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.

The figure below is a great reference for skeletal muscle fiber subtypes:

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 cat_6xbs and the lacY_6xbs 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 cat_6xbs and lacY_6xbs 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 cat_6xbs and lacY_6xbs 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 lacY_6xbs 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 lacY_6xbs transcript is found near the E. coli cell membrane, so it should be near phospholipids and proteins, making this the correct answer.

7) In the last paragraph, it is stated that chloramphenicol is a competitive inhibitor and we know that competitive inhibitors compete with the substrate at the active site to produce the following Michaelis-Menten curve: 

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. 

8) 

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 cat_6xbs 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.

 

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