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.

 

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