Albert Chen
Executive Director, Pre-Health Research and Development
Kaplan Test Prep
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BIOLOGICAL SCIENCES TEST 1 TRANSCRIPT
Passage I (Questions 1–7)
1. Choice C is the correct answer. According to the passage, myasthenia gravis is a skeletal muscle disease of the neuromuscular junction. And although the neurotransmitter ACh also acts on cardiac muscle, cardiac muscle is not regulated by the same division of the nervous system that regulates voluntary actions. And of the four answer choices, the only activity that is NOT under voluntary control is contraction of the heart. Walking, talking, and raising your arm are all actions that are under voluntary control; so choices A, B, and D are wrong. Therefore, choice
C is the right answer. Cardiac muscle has a different type of ACh receptor than the ones found in the neuromuscular junction. Nicotinic ACh receptors are found in the neuromuscular junction, while muscarinic ACh receptors are found in cardiac muscle. Since anti-ACh-receptor bodies produced by an individual afflicted with myasthenia gravis are specific for nicotinic ACh receptors, heart contraction is not affected. Again, choice C is the right answer.
2. Choice D is the correct answer. While an antigen is commonly thought of as a foreign molecule that elicits antibody production upon exposure to the host immune system, but when dealing with an autoimmune disorder, this is not the case. The immune system of an individual with an autoimmune disorder labels molecules normally regarded as "self" as foreign, and as a result, produces antibodies specific for these molecules. In other words, these molecules act an antigens. In myasthenia gravis, antibodies are inappropriately formed against the ACh receptors of the neuromuscular junction, which means that the ACh receptors are acting as antigens. Thus, choice D is the right answer.
Let's take a look at the other choices. Choice A, a substrate, is a substance that is acted upon by another compound, typically an enzyme. For example, glucose is the substrate of the enzyme glucokinase in the first step of glycolysis. Choice B, a competitive inhibitor, is a substance that mimics another substance, thereby fooling the body into reacting with it instead and inhibiting the typical response. Choice C, an inducer, is a compound that deactivates a repressor molecule, thereby promoting protein synthesis. So choices A, B, and C are wrong and choice D is the right answer.
3. Choice B is the correct answer. According to the passage, myasthenia gravis is an acute autoimmune disease. This is a disease in which the body's immune system mounts a reaction against its own tissue. The immune system produces an antibody that binds to ACh receptors in the muscle tissue. The antibody binds to the receptor, preventing ACh from binding, and thereby blocking the transmission of an action potential. By blocking action potential transmission, the anti-ACh-receptor antibody prevents muscle contraction. According to the passage, the enzyme that degrades ACh is acetylcholinesterase. Therefore, since the problem is with the ACh receptors, and not with the synthesis of ACh itself, it makes sense that neostigmine most likely improves muscle reactivity by inactivating acetylcholinesterase. Why? Because inhibiting acetylcholinesterase activity would increase the concentration of ACh, thereby increasing the opportunity for ACh to bind to its receptors and stimulate muscle activity. Therefore, choice B is the right answer.
Boosting the immune system, choice A, would increase the concentration of anti-ACh-receptor antibody.
This would mean that even more ACh receptors would be blocked, thus further reducing the frequency of nervous transmission to muscle tissue. Since we're told that neostigmine improves muscle reactivity, this would most definitely NOT be one of its mechanisms of action. Therefore, choice A is wrong. Choice C is wrong because if neostigmine inhibited ACh synthesis, this would only decrease the concentration of available ACh, thereby lowering muscle reactivity. Increasing the affinity of the anti-ACh-receptor antibody would further inhibit ACh binding, and muscle reactivity would further decrease. So choice D is also wrong. Again, choice B is the right answer.
4. Choice C is the correct answer. This question requires an understanding of the mechanics of impulse transmission, since no information on this subject is given to you in the passage. An action potential is a rapid change in the electric potential of a cell membrane that is capable of transmitting nervous signals. An action potential is generated by the depolarization of some point on the membrane. The depolarization and resultant action potential results in an increase in membrane potential from about
−
70 millivolts to
+
35 millivolts, due to the influx of positive sodium ions. The thing about an action potential is that it is an all-or-nothing response. A threshold potential must be reached for an action potential to be generated. If a stimulus is not strong enough to raise the potential to its threshold - no action potential. However, if the stimulus is strong enough, the resultant action potential will always be of the same magnitude, amplitude, and duration. The only variable is the frequency with which the action potentials are fired. Increasing the frequency increases the amount of ACh released into the synapse. Therefore, choices A, B, and D are wrong, and choice C is the right answer.
5. Choice D is the correct answer. According to the question stem, the mechanism of action for this drug is to occupy the anti-ACh-receptor antibody binding site before the antibody has a chance to reach the neuromuscular junction and bind to the ACh receptors there. Therefore, the drug would have to have a similar 3-dimensional structure to the ACh receptor, since the antibody is specific for the receptor. So, choice D is the right answer.
Anyway, you should have been able to eliminate choices A, B, and C right off the bat, and here's why. Since the antibody does not interact with ACh itself, the drug will not need to react with it either. Therefore, there is no reason
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Kaplan MCAT Biological Sciences Test 1 Transcript for the drug to be similar in 3-D structure, electrical charge, or amino acid composition to ACh. Again, choice D is the correct answer.
6. The correct answer is choice C. From the passage you know that ACh receptors are found on the motor end plate of a neuromuscular junction, not the nerve or the synapse. So whatever technique we select, it must be specific for examining the motor end plate, and not the other two structures of a neuromuscular junction. Based on that, choice B can be eliminated. The nerve is the site of ACh release. ACh normally traverses the synapse and binds to an
ACh receptor on the motor end plate. So studying the nerve will not do us any good. If we examine the motor end plate under a light microscope, choice A, we will only see the general shape of the motor end plate. A light microscope is not powerful enough to examine individual molecules on the surface of a plasma membrane. Even if it were, you wouldn't be able to conclude that the receptors you were looking at through the microscope were definitely
ACh receptors as opposed to some other type of receptor. Thus, choice A is incorrect. If we hybridize the motor end plate with radio-labeled anti-ACh-receptor antibodies, as in choice C, we will be able to determine the number of
ACh receptors there. Since anti-ACh-receptor antibodies are specific for the ACh receptors of the motor end plate, it will bind directly to the ACh receptors. And by binding to the ACh receptors, we will be able to determine the number of receptors by measuring the radioactivity bound to the motor end plate. Thus, choice C is the correct answer. Let's just look at choice D quickly. Hybridizing the motor end plate with radio-labeled acetylcholinesterase will not do any good, because this enzyme binds to acetylcholine and degrades it, it does NOT bind the ACh-receptor with great affinity. And although acetylcholine is bound to its receptor when acetylcholinesterase binds to it, the enzyme only remains at the receptor for a few microseconds. Thus, accurate results cannot be obtained, and choice D is incorrect.
Again choice C is correct.
7. Choice A is the correct answer. This is another question requiring outside knowledge, this time, of the neuromuscular junction. The sarcoplasmic reticulum is a specialized endoplasmic reticulum containing a store of calcium ions. The sarcoplasmic reticulum is found only in muscle cells. When a muscle is depolarized, the action potential that is generated stimulates the release of calcium ion from the sarcoplasmic reticulum. The calcium ions facilitate the sliding action of myosin and actin, causing contraction. While each of the other ions listed in the answer choices are present within muscle cells and regulate many cell functions, none of them is responsible for initiating contraction. Again, choice A is the correct answer.
Passage II (Questions 8–13)
8. The correct answer is C. UV spectroscopy involves the absorption of ultraviolet light by pi electrons. The longer the chain of conjugation, the longer the wavelength absorbed. 7-dehydrocholesterol has two conjugated double bonds, and Vitamin D
3
has three. Cholesterol is not shown in the passage, but its structure can be deduced from information in the passage. Since you are told that cholesterol is dehydrogenated to form 7-dehydrocholesterol, cholesterol must have less than two double bonds. Looking at the answer choices, you can see that choice C, Vitamin
D
3
, is the correct response.
9. The correct answer is D. Previtamin D
3
and Vitamin D
3
are structural isomers; they have the same molecular formula, but they have different structural formulas. In this case, they differ in the placement of a hydrogen atom. Answer choice A is incorrect because resonance structures differ only in the placement of electrons, not atoms.
Answer choices B, enantiomers, and C, diastereomers, are types of stereoisomers. In case you don't remember, stereoisomers differ from each other only in the spatial arrangement of their atoms. In other words, their structural formulas are the same, but the arrangement of the molecules groups are different. Since previtamin D
3
D
3
have different structural formulas, choices B and C can be eliminated. Again, the correct answer is D.
and Vitamin
10. The correct answer is D. You can see that the A ring is the first on the left. Be careful; the diagrams here show the A ring rotated in Previtamin D
3
compared to its orientation in 7-dehydrocholesterol. To see this, look at where the hydroxyl group lies on the ring in 7-dehydrocholesterol compared to where it is in Previtamin D
3
. You should now be able to see that the methyl group that was bonded to the A and B rings remains attached to the A ring when it is converted to Previtamin D
3
. As it is no longer attached to the B ring, choice C is incorrect. As far as answer choices A and B are concerned, you should be able to quickly count the number of carbon atoms and see that they are the same. Again, the correct answer is D.
11. The correct answer is B. The most obvious difference between these two molecules is the change in the methyl group of Previtamin D
3
to a methylene group in vitamin D
3
. This difference, one involving environment changes to protons, can be best detected by proton NMR. The methyl group would appear as a singlet with an area of three around 1.0 ppm, while the methylene group would appear as a singlet of area two around 5.5 ppm. Any changes in the peak positions of the hydroxyl group would be negligible in both IR and NMR, which eliminates Choices A and
C. The IR change in the methyl group, choice D, would be located in the fingerprint region of the spectrum: a complex region where peaks are difficult to identify. Again, the correct answer is B.
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Kaplan MCAT Biological Sciences Test 1 Transcript
12. The correct answer is D. As discussed in Question 9, the isomerization involves a hydrogen shift. The positions of the hydroxyl and methyl groups remain unchanged, making choices A, B, and C incorrect.
13. The correct answer is B. As can be seen in the passage, Vitamin D
3
has three double bonds and one hydroxyl group. Compounds with only these functional groups are commonly referred to as trienols, making choice B the correct response. Answer choice A, a dione, is a compound with two ketone groups. Answer choice C, a dienol, is a compound with two double bonds and an alcohol group. Answer choice D, a trienone, is a compound with a ketone group and three double bonds. Again, the correct answer is B.
Passage III (Questions 14–19)
14. The correct answer is choice B. To answer this question you need to have a basic understanding of immunology. From the passage you know that in graft-versus-host disease (GVHD), the immune cells from donor tissue attack incompatible host cells. GVHD typically occurs in transplant patients or people who receive blood transfusions. Anyway, if the donor immune cells attack the host cells, then this implies that donor antibodies must be binding to host antigens. Remember, a typical immune response involves the proliferation of antibodies that are specific for and attack foreign matter known as antigens. The antibodies recognize the antigen and bind to it. This antibody-antigen complex is then destroyed by other cells of the immune system, typically macrophages. So in
GVHD, the donor tissue recognizes the host tissue as foreign, and thus produces antibodies against the "foreign" cells of the host. These donor antibodies then bind to the host antigens. Therefore choice B is the correct answer.
Choice A is incorrect because it describes a situation in which the host tissue is rejecting the donor tissue.
Although the immune reaction is the same as in choice B, the passage describes the reaction as being the other way around in GVHD. Choice C is incorrect, because from the passage you know that the donor immune cells attack the host. If the donor cells switched surface antigens to become compatible with the host, you would not expect these cells to attack. Instead, you would expect the donor tissue to assimilate with the host tissue, and thus the patient would not suffer from GVHD. Choice D is also incorrect. Although you may not be completely familiar with the specific functions of T-cells and B-cells, you do know that in GVHD, the donor immune cells react against the host.
And as we just discussed, this means that the donor immune cells are producing antibodies. Therefore you would
NOT expect donor antibody production to be inhibited by host T-cells in a patient suffering from GVHD. In case you were wondering, in a TYPICAL immune response, T-cells INDUCE B-cells to proliferate into antibody secreting cells. Again, choice B is correct.
15. The correct answer is choice C. According to the passage, thalidomide has immunosuppressive properties.
What does this mean? It means that thalidomide inhibits the immune system. So to answer this question all you have to do is identify which cell type is involved in the immune system. Well, pancreatic beta cells are a type of cell found in the pancreas whose main function is to synthesize insulin. Insulin is a hormone that promotes the conversion of glucose into glycogen, thereby lowering blood glucose concentration. And the pancreas in general is a small gland that produces digestive enzymes and the hormones insulin and glucagon. Since these functions are not directly involved in the immune system, choice A is incorrect. Skeletal muscle cells are obviously the cells that comprise skeletal muscle tissue and are NOT involved in the immune system. So choice B is also wrong. Macrophages are a type of phagocytic white blood cell derived from bone marrow. Macrophages are important in both the inflammatory and immune responses, and you would therefore expect their function to be inhibited by an immunosuppressive drug such as thalidomide. Therefore, choice C is the correct answer. Gustatory cells are involved in the process of taste.
Thus, choice D is also incorrect. Again, choice C is the correct answer.
16. Choice C is the correct answer. To answer this question, you've got to remember a bit about human embryology. Ectoderm endoderm and mesoderm are the three primary germ layers. Therefore, the correct answer is the germ layer that gives rise to both the long bones and the heart, since we're told that thalidomide exposure results in deformities in both of these structures. Ectoderm gives rise to the skin, the lens of the eye, and the nervous system, so choice A is wrong. Endoderm gives rise to the digestive and respiratory tracts, as well as parts of the liver, pancreas, thyroid, and bladder; thus, choice B is wrong. Mesoderm gives rise to the musculoskeletal, circulatory, and excretory systems, as well as the gonads and connective tissue throughout the body. Therefore, choice C is the right answer.
The archenteron, choice D, is the name given to the cavity formed during gastrulation. Again, choice C is the right answer.
17. Choice A is the right answer. This is another outside knowledge question, this time testing your knowledge of the developmental events that occur during the different stages of human pregnancy. It is during the first trimester of pregnancy that the organ systems first appear, therefore choice A is the right answer. Choice B is wrong, because the organ systems don't fully mature until after birth. Choice C is wrong, because fetal circulation begins during the first trimester, as the heart and circulatory system development, and continues to get its oxygen from the placenta until birth. Choice D is wrong, because rapid brain growth occurs during the last trimester. Again, choice A is the correct answer.
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Kaplan MCAT Biological Sciences Test 1 Transcript
18. The correct answer is choice C. According to the passage, thalidomide can cause the absence or gross deformities in long bones. So basically all you need to do to answer this question is figure out which of the choices is a long bone. Human bones are of two types, flat bones and long bones. Flat bones include the skull bones, choice A, the ribs, choice B, and vertebrae, choice D. Flat bones contain red bone marrow, indicating high blood cell production, and are found where little movement is required. They usually perform protective functions. Since choices A, B and D are not long bones, they are all incorrect. Long bones include the bones of the hands, choice C, and the feet, arms and legs. Long bones contain yellow marrow, indicating low blood cell production. The long bones are typically involved in locomotion and motion. So, choice C is the correct answer.
19. Choice D is the correct answer. Since this question doesn't cover a topic discussed in the passage, nor is it something you're expected to be familiar with from introductory biology, you're expected to be able to infer why mice exposed to thalidomide during their gestation did not experience thalidomide's teratogenic effects. So let's take a look at the answer choices. There's no valid reason for you to conclude that teratogens have no effect on mice. A drug like aspirin, which reduces blood clotting, can cause damage to mouse embryos just as easily as to human embryos. That's probably why the scientists who did the initial thalidomide studies assumed that effects WOULD be seen in mice exposed to thalidomide. Therefore, choice A is wrong. Choice B, that mice have a gestation period of less than 45 days should not make a difference, because everything is relative. What I mean is, the critical period of the 45th to the
55th day of human gestation has a corresponding period in mouse gestation - say something like the 8th to the 13th day. What's important is what structures are developing in the fetus during the critical period, not the actual day of the gestation period. Since both mice and humans go through homologous stages of development, researchers only study the period of mouse development during which limb buds first appear. Choice D, that human genotypes differ from mouse genotypes, indicates that there is no guarantee that compounds that affect one will definitely affect the other. Although both humans and mice are mammals, their genotypes are too different to assume that all drugs, including a teratogen, will automatically affect them both in the same way. Although the developmental processes in the two species are similar, the presence of genetic differences accounts for the fact that thalidomide is teratogenic in humans but not in mice. Therefore, choice D is correct. In current drug trials, several species of animals are often studies, including one closely related to humans, such as chimpanzees. Choice C, that mice do not get leprosy, is irrelevant. Although we're told in the passage that thalidomide is used to treat one of the side effects of leprosy treatment, this is independent of thalidomide's harmful effects to developing human fetuses. So choice C is wrong.
Again, choice D is the right answer.
Discrete questions
20. Choice C is the correct answer. To answer this question you need to be familiar with the events that occur during the individual stages of mitosis. During metaphase, the centrioles have already migrated to the opposite poles of the cell, the chromosomes have aligned along the metaphase plate, and the centromeres of the chromosomes have already attached to the spindle fibers of the spindle apparatus. During normal cell division, metaphase is followed by anaphase, during which the spindle fibers shorten and thus pull the sister chromatids that comprise each chromosome apart, moving them towards opposite poles of the dividing cell. Of all the answer choices, the only one that would account for disruption at metaphase, is disruption of the spindle apparatus, and so choice C is correct. Nucleic acid synthesis and the actions of the enzyme DNA polymerase do not occur during metaphase; DNA polymerase synthesizes new DNA during the S phase of the cell cycle, which precedes mitosis. Therefore, choices A and B are wrong. If vinblastine inhibited cellular respiration, the rat embryo cells would have died soon after exposure, not just during metaphase, since cellular respiration is the cells' source of ATP, which is essential to cell existence. Thus, choice D is also wrong. Again, choice C is the right answer.
21. The correct choice is C. Optically inactive molecules are those that are achiral. Achiral molecules have a plane of symmetry and are identical with their mirror images. Keep in mind, that just because a molecule has chiral centers it is not necessarily chiral. It is possible that molecules of this type may have a plane of symmetry, rendering them optically inactive. Such is the case with meso compounds.
To answer the question, a good thing to do would be to draw out choices A and C as stick figures, this will make it easier to see if they contain a plane of symmetry. Anyway, you should be able to see that choice C does indeed have a plane of symmetry, making it the correct response. Choices A, B, and D are incorrect since each is a carbon bonded to four different substituents, you should know that this means that they are chiral and optically active.
Again choice C is the correct response.
22. The correct answer is choice B. To answer this one you've got to be familiar with the products and byproducts of both glycolysis, aerobic respiration, and anaerobic respiration. During aerobic respiration, glucose is broken down into pyruvate; the pyruvate enter the mitochondria where it is converted into acetyl CoA; acetyl CoA then enters the Krebs cycle. The NADH and FADH
2
molecules produced during this entire process pass their electrons into the electron transport chain where oxygen is the final electron acceptor. This entire process produces
36 molecules of ATP per molecule of glucose. So, when a tissue goes into oxygen debt, the electron transport chain ceases to function. In fact, all operations within the mitochondria that require NAD
+
or FAD cease to function because these molecules can only be regenerated in the mitochondria by the transfer of electrons to molecular oxygen.
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Kaplan MCAT Biological Sciences Test 1 Transcript
Oxygen debt often occurs in muscle tissue during strenuous exercise, when the oxygen supply cannot keep up with the mitochondria's oxygen needs. This leads to a buildup of NADH, preventing the conversion of pyruvate into acetyl
CoA and preventing the Krebs cycle from operating. Since the Krebs cycle and subsequently the electron transport chain provide the majority of the ATP generated during respiration, a new energy production system must be utilized.
Thus, the body begins to operate through anaerobic respiration. This leads to a increased concentration of the converted to lactic acid in order to regenerate NAD
+
for another round of glycolysis. This process produces only 2 molecules of ATP per molecule of glucose. Since lactic acid is an organic acid, increases in lactic acid concentration will cause a decrease in pH, choice C. Since anaerobic respiration is 18 times less efficient than aerobic respiration in terms of ATP synthesis, an the concentration of ATP decreases, choice D. While pyruvate is normally consumed in the Krebs cycle, it is also consumed by anaerobic respiration, so that during the course of oxygen debt there will NOT be a buildup of pyruvate. So from this review, during oxygen debt the concentration of lactic acid will increase; the concentration of ATP and pyruvate will decrease, and the pH will also decrease. Therefore, choice B is the only answer choice that does not occur during oxygen debt. Thus choice B is the correct answer.
23. The correct choice is C. This question applies Brønsted acid-base theory to organic compounds. According to this theory of acids/base chemistry, the reactants and products form conjugate pairs--that is, a weak acid always forms a strong base and a strong base always forms a weak acid. In the reaction given, you can see that one conjugate pair consists of water (which in this case behaves as a weak acid) and hydroxide ion (which is strongly basic), while the other pair consists of a carbanion and acetylene. Acetylene is a weak Brønsted acid, so the corresponding conjugated base, the carbanion, must be very strong. Therefore, choice C is correct.
24. Choice B is the right answer. To answer this question, you need to understand what happens during the different phases of the heart cycle and determine the steps of CPR that are analogous to these phases. CPR is a technique used to restore and maintain a victim's circulation and respiration during heart failure. Manually compressing the victim's sternum squeezes blood into the victim's circulatory pathway. Compressing the sternum simulates the systolic, or contraction, phase of the heart cycle; the compression causes the atria to contract and pump blood into the ventricles, and then causes the ventricles to contract and pump blood into the arteries. So choice A is wrong. Systole is followed by diastole, which is the relaxation phase of the heart cycle, during which both the atria and the ventricles fill with blood. Thus, each time the rescuer relaxes the pressure on the sternum, the ventricles fill with blood, and so choice B is the right answer. Choice C is wrong, because cardiac muscle cannot be stimulated to depolarize merely by relaxing pressure exerted on the sternum. Choice D is wrong because the pH of the blood is not regulated by the contraction or relaxation of the heart. Again, choice B is correct.
Passage IV (Questions 25–28)
25. Choice A is the right answer. The Hardy-Weinberg equilibrium describes the equilibrium that exists between the different alleles of a particular gene locus of a particular gene pool. There must be only two alleles for the locus in question, and the five conditions listed in the question stem must be met. Okay, let's see which of these conditions are met by our population of guppies. I hope you didn't waste your time analyzing conditions I and II, since they don't even appear in any of the answer choices. But for your information, condition I can be ruled out, because the population is not very large; at the start of the experiment there are only 500 guppies. Condition II is not upheld either, because there must have been variation in the gene pool for natural selection to have worked on. This is evidenced by the fact that the life cycle of the guppies evolved during the course of the experiment. Conditions IV and V are also NOT upheld in the guppy experiments. The experiments begin with the artificial "migration" of 500 guppies from a river to a tributary, and we know that all genes are NOT equally successful at reproducing themselves in this population. Those guppies that have the genes for early maturation and reproduction produce offspring that don't survive to reproductive age because they're preyed upon by the predator fish in the tributary. Since conditions
IV and or V appear in choices B, C, and D, so we know that these choices must be wrong and choice A must be the right answer. In fact, since we aren't told otherwise in the passage, you are meant to assume that mating is random within this small population, and so condition III is upheld. Thus, choice A is the right answer.
26. Choice D is the right answer. Let's just run through the answer choices and define them. Choice A, genetic drift, refers to the chance variation in gene frequencies seen in one generation to the next. Choice B, gene flow, is the movement of genes into and out of the population, for example, by migration. Choice C, hybrid breakdown, refers to the case where the F
2
generation hybrid offspring, produced by way of an interspecies mating two generations ago, are inviable or infertile, even though their hybrid parents were not. Choice D, natural selection, refers to the survival of those individuals and their offspring best adapted to their environmental conditions. Thus, if you were familiar with all of these terms, you would have known that the term genetic shift, as used in the passage, refers to natural selection.
Those guppies that matured later and had fewer, larger offspring were better adapted to their life in the tributary than guppies that matured early and had many, smaller offspring.
Even if you didn't know the definitions of all of these terms, you still should have been able to answer the question by inferring from the information in the passage. You're told that the guppies have adapted to their new environment in the tributary in a way that enhances their reproductive success. This is a pretty good example of
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Kaplan MCAT Biological Sciences Test 1 Transcript natural selection, if you ask me. And from this you should have been able to deduce that genetic shift, as used in the passage, means roughly the same thing as natural selection, and so choice D is the right answer.
27. Choice A is the correct answer. If two characteristics are always found together in a species, then it's usually safe to assume that the genes governing these characteristics are located closely enough on the same chromosome that they are unlikely to get separated during a crossover event during meiosis. This is known as genetic linkage, and so choice A is right. If you're not too clear on the issue of meiosis and the concept of recombination, now's a good time for you to review these topics in the Reproduction chapter of the Biology Home Study Book. Let's take a look at the wrong choices. Choice B, independent linkage, is a nonsense term made up by the test writer, combining genetic linkage and independent assortment, so choice B is wrong. Choice C, recombination frequency, refers to the tendency of genes to recombine during meiosis, and is a function of the physical distance between genes on the length of a chromosome. Linked genes have a really, really low recombination frequency. Choice D, codominance, refers to an allelic system, in which the characteristics associated with both alleles are phenotypically expressed in heterozygous individuals. An example of such as system is the ABO human blood groups. Even if you weren't familiar with the definitions for these terms, you should have at least been able to narrow it down to the two choices containing the word linkage, since the question stem contains the phrase "always inherited together", which is kind of what the word linkage implies. Again, choice A is the right answer.
28. Choice A is the right answer. First of all, we can deduce that BOTH of the parents have homozygous genotypes for the traits in question. How do we know this? Because, first of all, we're told that the genes for small size and late maturation are dominant, and that the small, later-maturing guppy in the cross is homozygous for both traits. Assuming that both of these traits have only two alleles, since we're not told otherwise, this means that the other traits, large size and early maturation, must be recessive. For an individual to exhibit an autosomally inherited recessive trait, it must have two copies of the recessive allele. Thus, a guppy exhibiting both of these traits must be genotypically homozygous for both traits. Well, we now have enough information to predict the results of a cross between these two guppies. Since both parents are doubly homozygous, the F
1
generation will be heterozygous for both traits. The F
2
generation is the result of crossing two guppies from the F
1
generation, which we know to be double heterozygotes. In such a cross, the probability of being homozygous for either small or large size is 1/4, and the probability of being heterozygous is 1/2. Likewise, the probability of being homozygous for either early or late maturation is also 1/4 and the probability of being heterozygous is 1/2. Since we're told that the predator fish preys on guppy young in the tributary, it is advantageous for the guppy to mature late and produce large offspring. The probability of being homozygous for both traits is 1/4 times 1/4, which is 1/16 or 6.25%. We multiply the two probabilities because, according to statistical law, the probability of two independent events occurring simultaneously is equal to the product of their individual probabilities. If you're uncomfortable with all of this statistics stuff, you could have drawn a Punnett square for a dihybrid cross, and counted the number of squares with the desired genotype.
However, this is an extremely time consuming process, since a dihybrid cross has 16 squares. Again, choice A is the right answer.
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