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MCAT Biology and Biochemistry Practice Axilogy MCAT Prep Amareen Dhaliwal

Copyright © 2016 Axilogy Consultants LLC. All rights reserved. ISBN:1534971327 ISBN-13: 978-1534891883

DEDICATION This book is dedicated to the aspiring physicians who are working day and night to make their dreams a reality. We wish you all the best in your endeavors and hope you are able to help the people who need you the most. Remember through all of this preparation, you will come out stronger, brighter, and more capable of achieving your dreams.

MCAT Biology and Biochemistry Practice

CONTENTS Introduction

1

1

Exam One

2

2

Exam One Answers

35

3

Exam Two

71

4

Exam Two Answers

100

5

Exam Three

129

6

Exam Three Answers

159

7

Exam Four

190

8

Exam Four Answers

217

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BIOLOGY AND BIOCHEMISTRY PRACTICE

Introduction Dear Future Doctor, We want to mention a few brief tips that will help make our practice workbooks as useful as possible. 1. Time yourself (59 minutes per exam) only if you have finished your General Chemistry, Physics, Organic Chemistry, and Biochemistry Content review before using this practice 2. If this is your first time doing section exams, do not time yourself and make sure to check answers after every question and/or passage so you absorb any mistakes as fast as possible 3. Our exams are harder than the real MCAT. We believe in the motto “train hard, fight easy” (so we want to give you difficult problems that you can learn from and apply on practice tests). Be aware of this before using our tests to determine your probable MCAT score :). 4. General “Strategies” are not for everyone. While our tutor program develops personalized strategies for each student, it still takes time to adjust these. The most commonly used strategies include “mapping” by linking common themes and parts of the passage on your note paper (like a family tree!). To learn how to read faster, physically cross out jargon and useless words on JUST the first few passages/questions/answers, then stop (you can’t do this on the exam!), after time your brain will do it on its own). Highlighting on the real exam slows students down but pacing with a cursor is helpful so we recommend tracking with your finger. Use a pencil and 2 sheets of printer paper for each exam. If you find yourself drifting off, label 1-15, 16-30, 31-45, and 46-59 on the 4 sides of paper. Put a :) or check on q’s you are sure about. Put A, B, C, D, and put a fraction next to the answer choices you are unsure about to weigh how “correct”they might be (this helps prevent second guessing). If your score is around 125-126 on practice tests and has not moved, try going through an entire exam using only your “first choice” (don’t change answer after your first decision unless 100% sure) and see if the score improves. If it does, listen to your “gut feeling”. If not, you make need to refresh on content :). 5. If you really want to challenge yourself: Finish all 4 exams in 1 day. (Yes, do it). We sometimes ask our students to challenge themselves by completing 2 full-lengths in a day. Doing more than 1 section back-to-back increases endurance for the section. 6. The last thing I want to say is: Be Happy. You have worked SO HARD to get to study for this test. This test is NOT the monster people might have called it. It is your golden ticket to medical school. This is the only way schools can compare you to anyone in the world. And IT IS POSSIBLE for you to do well. Don’t doubt yourself or get stressed! This is the only exam you can reschedule and take all the time in the world to prepare for. Listen to yourself, take the exam when you are ready, and practice as much as you need. If you can study for this exam happily and stress-free (and without using the fear of “not becoming a doctor” as a motivator) you will find this test to be much easier overall :). All the best, Amareen Dhaliwal CEO/Founder of Axilogy Test Prep

1

Exam One 1. "If you want to achieve greatness stop asking for permission." --Anonymous 2. "Things work out best for those who make the best of how things work out." --John Wooden 3. "To live a creative life, we must lose our fear of being wrong." --Anonymous 4. "If you are not willing to risk the usual you will have to settle for the ordinary." --Jim Rohn 5. "Trust because you are willing to accept the risk, not because it's safe or certain." -Anonymous 6. "Take up one idea. Make that one idea your life--think of it, dream of it, live on that idea. Let the brain, muscles, nerves, every part of your body, be full of that idea, and just leave every other idea alone. This is the way to success." --Swami Vivekananda 7. "All our dreams can come true if we have the courage to pursue them." --Walt Disney 8. "Good things come to people who wait, but better things come to those who go out and get them." --Anonymous 9. "If you do what you always did, you will get what you always got." --Anonymous 10. "Success is walking from failure to failure with no loss of enthusiasm." --Winston Churchill 11. "Just when the caterpillar thought the world was ending, he turned into a butterfly." -Proverb 12. "Successful entrepreneurs are givers and not takers of positive energy." --Anonymous 13. "Whenever you see a successful person you only see the public glories, never the private sacrifices to reach them." --Vaibhav Shah 14. "Opportunities don't happen, you create them." --Chris Grosser 15. "Try not to become a person of success, but rather try to become a person of value." -Albert Einstein 16. "Great minds discuss ideas; average minds discuss events; small minds discuss people." -Eleanor Roosevelt 17. "I have not failed. I've just found 10,000 ways that won't work." --Thomas A. Edison 18. "If you don't value your time, neither will others. Stop giving away your time and talents--start charging for it." --Kim Garst 19. "A successful man is one who can lay a firm foundation with the bricks others have thrown at him." --David Brinkley 20. "No one can make you feel inferior without your consent." --Eleanor Roosevelt

!2

Passage 1 (Questions 1 - 5)! Preliminary attempts to characterize a new human protein, XBR, have suggested that it may influence the expression of specific genes, including the gene!pyx. However, it was initially unclear whether this regulation was mediated via downstream transcription factors or direct interaction. Experiment 1! An electrophoretic mobility shift assay was used to determine whether XBR directly interacts with! pyx! (Figure 1). Different amounts of XBR were incubated with the DNA promoter region of!pyx!and a middle section of the ampicillin resistance gene as a control. Each DNA fragment was labeled with a radioisotope, was identical in length (250 bp), and contained one promoter, as well as a transcriptional and translational start site. !

! Figure 1!Interaction of XBR with!pyx! promoter and a fragment of the ampicillin resistance gene (amp) in an electrophoretic mobility shift assay ! Experiment 2! Interaction of XBR with the!pyx!promoter region was studied!in vitro!using a β-galactosidase assay (Figure 2). Two plasmids were created, one that carried the β-galactosidase gene controlled by the! pyx! promoter, and one with the XBR operon connected with a! lacUV5! promoter. The!pyx! promoter plasmid was transformed into!E. coli! cells, both with and without the additional co-transformation of the XBR operon plasmid. Expression of the! lacUV5! promoter is regulated by the! lacI repressor and can be induced with IPTG. Adding IPTG to the!E. coli! cultures containing the XBR plasmid induced the expression of XBR. To measure β-galactosidase expression, the!E. coli! cultures were given a solution that disrupted the cell membranes, but left the β-galactosidase intact. Samples were then treated with a synthetic color-reporting compound called ONPG, which was cleaved by βgalactosidase to yield a yellow compound. Activity of specific β-galactosidase is given in Miller units, a standardized measurement that quantifies β-galactosidase using ONPG. !

!3

! Figure 2 In vitro interactions between XBR and the!pyx! promoter region as detected with a β-galactosidase assay (Note: Since k2! zygote -> cleavage -> blastula (hollow ball of cells) > gastrulation: three layers (the germ layers) which differentiate into all adult tissues and organs.! ! Consider the following image.! !

Question!35!! Immediately prior to gastrulation, which of the following events take place? A.!The sperm penetrates the corona radiata and zona pellucida so that fusion of the gametes occurs. B.!The secondary oocyte becomes a mature ovum by completing its second meiotic division. C.!!Implantation in the endometrium occurs. D.! Rapid mitotic divisions result in daughter cells, smaller than their parent cells, being formed. The blastula precedes the formation of the gastrula in which, of course, the germ layers of the embryo form. However, gastrulation occurs after endometrial implantation. Question!36!! After sexual maturation, the primordial germ cells in the testes are initially called: A.!!spermatids and are haploid. B.!!primary spermatocytes and are haploid. C.!!primary spermatocytes and are diploid. D.!!spermatogonia and are diploid. You must be familiar with gametogenesis in males to answer this question. Note that the primordial germ cell is the spermatogonium (oogonium in females) and, along with the primary spermatocyte (cf. oocyte), is diploid (2n). Next: a reduction division, the chromosome number is reduced from diploid (46 chromosomes) to haploid (23 chromosomes). This is also known as Meiosis I (= first meiotic division or first meiosis), and it eventually produces germ cells (n), which can unite forming a diploid (2n) zygote.

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Question!37!! In the experiment described in the passage, which of the following materials in the germ plasm would likely be most functional in inducing germ cell function? A.!!Nuclear DNA. B.!!Mitochondria. C.!!RNA and protein molecules. D.!!Microtubules. The passage states that non-genetic material in the germ plasm is responsible for germ cell formation. This rules out nuclear DNA as an option. Mitochondria (energy) and microtubules (structural) are unlikely answers as they have very limited roles to actively change their outside environment. RNA and protein molecules can actively change the environment around them by aiding in the production of proteins, through functioning as enzymes (catalyzing reactions), as structural proteins, etc. Thus they are the most likely to induce germ cell function. Question!38!! Which of the following experimental results would contradict the conclusion made from the experiment in the passage? A.! ! The only subunits detected after digesting the germ plasm were non-uracil containing nucleotides. B.!!Microinjection of the midportion of a developing egg with germ plasm resulted in germ cell production at an abnormal site. C.! !Microinjection of proteases into the posterior end of the syncytial blastoderm inhibited the production of germ cells. D.! !The irradiated egg, which produced a sterile fly, demonstrated no post-radiation genetic abnormalities. The conclusion drawn from the experiment is that non-genetic material induces germ cell formation. You must identify the experimental results that would suggest that this conclusion is incorrect and that genetic material actually! does! have a role in germ cell formation. Option!B!is simply the same experiment as that in the passage but with the germ plasm injected at a different site. The results are the same, so it does not contradict the conclusion made in the passage. Option! C.! states that breaking down proteins (proteases) interferes with the induction of germ cell formation. This does not contradict the conclusion made since proteins are not genetic material. Option! D! shows that the interference in germ cell formation in the irradiated fly was not due to genetic mutations, which further supports the conclusion given in the passage. Option!A!indicates that the only thing present in the germ plasm appears to be genetic material (non-uracil containing nucleotides infers DNA but excludes RNA). Since it was the germ plasm that induced germ cell formation in the experiment, genetic material is implicated. This is the only evidence that contradicts the conclusion made in the passage. Question!39!! The mechanism by which blastomeres differentiate into germ cells is referred to as: A.!!induction. B.!!determination. C.!!specialization. D.!!differentiation. First off, it is important to know that a blastomere represents the first week of cell replication. The point at which a blastomere is committed to becoming a germ cell is different from the point at which the germ cell actually starts to function as a germ cell (i.e. producing cell-specific proteins). Determination is the point at which a cell is committed to becoming a particular type of cell, although it may not display any specific characteristics that would yet identify it as a specific type of cell. After determination, a cell will differentiate into a particular type of cell, and the fully differentiated cell is called specialized. Determination is the crucial point at which the fate of the blastomere is decided.

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Question!40!! During strenuous exercise, the NADH formed in the glyceraldehyde-3-phosphate dehydrogenase reaction in skeletal muscle must be reoxidized to NAD+! if glycolysis is to continue. The key reaction involved in the reoxidation of NADH is most clearly outlined by which of the following? A.!!glucose 6-phosphate → fructose 6-phosphate B.!!dihydroxyacetone phosphate → glycerol 3-phosphate C.!!isocitrate → α-ketoglutarate D.!!pyruvate → lactate NADH “must be reoxidized”, thus NADH is itself a reducing agent, and NAD+! is an oxidizing agent (these are expressions that you will likely see on the new MCAT). The process recycles NAD+!during anaerobic respiration (fermentation) implied by the strenuous exercise (the need for a ‘quick’ biochemical pathway to acquire energy; and confirmed by the use of the word ‘glycolysis’ in the question stem). In fermentation, NADH is used to reduce pyruvate to lactate. Question!41!! The biosynthesis of a pentasacharide of the monomer D-glucose (C6H12O6) would be expected to have a molecular formula consistent with which of the following? A.!!C30H52O26 B.!!C30H58O29 C.!!C30H50O25 D.!!(C6H12O6)5 Question!42!! When a dilute solution of formaldehyde is dissolved in! 18O-labeled water and allowed to equilibrate,! 18O incorporation occurs thus indicating the existence of the primary product. Which of the following compounds best represents this product of this!18O exchange?

A.! B.! C.! D.! The first step is to recognize that the polymerization of monosacharides in biological systems typically occurs through condensation reactions (the loss of water; typically it is a glycosidic linkage formed by a condensation reaction between the hydroxyl group of the first-position carbon and the hydroxyl group of the fourth-position carbon of each monosaccharide unit). Taking the analogy of a peptide: a dipeptide has 1 peptide bond between 2 amino acids. A pentapeptide logically has 4 peptide bonds. Thus there would be 4 condensations (dehydrations) creating 4 glycosidic bonds releasing 4 water molecules (4 H2O = 8 H + 4O). Thus (5 x C6H12O6) – (8H + 4O) = C30H52O26.

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Question!43!! Which of the following is the first step in the polymerase chain reaction (PCR)? A.!!Denaturation B.!!Cooling C.!!Primer extension D.!!Annealing This question begins with your ability to write the molecular structure for formaldehyde (a carbonyl group with two hydrogens attached to carbon!AKA methanal). Next, let's treat formaldehyde with water. Draw the mechanism on your scratch paper. The δ-!oxygen in water is attracted to the δ+! carbonyl carbon. The pi bond breaks and the free electrons go to oxygen. The attached water molecule now has three bonds (two H's and one carbon) and thus is positively charged. Electronegative oxygen is not happy! Oxygen pulls electrons to itself from its bond to H, thereby kicking off the proton H+. The proton is attracted to the oxygen with the free electron pair. Our product is a!diol!(=!2! alcohol! groups; IUPAC = methanediol). Note that the mechanism is the same as for hemiacetal formation, where R=H, R'=H, and R'''=H !

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Passage 7 (Questions 44 - 49)!! Question!44!! A pathogenic fungus is more capable of growth and reproduction on its native population of its sole host, the wild hog peanut, than on plants from other populations of the same species. It is reasonable to conclude that: A.! !the fungus, in this instance, was capable of more rapid adaptation to its host than vice versa. B.! !the fungus, in this instance, was capable of more rapid adaptation to all populations of the host species than vice versa. C.! !the host, in this instance, was capable of more rapid adaptation to the fungus than vice versa. D.! !all populations of the host species were capable of more rapid adaptation to the fungus than vice versa. PCR is often used to amplify a single copy or a few copies of a piece of DNA to several orders of magnitude, generating thousands to millions of copies of a particular DNA sequence. PCR begins with denaturation, then annealing the primer (stabilized by increased G/C content) and then primer extension. Aside: Practical uses include forensics and paternity testing. Question!45!! The passage suggests that one result of interspecific interactions might be: A.!!genetic drift within sympatric populations. B.!!genetic drift within allopatric populations. C.!!genetic mutations within sympatric populations. D.!!genetic mutations within allopatric populations. The fungus is better adapted to live on one population of the host species than another. In this case the fungus displays more rapid adaptation than the host since the fungus is better able to live on this host than others. The fungus did not display more rapid adaptation with respect to all host populations, as its ability to grow and reproduce on other populations of hosts is limited. Question!46! According to Fig. 1, the experiment showed that over time: A.!!coevolution caused a decrease in both the host and parasite populations. B.! ! coevolution caused both a decrease in fluctuation of the host and parasite populations, and a lowered density of the parasite population. C.!!coevolution caused a marked increase in the fluctuation of only the host population, and lowered the density of the parasite population. D.! ! coevolution caused a decrease in the population density of the parasite population but caused a marked increase in the density of the host population. The passage discusses evolution within two different species. Within a species, the passage distinguishes between "local" populations and "other" populations, which suggests that the populations live apart (=allopatric). Since the local population evolves differently (i.e.! "more capable of attacking the host . . ."),!genetic drift!may be implicated. Recall that genetic mutations are usually either negative or neutral with regard to the organism's survival.

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Question!47!! The control in the experiment likely consisted of: A.! ! members from different populations of the host and parasite species used in the experimental group, that had a short history of exposure to one another. B.!!members of the host and parasite species used in the experimental group, that had a long history of exposure to one another. C.! ! members of the host and parasite species used in the experimental group that had no history of exposure to one another. D.! ! members from different populations of the host and parasite species used in the experimental group, that had a long history of exposure to one another. By look...