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Your genes play an important role in your health, but so do your behaviors and environment, such as what you eat and how physically active you are. Epigenetics is the study of how your behaviors and environment can cause changes that affect the way your genes work. Unlike genetic changes, epigenetic changes are reversible and do not change your DNA sequence, but they can change how your body reads a DNA sequence. Gene expression refers to how often or when proteins are created from the instructions within your genes. While genetic changes can alter which protein is made, epigenetic changes affect gene expression to turn genes “on” and “off.” Since your environment and behaviors, such as diet and exercise, can result in epigenetic changes, it is easy to see the connection between your genes and your behaviors and environment. How Does Epigenetics Work?Epigenetic changes affect gene expression in different ways. Types of epigenetic changes include: DNA MethylationDNA methylation works by adding a chemical group to DNA. Typically, this group is added to specific places on the DNA, where it blocks the proteins that attach to DNA to “read” the gene. This chemical group can be removed through a process called demethylation. Typically, methylation turns genes “off” and demethylation turns genes “on.” Histone modificationDNA wraps around proteins called histones. When histones are tightly packed together, proteins that ‘read’ the gene cannot access the DNA as easily, so the gene is turned “off.” When histones are loosely packed, more DNA is exposed or not wrapped around a histone and can be accessed by proteins that ‘read’ the gene, so the gene is turned “on.” Chemical groups can be added or removed from histones to make the histones more tightly or loosely packed, turning genes “off” or “on.” Non-coding RNAYour DNA is used as instructions for making coding and non-coding RNA. Coding RNA is used to make proteins. Non-coding RNA helps control gene expression by attaching to coding RNA, along with certain proteins, to break down the coding RNA so that it cannot be used to make proteins. Non-coding RNA may also recruit proteins to modify histones to turn genes “on” or “off.” Example: Study of newborn vs. 26-year-old vs. 103-year-old DNA methylation at millions of sites were measured in a newborn, 26-year-old, and 103-year-old. The level of DNA methylation decreases with age. A newborn had the highest DNA methylation, the 103-year-old had the lowest DNA methylation, and the 26-year-old had a DNA methylation level between the newborn and 103-year-old (1). Example: Smokers vs. non-smokers vs. former smokers Smoking can result in epigenetic changes. For example, at certain parts of the AHRR gene, smokers tend to have less DNA methylation than non-smokers. The difference is greater for heavy smokers and long-term smokers. After quitting smoking, former smokers can begin to have increased DNA methylation at this gene. Eventually, they can reach levels similar to those of non-smokers. In some cases, this can happen in under a year, but the length of time depends on how long and how much someone smoked before quitting (2). If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. MCAT Content / AAMC MCAT Practice Exam 3 Bb Solutions AAMC FL3 BB [Web]Exam 3 Biology/Biochemistry Section Passage 1 1)
2) According to the passage, muscles that have similar diameters will also have similar strengths.
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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.
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.
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.
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.
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!
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.
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.
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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.
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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.
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.
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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.
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.
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.
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.
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.
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)
30) Enzymes are a very high yield topic on the MCAT; make sure you’re comfortable with enzyme kinetics.
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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)
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.
35) To answer this question, we will use the results from hypoxic conditions shown in Fig. 3:
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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.
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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.
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.
Exam 3 Biology/Biochemistry Section Passage 8 41)
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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:
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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)
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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.
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)?
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.
52) According to the passage, SSRIs work by blocking presynaptic neuron reuptake of serotonin. Knowing this, let’s move on to the answer choices.
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.
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.
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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.
Exam 3 Biology/Biochemistry Section Discretes 57)
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)
What is most likely the identity of protein X?What is the most likely identity of protein X ? An enzyme that participates in the degradation and recycling of cell components.
Which of the following best explains how molecules such as O2 and CO2 can move across the membrane of a cell quizlet?Which of the following best explains how molecules such as O2 and CO2 can move across the membrane of a cell? Small nonpolar molecules can diffuse through membranes.
Which of the following best explains how higher concentration of nitrogen and phosphorus contribute to eutrophication?Which of the following best explains how higher concentrations of nitrogen and phosphorus contribute to eutrophication? Algae require nitrogen and phosphorus to build macromolecules, so higher concentrations of these nutrients can result in algal blooms.
Which of the following describes the mechanism by which a plant stem grows toward light?"Even mature plants bend toward the strongest light. They do this by elongating the cells of the stem on the side that is farthest from the light. This type of light-oriented growth is called phototropism," explains Prof.
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