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MCAT Content / AAMC MCAT Practice Exam 3 Bb Solutions

AAMC FL3 BB [Web]

Exam 3 Biology/Biochemistry Section Passage 1

1)

  1. Before we consider the relative amount of work performed by each person, it’s important to note that work is independent of time. Work is the force exerted over some distance. Right off the bat this answer choice is a poor answer choice. On to the work: if work is the force over some distance and the two individuals lift the same amount of weight, they are exerting the same force (weight=force=mass*acceleration). However, the question stem says that individual M has longer limbs, which suggests they are taller than individual N. If both individuals lift the weight from the ground to their shoulders, the taller individual, M, is exerting the same force over a longer distance, and is doing more work. This answer is incorrect because it says the two individuals do the same amount of work and because it tries to justify the amount of work done using time.
  2. As noted above, individual M has longer limbs and is presumably taller than individual N, so they move the same weight over a longer distance and do more work than individual N, not less. Work and distance are positively proportional.
  3. This is correct. There is a positive relationship between distance and work performed. When the weight is moved over a shorter distance as is done by the individual with shorter limbs, the amount of work performed is less.
  4. The speed at which the muscle contracts, which takes time into account, is relevant for the power produced. The question is asking about the work performed. Answer choice C is the best answer.

2) According to the passage, muscles that have similar diameters will also have similar strengths.

  1. Based on the passage, it is correct that the two fibers in the question stem with similar diameters will produce similar forces. However, recall that power is the force produced multiplied by the velocity at which that force is generated. As the name implies, fast-twitch fibers contract at faster velocities than slow-twitch fibers. If the force is the same and the velocity is greater for fast-twitch fibers, then the fast-twitch fibers generate more power.
  2. This contradicts the passage which says similar muscle diameters produce similar strengths and strength is the production of force.
  3. As noted in the explanation for answer choice A, the power produced by fast-twitch fibers is greater than the power produced by slow-twitch fibers because the two will produce the same force when they are the same diameter, but the fast-twitch fibers do so with a greater velocity. This is the correct answer.
  4. Like answer choice B, this directly contradicts the passage. Muscles of similar diameter should produce similar strengths and forces. Answer C remains the best answer.

3)

  1. The fermentation of glucose to lactic acid is a type of anaerobic metabolism. This would be seen, making it incorrect for a LEAST question.
  2. The electron transport chain relies on oxygen to function. Under anaerobic conditions, the ETC will be off so phosphorylation of ADP would not occur. This is the correct answer.
  3. Anaerobic glucose metabolism produces less ATP than aerobic glucose metabolism, so more glucose needs to be metabolized to meet the same energy needs. Glucose will need to be mobilized from its glycogen storage form. Depletion of these glycogen stores could certainly occur under anaerobic conditions.
  4. Lactic acid fermentation under anaerobic conditions will, as the name suggests, generate acidic products which will acidify the cytoplasm. Answer choice B is the only answer which would not be seen under anaerobic conditions.

4) Power is equal to work performed divided by the time over which it is performed. The question stem specifies that the duration of contraction is the same, so time is constant across all four points. This means that we can estimate the work by looking at the power curve. The work performed will be lowest when power is lowest which will be when either the force or velocity approach zero.

  1. At point A, the power curve is virtually zero. Given a constant time, this means that work is also approaching zero. We’re off to a great start.
  2. The power produced at point B is greater than the power produced at point A. /the time of contraction is held constant, the power can be used to directly estimate the work performed. Since the power is greater at point B, the work performed is also greater at point B.
  3. Point C corresponds to the greatest power which, given our approximation, corresponds to the greatest work done by the muscle, not the least.
  4. Like point B, point D corresponds to a greater power than point A meaning it also corresponds to a greater work. Point A corresponds to the least amount of work done by the muscle.

5) One way to answer this question is by using the final statement in the passage about slow-twitch fibers, that they are adapted to increase oxidative capacity. An answer choice that would promote increased oxidative capacity would correspond to slow-twitch fibers and would be incorrect. Another approach to this question is to find an answer that corresponds to greater contractile strength for fast-twitch fibers.

  1. Greater Ca2+-pumping capacity would improve contractile strength which corresponds to fast-twitch fibers. Remember that the movement and subsequent binding of calcium to troponin is what allows for the binding of actin and the myosin heads, ultimately producing contraction. Also, it does not directly correlate with increased oxidative capacity which would be a marker for slow-twitch fibers. This is a great answer choice.
  2. Increased capillary density would promote blood flow and oxygen delivery to the slow-twitch muscle fibers, thereby improving oxidative capacity. The question stem asks for a property of fast-twitch fibers so this is incorrect.
  3. More mitochondria correspond to greater oxidative capacity for slow-twitch muscle fibers.
  4. Like answer choices B and C, this answer choice would improve oxidative capacity for ATP production in slow-twitch fibers. Only answer choice A would be a property of fast-twitch muscle fibers.

 

Exam 3 Biology/Biochemistry Section Passage 2

6) Below you will find a flowchart to help you determine the most likely mechanism of inheritance when looking at a pedigree. Please note that it was created with the pedigree from this passage in mind and as such is not exhaustive.

  1. This aligns with our pedigree analysis and mechanism of inheritance flowchart. Males are not truly disproportionately affected (see above) so autosomal is more likely than sex-linked. Also, two affected parents had unaffected offspring which means that it is a dominant mechanism of inheritance. If it were recessive, II-1 and II-2 would only be able to pass down recessive alleles and all of their children would be homozygous recessive and affected. Because III-1 and III-3 are not affected, the parents must both be heterozygous and only have one copy of the dominant allele.
  2. As noted in the explanation for answer choice A, this is not possible because two unaffected offspring are born to two affected parents.
  3. This would be more likely if males were disproportionately affected. Before you yell “wait a moment!” at your screen, note that both generations II and III have mostly male offspring to begin with. Also, II-2 “married” into the family; they are not offspring of I-1 and I-2 so they should not be included in our rough assessment of the number of males versus females affected. Because the male-to-female ratio is already biased at birth, we cannot say that males are disproportionately affected.
  4. Females are affected indicating that it cannot be Y-linked; only males have Y chromosomes. A is the correct answer choice.

7) Take a look at the 9th band from the top in the normal and then in the tumor gel. The original nucleotide was T in the normal tissue and C in the tumor tissue.

  1. Both thymine (normal) and cytosine (mutation/tumor) are pyrimidines. This is the opposite of what we’re looking for.
  2. While cytosine is indeed a pyrimidine, thymine is not a purine so the first half of this answer is incorrect.
  3. Thymine is indeed a pyrimidine, but so is cytosine. The second half of this answer is incorrect.
  4. This is it; both thymine (normal) and cytosine (mutation/tumor) are pyrimidines making the mutation a pyrimidine to pyrimidine mutation.

8) Halfway through the second paragraph, the author notes the amino acid substitution is an arginine for a histidine. The question asks for the amino acid that is replaced so the correct answer should be the structure of histidine. You’ve heard this before but if you haven’t yet memorized the structures of the amino acids, add that to your MCAT to-do list!

  1. This is the structure for proline, not histidine. Don’t let the ring confuse you.
  2. Not quite, if we see this structure we should think of tyrosine.
  3. Here’s histidine folks! This is the correct structure.
  4. This is phenylalanine, not histidine. Answer choice C it is.

9) The second sentence of the passage states that in HGF, a disulfide bond joins the α subunit to the β subunit. The correct amino acid should be able to form a disulfide bond.

  1. Alanine does not contain a sulfur atom to form a disulfide bond.
  2. Cysteine contains a sulfur atom and is capable of forming a disulfide bond.
  3. Don’t let the single letter abbreviation fool you; serine does not contain any sulfur atoms and cannot form disulfide bonds.
  4. Tyrosine cannot form disulfide bonds; only cysteine can form the bond that joins the α and β subunits of HGF making answer choice B the correct answer.

 

Exam 3 Biology/Biochemistry Section Discretes

10) Convergent evolution occurs when fairly unrelated organisms evolve to meet similar environmental needs and is associated with analogous structures, like the wings of a bat and butterfly. The bat and butterfly are distant relatives but both needed the ability to fly in their environments.

  1. Dolphins and sharks are distant relatives, specifically from different taxonomic families. They had to evolve similar characteristics in order to survive in similar environments, so this answer choice is looking good.
  2. Sheeps and mountain goats are a bit more closely related than dolphins and sharks and differ at the level of genus, so answer choice A is a better answer.
  3. Polar bears and pandas belong to the same family, so like answer choice B, are more related than dolphins and sharks making it a poor answer choice.
  4. These two are the most related out of all of the answer choices. For convergent evolution, we are looking for the least related animal pairing with similar traits making A the best answer. Now, you may be asking yourself, “am I supposed to know the family, genus and species of random animals for the MCAT?!” The answer is, thankfully, no. Without knowing the family and genus of each animal in these answer choices, you can probably tell that sheeps and goats are closely related, and that different types of bears are also closely related (probably why the MCAT test writer called it a panda bear instead of simply panda). You might also recognize that dolphins and sharks are less related; they have more visual and physical differences than some of the other answer choices. The MCAT is not about sitting down and memorizing a never-ending list of facts. Being able to critically reason your way through questions where you don’t have the exact answer, such as this one, is an invaluable skill for test day. Keep working on it, you’ve got this!

11)

  1. In this question stem, males have two copies of the R chromosome and females have one R and one S chromosome. The imbalance stems from the different numbers of R and S chromosomes. To compensate for the potential imbalance, either the males will need to have one less active R chromosome, have a form of the active S chromosome, or females would have to have another active R chromosome. It is unlikely that one of the sexes would gain an additional chromosome, but inactivating the “extra” R chromosome in males would mean each sex only has one active R chromosome, which would compensate for the imbalance.
  2. Doubling the transcription of the S chromosome in females would be like having another active S chromosome in females. However, males would still have no S chromosomes so the imbalance would be magnified, not compensated.
  3. Inactivating the only R chromosome in females would be like removing the R chromosome and would worsen the imbalance; males would have two functional R chromosomes and females would have none.
  4. Like answer choice B worsened the imbalance, doubling the transcription of the R chromosome in males would be the equivalent of having more active R chromosomes and would worsen the imbalance because women would still only have one. Answer choice A is the only answer choice that compensates for the imbalance instead of worsening it.

12)

  1. The stress axis would be better aligned with the HPA axis where the hypothalamus secretes corticotropin releasing hormone, the anterior pituitary releases adrenocorticotropic hormone and the adrenal cortex releases cortisol. Gonadotropin-releasing hormone is not a part of this axis so this axis should remain intact.
  2. In the growth hormone axis, the hypothalamus releases growth hormone releasing hormone which signals to the pituitary gland to release growth hormone. Like answer choice A, this axis should remain intact.
  3. The hypothalamus releases thyrotropin releasing hormone, the pituitary releases thyroid stimulating hormone and the thyroid gland responds by releasing the thyroid hormones T3 and T4. This axis should still be functional.
  4. This is the endocrine axis that will be disrupted in individuals with Kallman Syndrome. The hypothalamus normally secretes gonadotropin releasing hormone which tells the pituitary gland to release luteinizing hormone and follicle stimulating hormone to signal the gonads to produce their respective hormones. If the neurons that produce gonadotropin releasing hormone fail to migrate to the hypothalamus, then the release of every hormone downstream in the pathway will be compromised. This disrupts the reproductive axis because the gonads do not receive the central signal, making this the correct answer.

13)

  1. Primase and not ligase is the enzyme that lays down the RNA primer. Ligase joins the Okazaki fragments formed during replication of the lagging strand.
  2. As noted above, primase lays down the RNA primer on the lagging strand but is complementary to the DNA strand. It does not replace any of the original nucleotides.
  3. Helicase “unzips” the strands of DNA at the beginning of replication. This occurs at the origin of replication. This is the best answer.
  4. Topoisomerase prevents the re-coiling of the DNA strands during replication by addressing the DNA strands’ tendency towards supercoiling. As noted in answer choice A, ligase is the enzyme responsible for joining the Okazaki fragments.

 

Exam 3 Biology/Biochemistry Section Passage 3

14) The second paragraph states that cFLIP is structurally similar to caspase-8 but does not have catalytic activity. It also binds to FADD at the same (active) site as where caspase-8 binds.

  1. According to the passage, cFLIP binds to FADD at the same site as the Fas ligand caspase-8. The passage also states that TNF signaling does not depend on FADD. cFLIP is part of an FADD-dependent pathway and TNF is a non-FADD-dependent pathway so this answer choice is incorrect.
  2. cFLIP lacks catalytic activity meaning it will not activate the apoptosis pathway when it binds to FADD at the same site as the catalytically active caspase-8. Since it binds at the same site, overexpression of cFLIP will mean less caspase-8 binds to FADD and less apoptosis will occur.
  3. As noted in the explanation for answer choice A, TNF works through a non-FADD-dependent pathway and cFLIP signaling is through the FADD pathway so this answer choice is incorrect.
  4. As noted in the explanation for answer choice B, the correct answer, cFLIP is not catalytically active so it will not promote apoptosis. When it is overexpressed, there should be less apoptosis, not more.

15)

  1. Unlike caspase-8, cFLIP is catalytically inactive, so it is missing the catalytic activity that caspase-8 possesses. According to the first paragraph, caspase-8 is a protease. Proteases cleave peptide bonds and use water to do so, meaning they possess hydrolase activity. Oxidoreductases change the oxidation state of substrates via reduction and oxidation reactions.
  2. Lyases cleave bonds via mechanisms other than water. Peptide bonds are cleaved using water so while tempting, this answer choice is not the best answer.
  3. Isomerases catalyze isomerization reactions and alter the chemical structure of its substrates while maintaining the same chemical composition.
  4. Hydrolases cleave bonds using water, like proteases cleave peptide bonds using water. This is the correct answer.

16)

  1. When explaining experiment 1, the author notes that p38 phosphorylation indicates apoptosis. A cell increasing in size indicates that the cell is getting ready to multiply. This is the opposite of what we would expect of a cell that is set to die or apoptose.
  2. DNA replication is a marker of cell growth, the opposite of apoptosis.
  3. Like the answer choices above, this answer choice is the opposite of what we would expect in preparation for or during apoptosis.
  4. Here we go, undergoing arrest of growth is a much better indication of upcoming apoptosis, or cell death.

17) Most of the proteins described in the passage promote apoptosis. Cancerous cells are able to avoid apoptosis so the correct answer will either not promote apoptosis and/or actively prevent apoptosis.

  1. ccFLIP lacks the catalytic activity to activate the FADD pathway which, when activated, promotes apoptosis. Therefore, cFLIP does not promote apoptosis and because it binds at the same site as the catalytically active caspase-8, prevents the activation of the FADD pathway and decreases the amount of apoptosis that occurs.
  2. Caspase-8 is catalytically active and activates the pro-apoptotic FADD pathway. Overexpression would increase apoptosis.
  3. FADD promotes apoptosis, the opposite of what we’re looking for in cancerous cells.
  4. The second sentence of the passage states that Fas stimulation recruits FADD which is a death domain protein. Like answer choices B and C, this is the opposite of what we expect. Cancerous cells are able to avoid apoptosis but this answer choice promotes cell death.

 

Exam 3 Biology/Biochemistry Section Passage 4

18) The second paragraph states that cgr1 and cgr2 are produced in E. lenta.

A. Eukaryotes produce separate mRNA transcripts for different genes. These are known as monocistronic mRNA transcripts. E. lenta is a prokaryote and should produce polycistronic mRNA.

B. The answer choice correctly describes the “anatomy” of an operon as expected in prokaryotes like E. lenta.

C. The transcript produced from an operon does not undergo alternative splicing; transcription and translation occur simultaneously. Alternative splicing is a marker of eukaryotic transcription.

D. Like answer choice A, this answer describes a setup more akin to monocistronic mRNA transcripts which are produced in eukaryotes, not prokaryotes. Answer choice B is the correct answer.

19) A positive inotrope increases the force of contraction and contractility which is mediated by intracellular calcium levels. An increase in intracellular calcium will positively affect contractility.

  1. Roman numeral II explicitly mentions an increase in intracellular calcium; it will increase contractility. We would need to evaluate the other roman numerals before deciding if only roman numeral II is true.
  2. Roman numeral III, an increase in calcium storage in the sarcoplasmic reticulum, will mean that more calcium is available for release intracellularly. This will increase contractility. Because both roman numerals II and III are true, neither answers A nor B are correct.
  3. Because roman numerals II and III are both correct, this answer choice must be incorrect.
  4. I, II, and III– This is the only answer that shows that both roman numerals II and III are correct meaning that this is the correct answer and that roman numeral I must also be correct. On test day we would select this answer and move on. We’re not yet at test day so let’s evaluate roman numeral I as well. Decreasing transport of calcium to the extracellular environment means that there will be more calcium in the intracellular environment which will increase contractility. Roman numeral I is therefore correct as well.

20)

  1. Simple diffusion does not require proteins or energy. The passage begins by noting that digoxin, which is an example of a cardiac glycoside, exerts its effects by inhibiting the Na+K+ATPase. The sodium-potassium pump requires the use of both proteins and energy in its membrane transport.
  2. Facilitated diffusion does not require energy for membrane transport. This answer is incorrect.
  3. The sodium-potassium pump is a type of primary active transport because it uses ATP (hence the name ATPase) to move sodium and potassium against their concentration gradients.
  4. Secondary active transport uses energy stored in gradients produced by primary active transport. Na+K+ATPase uses ATP to move sodium and potassium. It is a type of primary active transport.

21) Dihydrodigoxin is the inactivated metabolite produced by the reduction of digoxin. This means that the highest level of dihydrodigoxin will be present when there was a lot of digoxin for the E. lenta cells to metabolize. The second paragraph notes that cgr1 and cgr2 are upregulated in E. lenta when digoxin is present. The greater the presence of digoxin and the greater metabolism of digoxin, the greater the cgr1 and cgr2 expression and the higher the levels of the dihydrodigoxin metabolite respectively.

  1. Data set 2 has the greatest cgr2 expression which means it would have the highest dihydrodigoxin levels.
  2. Data set 4 has about half the cgr2 expression of data set 2 so the dihydrodigoxin levels will be lower.
  3. Data set 6 has comparable cgr2 expression to data set 4. It will also have lower dihydrodigoxin levels than data set 2.
  4. Data set 8 has the lowest levels of cgr2 expression out of all of the answer choices; it will have the lowest levels of dihydrodigoxin.

22) Digoxin inhibits the Na+K+ATPase so the absence of digoxin will allow the sodium-potassium pump to move sodium and potassium across the cell membrane.

  1. This is incorrect. The sodium-potassium pump moves three sodium ions out of the cell and two potassium ions into the cell, not one potassium ion and two sodium ions.
  2. This is the opposite of answer choice A but is also incorrect. Three sodium ions leave the cell and two potassium ions come into the cell.
  3. This answer is also incorrect; it is the opposite of the correct answer in that three sodium ions exit the cell and two potassium ions enter the cell.
  4. This is the correct answer. The sodium-potassium pump moves three sodium ions out of the cell and moves two potassium ions into the cell.

23)

A. This is supported by the experimental data. Group 1 in Table 1 corresponds to the DSM2243 strain and group 2 corresponds to the FAA 1-3-56 strain. The DSM2243 strain has a higher cgr ratio. In Fig. 1, the DSM2243 strain produces greater reduction of digoxin compared to FAA 1-3-56 strain. Therefore the higher cgr ratio is correlated with greater digoxin reduction and excretion.

B. Per Fig. 1, the DSM2243 strain has a higher percentage of reduction than the FAA 1-3-56 strain, the opposite of this answer choice. This must be the correct answer to this LEAST supported question.

C. As shown above, the FAA 1-3-56 strain has a lower cgr ratio and the same strain has lower reduction (and lower digoxin inactivation) than the DSM2243 strain. This answer choice is supported by the experimental data.

D. The abundance of the cgr operon (cgr ratio) is correlated with the reduction and inactivation of digoxin (% reduction in Fig. 2). This means that the abundance of the cgr operon could predict the bioavailability (approximated by the percentage that remains active) of digoxin. This is supported by the experimental data; answer choice B is the correct answer.

 

Exam 3 Biology/Biochemistry Section Passage 5

24) The last sentence of the first paragraph states that HDACs and HATs are responsible for modifying basic residues. The correct answer will be a basic amino acid and will have a positively charged side chain.

  1. This amino acid is acidic because it has a negatively charged deprotonated carboxylic acid. This is aspartate.
  2. This amino acid is methionine; it is not a basic amino acid.
  3. While polar, this amino acid is not a basic amino acid; it is asparagine.
  4. This amino acid is basic and has a positively charged amino acid. It is lysine and it can be modified by HDACs; this is the correct answer.

25) The first paragraph notes that HDACs counter the effects of HATs. Histone acetyltransferases promote transcription by decreasing the attractive interaction between lysine residues in the histones and DNA. If HDACs counter the effects of the histone acetyltransferases, they must inhibit transcription, not promote it.

  1. This is aligned with the effect of histone acetyltransferases (HATs) and the opposite of histone deacetyltransferases (HDACs) which promote and inhibit transcription respectively.
  2. This is the correct answer. HDACs promote the attraction of the histones and DNA which will condense chromatin and inhibit transcription.
  3. This is incorrect because HDACs decrease transcription, they do not increase transcription.
  4. This would promote transcription, not inhibit it. In general, if the effect is to open and expose a chromosomal region, or loosen the packing of nucleosomes, this will increase translation. If the effect is to cause the nucleosomes to pack more tightly together, transcription will decrease. Take a look at the following content page for chromatin structure regulation: https://jackwestin.com/resources/mcat-content/control-of-gene-expression-in-eukaryotes/regulation-of-chromatin-structure

26)

A. The second paragraph begins by saying that βOHB) is a major energy source when exercising or during starvation. When someone first starts fasting, glycogenolysis and gluconeogenesis are initiated to maintain blood glucose levels and provide energy. During prolonged fasting, such as when the body perceives starvation, fatty acid oxidation and ketogenesis are engaged to provide sustained energy and βOHB is produced. This means that βOHB will increase and HDACs will be inhibited, correctly addressing the question stem.

B. This will not increase the levels of βOHB. The pentose phosphate pathway is of note because no ATP is used or produced. This means that it wouldn’t be especially useful during times of energy depletion the way that other energy-producing pathways are useful. See below for a review of the PPP.

C. While gluconeogenesis will be upregulated during fasting, it will not directly increase levels of βOHB unlike fatty acid oxidation and ketogenesis.

D. Like many of the other answer choices, the Cori cycle does not produce βOHB. Only answer choice A directly affects βOHB levels in living organisms during prolonged fasting.

27) Per the passage, the amino acids subject to metal-catalyzed carbonylation are proline, arginine, lysine and threonine.

  1. Lysine and arginine are basic amino acids. This answer choice is incorrect for this CANNOT question.
  2. Arginine, lysine and threonine are hydrophilic amino acids. Polar amino acids will be hydrophilic.
  3. This is the correct answer, none of the four amino acids mentioned in the passage are acidic. The acidic amino acids are aspartic acid and glutamic acid.
  4. Proline is classified as a hydrophobic amino acid so this answer is incorrect.

28)

A. The data does not note all of the potential target proteins so we cannot confidently say this. The only target other than histone described in the passage is tubulin. We don’t know if there are “many” others.

B. Let’s take a look at the curve for AcTubK40 in Fig. 1 to answer this question. The only non-histone target described is tubulin, and the anti-AcTubK40 antibody asked about in the question stem only binds to acetylated tubulin. The results in the figure below are obtained after treatment with βOHB. The  βOHB would inhibit deacetylase activity, which would mean more acetylation. However, we see that the line representing the anti-AcTubK40 antibody does not change with increasing βOHB concentration. The βOHB must not be inhibiting the HDACs which would explain why the acetylation level does not increase and stays the same. This is the correct answer.

C. The researchers treated the cells with βOHB; the βOHB had to get into the cell, presumably through the membrane and via the cytoplasm before reaching its target HDACs in the nucleus.

D. The passage doesn’t actually tell us what HDACs butyrate inhibits so we cannot compare the two.

 

Exam 3 Biology/Biochemistry Section Discretes

29)

  1. The somatic nervous system is associated with voluntary control. The adrenal medulla is responsible for the synthesis and secretion of epinephrine and norepinephrine in response to stress. This is not under voluntary control making this answer choice incorrect.
  2. The sympathetic nervous system regulates the “fight or flight” response and is activated during times of stress. The adrenal medulla secretes epinephrine and norepinephrine in response to stress; this is a good answer choice!
  3. The parasympathetic nervous system or branch of the autonomic nervous system is responsible for the baseline “rest, digest and sex” signaling. The adrenal medulla responds to stress and is not a part of the parasympathetic nervous system.
  4. As noted above, the adrenal medulla secretes hormones in response to stress and is a part of the sympathetic branch of the autonomic nervous system but not a part of the parasympathetic branch.

30) Enzymes are a very high yield topic on the MCAT; make sure you’re comfortable with enzyme kinetics.

  1. Allosteric inhibition requires the inhibitor to bind at a site other than the active site. Binding in place of the substrate implies that it is still at the active site.
  2. This is a marker of competitive inhibition at the active site, not allosteric inhibition.
  3. While “noncompetitively” is tempting given that noncompetitive inhibition and allosteric inhibition are sometimes used interchangeably, this particular answer choice is incorrect because it notes that the inhibitor is still binding at the active site.
  4. This is the correct answer; it is the only one that has the inhibitor binding at a place other than the active site which is the definition of allosteric inhibition. If you have any doubts about enzyme inhibition, there are two content pages you might want to review: https://jackwestin.com/resources/mcat-content/control-of-enzyme-activity/inhibition-types and https://jackwestin.com/resources/mcat-content/enzymes/inhibition

31)

  1. NAD+ is a substrate and not a product of the Krebs cycle. Recall that NAD+ is the oxidized form of NADH and that the reduced form NADH is what serves as the electron carrier that feeds into the electron transport chain to produce energy.
  2. As noted above, NAD+ is not a source of energy and is not a direct product of the Krebs cycle.
  3. This is a solid answer choice. Both ATP and NADH are products of the Krebs cycle and are direct and indirect suppliers of energy respectively. ATP can be used directly in many cellular processes while NADH donates its electrons to the electron transport chain where a series of redox reactions creates a proton gradient that is used for the production of ATP.
  4. Both NADH and ATP are products of the Krebs cycle. However, NADH does not directly supply energy; NADH feeds its electrons into the electron transport chain to produce ATP which then serves as a direct energy supplier. Answer choice C remains the best answer. See below for an overview of the connection between the Krebs cycle and the ETC.

32)

A. This is correct. The approximate concentration of solutes in blood with ¼ of the osmotic pressure of ocean water will be ¼ the concentration of the ocean water.

B. This would be true if the osmotic pressure of the blood were 14atm. Note that the solutes are approximated by the same molecule, NaCl so the van’t Hoff factor remains the same. There is no need to multiply by 2.

C. The osmotic pressure is directly proportional to the concentration of solutes. A lower osmotic pressure should correspond to a lower solute concentration, not a higher one.

D. As in answer choice C, the solute concentration should be lower than that found in ocean water, not greater. Answer choice A is the correct answer.

 

Exam 3 Biology/Biochemistry Section Passage 6

33)

  1. Let’s use the last sentence of the first paragraph to answer this question: it says that hypoxia attenuates or decreases oxidative phosphorylation, the processes that occur in the mitochondria, to conserve oxygen. This means that the citric acid cycle will be downregulated, not induced.
  2. This is correct. Recall that lactic acid fermentation consumes NADH which donates electrons to pyruvate in order to form lactate and regenerate NAD+ for use in glycolysis. This will maximize energy production under hypoxic or increasingly anaerobic conditions.
  3. Like the citric acid cycle, the electron transport chain will be downregulated and not induced in order to minimize oxygen consumption. Remember that the passage says that oxidative processes (which includes the ETC) are downregulated in response to hypoxia.
  4. Glycogenesis is the synthesis of glycogen, not to be confused with glycolytic energy production which refers to glycolysis. It will not directly regenerate NAD+.

34) Figure 1 demonstrates the positive effect of hypoxia on phosphoglucose isomerase induction. If you’ve already memorized the enzymes of various metabolic pathways, great! If not, that’s okay too! You should definitely have some of the major enzymes memorized (think hexokinase and succinate dehydrogenase) but you can reason your way through many of the other enzyme names. This is a great skill to have for test day whether or not you have the enzyme names memorized. Phospho- represents a phosphorylated compound, glucose is the phosphorylated molecule, and an isomerase rearranges the bonds within a compound. That means that phosphoglucose isomerase takes a phosphorylated glucose and rearranges its structure.

  1. This is another, arguably less often used, name for the electron transport chain. Phosphorylated glucose is not a part of this metabolic pathway.
  2. Phosphoglucose isomerase catalyzes the second step of glycolysis and rearranges the phosphorylated glucose (glucose-6-phosphate) to form fructose-6-phosphate making this the correct answer.
  3. While phosphorylated glucose is indeed a part of the glycogen synthesis pathway, it is added to existing glycogen residues, not isomerized.
  4. Phosphorylated glucose is not a part of the citric acid cycle. Answer choice B remains the best answer.

35) To answer this question, we will use the results from hypoxic conditions shown in Fig. 3:

  1. Oxygen consumption increased when the NDU gene was knocked out. We would expect the overexpression of the gene product to have the opposite effect.
  2. This is the best answer. The knockout of NDU significantly increased oxygen consumption, so overexpression is likely to decrease oxygen consumption.
  3. The NDU knockout did not have a statistically significant impact on Complex IV activity compared to the wild type control. Overexpression of the protein is unlikely to affect Complex IV activity.
  4. Like answer choice C, this answer incorrectly assumes that the NDU gene expression will have an effect on Complex IV activity. The data showed no significant impact of NDU knockout on Complex IV activity.

36)

  1. Like GAPDH and actin, tubulin is often used as a housekeeping gene and loading control because it is constitutively expressed. Tubulin expression should not change under hypoxic conditions so ensuring the presence of and roughly equal amounts of tubulin in the gel means that each lane was loaded correctly. If a lane does not show tubulin, or if the lanes show immense differences in the relative concentration of tubulin, then the results are compromised due to poor loading. The use of tubulin allows researchers to better assess protein levels making this the correct answer.
  2. Tubulin neither controls nor is controlled by NDU expression; there is no evidence of this in the passage.
  3. Tubulin levels are stable irrespective of hypoxic status; tubulin is a cytoskeletal protein that is constitutively expressed. Hypoxia should not affect the cell’s cytoskeleton.
  4. Tubulin expression is not controlled or regulated by NDU; there is no mention of this in the passage. As noted in the correct answer, A, tubulin is serving as a loading control.

 

Exam 3 Biology/Biochemistry Section Passage 7

37) The first paragraph indicates that the HIF binding sequence is CCCCGGGC. Notice that the first and last nucleotides are not complementary and are thus not a part of the palindrome. Recall that a palindromic sequence is a sequence where both DNA strands have the same sequence when read from 5’ to 3.’ Another way to think about this is that the complement of the sequence is the same as reading the sequence in reverse. The palindromic sequence is six nucleotides long and is CCCGGG.

  1. A four base recognition sequence is smaller than the six nucleotides that make up the palindromic sequence so the restriction enzyme should be able to recognize the HIF binding sequence.
  2. We already established that roman numeral I is correct so this answer choice is incorrect.
  3. An eight base recognition sequence is too long for a six nucleotide palindrome. If you selected this answer, you might have missed that the first and last nucleotides in the sequence are not complementary and therefore are not a part of the palindromic sequence.
  4. Time to evaluate roman numeral II. A six base recognition sequence would be able to recognize the six nucleotide palindromic sequence so both roman numerals I and II are correct making this the correct answer. Restriction enzymes can be really tricky for a lot of students so we’ve broken down the topic here: https://jackwestin.com/resources/mcat-content/recombinant-dna-and-biotechnology/restriction-enzymes

38)

  1. Succinate dehydrogenase is not found in Complex I of the electron transport chain. The corresponding Complex I enzyme is NADH reductase.
  2. This is correct, complex II uses succinate dehydrogenase. It is the only enzyme involved in both the citric acid cycle and the electron transport chain.
  3. SDH is found in complex II and not complex III.
  4. SDH is found in complex II and not complex IV.

39) According to the passage, succinate competitively inhibits HIF hydroxylase which normally targets HIF for degradation. An enzyme that increases the levels of succinate will inhibit the degradation of HIF by HIF hydroxylase and result in increased levels of HIF.

  1. Decarboxylating succinyl-CoA, the precursor to succinate, will produce an entirely different molecule with only three carbons; succinate has four carbons.
  2. This is the correct answer. Succinyl-CoA synthetase is responsible for the synthesis of succinate from succinyl-CoA. Overexpression of succinyl-CoA synthetase will increase succinate levels, increase the inhibition of HIF hydroxylase thereby decreasing the degradation of HIF and ultimately result in increased levels of HIF.
  3. Succinate dehydrogenase catalyzes the conversion of succinate to produce fumarate. Test day tip: the first term in the name of an enzyme is often its substrate. This is not a hard and fast rule but can be useful nonetheless. The overexpression of an enzyme that uses succinate as its substrate would decrease the levels of succinate as it is converted to another compound ultimately decreasing HIF levels.
  4. Like the enzyme above, succinate carboxylase would catalyze a reaction in which succinate is a reactant, or substrate. If an enzyme that uses succinate as its substrate were overexpressed, succinate levels would go down and HIF degradation would rise. Only succinyl-CoA synthetase would cause an increase in HIF levels.

40) The second paragraph is the key to understanding the hereditary transmission of SDH-linked paraganglioma: according to the passage it only segregates with the SDH-linked paraganglioma after paternal transmission of the mutant allele. There is no indication that males and females are unequally affected and the mutation segregates exclusively with the paternal lineage so something specific to the fathers must be at play.

  1. Imprinted genes are parent-specific based on inherited epigenetic modifications. For example, if the gene is methylated and “off” in the parent, it will be “off” in the offspring. If the transmission is exclusive to the paternal lineage and affects both sexes, an imprinted gene would be a great way to account for the differential transmission of the SDH-linked paraganglioma.
  2. A Y-linked gene would only affect the male offspring. The passage gives us no reason to believe only male offspring are affected.
  3. An X-linked gene could affect both males and females, but it would disproportionately affect males and would not be exclusive to paternal transmission.
  4. Tumor suppressor genes are not transmitted in a parent-specific manner. Only answer choice A correctly identifies a parent-specific transmission that could explain transmission exclusively via the paternal lineage.

 

Exam 3 Biology/Biochemistry Section Passage 8

41)

  1. This answer is incorrect because it implies that protein synthesis is random and that amino acids can be replaced during translation. Proteins are systematically synthesized from N terminus to C terminus; the random allocation of radioactive amino acids is inconsistent with this systematic synthesis.
  2. This is a good answer. The question stem notes that protein synthesis was under way, or already initiated, when the radioactive amino acids were added. This suggests that at least the N terminus had been translated, and possibly the middle portion of the protein as well. We have no way of knowing exactly how much of the protein had already been translated from the question stem alone, but it’s fair to say that at least the C terminus would contain radioactive amino acids while the N terminus would not.
  3. For only the middle of the protein to be free of radioactive amino acids, the center would have had to have been translated before the addition of the radioactive amino acids and the ends afterwards. However, we know that the middle of an mRNA transcript is translated after the N terminus so this answer choice is incorrect.
  4. This is the opposite of answer choice C and is also incorrect. For this to be true, proteins would need to be synthesized from the outside in with both the N and C termini synthesized at roughly the same time. We know this is not true and that proteins are synthesized in the following order: N terminus -> middle -> C terminus. In the absence of an answer choice that takes into consideration that possibly both one end and the middle have radioactive amino acids, answer choice B remains the best answer.

42)

  1. This answer is tempting because it is the same as the curve shown in Fig. 1 for which hypothesis 2 was proposed. However, this is the measurement of radioactivity within the cells. The question stem asks for the antibody–Protein X precipitate in the extracellular growth medium. If hypothesis 2 is correct and the radioactive mature form of Protein X is being secreted from the cell, the extracellular radioactivity should increase with time, not decrease. There should also be a delay in the increase because the mature form of Protein X needs to be produced before its secretion.
  2. As noted above, the key to this question is that the question stem asks for the radioactivity in the extracellular growth medium if hypothesis 2 is correct. Hypothesis 2 states that the fall of radioactivity levels is the result of the secretion of a mature form of Protein X. This decline refers to the decline of radioactivity within the cell. The extracellular radioactivity should increase with the passage of time as more of the mature Protein X is secreted out of the cell. This answer choice shows the radioactivity decreasing, not increasing.
  3. While this answer choice correctly shows the extracellular radioactivity increasing (see the explanations for answer choices A and B for more on this if you’re unsure), it lacks the delay we would expect. Remember that hypothesis 2 notes that the decline occurs when a precursor is modified to a mature form and it is this mature form that is secreted by the cell. Some time will inevitably pass as the mature form of Protein X is made, so some time needs to pass before the radioactivity begins to increase. Also, the radioactivity within the cell (Fig. 1 in the passage) does not begin to decline until about 1200s, so the corresponding extracellular increase in radioactivity should also be at about 1200s.
  4. This is the correct answer. As noted in the explanations above, as the cellular radioactivity decreases, the extracellular radioactivity should increase assuming hypothesis 2 is correct and the mature form of Protein X is secreted from the cell. Furthermore, there should be a delay in this increase in radioactivity that begins at and corresponds with the time at which cellular radioactivity begins to decline, about 1200s. This increase should be logarithmic because the decrease in Fig. 1 is an exponential decrease or decay curve.

43) We’ve gone ahead and annotated a simplified version of Fig. 1 to help visualize the “half life,” specifically the time during which half of the radioactive Protein X is lost from the cell:

  1. As noted above, this is the correct answer. Half of the radioactive Protein X is lost after about 100 seconds.
  2. About ¾ of the radioactive Protein X is lost after 200 seconds, ½.
  3. Much more than half of the radioactive Protein X is lost from the cell by 300 seconds.
  4. Almost all of the radioactive Protein X has been lost from the cell after 500 seconds.

44)

  1. This curve shows a flattening of the initial upward slope at 45°C compared to 37°C. This would indicate a decrease in the rate or protein synthesis, not an increase.
  2. This graph appropriately depicts the rate of protein synthesis, or the amount of increase in radioactivity given the passage of time, increasing at 45°C compared to 37°C.
  3. The graph on the left shows a decrease in total protein synthesis or radioactivity at 45°C compared to at 37°C. The question stem asks about the rate which would be analogous to the upward linear progression of the first part of the curve.
  4. Here the maximum amount of radioactivity and protein synthesis increases at 45°C but the question stem asks about the rate of protein synthesis. Answer choice B is the best answer.

 

Exam 3 Biology/Biochemistry Section Discretes

45)

A. This is correct. RNA viruses contain transcriptases, namely reverse transcriptase, that can use the viral RNA transcript to make cDNA or complementary DNA that can be recognized by the host for replication. Another name for reverse transcriptase that you might see on test day is RNA dependent DNA polymerase because it depends on RNA and will produce DNA.

B. RNA viruses may only possess RNA as their genetic material, but their hosts contain DNA. Remember that the central dogma of biology for living organisms is DNA -> RNA -> protein as shown in action below.

C. RNA viruses code for or carry their own polymerases that recognize the RNA template to produce DNA. They do not alter the host’s own polymerases.

D. The key here is that these viruses contain RNA as their genetic material; how can the RNA instructions be used for viral replication? Even if the viral RNA stimulated the transcription of host genes, the problem of converting the RNA genetic material into clear instructions would remain. On the other hand, answer choice A and the presence of reverse transcriptase would allow for the viral RNA instructions to be converted to DNA instructions that the host can use to replicate the virus.

46)

  1. The collecting duct is responsible for concentrating the filtrate and future urine via aquaporin-mediated water reabsorption and secretion. No proteins are reabsorbed at this location.
  2. The distal convoluted tubule is primarily responsible for additional salt (NaCl), bicarbonate and water reabsorption, and proton, potassium and ammonia secretion; no proteins here.
  3. The glomerulus normally prevents large molecules, including proteins, from making it into the filtrate. If the glomerulus is compromised or damaged, the proteins from the blood can make their way into the filtrate and produce proteinuria or protein in the urine. This is the correct answer.
  4. The loop of Henle and its countercurrent multiplier allow for the reabsorption of water in the descending limb and the reabsorption of salt in the ascending limb. Proteins are neither secreted nor reabsorbed in the loop of Henle. The glomerulus is the only answer choice that regulates the movement of proteins in the nephron.

47)

  1. Decreasing the concentration of the substrate would decrease the rate at which the enzyme reacts, not increase it.
  2. The optimal pH at which an enzyme functions is dependent on the enzyme. Just think about the difference in pH between the stomach, the blood and the small intestine! At least one type of enzyme can be found in each location, highlighting the varied pHs at which enzymatic activity is optimized.
  3. Increasing the activation energy increases the energetic barrier for the reaction and would decrease the rate of the enzymatic reaction.
  4. This is correct; enzymes are typically most active at a temperature of 37°C (average body temperature). If the temperature gets too hot or too cold, the enzymes will lose their enzymatic activity.

48)

  1. Smooth muscle is muscle that is under involuntary control and forms the walls of hollow organs and blood vessels. The way I like to remember this is that smooth, hollow and blood all have two “o”s. Smooth muscle can be found in the lung airways, but it does not secrete mucus. For a more in-depth review of muscle structure and contraction, visit: https://jackwestin.com/resources/mcat-content/muscle-system/muscle-structure-and-control-of-contraction
  2. Epithelial cells can be found in the luminal aspect of internal organs including the lungs. There are many types of epithelial cells: squamous, cuboidal, columnar, transitional and Goblet cells. Goblet cells in particular secrete mucus that can trap irritants and smaller foreign bodies. These Goblet epithelial cells can be found in the lungs to trap and inhibit microbial infections making this the correct answer. It should also be noted that the cilia in the bronchi will move the debris and mucus up and out of the lungs in a process called mucociliary clearance.
  3. The nervous system may provide signals to other endocrine and exocrine organs to secrete their respective hormones, but they themselves secrete neurotransmitters. Neither of these correspond to the secretion of mucus in the lungs.
  4. Connective tissues include the blood, fat, bone and cartilage. None of these will secrete mucus. Answer choice B, epithelial cells, is the only one that mentions a cell type responsible for secreting mucus. More information on connective tissue cells can be found at https://jackwestin.com/resources/mcat-content/tissues-formed-from-eukaryotic-cells/connective-tissue-cells

 

Exam 3 Biology/Biochemistry Section Passage 9

49) The first paragraph ends by saying that SSRIs block the normal reuptake of serotonin into presynaptic neurons. This means that the usual way that serotonin signaling is terminated is by reuptake.

  1. While this could affect serotonin’s ability to bind and activate the receptor if true, the passage does not mention any changes to postsynaptic receptor affinity for serotonin. As noted above, the usual mechanism of termination involves serotonin reuptake.
  2. By definition, mutant enzymes would be deviations from the normal or wild type versions of the enzyme. Additionally, the question stem asks for ways in which the action of serotonin on postsynaptic receptors is terminated. Once the serotonin is released into the synaptic cleft, altering the synthesis of serotonin will not affect the serotonin already released. The usual mechanisms of neurotransmitter signalling termination are neurotransmitter degradation or reuptake. The passage explicitly mentions reuptake so the correct answer should have some mention of reuptake.
  3. This is the best answer; it aligns with the mechanism described in the passage. Transportation back into the presynaptic terminal is another way of saying that the presynaptic neuron reuptakes the serotonin.
  4. While this could terminate signaling if it were true, like answer choice A there is no mention of this in the passage. There are questions where an answer choice that could be true is an appropriate answer, but for questions that begin with “based on the passage,” we want to limit ourselves to what’s explicitly stated, shown or implied in the passage as much as possible. Answer choice C is the best answer because it incorporates information mentioned in the passage.

50) After reading the question stem you might be tempted to look back at the passage and see if you missed something; where did these 7 out of the 9 that didn’t respond to treatment come from? I was tempted too! This question teaches us how important it is to read the passage carefully the first time and make sure we trust that we read the passage well. This question is providing us with a “what if” scenario and wants us to apply our understanding of the passage to a new scenario (kind of like a reasoning beyond the text question in CARS…). If you read a question stem on test day and your first thought is something along the lines of “wait, what?! I don’t remember reading anything about that!” pause before jumping back into the passage. Do you know where to look or will you be blindly skimming the entire passage again and cost yourself precious time? Ideally you should be able to pinpoint a related excerpt in the passage to help you answer these types of questions.

According to Fig. 2, the mutation in question significantly decreased serotonin synthesis to just under a quarter of serotonin synthesis in cells with the wild type hTPH2. How would decreased serotonin synthesis affect the action of selective serotonin reuptake inhibitors (SSRIs)?

  1. This answer choice is extrapolating too far beyond the passage. There is no mention of a mutation in the postsynaptic serotonin receptor, only in the rate-limiting enzyme in serotonin synthesis. The ideal answer will incorporate the effect of the mutation as described in the passage with the hypothetical scenario described in the question stem.
  2. This is a great answer because it directly ties in the effect of the hTPH2 variant on serotonin synthesis with the mechanism of action of SSRIs. Recall that Fig. 2 showed that cells with the mutant enzyme only synthesized 25% as much serotonin as the wild type. If only ¼ the normal amount of serotonin is released into the synaptic cleft to begin with, preventing the reuptake of such a small amount of serotonin (mechanism of action for SSRIs) won’t have much effect. This is the correct answer.
  3. The question stem specifically asks about patients with the R441H hTPH2 allele and, per the second paragraph, this mutation affects the rate-limiting enzyme of serotonin synthesis. There is no mention of the enzyme degrading the SSRIs making this answer choice incorrect.
  4. This is the opposite of what we would expect. The patients with the R441H hTPH2 allele produced enzymes that significantly decreased the production of serotonin and the first sentence in the passage notes that reduced serotonin can cause depression. If the postsynaptic serotonin receptors were constitutively activated, it would mimic the signaling that would occur if there were a lot of serotonin in the synaptic cleft, not low levels of serotonin. Only answer choice B ties together the role of allele in question on serotonin synthesis with the mechanism of action of SSRIs making it the correct answer.

51) The inhibition of an enzyme will cause a relative decrease in the concentration of the products and an increase in the concentration of the reactants as the reaction slows down. (Note that this does not violate Le Chatelier’s principle because this is a temporary or biologically relevant change in concentrations due to the slowing of the reaction. The equilibrium concentrations will remain the same and thermodynamic favorability is unchanged.) Below we have noted which reactants would build up with the inhibition of each of the enzymes mentioned in the passage.

  1. As noted above, inhibition of aldehyde dehydrogenase would cause a build up or increased levels of 5-hydroxyindoleacetaldehyde, the product of serotonin metabolism. Let’s see if there’s a better answer that directly impacts serotonin levels.
  2. Inhibition of monoamine oxidase A would prevent the metabolism of serotonin to 5-hydroxyindoleacetaldehyde which would cause levels of serotonin to increase. According to the passage, reduced levels of serotonin can cause depression in some people, so increasing the levels of serotonin by preventing its breakdown should have the opposite effect and relieve depression in some folks. Irreversible inhibition will amplify this effect making this a great answer.
  3. Inhibition of L-Aromatic amino acid carboxylase will decrease the catalysis of 5-hydroxytryptophan to serotonin, directly decreasing the levels of serotonin and possibly worsening depression instead of relieving it.
  4. Inhibition of tryptophan-5-hydroxylase will prevent the enzyme from catalyzing the conversion of L-tryptophan to 5-hydroxytryptophan. The passage notes that this is the rate-limiting enzyme and step in serotonin synthesis. If this reaction does not occur, the direct serotonin precursor will not be formed, causing the synthesis of serotonin to grind to halt and lowering the levels of serotonin, potentially worsening depression instead of alleviating it.

52) According to the passage, SSRIs work by blocking presynaptic neuron reuptake of serotonin. Knowing this, let’s move on to the answer choices.

  1. There is no mention of the postsynaptic neuron absorbing the serotonin in the description of the mechanism of action for SSRIs. Furthermore, the blocking of reuptake of serotonin will cause serotonin to stay in the synaptic cleft, not be absorbed into another neuron.
  2. Like answer choice A, the sentence describing how SSRIs work does not address this. The SSRIs block reuptake, they do not promote the secretion of serotonin.
  3. The mutation in hTPH2 mentioned later in the passage affects the synthesis of serotonin but the SSRIs work by decreasing reuptake of serotonin into the presynaptic neuron. It is important to keep the different effects of the mutations and SSRIs separate when answering the question.
  4. This is correct; if the serotonin is not taken back up by the presynaptic neuron in the presence of SSRIs, the serotonin will spend more time in the synaptic cleft. This prolonged exposure of the postsynaptic neuron to serotonin is what causes the relief in depression for some individuals.

 

Exam 3 Biology/Biochemistry Section Passage 10

53) To understand why the boy lost weight, let’s review the end of the second paragraph: the polypeptide damages the lining of the small intestine and promotes the atrophy of intestinal villi which contributes to malnutrition. The first part notes that the deaminated form of the gluten polypeptide causes damage in the small intestine. The correct answer should be related specifically to the small intestine. The second part is even more specific; it is the villi in the small intestine that are damaged, and this damage causes malnutrition. Thus, the correct answer should mention the small intestine and have something to do with the primary function of the villi.

  1. This answer choice is incorrect because it focuses on the large intestine instead of the small intestine. Additionally, the primary function of the villi found in the small intestine is to increase the surface area across which nutrient absorption can occur. The large intestine is responsible for water reabsorption and the formation of stool. It contains invaginations instead of villi and contains our “gut microbiome” or bacterial flora. For more on the large intestine, visit https://jackwestin.com/resources/mcat-content/digestive-system/large-intestine
  2. An increase in enzyme secretion and excess digestion of nutrients would mean there are more nutrients available for absorption. This is the opposite of what would be expected; in order for the young boy to lose weight, he would need to absorb less nutrients, not more. This is also incorrect.
  3. This is the best answer because it has both key components: the answer is focused on the small intestine, noted in the passage as the site of damage, and it incorporates the primary function of the villi in the small intestine, increasing surface area for absorption. If the villi are destroyed, the surface area available for nutrient absorption decreases, less nutrients are absorbed, malnutrition develops and weight loss occurs. The content page for the small intestine is: https://jackwestin.com/resources/mcat-content/digestive-system/small-intestine
  4. The primary source of digestive enzymatic activity comes from the pancreatic juices released into the small intestine. The small intestine secretes and contributes additional proteases, disaccharidases and lipase. That said, the main function of the villi is to increase the surface area available for absorption. Answer choice C is a better answer.

54) Let’s take a look at the two main pressures within capillaries, the hydrostatic and the osmotic pressures:

In the arterial side of the capillary bed, the hydrostatic pressure is greater than the oncotic pressure and fluid moves out and into the interstitial space/tissues. However, on the venous side of the capillary, there is less fluid relative to the arterial side but the large plasma proteins that cannot cross the capillary wall are still present. The same amount of proteins with less fluid surrounding them will draw fluid back into the capillary as osmotic pressure in will be greater than hydrostatic pressure out. The net movement of fluid out of the arterial capillary and into the tissue due to hydrostatic pressure and the net movement of fluid into the venous capillary due to the osmotic pressure are for the most part balanced. However, there is some discrepancy between the two net pressures and not all of the fluid re-enters the capillaries. The lymphatics drain this fluid into the venous circulation to prevent edema.

  1. The osmotic pressure is the pressure that draws fluid into the capillary. If there are less proteins in the blood, there will indeed be a drop in the osmotic pressure as there would be fewer solutes in the blood to pull in fluid. However, we noted that the osmotic pressure is responsible for bringing fluid out of the tissue and into the capillaries. If there is less osmotic pressure and less “pulling” of the fluid back into the capillary, there would be an increase in fluid remaining in the tissue, not a decrease.
  2. As noted above, the osmotic pressure would decrease with fewer proteins in the blood because the solute concentration that pulls fluid in will decrease.
  3. This is the correct answer. Fewer plasma proteins will decrease the osmotic pressure. This means less fluid will move out of the interstitial space/tissues and into the capillaries, resulting in more fluid remaining in the tissues. Pre-medical school fact: this mechanism is part of one of the two main hypotheses regarding the cause of edema! If the osmotic pressure drops too low, the fluid left in the tissues can exceed the lymphatic system’s ability to drain the fluid back into the venous circulation and cause the accumulation of fluid in the interstitial space, most often the lower extremities.
  4. Like answer choice B, this answer is incorrect because fewer proteins in the blood would mean a lower effective solute concentration and a lower osmotic pressure, not an increased osmotic pressure.

55)

  1. If there is an excess of unabsorbed fats, there will be a disruption of normal electrolyte and water reabsorption. The sodium absorption that is usually stimulated by short chain fatty acids and the bile salt reabsorption that occurs with each meal will decrease. The excess solutes in the intestine that would otherwise have been absorbed increase the osmotic pressure, and there will be increased fluid in the intestines which in turn promotes diarrhea.
  2. The presence of additional unabsorbed solutes in the intestines will increase osmotic pressure, not decrease the osmotic pressure.
  3. An increase in osmotic pressure within the intestines would mean more water is remaining in the intestines, promoting diarrhea and not constipation. If the question were asking about a scenario that would result in less fluid in the intestinal lumen, then constipation would be an appropriate expectation.
  4. As noted above, the presence of additional unabsorbed solutes in the intestinal lumen will increase the osmotic pressure compared to if the fats had been appropriately absorbed.

56) The three key players in bone homeostasis that you should be familiar with on test day are osteoblasts, osteocytes and osteoclasts. OsteoBlasts are responsible for Building bones. They are responsible for the deposition of the calcium, collagen and protein matrix that make up bones. Osteocytes are osteoblasts that have become a part of the bone matrix and are involved in cell communication and mechanical sensing. These cells are recognizable due to their dendritic processes. OsteoClasts Consume or Chew bone; they break down and reabsorb the matrix. In order for the body to mobilize the calcium stores present in bones and increase blood calcium levels as effectively as possible, 1) bone growth should be downregulated to minimize calcium deposition and use, and 2) bone should be broken down to release and reabsorb calcium.

  1. This would minimize blood or serum calcium levels, not maximize them. The activation of osteoblasts would promote the consumption of calcium and the inactivation of osteoclasts would decrease calcium release and reabsorption.
  2. While increasing osteoclast activity would promote the release of calcium, the increased osteoblast activity would increase the use of calcium in the deposition of new bone matrix. This will not maximize the increase in blood calcium.
  3. The first part of this answer choice is on the right track; decreasing osteoblast activity will slow the use of calcium thereby allowing more calcium to remain in the blood. However, decreased osteoclast activity will mean less bone is being broken down and less calcium is being released and reabsorbed. Osteoclast activity should be increased in order to mobilize calcium.
  4. This is the correct answer because it is the most effective way that the body can use bone to increase serum calcium levels. The decrease in osteoblast activity means that less calcium is being deposited into the bone matrix so more remains in the blood, and increased osteoclast activity means more bone is being broken down and more calcium is released and reabsorbed into the blood. Both changes in activity will increase blood/serum calcium.

 

Exam 3 Biology/Biochemistry Section Discretes

57)

  1. Translational control would cause a change in the translation of the mRNA to protein and would change the number of enzymes produced and observed thereby contradicting the question stem.
  2. According to Le Chatelier’s principle, glutamine degradation and the corresponding decrease in product concentration would shift the reaction forward to restore equilibrium. This is inconsistent with a decrease in the activity of glutamine synthetase.
  3. Transcriptional control would alter the number of transcripts produced, which would affect the translation of the enzyme and like translational control, would change the number of enzymes produced. The question stem specifies that the number of enzyme molecules remains constant.
  4. Feedback inhibition relies on a component of the pathway affecting either earlier steps in the pathway (most types of feedback inhibition) or later (in the case of feedforward inhibition). Glutamine synthetase’s activity is decreasing as the concentration of its product, tryptophan, increases. If tryptophan is inhibiting the enzyme, then the enzyme is under feedback inhibition. The product “feeds the signal back” to the enzyme. This is the correct answer.

58) The drug binds reversibly at the active site, meaning it is directly competing with the substrate at the active site.

A. Lowering the activation energy is a characteristic of enzymes and speeds up the reaction. An inhibitor neither speeds up the reaction nor makes it more likely. This is not the right answer.

B. An example of feedback inhibition would be the product inhibiting the enzyme, not a drug that is competing with the substrate for the active site.

C. This is correct. As shown below, the inhibitor that binds at the active site is engaging in competitive inhibition. Make sure that you’re comfortable with the main types of enzyme inhibition on test day.

D. Noncompetitive inhibitors bind the enzyme at a site other than the active site.

59)

  1. Bowman’s capsule is indeed the first part of the nephron through which the filtrate will pass; blood is filtered into Bowman’s space to generate the filtrate. However, the loop of Henle comes after the proximal tubule, not before it.
  2. The proximal tubule comes after Bowman’s capsule and is responsible for the majority of the reabsorption of nutrients in the nephron. We don’t need to look any further; this answer choice is incorrect.
  3. The collecting duct is the final component of the nephron through which the filtrate will pass, not the first. No point in reading the rest of the answer.
  4. This is the correct answer. The filtrate will first pass through Bowman’s capsule and enter the proximal tubule; most of the reabsorption that will occur in the nephron occurs here. Next comes the loop of Henle and the countercurrent multiplier, the distal tubule where sodium is reabsorbed and water follows, and lastly the collecting duct to produce concentrated filtrate to be urinated.


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