-of-respiration-processes-2. You should also review glycolysis and the citric acid cycle if you are not yet comfortable with glucose metabolism. 

29)

1. The bacteria that were unable to digest the polysaccharide died. – If bacteria were dying, there would be a decrease in the number of bacterial cells on the y-axis, but there is no such decrease.
2. The digestive enzymes for the polysaccharide had to be transcribed and translated. – This would make sense. When the polysaccharide isn’t around, the cells have no reason to make the enzymes needed to digest the polysaccharide. However, when the polysaccharide is introduced to the bacterial colony, the cells have to transcribe and translate the appropriate enzymes to digest it. This is not an immediate process and takes time, hence the delay before we see an increase in the number of bacteria.
3. The hydrolysis of fatty acids is a slow process. – Be very careful here, a polysaccharide is a carbohydrate and not a lipid. The hydrolysis of fatty acids is not relevant to this question. 
4. The polysaccharide directly inhibited bacterial fission. – If the polysaccharide directly inhibited bacterial fission, we would not see the eventual increase in the number of bacterial cells. Only answer choice B correctly explains the delay before the increase in the number of bacteria present.

30)

1. In euchromatin – I used to confuse euchromatin and heterochromatin quite frequently when I was studying for the MCAT. How can we remember the differences between the two? Well, euchromatin says “eu/you are ready to go!” Euchromatin is more loosely packed and is transcriptionally active. Heterochromatin says “hold up, we’re highly condensed here!” Heterochromatin is tightly packed which causes it to stain darker and is associated with silenced genes. If GAPDH is continuously expressed, it could very well be located in a transcriptionally active region of euchromatin. 
2. In a telomere – Telomeres are protective regions of DNA that prevent the degradation of important, transcriptionally active DNA. Telomeres shorten with progressive chromosomal replication. We sure hope that a gene that is continuously expressed isn’t located in a region that keeps getting degraded… 
3. In heterochromatin – Heterochromatin says, “hold up, we’re highly condensed here!” We would expect silenced and suppressed genes to be present in heterochromatin, not genes that are always on like housekeeping genes.
4. In a centromere – Centromeres are actually heterochromatin! Only answer choice A represents a region that is conducive to constant transcription of a housekeeping gene like GAPDH. See the figure below for a visual representation of these structures.

31)

1. Gln – Glutamine contains two nitrogen atoms and is neutral! You might be tired of hearing this, but it cannot be overstressed—know the amino acids and their structures (among other characteristics). Add them to your study list if needed.
2. Lys – Lysine does have two nitrogen atoms, but alas, it’s positively charged and not neutral.
3. Tyr – While neutral, tyrosine only contains one nitrogen atom, not two.
4. Arg – Arginine contains four nitrogen atoms and is positively charged at a pH of 7. Answer choice A is the correct answer.

Sample Test Biology/Biochemistry Section Passage 6

32)

1. uptake by erythrocytes. – The passage lets us know that P. falciparum binds and infects erythrocytes. If this uptake were reduced, the infectivity of P. falciparum would decrease which would not benefit the parasite. The question specifically asks for an answer that would benefit the parasite.
2. sensitivity to jasplakinolide. – The last paragraph discusses the effect of jasplakinolide on which var genes are expressed but does not discuss the effect of var gene switching on jasplakinolide. The two are not interchangeable so there should be a better answer choice. 
3. overall adhesiveness to blood vessel walls. – According to the first paragraph, “PfEMP1 enhances P. falciparum survival by binding infected erythrocytes to endothelium.” If binding to the endothelium increases survival, reducing the adhesiveness to blood vessels would decrease survival which would be unfavorable for P. falciparum. Next answer choice.
4. elimination by the human adaptive immune system. – The second paragraph lets us know that the var genes switch and that “the antigenicity of the specific PfEMP1 on the erythrocyte surface, switches throughout an infection.” If the antigens presented are changing, then the antibodies that have already been produced won’t bind and neutralize the new PfEMP1 antigens. This will reduce the effectiveness of the immune system and its elimination of the parasite. Quick recap: var gene switching –> change in antigens presented –> old antibodies don’t bind –> immune system is confused until new antibodies are made and parasite elimination is reduced. This is the best answer.

33)

1. Microtubules – From the last paragraph, we know that “Perturbation of normal actin with jasplakinolide, an actin stabilizer, results in movement from the heterochromatin clusters and activation of previously silent var genes.” This implies that actin is involved in the transport of the var genes in order to alter activation status. Actin filaments are also known as microfilaments which are different from microtubules.
2. Microfilaments – As noted above, the passage points to the role of actin, or microfilaments in var gene transport and activation. This is it!
3. Intermediate filaments – Intermediate filaments are a diverse group that are neither microtubules nor microfilaments. The most likely example you’ll see on test day is keratin.
4. Sarcomere thin filaments – Sarcomeres are not found in erythrocytes. We’ll stick with microfilaments, answer B.

34) The question asks about PfEMP1-mediated characteristics, so let’s refresh our memory on what’s said in the passage: “PfEMP1 enhances P. falciparum survival by binding infected erythrocytes to endothelium.” Erythrocytes binding to endothelium means these red blood cells are collecting on the lining of the blood vessels.

1. Blood vessel blockage – As the RBCs accumulate on the endothelium of the blood vessels, they will begin to block flow through the blood vessels. This answer choice directly ties the role of PfEMP1 to a possible symptom.
2. Low red blood cell count – The most likely cause of a low RBC count would be if the RBCs were being destroyed. The passage specifies that PfEMP1 promotes the binding of RBCs to endothelium but makes no mention of RBC destruction so answer choice A remains the better answer.
3. Recurring fever and chills – Similar to answer choice B, we would expect fevers and chills if the parasite were causing rupture or destruction of the RBCs. The connection between binding of RBCs to the endothelium and blood vessel blockage is clearer than a connection between that binding and destruction of RBcs.
4. Sickle cell hemoglobin-mediated malaria resistance – Sickle cell trait would cause greater turnover of the RBCs, making malaria infection more difficult. We have no passage-based reason to believe that PfEMP1, which increases binding of RBCs to endothelium, would increase sickle cell mediated erythrocyte turnover. Answer choice A is the clearest connection of all of the answer choices.

35) The last sentence of the first paragraph specifies that the binding of infected erythrocytes in brain venules causes cerebral malaria. The question is asking what comes after venules. 

1. Arterioles – arterioles come after arteries and before the capillaries. The venules are past the capillary bed, so this is not the correct answer. 
2. Veins – As noted above, the venules drain into the larger veins to return blood to the heart. If the erythrocyte lost its adhesion, it would move to the next blood vessel after the venules which are the veins.
3. Capillaries – The capillaries are immediately before the venules, not after so this is in a sense the opposite of what we’re looking for.
4. Arteries – The arteries move blood away from the heart. The erythrocyte would need to travel out of the venules, into the veins, to the heart and the associated pulmonary vasculature and then back to the heart before entering the arteries. The correct answer is answer choice B.

Sample Test Biology/Biochemistry Section Passage 7

36)

1. Fewer signals of the weakly perceived color are sent to the brain. – If there are fewer pigment molecules available to absorb light, then fewer stimuli are being processed and fewer signals will be sent to the brain. 
2. A normal number of signals of the weakly perceived color is sent to the brain, but each signal is less intense. – This is incorrect; the question stem specifies that there are fewer pigment molecules to absorb the light which means fewer signals are sent to the brain.
3. The signals for the weakly perceived color are sent to the brain via other, more plentiful, pigments. – There is no mention in the passage of collateral or alternative pigment signaling. If the pigment molecules are compromised, so is the signaling for that pigment.
4. Light molecules stimulate the cells to make additional pigment. – There is no indication in the passage that light stimulates pigment synthesis. Answer choice A is the best answer.

37) The first paragraph primes us with the following: “Normal color vision in humans is trichromatic…absorb either green, red, or blue light.” If there are three colors that determine color vision and three colors with absorption curves, then we can determine which color corresponds with which curve to answer the question. 

1. Red colors appear more green than normal. – As seen above, when the color of interest is red, there is still some absorption of green (see rightmost highlighted region). When we’re looking at green (the leftmost highlighted region) there is greater absorption of green than red and at times even more blue absorption. If the phenotype is red pigment absent as in this question stem, imagine what would happen if the red curve were not present. The “red” highlighted region would still absorb green light, so the colors that would normally be seen as red will now be seen as more green. This answer choice best represents color perception with red pigment absent.
2. Green colors appear more red than normal. – This is the opposite of what would happen. As noted above, imagine for “red pigment absent” that the red curve did not exist. The “green” region, highlighted region to the left, would appear less red than normal as red would not be absorbed.
3. Both red and green colors appear more blue than normal. – Blue light absorption overlaps with green light absorption but not with red light absorption. If the red curve is removed, only green light is absorbed in the red region (rightmost highlight) with no contribution from blue.
4. Neither red nor green colors are perceived. – Nothing is wrong with the curve for green light absorption; it is the curve for the red light absorption that has been compromised. Green can still be perceived making A the correct answer.

38) The best approach to this question is to review what the passage says about the different alleles, also summed up in Table 1, and then determine the combination that allows the mother to have normal vision but have a son with green pigment absent.

“The allele for normal color vision (+) is dominant over the allele for color weakness (′), which is dominant over the allele for extreme color weakness (″), which is dominant over the allele for lack of pigment (–).”

1. G+R+ / G–R– – While the G+R+ allele would account for the mother’s normal vision, the G–R– allele would be both green and red pigment absent, not just green pigment absent as we’re looking for in the son.
2. G′R′ / G′R′ – These alleles are neither normal vision nor green pigment absent phenotype. 
3. G+R+ / G–R+ ­– The G+R+ dominant allele would allow the mother to have normal vision and the son that inherits G–R+ will be green pigment absent. This is the correct allele pairing.
4. G″R″ / G+R+ – Like answer choice A, this answer accounts for the mother’s normal vision but not the son’s green pigment absent phenotype which should be G–R+ leaving C as the best answer.

39) The first sentence states that the visual pigments absorb green, red, or blue light. The correct answer will involve a cell type that absorbs light and can pass along signals to the brain.

1. Neurons in the visual center of the brain – The occipital lobe houses the visual processing center of the brain, but it does not directly absorb light.  
2. Neurons making up the optic nerve – The neurons in the optic nerve transmit visual signals but do not directly absorb light.
3. Pigment cells of the iris – The pigment of the iris, what we colloquially refer to as someone’s eye color, does not determine our ability to perceive color. 
4. Visual receptor cells in the retina – The photoreceptors and cones in the retina absorb light and transmit signals to the brain for processing. This is the best answer.

Sample Test Biology/Biochemistry Section Passage 8

40)

1. Mitochondria – The third paragraph notes that “BAT is the primary site of nonshivering thermogenesis (heat production) in rodents.” If BAT is involved in heat production, it will need a lot of mitochondria to produce the energy required for heat production. Looking good.
2. Rough endoplasmic reticulum – The rough ER is key for protein production, not so much for energy production.
3. Golgi apparatus – The Golgi apparatus is involved in the processing and transport of proteins and lipids. Mitochondria are more directly related to energy and heat production.
4. Lysosomes – Lysosomes are important for cellular digestion and recycling. The key characteristic of BAT is its heat production which requires energy. Lysosomes are not directly involved in energy production the way mitochondria are so we’ll keep answer choice A.

41) A control is a baseline against which researchers can compare an intervention. A negative control is when researchers do not perform the intervention or do not change the variable of interest. 

1. I and II only – The sham operation is indeed one of the controls because it represents when the pineal gland was not removed and allows the researchers to have something to compare the results for when the pineal gland was removed. If the researchers are interested in what happens when the pineal gland is removed, the pinealectomy is the intervention and not a control. Only half of this answer is correct.
2. I and III only – The sham operation is a control because the intervention, the removal of the pineal gland to observe the effects, is not performed. Because the control is when the researchers do not change the independent variable or do not perform the intervention, keeping the hamsters at room temperature serves as the control. The researchers want to know what happens when the hamsters are placed in a hot environment, so a “normal” or baseline environment at a normal temperature provides the researchers a reference point they could later use to interpret the results. This is the best answer because both roman numerals are controls.
3. II and IV only – As noted above, the pinealectomy is not a control, but the intervention of interest. Also, the exposure to heat is what the researchers are manipulating and changing so they can observe the difference. They actively changed the temperature to determine its effect—this is not a control, but also a variable of interest/intervention.
4. III and IV only – While normal temperature is a control, the hot temperature is the intervention.

42) To answer this question, we’ll need to review the results shown in Table 1; see below for a brief interpretation. 

1. Increased thermogenesis, because BAT mass was greater in Normal/Px than in Hot/Px hamsters – This answer choice compares the change in temperature and not the effect of the pinealectomy because it specifies “Px” or having undergone the pinealectomy in both Normal/Px and in Hot/Px. In order to determine the effect of the pineal gland, we would need to compare having undergone the pinealectomy “Px” and not having undergone the pinealectomy “sham” making this answer choice incorrect.
2. Decreased thermogenesis, because BAT mass was greater in Hot/Sham than in Hot/Px hamsters – This answer choice is super tempting because 177mg is greater than 161mg, but the lack of the cross/letter t indicating significance means that this result is likely due to random chance and that the results are, for all intents and purposes, equivalent.
3. No effect, because BAT mass was equivalent in Normal/Sham and Normal/Px hamsters – This is the best answer choice because none of the differences in BAT mass are statistically significant. If the results are not statistically significant, they are most likely due to chance and are not the result of the intervention. If having the pineal gland removed doesn’t affect the BAT mass, then the pineal gland does not affect the hamsters’ BAT-mediated thermogenesis.
4. No effect, because BAT mass was equivalent in Normal/Px and Hot/Px hamsters. – Like answer choice A, this answer choice incorrectly compares two groups of hamsters that received the pinealectomy. The differences in the two are the result of temperature change, not the effects of the pineal gland.

43) Once again, to answer this question we’ll need the results shown in Table 1. A quick skim of the answer choices lets us know to focus our analysis on the pituitary and testes weights.

In this new experiment, the administration of pineal gland extract (Pgex) is like having the pineal gland intact (the sham/control) and saline (Psal) is like not having the pineal gland (pinealectomy/Px).

1. Pituitary weights of Pgex/Hot hamsters are less than those of Psal/Hot hamsters. – Another way of thinking about this is that Sham/Hot pituitary weight is less than Px/Hot pituitary weight. This agrees with our note in blue above and supports the results from the original experiment making it an incorrect answer choice.
2. Pituitary weights of Pgex/Hot hamsters are greater than those of Pgex/Normal hamsters. – Once again rewriting the new experiment to align with the original experiment, this answer choice says the Sham/Hot hamsters have greater pituitary weights than Sham/Normal hamsters. According to the original experiment, no significant difference was observed with temperature change, but that doesn’t make it impossible, just very unlikely. This answer choice doesn’t actively support or validate the original experiment, but it also doesn’t directly contradict it. Going from no change to greater than is not the same as greater than to less than. Let’s see if we can find an answer choice that directly contradicts the original experiment.
3. Testes weights of Pgex/Hot hamsters are less than those of Pgex/Normal hamsters. – This answer choice is analogous to Sham/Hot having lower testes weights than Sham/Normal. According to the original experiment, the hot temperature decreased testes weight in the sham and pinealectomy conditions so this answer choice supports and validates the original results. 
4. Testes weights of Pgex/Hot hamsters are greater than those of Psal/Hot hamsters. – The testes weights of Pgex/Hot hamsters will resemble those of the Sham/Hot hamsters and the Psal/Hot hamsters will resemble the Px/Hot hamsters. In the original experiment, Px/Hot had significantly greater testes weights than Sham/Hot, the opposite of what is seen in this answer choice. This answer directly contradicts the results of the original experiment and will therefore NOT validate the original experiment. Because it most clearly contradicts the original experiment, it is the correct answer.

Discrete Questions

44) A paramecium is a type of protozoa that is a single-celled organism. Even if we didn’t know this, we could still work through this question! 

1. reproduce by mitosis. – Single-celled eukaryotes undergo mitosis, as do the cells that make up different tissues. If we weren’t sure what a paramecium was, we would take a mental note that this answer is characteristic of eukaryotes and keep going. 
2. have subcellular organelles. – Like the answer choice above, this answer choice is characteristic of eukaryotes which contain membrane-bound organelles within the cell. Knowing that paramecia are eukaryotes makes it easier to rule out this answer choice because both paramecium and tissue cells share this characteristic. If we didn’t know that paramecia are eukaryotes, we could still rule out this answer choice because it’s the second characteristic of eukaryotes in the answer choices. This means that if paramecium had not been a eukaryote, there would be two right answers which would not happen! We would then start looking for an answer that is not a basic characteristic of eukaryotes.
3. are capable of extended independent life. – While the single-celled paramecium is capable of living an extended and independent life, its more extroverted cousin, an individual cell making up tissue, needs its neighbors, signals and the like to maximize its lifespan. Think about all of the integrated signals in the body that promote life! It sure would be hard to mimic all of that complexity by isolating a single cell from a tissue. This must be the right answer then.
4. can metabolize nutrient molecules. – If a cell could not metabolize nutrients, it would not live. We’ll stick with answer choice C because it’s the only answer that applies to only one of the two cells mentioned in the questions stem.

45)

A. Pyruvic acid – Pyruvate is the end product of glycolysis and does not require oxygen, so it will be produced in a significant amount. The question asks for something that will not be produced in a significant amount so this answer choice is incorrect.

B. Glucose-6-phosphate – G6P is produced in the first step of glycolysis so, like pyruvate, plenty of G6P will be produced.

C. Lactic acid – Lactic acid is the product of lactic acid fermentation which is an anaerobic fermentation. Lots of lactic acid will be produced.

D. Acetyl-CoA – when oxygen is present, pyruvate is brought into the mitochondria and is converted to acetyl-CoA. But, if there is a complete absence of oxygen as described in the question stem, this will not occur, and acetyl-CoA will not be produced in a significant amount making it the correct answer.

Below is a graphic representation of the different fates of pyruvate depending on oxygen status:

46)

1. 11% – This percentage is smaller than any of the percentages for the individual nitrogenous bases.
2. 23% – This is the percentage of adenine or thymine.
3. 27% – This is the percentage of both guanine and cytosine individually. Looking good!
4. 46% – This is the combined percentage of adenine and thymine.

For more on base pairing, take a peek at the base pairing content page: https://jackwestin.com/resources/mcat-content/nucleic-acid-structure-and-function/base-pairing-specificity-a-with-t-g-with-c

47)

1. a hydroxyl group. – While there is indeed an -OH attached to the head of a fatty acid, it is attached to a carbonyl. A -COOH group is better described as a carboxyl group, and not simply a hydroxyl group.
2. a carboxyl group. – Fatty acids contain a long hydrocarbon tail and a carboxylic acid on the other end, meaning the carboxyl group makes up the fatty acid head. The content page describes fatty acids in more detail: https://jackwestin.com/resources/mcat-content/metabolism-of-fatty-acids-and-proteins/description-of-fatty-acids-bc
3. a phosphate group. – While a phospholipid would contain a phosphate group, a simple fatty acid does not.
4. an amino group. –  A phosphatidylethanolamine, a type of phospholipid, has an amine group hence its name. This does not apply to fatty acids. 

Sample Test Biology/Biochemistry Section Passage 9

48)

1. HCO3– in the RBCs. – The generation of bicarbonate is dependent on carbonic anhydrase. If a carbonic anhydrase inhibitor is given, the reaction will slow and cause a buildup of the leftmost reactants, not the products.
2. H2CO3 in the RBCs. – For the same reason described above, there will be a buildup of the main reactants, not the downstream products carbonic acid and bicarbonate.
3. CO2 in the tissues. – If a carbonic anhydrase inhibitor were administered, the forward reaction shown in equation 1 would still occur, but much more slowly (remember that enzymes catalyze reactions and speed them up, they don’t alter the thermodynamics). This would result in a buildup of the reactant CO2 which is produced in the tissues.
4. H2O in the tissues. – Water is involved in many reactions and is not dependent on the bicarbonate buffer system for its removal. CO2 in the tissues is a better answer choice.

49)

1. Veins, because the HCO3– concentration is higher in veins than in arteries. – The passage says that following the release of bicarbonate ions, “The resulting ionic imbalance in the RBCs is equalized by the rapid movement of chloride (Cl–) ions into the RBCs from the plasma.” This means that an increase in Cl– would follow an increase in HCO3–. The bicarbonate buffer system demonstrates that an increase in CO2 would produce an increase in HCO3–. Thus, an increase in carbon dioxide ultimately leads to an influx of chloride anions into the RBCs. Carbon dioxide concentration is greater in the systemic veins because it is a metabolic waste product and has not been expelled during gas exchange. In the systemic veins: higher CO2 concentration –> generation of more HCO3– –> influx of Cl– into RBCs.
2. Veins, because there are fewer RBCs in veins than in arteries. – There is no systematic difference in the number of RBCs in the veins versus the arteries. The circulatory system is a closed loop so the same number of RBCs that pass through the arteries must pass through the veins.
3. Arteries, because the HCO3– concentration is higher in arteries than in veins. – The concentration of carbon dioxide is greater in the systemic veins, not the systemic arteries. There will be less HCO3– in the arteries, not more.
4. Arteries, because there are fewer RBCs in veins than in arteries. – As mentioned in the explanation for B, the number of RBCs traveling through the arteries is the same as the number of RBCs traveling through the veins. Answer choice A is the correct answer.

50) To understand the relationship between PCO2 and acidity, we can head back to Equation 1.

1. high, because Equation 1 will proceed to the right. – According to the equation above and Le Chatelier’s principle, increasing the concentration of a reactant, CO2, will cause a rightward shift. On the righthand side of the equation there are protons. As proton concentration increases, pH decreases, and the environment becomes more acidic. Thus dissociation, which occurs more readily in an acidic environment, will increase when the concentration of CO2 as measured by PCO2 is high.
2. high, because Equation 1 will proceed to the left. – As noted above, an increase in the concentration of a reactant will cause a rightward shift to restore equilibrium, not a leftward shift.
3. low, because Equation 1 will proceed to the right. – If the concentration of a reactant decreases, the reaction will proceed in the direction of the reactants to restore equilibrium, so to the left.
4. low, because Equation 1 will proceed to the left. – While a decrease in the concentration of a reactant will indeed cause the reaction to proceed to the left, a leftward shift in the bicarbonate buffer system will cause the consumption of protons and will increase the pH, creating a more basic environment thereby decreasing dissociation. Answer choice A is the answer that correctly describes the relationship between PCO2 and environmental acidity to maximize dissociation.

51)

1. increasing the production of CO2 in the RBCs. – The narrowing of the capillaries will not cause RBCs to produce CO2. Our focus should be on how the narrowing of a blood vessel will affect the capillary-alveolar interface and how this could affect the exchange of gas.
2. allowing RBCs to have direct contact with alveoli. – RBCs should not have direct contact with alveoli; this would be the result of a breakdown of the capillary-alveolar interface and would allow blood into the lungs.
3. giving maximum exposure of each RBC to diffusing gases. – Forcing the RBCs to move in a single file through the capillaries would allow all RBCs to come into contact with the capillary-alveolar interface and participate in gas exchange. Think of it this way, if you have a giant hose, most of the fluid would be located in the center and very little would actually be in contact with the hose itself. If the hose is narrowed, more of the fluid will come into contact with the edges of the hose. The same applies for RBCs forced into a single file within the capillaries.
4. making Hb available for CO2 but not O2 to bind. – Having to pass through capillaries in a single file will not change their affinity for O2 or CO2. In case you were confused regarding affinity, it is classically CO that Hb will preferentially bind over O2. Answer choice C is the best answer.

Sample Test Biology/Biochemistry Section Passage 10

52)

1. Centriole – Centrioles are important for cell division and make up the microtubule organizing centers or MTOCs. They are not involved in retrograde trafficking. 
2. Plastid – Plastids are specific to plants and algae which are eukaryotes. They are involved in nutrient creation and storage, and may contain photosynthetic pigments.
3. Nucleolus – The nucleolus is located within the nucleus and is not involved in retrograde trafficking.
4. Lysosome – Of the answer choices provided, only lysosomes are involved in retrograde transport with the help of dynein proteins. By allowing STx to go directly from the early endosome to the Golgi, subunit B allows the toxin to bypass lysosomes.

53)

To answer this question, we need to know the properties of alanine and proline. Alanine is a small hydrophobic amino acid, while proline is polar and disrupts alpha helices. Subunit A is introduced as “an α-helical 293-amino acid subunit A” which means the substitution of alanine to proline would disrupt the α-helical structure.

1. lack of retrograde trafficking. – Subunit B is the subunit involved in retrograde trafficking. Disrupting the structure and function of subunit A should not affect the function of subunit B.
2. increased hydrogen bonding. – Proline does not have any more nitrogens, oxygens or fluorines than alanine. Not every polar substance is capable of H-bonding. This is not an effect of the above substitution. 
3. loss of secondary structure. – Here we have it; an α-helical structure is disrupted by proline kinks and α-helices are a type of secondary structure.  
4. higher catalytic activity. – Disrupting the structure of the subunit would decrease catalytic activity, not increase it. Answer choice C is the correct answer.

54)

1. GPP130 is also a target of several common antibiotics. – While tempting because GPP130 is a host protein, it’s important to recall the role of GPP130 in infection: “GPP130, a host membrane protein that cycles between the Golgi apparatus and endosomes, might assist STx in evading degradation.” If a host protein that aids in STx evading degradation were targeted, it would actually decrease the burden of infection. STx would not be able to evade the immune system and would be destroyed. This would be a favorable effect of taking antibiotics, not a reason to withhold them.
2. the bacterial lysate will disrupt host translation. – This is true. The author states that the STx protein “halts protein synthesis” in the host. If translation is halted in the host when the E. coli is lysed, the host cell would be negatively affected. The compromise of the host in addition to the lysis of the E. coli would be a contraindication for the administration of the antibiotic. An ideal therapeutic treatment will address the threat (in this case the STx) without harming the host.
3. STx confers antibiotic resistance once inside the host cell. – This could also be a good answer choice—the potential for antibiotic resistance makes physicians less likely to use a particular antibiotic. However, this is not supported in the question stem which specifies that the antibiotics are able to lyse the E. coli. Also, this is not as specific concerning the STx protein which is the focus of the passage.
4. antibiotics are not effective against viruses. – E. coli are bacteria, not viruses. This answer choice does not actually address the question of why antibiotics that are known to lyse E. coli are not given. Answer choice B is the correct answer.

55) The end of paragraph tells us that “GPP130…might assist STx in evading degradation and reaching the endoplasmic reticulum as it binds the toxin with a Kd of 25 mM.” For the novel compound to be effective, it needs to bind STx more readily than GPP130, which means it needs to bind at a lower concentration and a lower Kd. 

1. 0.25 mM – This is lower than the 25mM Kd for GPP130. Looking promising.
2. 2.5 mM – This is also lower than the Kd given for GPP130. However, it’s a higher concentration than 0.25mM and the question is asking for the most effective drug. The more effective of the two is the one that binds at the lower concentration, so we’ll keep A so far.
3. 25 mM – This is the same Kd as GPP130 and unlike the two answer choices above, would not provide any binding advantage over GPP130.
4. 250 mM – This concentration is higher than the one used as the Kd for GPP130. A drug with a Kd of 250mM would be less likely to bind STx; we’ll go with answer A.

56) The first paragraph states that “The STx protein is comprised of two subunits… Subunit B mediates retrograde trafficking from the cell surface to the endoplasmic reticulum, after which subunit A is cleaved into the catalytic A1 domain.” The enzyme cleaves the protein to activate subunit A.

1. Endonuclease – Endonucleases cleave DNA and RNA bonds, specifically between the 5’ and 3’ ends/the “middle”. 
2. Exonuclease – Exonucleases, like endonucleases, cleave phosphodiester bonds between nucleotides. Unlike their endo- counterpart, exonucleases cleave at the 5’ and 3’ ends and work their way in.
3. Ribonuclease – Ribonucleases are specific to RNA. We need an enzyme that will cleave peptide bonds in a protein.
4. Protease – A protease cleaves the peptide bonds between amino acids within a protein and is the only answer choice that is specific to cleaving proteins which is what allows for the activation of subunit A.

Discrete Questions

57) On test day, we would quickly scan the answer choices and realize that we just need to determine whether CAP and LacI are transcription activators or repressors.

Row 1: Both proteins are bound with no lacZ transcription. At least one of the proteins must be a repressor because if both proteins bound were activators, there would be lots of transcription. 

Row 2: Here, CAP is bound and LacI is not bound—the release of LacI has allowed for lots of transcription (+++) which means LacI has to be a repressor. Removing an activator is not going to increase transcription, it would decrease it. This row also tells us that CAP is an activator. If CAP were also a repressor, there would not be lots of transcription. The first two rows together tell us that LacI is a repressor and CAP is a transcription activator protein.

Row 3: Here CAP is not bound and LacI is bound. The absence of an activator and the presence of a repressor should produce no transcription, and indeed that’s what is shown! This supports our interpretation of row 2.

1. CAP is a transcription activator, and LacI is a transcription repressor. – This aligns with the interpretation we settled on above. When CAP is present and LacI is absent, transcription occurs.
2. CAP is a transcription repressor, and LacI is a transcription activator. – This is the opposite of the correct answer. When only CAP is present, transcription occurs (activator) but when LacI is present, no transcription occurs (repressor).
3. CAP is a transcription activator, and LacI is a transcription activator. – When LacI is present, no transcription occurs indicating that LacI is a transcription repressor and not a transcription activator.
4. CAP is a transcription repressor, and LacI is a transcription repressor. – When only CAP is present, transcription occurs so it is a transcription activator and not a transcription repressor. Answer choice A is the correct answer.

58) Once again there’s a question on amino acids, but this time it’s asking about decreasing entropic penalty. To better understand this question, let’s imagine a folded protein with two main regions, a hydrophobic core, and a hydrophilic outside. Would the protein spontaneously unfold to expose the hydrophobic amino acids to the aqueous environment? No, that wouldn’t be too favorable. However, the folding of the protein is favorable. This question causes a lot of students distress because “entropic penalty” is often an unfamiliar term. Breaking it down, we know that entropy is loosely the disorder of a system, and a penalty is usually a price paid for something unfavorable. Putting the two together, along with our understanding of the favorability of protein folding, we can appreciate the following:

-entropic penalty: price paid for an unfavorable ordering of a system

-decrease in entropic penalty: decreasing the energetic price for the folding, so a switch between amino acids with similar characteristics such as valine to leucine

-increase in entropic penalty: increasing the energetic price for folding, so swapping amino acids with different characteristics such as alanine to aspartate

1. Ile to Asp substitution at a buried site – Isoleucine is a nonpolar amino acid, while aspartate is a negatively charged amino acid. Assuming the buried site is hydrophobic, swapping a nonpolar amino acid for a charged amino acid would increase entropic penalty—this is an unfavorable change.
2. Leu to Thr substitution at a surface-exposed site – Leucine is a nonpolar amino acid and threonine is polar uncharged. At a surface-exposed site, which would be polar, amino acids should be hydrophilic or polar. Swapping a nonpolar amino acid for a polar one at a polar site is favorable and decreases entropic penalty.
3. Gly to Pro substitution in a flexible loop – Glycine is nonpolar and has a very small R group while proline is polar and incorporates the backbone into its side chain ring. If the region is for a flexible loop, swapping an amino acid with a small side chain for one with a bulky, fixed conformation would be unfavorable and increase entropic penalty.
4. Arg to Tyr substitution at a surface-exposed site – Arginine is positively charged, and tyrosine is aromatic and nonpolar. Changing a charged amino acid in a polar region for a nonpolar acid in a polar environment is energetically unfavorable and increases entropic penalty. Answer choice B is the only answer choice that correctly describes a decrease in entropic penalty.

59) We’re ending the bio/biochem section with a very straightforward question; either you know the content or you don’t. On test day, we’ll devote less time to these types of questions. As shown below, the endoderm gives rise to epithelial linings, endocrine tissue and many of the internal organs. If you’ve had trouble memorizing the different structures that each of the germ layers gives rise to, hopefully this can help:

Ectoderm: The ectoderm gives rise to the structures people are often attracted to—the epidermis and hair, the mouth and teeth (and the anus), the eyes (cornea and lens), the nervous system (because we care about the mind!) and the sensory receptors that help us interact with the world around us. It also gives rise to the adrenal medulla and the pineal and pituitary glands. When in doubt, what are people often attracted to?

Mesoderm: The mesoderm is all about movement, go, go, go! The mesoderm provides the skeletal system, muscles and excretory system, the circulatory and lymphatic systems, the reproductive system and adrenal cortex. Mesoderm ~ movement of something.

Endoderm: When I think endoderm, I think of smooth and bumpy twins, lining the “insides.” Here we have the epithelial linings of the digestive and respiratory system, the lining of smooth structures like the urinary and reproductive systems, the bumpy endocrine tissues and other internal organs.

1. Mouth – Using our quick and dirty categorizations above, the mouth is something people are attracted to, so it’s derived from the ectoderm, not the endoderm. This is a NOT question so this is a good answer choice.
2. Bronchi – The bronchi are part of the respiratory system and are indeed derived from the endoderm. This is not a good answer choice. 
3. Bladder – The bladder is one of those “smooth” linings so it’s endoderm-derived and an incorrect answer.
4. Stomach – Like the bladder, the stomach is one of those endoderm derivatives. Only A is not primarily derived from the endoderm and is the correct answer.

For a more extensive review of the germ layers, check out https://jackwestin.com/resources/mcat-content/embryogenesis/major-structures-arising-out-of-primary-germ-layers