Official Guide Biology/Biochemistry Section Passage 1
1) To answer this question, we need to figure out which muscle fiber subtype is represented by culture C. To do that, we need to use Table 1 (shown below) to assign muscle fiber subtypes to cultures A and B, and whichever muscle fiber subtype remains is culture C.
Culture A: is
- “white when visualized with a microscope” –> fast glycolytic twitch muscle fibers (IIb or IIx) have very little myoglobin and mitochondria, so they appear whiter than their slow twitch counterparts
- “When an electrical current was applied to the culture medium, the concentrations of lactate and H+ ions in that medium increased” –> the presence of lactate in culture A indicates the muscle is using lactic acid/anaerobic fermentation to meet its metabolic needs
- “The pH of the medium changed from 7 to 5 after 2 minutes of continuous, electrically-stimulated contractions.” –> the pH changed in response to the fermentation very quickly, so the muscle fiber cannot tolerate prolonged use or it becomes too acidic
- Culture A must be type IIx because it is for short-term anaerobic use
Culture B: is
- “red in appearance and had a high mitochondrial density” –> this muscle fiber subtype has plenty of myoglobin to bring in oxygen and has a lot of mitochondria to make use of that oxygen for aerobic metabolism –> high oxidative capacity
- “numerous intact capillaries, and abundant intracellular triglyceride deposits” –> glucose is not the primary form of energy storage so glycolytic capacity is low
- “slow contraction phase…very slow rate of ATP utilization” –> Culture B is slow so it has low power (power=force/time) and because it uses ATP more slowly, will have a longer duration
- Culture B is muscle fiber subtype I because it exhibits long duration aerobic metabolism and has low glycolytic capacity
If Culture A is type IIx and Culture B is type I, Culture C must be type IIa. We will look at type IIa in the table when deciding between the answer choices, remembering that the goal is to find an answer choice that does not describe Culture C.
A. a fast rate of muscle contraction – type IIa is “moderately fast” which is still fast. Let’s see if we can find a better answer choice.
B. the ability to engage in oxidative and anaerobic respiration – type IIa is used for long-term anaerobic respiration, but it also has high oxidative capacity which means it can engage in both types of respiration.
C. the presence of medium-sized motor units – the table shows that IIa muscle fibers have medium-sized motor neurons.
D. low densities of mitochondria and capillaries – as noted with answer choice B, the table shows that type IIa fibers have high oxidative capacity. In order to have high oxidative capacity, the muscle fibers would need plenty of mitochondria and capillaries to bring oxygen to the muscle fibers. These characteristics of IIa fibers are the opposite of answer choice D. This question is an EXCEPT question which makes answer choice D correct.
2) Which steps involved in the contraction of a skeletal muscle require binding and/or hydrolysis of ATP?
I. Dissociation of myosin head from actin filament
II. Attachment of myosin head to actin filament
III. Conformational change that moves actin and myosin filaments relative to one another
IV. Binding of troponin to actin filament
V. Release of calcium from the sarcoplasmic reticulum
VI. Reuptake of calcium into the sarcoplasm
On test day, we should skim the roman numerals included in each answer choice and narrow down which roman numerals to assess. Each of the answer choices below includes III which means it must be true, and none of the answer choices include V so it must be false. Next, we would look at a roman numeral found in half of the answer choices. While we’re acknowledging the “roman numeral trick,” it doesn’t help us with this question because I, II, IV and VI are all included in half of the answer choices.
A. I, II, and III only – Looking at the figure above, we can see that the dissociation of the myosin head from the actin filament requires ATP so I is correct. On the other hand, it is not required for attachment of the myosin head so II is false.
B. II, III, and IV only – As noted above, II is false so answer choice B is also incorrect.
C. I, III, and VI only – We’ve established that I and III are both true, so we only have to evaluate VI. Reuptake of calcium from the sarcoplasmic reticulum will move calcium against its concentration gradient and require energy. All three roman numerals are correct!
D. III, IV, and VI only – The binding of troponin to actin does not require energy so answer choice D is incorrect and we’ll stick with answer choice C.
3) To answer this question, we need to know the role of acetylcholine within the medium of interest, that is muscles. Acetylcholine is a neurotransmitter released at neuromuscular junctions by the presynaptic neurons that binds to nicotinic receptors on the muscle fiber. This allows for the influx of sodium, the release of Ca2+ from the sarcoplasmic reticulum and ultimately muscle contraction. Between this question and the previous question, we see knowing the mechanisms of muscle contraction are important for test day. If you haven’t already studied muscle contraction, jot down the topic for future review.
A. depolarization of the cell membrane that resulted in contraction – the influx of sodium will depolarize the cell and the ultimate release of Ca2+ will cause muscle contraction. We’re off to a good start.
B. repolarization of the cell membrane that resulted in relaxation – the influx of Na+ will cause the cell to become more positive compared to baseline. Repolarization implies that we’re below baseline and are becoming more positive to return to the baseline voltage so this answer choice is incorrect. We also know that acetylcholine causes muscle contraction, not relaxation.
C. hyperpolarization of the cell membrane that resulted in contraction – hyperpolarization indicates that the membrane voltage is becoming more negative, the opposite of depolarization making this answer choice incorrect.
D. depolarization of the cell membrane that resulted in relaxation – the release of Ca2+ by the sarcoplasmic reticulum following acetylcholine release causes muscle contraction, not muscle relaxation.
4) Quoting paragraph 2, when culture A was stimulated, “the concentrations of lactate and H+ ions in that medium increased.” The presence of lactate indicates that culture A was using lactic acid fermentation to produce energy.
A. pyruvate – As shown in the figure above, pyruvate accepts the electrons from NADH and is reduced during lactic acid fermentation.
B. oxygen – Oxygen is the terminal electron acceptor in aerobic respiration, but not in lactic acid fermentation which is an anaerobic process.
C. NAD+ – NAD+ is a product of NADH giving up its electrons and reducing pyruvate; it does not accept any electrons.
D. water – Water is not involved in lactic acid fermentation. Water is a product of aerobic metabolism after oxygen accepts the electrons at the end of the electron transport chain. Answer choice A is the correct answer.
Official Guide Biology/Biochemistry Section Passage 2
5) To ensure that the MS2-GFP fusion protein isn’t influencing where the cat6xbs and the lacY6xbs transcripts localize, we need to see where the fusion protein goes when it is by itself.
A. Determination of expression of MS2–GFP in cells that lacked the 6xbs insertion upstream of the cat and lacY genes – This would isolate the fusion protein and determine where it goes when not bound to the cat and lacY genesis, so when it is by itself. This is likely to be the correct answer.
B. Use of E. coli cells that expressed only MS2 instead of the MS2–GFP fusion protein – If only MS2 were expressed, we would not be able to see where the protein localized because the GFP component of the fusion protein is what allows for fluorescence and visualization.
C. Insertion of the 6xbs region upstream of both the cat and lacY genes in the same cells – If instead of removing the individual cat6xbs and lacY6xbs regions we included both of them at the same time, we would complicate the experiment and make the results less clear. We wouldn’t be any closer to understanding where the MS2-GFP protein goes when by itself and the cat6xbs and lacY6xbs regions would compete with one another to localize to the cytoplasm and cell membrane respectively.
D. Determination of expression of both MS2 and GFP as separate proteins rather than as a fusion protein – Like answer choice B, we would not be able to see where the MS2 protein goes without the GFP component attached.
6) In the last sentence of the first paragraph, the author states that the “the lacY6xbs transcript was localized near the cell membrane” and the cell membrane is in E. coli, identified in the first sentence of the same paragraph. The correct answer choice will include components of a prokaryotic cell membrane.
A. Proteins and glycolipids – While prokaryotic cell membranes contain protein, they do not contain glycolipids. Glycolipids are synthesized in the Golgi apparatus, but prokaryotes do not contain Golgi because they do not contain membrane-bound organelles.
B. Glycolipids and sterols – Sterols are synthesized in the endoplasmic reticulum. E. coli is a prokaryote that does not contain this membrane-bound organelle, so we would not expect to see sterols like cholesterol in the cell membrane.
C. Sterols and phospholipids – As noted above, sterols like cholesterol are not found in the cell membrane of E. coli.
D. Phospholipids and proteins – Phospholipids and proteins are certainly found in the cell membranes of E. coli and are in fact key and the major components of cell membranes in general. The lacY6xbs transcript is found near the E. coli cell membrane, so it should be near phospholipids and proteins, making this the correct answer.
A. Vmax decreases, and KM increases. – A competitive inhibitor does not decrease Vmax. Imagine a game of musical chairs, and there is only one chair left. If there is only one of you (the substrate) and one of your friends (the inhibitor), then it will be harder for you to win the game. However, if you clone yourself and now there are 20 of you/substrate molecules, and still only one friend/inhibitor, you can easily displace your friend and win the game. Similarly, a competitive inhibitor can be overcome by increasing the substrate concentration, so Vmax is unchanged.
B. Vmax decreases, and KM remains unchanged. – As noted above, Vmax remains unchanged.
C. Vmax remains unchanged, and KM increases. – Now let’s think about the Km: a low Km corresponds to a higher affinity, and a higher Km corresponds to a lower affinity; the two are inversely related. Using the same musical chairs analogy: if you/the substrate are the only one fighting for the chair, you’re very likely to sit in the chair and win the game. However, when your friend/the inhibitor comes to challenge you for the chair, you’re less likely to sit in the chair than when you were alone, so your affinity for the chair decreases. As affinity decreases, Km increases. This must be the correct answer.
D. Vmax remains unchanged, and KM decreases. – As noted above, Km increases, it does not decrease. On test day we select answer choice C and keep going.
A. Diffusion across the cytoplasm – diffusion across the plasma is quite slow. If the transcript has a short half-life, it may well degrade before reaching its destination. We need something faster.
B. Transport via attachment to the mitotic spindle – Tempting, except E. coli, the organism used in this experiment, does not undergo mitosis, but rather a similar process called binary fission. Prokaryotes do not have a nucleus, so they do not undergo mitosis and do not form mitotic spindles. The ultimate goals of binary fission and mitosis are also distinct. Binary fission is a form of asexual reproduction, whereas mitosis is centered on growth. Note that keeping track of the cell line or organism used in an experiment is incredibly important on test day. There are many questions that require you to distinguish between prokaryotes and eukaryotes—know the differences and use them to your advantage. When reading a bio/biochem passage, ask yourself if the researchers are working with prokaryotes or eukaryotes.
C. Active transport along cytoskeletal filaments – Prokaryotes contain cytoskeletal filaments (think structure and transport), and moving transcripts using this active method will be much faster than diffusion. The speed of transport is important here because the bgIF transcript has a short half-life. The faster the transport, the more likely the bgIF transcript makes it to the membrane without degrading.
D. Transport from the endoplasmic reticulum in vesicles – Remember that we’re working with prokaryotes, which do not contain membrane-bound organelles, including the endoplasmic reticulum. Answer choice C is the correct answer.
9) The question stem specifies that chloramphenicol does not inhibit translation in cells containing the cat6xbs plasmid, so we’ll need to remind ourselves of what the plasmid contains. Looking back at the passage: “The gene cat encodes the soluble protein chloramphenicol acetyltransferase” and 6xbs is “the binding sequence (6xbs) for the phage MS2 coat protein.” The 6xbs component doesn’t seem to have anything to do with chloramphenicol, but the cat gene does! Let’s break down the name chloramphenicol acetyltransferase. The -ase ending lets us know it’s an enzyme, the “transferase” tells us it’s transferring a functional group from one substrate to another and the “acetyl” lets us know what that functional group is. Including “chloramphenicol” before the enzyme type tells us that chloramphenicol is the substrate. This enzyme is likely inactivating chloramphenicol by transferring an acetyl group. On to the answer choices.
A. Increase the chloramphenicol concentration. – If the cat gene is producing an enzyme that inactivates chloramphenicol, adding more chloramphenicol won’t help because the enzyme will just inactivate it.
B. Increase the chloramphenicol incubation time. – Again, the chloramphenicol is being inactivated by an enzyme. Changing the incubation time will not stop its inactivation.
C. Alter the incubation temperature by a few degrees. – While increasing incubation temperature by a few degrees can improve translation (depending on the new temperature), it will not stop chloramphenicol acetyltransferase from rendering chloramphenicol ineffective and will not inhibit translation. We need to find a better answer choice; hopefully answer choice D is correct.
D. Use an alternate antibiotic. – If chloramphenicol is not functional, it should be replaced with another antibiotic, so this is the correct answer choice. Note that we are not explicitly told in the passage that chloramphenicol is an antibiotic, but we can assume this to be the case because we are told that it inhibits translation in E. coli which would ultimately cause the cell to die. On test day, we will need to use process of elimination to rule out answer choices A-C, and then we’ll need to connect the dots and identify chloramphenicol as an antibiotic.
Official Guide Biology/Biochemistry Section Passage 3
10) Before we look at the answer choices, we need to identify the three cell types: “Infected RBCs adhere to platelets, endothelial cells, and other mature RBCs.” We need to find an answer choice that applies to at least two of these—platelets, endothelial cells and RBCs.
A. They have nuclei. – platelets are cell fragments derived from megakaryocytes and do not contain nuclei. Mature red blood cells have lost their nuclei. Only endothelial cells have nuclei. This answer choice doesn’t meet the requirements.
B. They are cell fragments. – Endothelial cells and RBCs are both cells, only platelets are cell fragments. This cannot be the correct answer.
C. They are bone marrow–derived. – Both RBCs and platelets are derived from the bone marrow. Red blood cell precursors, or erythrocyte precursors, are made in the bone marrow. Megakaryocytes, the cells that platelets are derived from, are also made in the bone marrow.
D. They are connected by tight junctions. – Only endothelial cells are connected by tight junctions. The individual RBCs and platelets need to be mobile in the blood in order to perform their functions. Answer choice C is the only answer that meets the criteria for this question.
11) We’ll need Fig. 2 to help us answer this question.
Focusing on the bottom of Fig. 2, we see that SET/PHD has strong binding (++) to unmodified histone H3, and that SET/ΔPHD has poor binding (-) to unmodified histone H3. What does the delta stand for then? The ΔPHD indicates that the PHD domain has been removed, supported by the fewer amino acids (581 amino acids vs 449 amino acids) to the right of the protein schematic. Moving on to the answer choices, we know that removing the PHD domain decreases the function and binding of the PfSET10 protein.
A. the SET domain. – Fig. 2 showed that the binding function is determined by the presence of the PHD domain, not the SET domain. The SET/ΔPHD protein only has the SET domain, and the SET domain is not enough to preserve binding function, so it’s not going to be the binding domain.
B. the PHD domain. – This aligns with our analysis of Fig. 2; the presence or absence of PHD is what affects the binding function which supports PHD being the binding domain.
C. the N-terminal domain. – Fig. 2 shows us that PHD, the domain that we identified as influencing binding, is located in the center of the protein, not the N-terminus. PHD is the thin black rectangle next to the larger black rectangle SET domain. PHD is highlighted above.
D. the C-terminal domain. – Fig. 2 shows us that PHD, the domain that we identified as influencing binding, is located in the center of the protein, not the C-terminus. We’ll stick with answer choice B.
12) The question is asking us to identify which protein needs to exit the parasite. Let’s take a look at the options.
A. PfEMP1 – In paragraph 1, we’re told that “transmembrane protein PfEMP1 variants…localize to the RBC plasma membrane and bind to endothelial cells.” Since the protein is on the RBC plasma membrane, it must somehow leave the parasite itself. Similarly, the fact that it binds to endothelial cells means that PfEMP1 needs to travel beyond and outside of the parasite. This answer choice is promising!
B. PfSET10 – From the passage we know that PfSET10 is P. falciparum histone lysine methyltransferase. We also know that it is associated with histones (the name of the protein tells us that it interacts with the lysine in histones) and with the var genes. It should then be located in the nucleus and we have no reason to believe that it would leave the nucleus.
C. Histone H3 – The passage says “Histone H3 on the active var promoter” indicating that histone is associated with the genetic material in the nucleus of the parasite. We have no reason to believe that it would need to leave the nucleus.
D. Hemoglobin – This is not a parasite protein and therefore a distractor. Hemoglobin is found in the RBCs that the parasite infects. Answer choice A is the only one that was satisfying so we’ll select it and move on to the next question.
13) The question is asking us to consider the relationship between PfSET10 and the var genes. This is a very general question so it’s a bit more difficult to predict an answer, however, we’ll bookmark Fig. 1 for reference if we need it.
A. allows active and silent var genes to colocalize in the nucleus. – The passage states “expressed var gene is located in a different place in the periphery of the nucleus than are silent var genes” so they are in different parts of the nucleus and do not colocalize.
B. marks the chromatin of the active var promoter for reexpression after mitosis. – Here we’ll want to go back to Fig. 1 and refresh our memory as to the relationship between PfSET10 and the var genes. On the y-axis, we are given the distance between the var gene and PfSET10. The “on” gene is active and the “off” gene is inactive. For both var gene 1 and var gene 2, we see that the on/active gene is closer to the PfSET10. The passage also says that “P. falciparum cells contain the most PfSET10 when the intraerythrocyte parasites are in an actively dividing life cycle phase.” If there is more PfSET10 when the parasites are actively dividing (undergoing mitosis) and PfSET10 is closer to the active var gene, then it makes sense for the PfSET10 to be marking the active gene for reexpression, or to make sure it gets turned on again, following mitosis. This answer choice looks good but we’ll keep reading.
C. marks the chromatin of a silent var promoter to be expressed after mitosis. – In Fig. 1 we saw that the PfSET10 was closer to the on/active var gene and further from the off/inactive var gene. This means that it should be marking active, and not silent var genes. We’ll stick with answer choice B.
D. marks the chromatin of multiple var promoters for simultaneous expression. – The passage tells us that “Only one var gene is transcribed at a time over multiple mitotic generations” so the “simultaneous” in this answer choice makes it incorrect. Answer choice B is then the correct answer.
A. Purines – purines are not negatively charged
B. Pyrimidines – pyrimidines are not negatively charge
C. Deoxyribose – deoxyribose is not negatively charged
D. Phosphate groups – phosphate groups are negatively charged! By identifying the properties of lysine, we made this question a quick one.
Official Guide Biology/Biochemistry Section Passage 4
15) This question is a pseudodiscrete question. The question is asking us to identify what both the liver and platelets are involved in.
A. cholesterol synthesis – The liver is the primary site for cholesterol synthesis, but platelets do not work together with the liver during cholesterol synthesis.
B. glucose metabolism – We know from our extensive carbohydrate metabolism review that the liver is involved in glucose metabolism. The liver is perhaps best known for controlling glucose levels via glycogen storage and breakdown to facilitate appropriate glucose metabolism. Additionally, the liver is the primary site for gluconeogenesis. However, platelets are not involved in glucose metabolism.
C. blood clotting – When bleeding begins, platelets aggregate together at the site of bleeding and are activated, becoming a plug that helps stop the bleeding. The other major clotting factors involved in recruitment and synthesis of the fibrin clot are produced by the liver. Both the liver and platelets work together during blood clotting. This is also the only answer so far that has a major contribution from platelets!
D. fat digestion – Like answer choices A and B, this answer choice applies to the liver, but not to platelets. The liver produces bile which is then stored in the gallbladder and is responsible for emulsifying fats during digestion. Answer choice C is the only one that has a clear role for platelets and therefore the only one that involves both the liver and platelets working together.
A. transmembrane serotonin transporters – In the second paragraph, we’re told that the serotonin carried by the platelets “is synthesized from tryptophan and secreted by endocrine cells in the lining of the gastrointestinal tract.” Platelets have to pick up and carry the serotonin made and then secreted by the endocrine cells. A transmembrane transporter is a good way to ensure that the serotonin can be picked up and carried by the platelet at a moment’s notice. This answer choice is incorrect for a LEAST question.
B. ribosomes – Platelets are cell fragments derived from megakaryocytes. While they do not contain a nucleus, they do contain organelles, including ribosomes. If you ever forget what’s inside a platelet, imagine a large cell (megakaryocyte) and pinch off a piece of the membrane with cytoplasm inside. What you find in the cytoplasm is what you’ll find in a platelet. Platelets do contain ribosomes, so this answer choice is incorrect.
C. serotonin – The first sentence of the second paragraph begins with: “Platelets carry 95 percent of blood serotonin.” Platelets contain serotonin.
D. Ki67 – As mentioned above, platelets do not contain nuclei. The passage also tells us that the Ki67 protein “is detected exclusively in the nuclei of proliferating cells.” If there’s no nucleus, there shouldn’t be Ki67. This is a least question and we should not find Ki67 in platelets so this is the correct answer.
17) To know where the serotonin receptors are located, we need to identify the general properties of serotonin. Is serotonin as a whole polar or nonpolar? This will help us decide whether serotonin will be able to cross the plasma membrane without help. Looking at serotonin, we see an alcohol and an amine group. These groups are polar and by making serotonin as a whole polar, would prevent it from freely crossing the plasma membrane.
A. in the nucleus – If the receptor were in the nucleus, serotonin would have to be able to cross the plasma membrane, so this answer choice is incorrect.
B. in the cytosol – Like the nucleus, for the serotonin receptor to make it into the cytoplasm, it would need to be able to cross the membrane without help, which the polar groups do not allow. Onward!
C. embedded in the mitochondrial membrane – Don’t let them trick you… The mitochondrial membrane, while still a membrane, is located inside of the cell and past the cell’s plasma membrane. Like the answer choices above, serotonin would have to cross the plasma membrane first. Let’s hope the last answer choice is a better fit!
D. embedded in the cell membrane – This answer choice is a better fit! The alcohol and the amine on serotonin make it a polar molecule and would prevent it from crossing the plasma membrane. As such, the serotonin receptor must be on the plasma membrane.
18) To answer this question, we’ll need to revisit Fig. 2, shown above. What we see is that the addition of a 5-HT2A/2B antagonist decreases the number of Ki67-positive hepatocytes. An antagonist prevents the stimulation of the receptor it binds. The two receptors were associated with little proliferation when prevented from activating, so 5-HT2A and 5-HT2B are likely associated with increased proliferation when activated. Let’s see what the answer choices hold.
A. Administration of 5-HT2A receptor agonist resulted in reduced Ki67 staining. – While this answer choice is tempting at first glance because it describes what we see in Fig. 2 (decreased staining), it incorrectly attributes this to a 5-HT2A agonist, but what was used in the experiment was an antagonist. An agonist activates and stimulates a receptor, while an antagonist prevents the receptor from being activated. This answer choice is the opposite of what we want.
B. RNA for seven different receptor subtypes was detectable in naive liver tissue. – The question is then asking about 5-HT2A and 5-HT2B specifically. Seeing RNA for other receptor subtypes does not give us information about the two receptors we’re asked about. If an answer choice doesn’t actually address or answer the question stem, it’s incorrect.
C. Up-regulation of 5-HT2A and 5-HT2B was observed during periods of peak hepatocyte proliferation. – We said that in Fig. 2, the lack of activation of 5-HT2A/2B was associated with little Ki67 staining which was used as a proxy for hepatocyte proliferation. This led us to believe that activation could be associated with increased staining and hepatocyte proliferation. This answer choice captures that well—if 5-HT2A and 5-HT2B are upregulated, they are activated and associated with hepatocyte proliferation/staining. We’ll take a look at D just in case, but C is looking good.
D. Administration of 5-HT2C and 5-HT3 receptor antagonists reduced the number of Ki67-positive cells. – Like answer choice A, this answer is tempting because it matches the visual in Fig. 2 and also correctly states that an antagonist was used. Pause and reread that answer choice out loud. What happens to many of us is that we read this answer choice too quickly and don’t realize that it’s talking about receptors 5-HT2C and 5-HT3, and not 5-HT2A and 5-HT2B. Unfortunately, this answer choice doesn’t help us understand the mediating role played by receptors 5-HT2A and 5-HT2B. We’ll stick with answer choice C.
19) This question is a pseudodiscrete question. We need to be very comfortable with the amino acids on test day, including knowing that serotonin is derived from tryptophan. What the question is really asking then, is to describe the side chain of tryptophan. The structures of the amino acids are shown below.
A. nonpolar, aliphatic – tryptophan does not have open carbon chains, aliphatic does not apply.
B. polar, uncharged – tryptophan is not polar as evidenced by the two fused aromatic rings.
C. aromatic – tryptophan is indeed aromatic; it has two fused rings that meet Huckel’s rule for aromaticity. While we may not need to use Huckel’s rule to determine the aromaticity of tryptophan, we should be familiar with Huckel’s rule and know which amino acids are aromatic.
D. negatively charged – tryptophan’s side chain is not negatively charged.
For those who are unfamiliar with Huckel’s rule: Huckel’s rule states that a planar cyclic molecule or planar ring with a p orbital at each atom within the ring and 4n+2 π electrons is aromatic, where n is an integer.
Official Guide Biology/Biochemistry Section Passage 5
20) We are not expected to have the structure of GABA memorized and the author was kind enough to provide the formulas for both glutamate and GABA (shown below). We need to determine what changes between glutamate and GABA. Looking at the very end of each formula, a carboxylic acid has been removed, with its hydrogen now on the originally deprotonated carboxylic acid/carboxylate. Going from glutamate to GABA then requires the loss of CO2.
A. Kinase – kinases phosphorylate their substrates. No phosphorylation has occurred.
B. Transferase – transferases transfer a functional group from one substrate to another. The conversion of glutamate to GABA involves the loss of carbon dioxide so no transfer is occurring.
C. Decarboxylase – decarboxylase catalyzes the loss of a carbon and produces CO2 in the process. This is the correct answer.
D. Dehydrogenase – dehydrogenase is involved in redox reactions but no redox reaction occurs in the conversion of glutamate to GABA. For redox reactions in organic molecules, look for changes in the number of hydrogen and oxygen bonds. For inorganic molecules, calculate the oxidation number. The bond broken in the above reaction is a C-C bond which would not constitute a redox reaction.
21) Questions that ask why some molecule/substrate/etc. was used in an experiment are asking you to assess the experimental design and explore what made the molecule of interest an appropriate measure or reagent. The first step before moving on to the questions is to revisit what the passage says about Tuj1 – “a neuron-specific class III β-tubulin.” Now let’s take a look at the options.
A. Tuj1 induces expression of the TauEGFP protein. – This answer choice implies a causative relationship that is not supported by the passage. If Tuj1 is a tubulin, it should be involved in cell structure, not controlling the expression of the fluorescent fusion protein.
B. Tuj1 is expressed in fibroblasts and neurons. – We are told that Tuj1 is “neuron-specific” so it should not be expressed in other cell types. We can also tell that this answer choice is incorrect because if it were true, it wouldn’t help the researchers distinguish between the fibroblasts that they started with and the neurons they were trying to make!
C. Tuj1 is an early marker of neural differentiation. – Here we have a component of neuron specificity which is important. We’re starting off on the right foot. If this statement were true and if Tuj1 is an early marker, that would be even better because the researchers would be able to assess their results early on. This is a good answer choice.
D. Tuj1 is present in embryonic and adult cells in culture. – While the researchers used embryonic fibroblasts in their experiment, we do not expect Tuj1 to be present in the fibroblasts, regardless of their embryologic status. Additionally, the focus of measuring Tuj1 is not on whether the source cell is embryonic or adult; the focus is on whether the cell produced is a neuron or not. Whether Tuj1 is present in embryonic and adult cells does not address the question of what cell type is produced because the source cell type remains constant. Answer choice C is a much better answer.
A. Ascl1 – Based on our figure analysis above, we settled on Ascl1 causing the greatest increase so this is the correct answer.
B. Brn2 – Brn2 does not decrease the number of neurons as much as Ascl1 when removed, so it should have a smaller effect when added.
C. Zic1 – Zic1 does not decrease the number of neurons as much as Ascl1 when removed, so it should have a smaller effect when added.
D. Olig2 – Olig2 does not decrease the number of neurons as much as Ascl1 when removed, so it should have a smaller effect when added.
For the neurons in the passage to be used to replace dopamine-deficient neurons in the brain, the neurons would need to produce dopamine. The passage tells us that the neurons created by the researchers “produced the excitatory neurotransmitter glutamate (–OOC-CH2-CH2-CH(NH2)-COOH), a few produced the inhibitory neurotransmitter γ-aminobutyric acid (GABA) (HOOC-CH2-CH2-CH2NH2).” There is no mention of dopamine, so the neurons would not be able to replace dopamine-deficient neurons.
A. Yes, because the cells obtained have the functional characteristics of nerve cells. – Having similar functionality is useful, but not enough to replace dopamine-deficient neurons in the brain. If they do not produce dopamine, they cannot help with dopamine deficiency.
B. Yes, because the cells obtained are similar to cells in the central nervous system. – Again, this is true and would be a good place to begin, but that’s not enough to help with dopamine deficiency. No dopamine, no help.
C. No, because the cells obtained may contain tumorigenic pluripotent cells. – We have no reason to believe that the neurons produced are prone to tumors, and we do not create any pluripotent cells in the process of neuronal differentiation.
D. No, because the cells obtained lack the correct neurotransmitter phenotype. – This is correct. The neurons synthesized and released glutamate and GABA, not dopamine. Glutamate and GABA are the wrong neurotransmitters, which explains why these cells would not be used in an animal model of Parkinson disease to replace dopamine-deficient neurons.
24) The question stem is asking us to identify a mechanism to increase the accuracy of our estimates, in other words to be more sure of the results we obtain.
A. Increase the magnification of the oculars used to define the field of view. – If the researchers increase the magnification, they will zoom in and see a smaller area. This would decrease the size of the regions observed and make the researchers more prone to inaccurate estimates of the total number of neurons produced in the entire petri dish.
B. Increase the number of visual fields counted per petri dish. – This amounts to increasing the number of samples taken. If the researchers count more samples, they can be more sure about the total amount of neurons in the petri dish. This represents an increase in accuracy.
C. Select visual fields from the central portion of the petri dish where cell density is highest. – Selecting samples from areas where the researchers know the cell density is highest will skew the results and cause the researchers to overestimate the number of neurons in the petri dish. This would decrease, not increase the accuracy.
D. Use the presence of green fluorescence to identify cells appropriate for quantification. – The cell line used in the experiment expressed the TauEGFP green fluorescent fusion protein. Thus, all the cells, neuron or not, would have to express the green fluorescent protein. If the researchers counted every cell that fluoresces, they would be including fibroblasts in that count which would bias their count and cause them to overestimate the number of neurons. Answer choice B is the only that would improve accuracy.
25) The question requires us to understand the fundamental principles behind enzymes. What do enzymes do, and what don’t they do? Enzymes speed up the forward and reverse rates of a given reaction. They do not change the amount of product created or the amount of reactant consumed, and do not affect any thermodynamic properties. Enzymes may bring the reactants in closer proximity to each other, create an environment more suitable for the catalytic reaction to take place, and stabilize the transition state of the reaction thereby decreasing the activation energy for the reaction. Below we have a reaction coordinate curve that will show what energetic component of a reaction is affected by an enzyme.
A. the substrate. – If the substrate were stabilized, it would lower the energy state of the reactants and increase the activation energy.
B. the product. – If the product were stabilized, the activation energy would remain unchanged, but the thermodynamic favorability of the reaction would increase. Enzymes do not affect the thermodynamics of a reaction.
C. the transition state. – This is correct! Enzymes stabilize the transition state to decrease the activation energy of a reaction. See the diagram above for more visual representation of this.
D. the equilibrium. – Enzymes don’t affect equilibrium, which is a result of the thermodynamic properties of a reaction. Answer choice C best captures the mechanistic advantage of enzymes.
Content page for thermodynamics: https://jackwestin.com/resources/mcat-content/principles-of-bioenergetics/bioenergetics-thermodynamics-1d
26) Ah, more thermodynamics, one of the most beloved topics on the MCAT. Like amino acids, this is a topic you should be intimately familiar with on test day. The key equation to consider for this question stem is ΔG=ΔH -TΔS. A negative ΔG corresponds to a spontaneous reaction, and the question stem is asking for an enthalpy and entropy pairing that always produces a spontaneous reaction. We need to determine what sign each should have to make ΔG more negative. Enthalpy (ΔH) is added in the equation, meaning it will have the same sign as the Gibbs free energy. To produce a negative ΔG, enthalpy should be negative. On the other hand, entropy (ΔS) is subtracted in the equation, meaning it should have the opposite sign as the Gibbs free energy, so it should be positive in order to make ΔG more negative.
A. A positive ΔH and a negative ΔS – This pairing is the opposite of what we’re looking for, and would correspond with a reaction that is never spontaneous.
B. A positive ΔH and a positive ΔS – This pairing is ambiguous, and will only produce a spontaneous reaction when the product of TΔS is greater than ΔH, that is at higher temperatures.
C. A negative ΔH and a negative ΔS – This pairing will only produce a spontaneous reaction when the product of TΔS is less than ΔH, or lower temperatures.
D. A negative ΔH and a positive ΔS – Here we have it folks, the pairing we’ve been searching for; both enthalpy and entropy contribute to a spontaneous reaction.
27) The key to this question is realizing that we’re being asked about a eukaryotic cell and not a prokaryotic cell; eukaryotic cells contain membrane-bound organelles and are capable of producing a wider variety of molecules. Below is a representative eukaryotic plasma membrane.
A. Phospholipid – This is the primary molecule in the cell membrane, hence the other name for the cell or plasma membrane, the phospholipid bilayer.
B. Cholesterol – Unlike prokaryotes, eukaryotes are able to produce cholesterol and other sterols in the endoplasmic reticulum. In eukaryotic cell membranes, cholesterol is responsible for stabilization. It will make the membrane more fluid at low temperatures and less fluid at high temperatures.
C. Glycoprotein – Unlike prokaryotes, eukaryotes are able to produce glycoproteins because they contain the Golgi apparatus. They are key for cell communication and recognition.
D. Peptidoglycan – Peptidoglycan is produced by bacteria and the thickness of the peptidoglycan layer is important for classifying Gram negative versus Gram positive bacteria. While prokaryotes contain peptidoglycan, eukaryotic cell membranes do not; this is the right answer.
28) The question stem indicates that an epitope binds a specific antibody, so the epitope-antibody ratio is 1:1. The more antibodies that bind to a protein, the stronger the immune response. The question stem also states that an epitope is a region on the surface of an antigen, so more antigens correspond to more epitopes, and the more epitopes, the more antibodies bind and the greater the immune response.
A. High molecular weight and increased number of epitopes – As noted above, more epitopes do produce a stronger immune response. Looking at the table, the molecule with the greatest number of epitopes also has the greatest molecular weight, while the molecule with the fewest epitopes has the lowest molecular weight. This means that high molecular weight is also associated with a more robust immune response. We like this answer choice.
B. High molecular weight and reduced number of epitopes – Reduced number of epitopes means less antibody binding and a weaker immune response.
C. Low molecular weight and increased number of epitopes – As discussed in answer choice A, epitope count is positively associated with molecular weight, so a low molecular weight corresponds to fewer epitopes, less antibody binding and a weaker immune response.
D. Low molecular weight and reduced number of epitopes – This answer choice is the opposite of that we’re looking for; a low molecular weight is associated with fewer antigens and epitopes, and a weaker response. Answer choice A produces the most robust response.
29) Increasing the volume of air that reaches the alveoli increases the gas exchange that occurs in the lungs. This means more O2 is brought into the body and bloodstream, and more CO2 is exhaled with each breath. What will the decrease in blood CO2 do to the pH of the blood? Enter the bicarbonate buffer system, a fan favorite of the MCAT because it connects acid-base chemistry, equilibrium and Le Chatelier’s principle, the respiratory system and the cardiovascular system all in one neat package. This buffer will also some up again in medical school, so tuck it in your back pocket for future reference.
CO2 + H2O <–> H2CO3 <–> H+ + HCO3– <–> 2H+ + CO32-
As CO2 decreases, according to Le Chatelier’s principle, the reaction will shift to the left to restore equilibrium. This means there will be fewer protons. pH=-log[H+], so a decrease in the proton concentration will increase the pH.
A. increase, because the neural mechanisms that remove acid from the blood will be activated.– The pH will increase, but it is the result of a shift in the bicarbonate buffer system, not neural mechanisms. Neural mechanisms are responsible for respiratory rate/breathing rate in response to changes in carbon dioxide concentration.
B. increase, because the partial pressure of CO2 in the blood will decrease. – This is true. The decrease in carbon dioxide concentration prompts a response via the bicarbonate buffer system that increases the pH of the blood.
C. decrease, because the affinity of hemoglobin for oxygen will be increased. – The pH of the blood will increase, not decrease. The pH is also determined by the concentration of carbon dioxide, not Hb binding affinity.
D. decrease, because the work associated with increased ventilation will come more O2. – If you were confused by the wording of this answer choice, you were not alone. At the very least, the work of breathing does not directly affect blood pH. Answer choice A is the correct answer.
30) This question is all about mitochondria and aerobic versus anaerobic processes. This means that the prime differentiating factor, the presence and use of oxygen, is most important.
A. Glycolysis – Glycolysis is an oxygen-independent process, meaning it can occur under both aerobic and anaerobic conditions in the cytoplasm. It doesn’t require oxygen, and it is not inhibited by oxygen. Both the aerobic bacteria and the anaerobic eukaryotic cells would be able to carry out glycolysis.
B. Citric acid cycle and electron transport – Both the citric acid cycle and the electron transport chain are oxygen-dependent processes and occur in the mitochondria. Because they require oxygen, only the aerobic bacteria which contained the mitochondria would be able to carry them out. The primitive anaerobic eukaryotes would not be able to carry out these functions, aligning with what the question stem requires.
C. Cell division – Cell division is not an oxygen-dependent process. Also, cell division is key for reproduction, so it is unlikely that a cell line would survive without dividing at some point.
D. Transcription and translation – Like cell division, transcription and translation are oxygen-independent processes and are crucial for cell survival. A cell line would not be able to survive without the two. Answer choice B is the only answer choice that describes functions not carried out by the primitive anaerobic eukaryotes.