10Biology discretes test w solutions

Biology Discretes Test Time: 30 Minutes Number of Questions: 30 This test consists of 30 discrete questions—questions that are NOT based on a descriptive passage These discretes comprise 15 of the 77 questions on the Physical Sciences and Biological Sciences sections of the MCAT MCAT BIOLOGY DISCRETES TEST DIRECTIONS: The following questions are not based on a descriptive passage; you must select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet I II III IV A B C D They are eukaryotic They possess ribosomes They possess cell walls They reproduce asexually I only I and 1I only I, II, and III only II III, and IV only The solid line in the figure below shows the course of the reaction A + B → C + D in the absence of its catalyst Which of the dotted lines would best represent the course of the same reaction in the presence of its catalyst? potential energy Which of the following are characteristic of animal and fungal cells, but NOT bacterial cells? reaction coordinate What is the function of a lysosome’s membrane? A It provides an acidic environment for the lysosome’s hydrolytic enzymes within the neutral environment of the cell B It is continuous with the nuclear membrane, thereby linking the lysosome with the endoplasmic reticulum C It is used as an alternative site of protein synthesis D The cytochrome carriers of the electron transport chain are embedded within it A B C D 4 A culture of red blood cells is grown on a nutrient medium containing dinitrophenol which is a poison that blocks the electron transport chain Under these conditions: A B C D ATP production will remain the same ATP production will decrease oxygen consumption will increase lactic acid production will increase GO ON TO THE NEXT PAGE as developed by Biology Discretes Test Down’s syndrome is a human genetic disorder that can be caused by the nondisjunction of chromosome 21 during meiosis During which phase of meiosis does nondisjunction occur? A B C D Interphase Prophase Anaphase Metaphase Brain cells of the housefly Musca domestica have pairs of chromosomes Therefore, it can be concluded that: A B C D the fly’s diploid number is 24 the fly’s haploid number is 12 the fly’s gametes contain chromosomes the fly’s gametes contain chromosomes Fetal lungs are supplied with only enough blood to nourish the lung tissue itself, since fetal lungs are nonfunctional prior to birth Obstruction of which of the following fetal structures would cause an increase in blood supply to fetal lungs? A B C D Aorta Ductus arteriosus Ductus venosus Pulmonary artery Which of the following pairs of differentiated tissue might be malformed if only one of the primary germ layers failed to properly differentiate during development? A B C D Brain and skeleton Muscle and kidneys Kidneys and brain Intestinal epithelium and kidneys Strontium is preferentially incorporated into growing long bone Therefore, if a child were exposed to strontium, where would the highest concentration of strontium most likely be found? A B C D In the cartilage lining the joints In the center of long bones Near the epiphyseal plates of long bones Evenly distributed throughout long bones 1 A patient with a peptic ulcer takes a large overdose of an antacid As a result, the activity of which of the following enzymes would be most affected? A B C D Trypsin Procarboxypeptidase Lipase Pepsin Pancreatic lipase is involved in the digestion of: A B C D starch protein fat cellulose Which of the following would trigger an increase in breathing rate? A B C D High blood CO2 partial pressure Low blood CO2 partial pressure High blood pH High blood O2 partial pressure Which of the following statements illustrates the principle of induction during vertebrate development? A The presence of a notochord beneath the ectoderm results in the formation of a neural tube B A neuron synapses with another neuron via a neurotransmitter C The neural tube develops into the brain, the spinal chord, and the rest of the nervous system D Secretion of TSH stimulates the secretion of the hormone thyroxine If a tracer substance is injected into a patient’s superior vena cava, which of the following structures would the tracer reach last? A B C D Right ventricle Left ventricle Pulmonary veins Left atrium GO ON TO THE NEXT PAGE KAPLAN MCAT Stenosis is a condition in which the leaves of a heart valve adhere to each other, decreasing the volume of blood flow through the valve A patient diagnosed with stenosis of the mitral valve would experience the greatest increase in blood pressure in his: A B C D left ventricle left atrium right atrium aorta If the victim of an automobile accident suffered isolated damage to the cerebellum, which of the following would most likely occur? A B C D Loss of voluntary muscle contraction Loss of sensation in the extremities Loss of muscular coordination Loss of speech Ingestion of the insecticide Parathion, which blocks acetylcholinesterase function, would cause a(n): All of the following thermoregulatory mechanisms are involved in heat conservation EXCEPT: A B C D piloerection perspiration shivering blood vessel constriction A decrease in postsynaptic depolarization B halt to all synaptic nervous transmission C increase in acetylcholine concentration in synapses D decrease in acetylcholine concentration in synapses 2 How many different types of gametes would be produced by an organism of genotype AabbCcDdEE, if all of the genes assort independently? An increased plasma osmolarity would most likely result in: A B C D excretion of dilute urine decreased water permeability in the nephron increased ADH secretion dehydration A patient’s anterior pituitary gland is removed due to the presence of a malignant tumor Plasma concentration of which of the following hormones would be LEAST affected following surgery? A B C D FSH Insulin Estrogen ACTH If a patient with high blood glucose levels (a diabetic) accidentally took an overdose of insulin, which of the following would most likely occur? A Convulsions due to decreased blood glucose concentration B Increased glucose concentration in urine C Dehydration due to increased urine excretion D Increased conversion of glycogen to glucose A B C D 10 16 The gene for color-blindness is X-linked If normal parents have a color-blind son, what is the probability that he inherited the gene for color-blindness from his mother? A B C D 25% 50% 75% 100% What is the sequence of linked genes D, E, F, and G, given the following recombinant frequencies? GE: 23% GD: 8% EF: 8% A B C D ED: 15% GF: 15% DF: 7% GDFE GFDE EFGD FGDE GO ON TO THE NEXT PAGE as developed by Biology Discretes Test In fruit flies, the gene for wing type is located on an autosomal chromosome The allele for wild-type wings is dominant to the allele for vestigial wings If a homozygous dominant male fly is crossed with a female with vestigial wings, what percentage of their female progeny are expected to have wild-type wings? A 0% B 25% C 75% D 100% In a particular population, for a trait with two alleles, the frequency of the recessive allele is 0.6 What is the frequency of individuals expressing the dominant phenotype? A B C D 0.16 0.36 0.48 0.64 Sexually reproducing species have a selective advantage over asexually reproducing species, because sexual reproduction: A cell is placed in a medium containing radioactively labeled thymidine If the cell undergoes two rounds of replication while in this medium, the radioactivity will appear: A B C D A B C D is more energy efficient creates novel genetic recombinations decreases the likelihood of mutations always decreases an offspring’s survival ability in both strands of all the DNA molecules in both strands of half the DNA molecules in one of the strands of every DNA molecule in both strands of half the DNA molecules; the other half of the DNA molecules will have one labeled strand Which of the following is an acceptable nitrogen base composition for double-stranded DNA? A B C D 31% A; 19% T; 31% C; 19% G 36% A; 36% U; 24% C; 24% G 48% A; 48% T; 52% C; 52% G 31% A; 31% T.; 19% C; 19% G What would be the sequence of mRNA transcribed from the DNA segment: 5’ -ACGTCA-3’ ? A B C D 5’ -TGCAGT-3’ 5’ -UGCAGU-3’ 5’ -UGACGU-3’ 5’ -UCGAGU-3’ END OF TEST KAPLAN MCAT ANSWER KEY: A 11 A 12 D 13 A 14 C 15 D C A B B 21 22 23 24 25 C B D A D 10 B C B A C 26 27 28 29 30 D D C D B D B B A C 16 17 18 19 20 as developed by Biology Discretes Test BIOLOGY DISCRETES TEST TRANSCRIPT The correct answer is choice A This question requires you to know the differences between your basic animal cell, fungal cell, and bacterial cell You should know by now that all animal cells (as well as plant cells) are eukaryotic and that all bacterial cells are prokaryotic, but what about fungal cells? Though they now have their own kingdom in the classification scheme, fungi used to be considered part of the Plant kingdom, which should tell you something about fungal cells: they’re EUKARYOTIC This means that choice D must be wrong because it doesn’t contain roman numeral I Fungi are heterotrophic organisms that come in all shapes and sizes; they can be either multicellular, like mushrooms, or unicellular, like yeast And like plant cells, fungi have cell walls, along with all the other eukaryotic organelles Bacteria also have cell walls, though they’re different in structure from plant cell walls and fungal cell walls Remember, however, that animal cells are NOT surrounded by cell walls Therefore, roman numeral III can be eliminated, which means that so can choice C As you know, animals reproduce sexually The primary means of reproduction in fungi, however, is asexual, via spores Bacteria reproduce asexually via binary fission; therefore, roman numeral IV is not a common characteristic of animal and fungal cells, and so it can be eliminated This doesn’t help you any since roman numeral IV only appears in choice D, which has already been eliminated So now you’re down to choices A and B Since we’ve established that both animal cells and fungal cells are eukaryotic, this means that both cell types have ribosomes the organelles responsible for translating a strand of mRNA into a protein Bacteria also synthesize proteins with ribosomes And even though prokaryotic ribosomes are structurally different from eukaryotic ribosomes, they are ribosomes nonetheless Therefore, roman numeral II must be wrong along with choice B, since this is a characteristic shared by ALL THREE cell types So, of the four characteristics, the only one shared by animal cells and fungal cells, but NOT by bacterial cells, is that they are both eukaryotic So, choice A, I only, must be the correct answer Choice A is the correct answer Lysosomes are membrane-bound sacs containing hydrolytic enzymes involved in intracellular digestion Lysosomes fuse with intracellular vesicles and digest their contents, such as proteins, polysaccharides, fats, and even nucleic acids The end results are breakdown products that can now be directly used by the cell Lysosomes also play a role in the recycling of cell organelles; lysosomes engulf and digest old organelles, releasing their component molecules into the cytosol for reuse The reason these hydrolytic enzymes are contained within the lysosome is that they are optimally functional at a pH of 5, which is acidic Like the stomach, which “accommodates” its enzymes by maintaining a pH of 2, the lysosome pumps in hydrogen ions from the cytosol to maintain an internal pH of The lysosomal enzymes would not function optimally in the neutral environment of the cytosol; the lysosome membrane enables the lysosome to maintain the acidic environment needed by its enzymes Therefore, choice A is the correct answer Choice B sounds nice, but it isn’t quite correct Although lysosomes, like the other membrane-bound organelles in the cell, are considered to be a part of the cell’s endomembrane system a system of membranes linked through direct contact or “communication” via tiny vesicles lysosome membranes are NOT in direct physical contact with the nuclear membrane and NOT provide a link between it and the endoplasmic reticulum The endoplasmic reticulum doesn’t need a link because it IS continuous with the nuclear membrane itself at certain points; thus, choice B is wrong Choice C is incorrect because it says that lysosomes are an alternate site of protein synthesis; the RIBOSOME is ALWAYS the site of protein synthesis, no matter what type of protein is being synthesized Lysosomes are involved in protein degradation, not synthesis Choice D is incorrect because the cytochrome carrier molecules of the electron transport chain the third stage of aerobic respiration are embedded in the inner membrane of the MITOCHONDRION, not the lysosome Again, choice A is the correct answer The correct answer is choice D This question tests your understanding of the function of a catalyst A catalyst is a substance, typically a protein, that increases the rate of a reaction by LOWERING THE ACTIVATION ENERGY OF THE REACTION The activation energy is the energy required by the reactants to get the reaction to go to completion A catalyst is neither altered nor consumed during the reaction Furthermore, a catalyst does not change the initial potential energy of the reactants, nor the final potential energy of the products In other words, a catalyst doesn’t change the delta G of a reaction In terms of the figure, since the solid curve corresponds to the reaction rate of the uncatalyzed reaction, the dotted line that corresponds to the catalyzed reaction must begin and end at the same points as the solid line Therefore, curves and can be immediately ruled out, and so choices A and C must be wrong Since the activation energy of the uncatalyzed reaction the solid curve correlates to the distance between the initial potential energy and the potential energy at the peak of the curve, the activation energy of the catalyzed reaction would necessarily be lower In other words, the distance between the peak potential energy and the initial potential energy would be lower Therefore, curve choice B is incorrect, and choice D curve is the correct answer The correct answer is choice A To answer this question you need to recall that red blood cells are enucleated cells that is, they don’t have nuclei Why? Because when red blood cells mature in the bone marrow, they lose their nuclei, mitochondria, and other membranous organelles According to the question stem, the poison dinitrophenol blocks the electron transport chain And you know from your introductory biology that the electron transport chain is embedded within the inner mitochondrial membrane However, since red blood cells lack mitochondria, they are not affected by drugs, such as dinitrophenol, that disrupt the electron transport chain Red blood cells derive their ATP anaerobically via glycolysis Therefore, ATP production in red blood cells poisoned with dinitrophenol will remain the same, and so choice B is wrong and KAPLAN MCAT choice A is correct Choice C is wrong because oxygen consumption must remain the same if ATP production remains the same Choice D is wrong because lactic acid buildup would occur only in an aerobic cell forced to revert to anaerobic metabolism in the absence of adequate oxygen supply It’s actually beneficial for an organism’s red blood cells to be anaerobic Red blood cells are the oxygen-carrying component of the blood If red blood cells were aerobic, they would require oxygen for metabolism, which would most likely be acquired from the very supply of oxygen that they transport This means that an aerobic red blood cell would deliver less oxygen to the tissues than an anaerobic red blood cell, which would be an extremely inefficient system Again, choice A is the correct answer The correct answer is choice C Before looking at the answer choices, let’s briefly review meiosis During prophase I, the DNA, which was already replicated during interphase, condenses into chromosomes Homologous chromosomes pair up during prophase I Homologous chromosomes are those that code for the same traits; humans have 22 pairs of homologous chromosomes and a pair of sex chromosomes During metaphase I, the homologous pairs align themselves at the center of the nucleus along the metaphase plate During anaphase I, the homologous chromosomes are pulled toward opposite poles of the cells by the spindle fibers attached to their centromeres During telophase I, the chromosomes reach the opposite sides of the cell, cytokinesis occurs, and two daughter cells are formed These daughter cells then undergo the second round of meiotic division During prophase II, the chromosomes move toward the equator of the cell During metaphase II, the chromosomes line up at the metaphase plate, and during anaphase II, the sister chromatids separate and move toward opposite poles of the cell During telophase II, cytokinesis occurs and a nuclear membrane reforms around each haploid set of chromosomes The end result is four daughter cells, each with a unique set of N chromosomes This question involves the nondisjunction of chromosome 21, which is known to be one of the ways in which Down’s syndrome occurs Nondisjunction occurs when a pair of chromosomes (either homologous chromosomes during meiosis I, or sister chromatids during meiosis II) fail to properly separate Since chromosomes normally separate during anaphase, the correct answer is choice C The correct answer is choice D All non-sex cells in a eukaryotic organism are known as autosomal, or somatic, cells and contain the diploid number of chromosomes, which is also referred to as 2N All sex cells, or gametes, contain the haploid number of chromosomes, or N, which amounts to half of the diploid number If the brain cells of the fly contain pairs of chromosomes, which is 12 chromosomes in total, then all of its other somatic cells must also contain 12 chromosomes as well Since the diploid number is the number of chromosomes in a somatic cell, this species of fly has a diploid number of 12 The only cells in the fly that are haploid are its gametes Therefore, the fly’s gametes must contain chromosomes, and so choice D is correct During fertilization, the diploid number is restored when two haploid nuclei fuse to become one cell the zygote Let’s take a look at the wrong answers Choice A is wrong because the fly’s diploid number is 12, not 24 Choice B is wrong because the fly’s haploid number is 6, not 12 Choice C is wrong because every sperm and egg of this fly species contain chromosomes, not Again, the correct answer is choice D The correct answer is choice B Answering this question gives us the perfect opportunity to review fetal circulation Fetal circulation differs from adult circulation in several important ways The major difference is that in fetal circulation, blood is oxygenated in the placenta (because fetal lungs are nonfunctional prior to birth), while in adult circulation, blood is oxygenated in the lungs In addition, the fetal circulatory route contains three shunts that divert blood flow away from the developing fetal liver and lungs The umbilical vein carries oxygenated blood from the placenta to the fetus The blood bypasses the fetal liver by way of a shunt called the ductus venosus, before converging with the inferior vena cava The inferior and superior venae cavae return deoxygenated blood to the right atrium Since the oxygenated blood from the umbilical vein mixes with the deoxygenated blood of the venae cavae, the blood entering the right atrium is only partially oxygenated Most of this blood bypasses the pulmonary circulation and enters the left atrium directly from the right atrium by way of the foramen ovale, a shunt that diverts blood away from the right ventricle and pulmonary artery The remaining blood in the right atrium empties into the right ventricle and is pumped to the lungs via the pulmonary artery Most of this blood is shunted directly from the pulmonary artery to the aorta via the ductus arteriosus, diverting even more blood away from the lungs This means that in the fetus, the pulmonary arteries carry oxygenated blood to the lungs, though this blood is by no means saturated with oxygen The blood that is delivered to the lungs is further deoxygenated there, because the blood unloads its oxygen to the fetal lungs, which need it for proper development Remember, gas exchange does not occur in the fetal lungs it occurs in the placenta The deoxygenated blood then returns to the left atrium via pulmonary veins Despite the fact that this blood mixes with the partially oxygenated blood that crossed over from the right atrium (via the foramen ovale) before being pumped into the systemic circulation by the left ventricle, the blood delivered via the aorta has an even lower partial pressure of oxygen than the blood that was delivered to the lungs Deoxygenated blood is returned to the placenta via the umbilical arteries Obstruction of the ductus arteriosus would cause an increase in blood supply to the fetal lungs Why? Because ALL of the blood pumped into the pulmonary arteries by the right ventricle would then have to flow through the lungs there would be no place else for it to go So choice B is correct Let’s look at the other choices If the aorta were obstructed, the blood supply delivered to the tissue would be greatly diminished, but this would not have an effect on the volume of blood delivered to the fetal lungs Thus, choice A is wrong Choice C is wrong, because obstruction of the ductus venosus would greatly increase the blood supply to the fetal liver, since all of the oxygenated blood from the umbilical vein would be forced as developed by Biology Discretes Test to pass through the liver before being sent to the heart Choice D is wrong, because obstruction of the pulmonary artery would result in a DECREASE, not an increase, in blood supply delivered to the fetal lungs Again, the correct answer is choice B The correct answer is choice B This is really a simple knowledge question either you knew which of the pairs of tissue are derived from the same primary germ layer or you didn’t Let’s go through each of the choices Choice A is incorrect because the brain, along with the rest of the nervous system, is an ectodermal derivative, while the skeleton is a mesodermal derivative Choice B is correct since both muscle and the kidneys are derived from mesoderm Choice C is incorrect because the kidneys are mesodermal in origin while the brain, as noted, is an ectodermal derivative Choice D is incorrect because the epithelial linings of both the digestive and respiratory tracts are endodermal in origin, while, as just noted, the kidneys arise from mesoderm it is important that you know which tissues derive from which embryonic germ layers, as this has always been a favorite MCAT discrete question topic A more detailed discussion of this topic can be found in the Embryology chapter of your Biology Review Notes Again, the correct answer is choice B The correct answer is choice A In terms of vertebrate development, induction is defined as the process by which a particular group of cells causes the differentiation of another group of cells Choice A is an example of induction: the group of cells that form the notochord induces the formation of the neural tube Other examples of induction in vertebrate development include the formation of the eye, where the optic vesicles induce the ectoderm to thicken and form the lens placode, which in turn induces the optic vesicle to form the optic cup, which in turn induces the lens placode to form the cornea and the lungs Let’s look at the other answer choices In choice B, while it IS true that a neuron synapses with another neuron via a neurotransmitter the chemical messenger of the nervous system this is NOT an example of induction Thus, choice B can be eliminated In choice C, while the neural tube DOES develop into the nervous system, it is not induced to so by another group of cells or tissue at least not as far as we know Therefore, choice C is also wrong Finally, choice D is also a true statement TSH, or thyroxin stimulating hormone, DOES stimulate the secretion of thyroxine, as its name implies However, this is not an example of induction Remember, just because an answer choice is a true statement does not mean it’s the correct answer It’s only correct if it answers the question that was asked Again, the correct answer is choice A 10 The correct answer is choice C This question is simply asking you where growth occurs in long bone The answer is: the epiphyseal plates The epiphyseal plates are regions of cartilaginous cells separating the shaft of the long bone, called the diaphysis, from its two dilated ends, called the epiphyses The epiphyseal plates are located at either end of long bone, which are referred to as the proximal and distal ends Since growth occurs only at the epiphyseal plates, the strontium would be incorporated near the plates Therefore, choice C must be the correct answer Choices B and D are incorrect because they incorrectly identify the location of long bone growth Choice A is incorrect because the question is only dealing with long bone growth; joints are what link two bones together they are not part of the long bones themselves, so you’re not going to find strontium there Again, choice C is correct 11 The correct answer is choice D An individual develops a peptic ulcer either when her gastric mucosa overproduces hydrochloric acid (HCl), or when her mucosal defenses are inadequate to protect the stomach mucosa from the normal concentration of HCl in the stomach The pH in the stomach is normally very low around due to this secretion of HCl This acidic environment is needed by the stomach enzyme pepsin, which works optimally at a pH of One way to treat a peptic ulcer is to take antacids, which neutralize the HCl in the stomach and thereby raise the gastric pH By raising pH, pepsin becomes nonfunctional Therefore, pepsin activity would be most affected by an antacid overdose, and so choice D is the correct answer By contrast, the contents of the small intestine are relatively alkaline, due to the secretion of bicarbonate, and the enzymes secreted into it function best at this higher pH Trypsin, procarboxypeptidase, and lipase are all products of the exocrine pancreas secreted into the small intestine, and function best in the alkaline environment there Therefore, they would not be affected by an antacid overdose, and so choices A, B and C are incorrect Trypsin and procarboxypeptidase are both protein-digesting enzymes, while lipase is involved in fat digestion Again, choice D is the correct answer 12 The correct answer is choice C Lipases are enzymes involved in lipid digestion Lipids are a category of nonpolar organic substances that include fats, oils, waxes, and steroids Therefore, pancreatic lipase is an enzyme involved in the digestion of fat If you didn’t know this outright, you might have been able to deduce it from prefix “lip” shared by lipase and lipid and liposuction Thus, choice C is the correct answer As for the other answers starch, choice A, is digested by amylase, which is secreted by both the salivary glands and the pancreas Protein, choice B, is digested by proteases such as pepsin, trypsin, and chymotrypsin, just to name a few Choice D, cellulose, is actually an indigestible molecule for most mammals Cellulose is the plant equivalent of starch, and vertebrates not have an enzyme capable of digesting it Mammals that consume a lot of cellulose-rich plant material, such as elephants, horses, and cows, have bacterial symbiotes in their guts that digest cellulose for them So, choices A, B, and D are all incorrect, and choice C is the correct answer 13 The correct answer is choice A When the partial pressure of the respiratory gases are abnormally high or low, chemoreceptors located in the medulla oblongata, on the carotid arteries, and on the aorta, detect and signal the KAPLAN MCAT respiratory centers in the medulla to modify the breathing rate The chemoreceptors are most responsible to changes in the concentrations of carbon dioxide and hydrogen ion; only extreme changes in oxygen concentration are relayed to the medulla To answer this question correctly, it’s important to understand that the concentration of hydrogen ion in the blood is directly proportional to the partial pressure of carbon dioxide in the blood Inside red blood cells, carbon dioxide combines with water to form carbonic acid, which then dissociates into bicarbonate ion and hydrogen ion And an increase in hydrogen ion concentration leads to a decrease in blood pH So if the chemoreceptors detected an increase in the partial pressure of CO2 in the blood, or an increase in the concentration of hydrogen ion, they would send a signal to the brain that would result in an increase in breathing rate Therefore, choice A must be correct and choices B and C must be wrong As for choice D, a high partial pressure of oxygen in the blood goes hand in hand with choices B and C In other words, choices B, C, and D would most likely trigger a decrease, not an increase, in breathing rate Again, choice A is the correct answer 14 Choice B is the correct answer This question is simply a matter of knowing the pathway traveled by blood in the heart Deoxygenated blood drains into the right atrium from both the inferior vena cava and the superior vena cava From the right atrium the blood flows into the right ventricle, which then pumps it to the lungs via the pulmonary arteries Carbon dioxide is exchanged for oxygen in the alveoli of the lungs Oxygenated blood is returned to the left atrium via the pulmonary veins From the left atrium the blood drains into the left ventricle, which pumps it into the aorta for circulation throughout the body Okay, getting back to our question Based on the previous discussion, if a tracer substance is injected into the superior vena cava, it would take the longest amount of time to reach the left ventricle Thus, choice B is the correct answer Remember, following injection, the tracer would first enter the right atrium, then drain into the right ventricle, choice A The blood with the tracer would travel to the lungs via the pulmonary arteries and return through the pulmonary veins, choice C, into the left atrium, choice D The tracer would then travel from the left atrium into the left ventricle, choice B Therefore, the tracer reaches the left ventricle last Again, choice B is the correct answer 15 The correct answer is choice B The key to answering this question hinges on our knowledge of the mitral valve The mitral valve sits between the left atrium and the left ventricle, preventing the backflow of blood into the left atrium when the left ventricle contracts In general, the term stenosis refers to a narrowing of the valve such that the volume of blood pumped through it decreased So in an individual suffering from stenosis of the mitral valve, blood flow from the left atrium into the left ventricle is impeded Therefore, the effect is a great increase in blood volume in the left atrium and a reduced net movement of blood from the left atrium into the left ventricle This means that there will be an increased left atrial pressure and decreased left ventricular pressure Therefore choice A is wrong and choice B is the correct answer Let’s take a quick look at the other choices Blood flows out of the left ventricle into the aorta, which branches off into the arterial system Since pressure in the left ventricle is reduced, you would also expect to see decreased pressure in the aorta as well Therefore, choice D is also incorrect The right atrium, where deoxygenated blood is returned by venous circulation, is too far removed from the mitral valve to be much affected by mitral valve stenosis Therefore, choice C is also incorrect Again, the correct answer is choice B 16 The correct answer is choice B Perspiration, or sweat, is the secretion of water, salts, and urea from the sweat pores of the skin As the sweat comes into contact with air, it evaporates, thereby cooling the skin Thus, perspiration is a thermoregulatory mechanism involved in cooling down the body, or heat dissipation, NOT heat conservation Let’s take a look at the wrong choices Choice A, piloerection, refers to a reflex contraction of the small muscles found at the base of skin hairs Contraction of these muscles causes the hairs to stand erect, forming an insulating layer that prevents the loss of body heat Piloerection in humans is commonly known as “goose bumps.” So, choice A is incorrect Choice C, shivering, is also wrong, because shivering is a thermoregulatory mechanism designed to conserve heat Shivering is an intensive, rhythmic involuntary contraction of muscle tissue that increases internal heat production Choice D is incorrect since the selective constriction of blood vessels at the skin’s surface reroutes blood flow from the skin to deeper tissues, minimizing the loss of body heat Again choice B, perspiration, is the correct answer 17 The correct answer is choice C To answer this question, you need to figure out which of the physiological responses listed in the answer choices would be the most effective method for the body to return plasma osmolarity to a normal level When plasma osmolarity rises above normal levels, it is detected by the osmoreceptors of the hypothalamus, which send signals to the pituitary gland that stimulate it to secrete ADH ADH, or vasopressin, acts directly on the nephron to increase its permeability to water, thereby increasing water re-absorption Increasing water re-absorption serves to decrease plasma osmolarity So, increased ADH secretion alleviates elevated plasma osmolarity and therefore, choice C is correct Let’s look at the other answer choices just to make sure Choice A, excretion of dilute urine, is incorrect since this would be a result of LOW plasma osmolarity or inhibition of ADH secretion by substances such as alcohol and caffeine A low solute concentration in the blood would decrease pituitary secretion of ADH, which in turn, would reduce water reabsorption, which would lead to the excretion of dilute urine Excretion of dilute urine serves to raise plasma osmolarity, not decrease it; thus, choice A is incorrect Choice D is wrong, due to its association with choice A Dehydration results from an excess excretion of dilute urine, known as diuresis, which can be caused by excessive loss of fluid and an increased plasma osmotic pressure Therefore, dehydration increases plasma osmolarity, and would therefore not be a likely result of increased 10 as developed by Biology Discretes Test plasma osmolarity Choice B, decreased water permeability in the nephron, is incorrect because, as mentioned before, increased plasma osmolarity would lead to increased ADH secretion, which in turn leads to increased water re-absorption And as we previously discussed, ADH increases water re-absorption by increasing water permeability of the nephron Therefore, decreasing water permeability would serve to increase plasma osmotic pressure, not return it to normal, and so choice B is also incorrect Again, the correct answer is choice C 18 The correct answer is choice B If the anterior pituitary gland is removed, then all of the hormones that it normally synthesizes and secretes would no longer be present in the bloodstream These hormones include luteinizing hormone (LH), follicle-stimulating hormone (FSH), adrenocorticotropic hormone (ACTH), thyroid-stimulating hormone (TSH), prolactin, and growth hormone Since the secretion of estrogen and progesterone is stimulated by the secretion of FSH and LH, the plasma levels of both these hormones estrogen and progesterone would also decrease if the anterior pituitary were removed Likewise, since ACTH stimulates the adrenal cortex to secrete glucocorticoid hormones, the levels of these hormones would also decrease Thus, choices A, C, and D are wrong Insulin, on the other hand, is secreted by the pancreas in response to high blood glucose levels, and would therefore be the LEAST affected by the removal of the anterior pituitary Thus, choice B is the correct answer 19 The correct answer is choice A Insulin is the hormone secreted by the beta cells of the pancreas in response to high blood glucose levels Insulin decreases blood glucose by stimulating cells to uptake glucose, and by stimulating the conversion of glucose into its storage form, glycogen, in liver cells and muscle cells An overdose of insulin can, and often does, lead to convulsions, due to the sharp decrease in blood glucose concentration that results Therefore, choice A is the correct answer Choice B is incorrect because the presence of glucose in urine is indicative of a high blood glucose concentration and is actually one of the symptoms of untreated diabetes A person with diabetes has high blood glucose levels because of an insufficiency in, or a complete lack of, insulin production, or because insulin-specific receptors in the body are insensitive to insulin As a result, in untreated diabetes, there is a high blood glucose concentration along with a high urine glucose concentration, since the kidneys aren’t equipped to re-absorb the excess glucose in the blood Controlled doses of insulin are used to alleviate the symptoms of diabetes, so clearly, insulin wouldn’t cause an increased glucose concentration in the urine, and so choice B is wrong Choice D is wrong, because as just discussed, one of the effects of insulin is to increase the conversion of glucose into glycogen, NOT vice versa Therefore, an overdose of insulin would not cause an increase in the conversion of glycogen back into glucose It is actually the hormone glucagon, also secreted by the pancreas, that stimulates the conversion of glycogen into glucose Diabetics DO become dehydrated if untreated because, in the kidneys, the excess glucose in the nephrons causes water to diffuse into the nephrons, not out of them, as it would under normal circumstances The net result is an increase in urine production and excretion Insulin would decrease urine excretion by decreasing blood glucose, and so choice C is also wrong Again, choice A is the correct answer 20 The answer is choice C The cerebellum is located in the hindbrain along with the pons and medulla All higher brain sensory neutrons and motor neurons pass through the hind brain The main function of the cerebellum is coordinating unconscious movement Hand-eye coordination, posture, and balance are all controlled by the cerebellum Therefore, damage to the cerebellum would most likely result in loss of muscle coordination, and so choice C is the correct answer Let’s look at the other choices briefly Loss of either voluntary muscle contraction, choice A, sensation in the extremities, choice B, or loss of speech, choice D, may be caused by damage to specific areas of the cerebrum The cerebrum is located in the forebrain and is divided into two hemispheres, the left and the right Each hemisphere is further subdivided into four lobes The cerebrum is responsible for the coordination of most voluntary activities, sensation, and “higher functions”, including speech and cognition Sensation of the extremities may also be controlled in part by the spinal cord Since choices A, B, and D are not under the control of the cerebellum, they are all incorrect Again, choice C is the correct answer 21 The correct answer is choice C Acetylcholine is a neurotransmitter, which causes depolarization of the postsynaptic membrane of one neuron when released by the presynaptic terminal of another neuron Acetylcholine is removed from the synapse by way of hydrolysis; that is, in the synapse, the enzyme acetylcholinesterase catalyzes the hydrolysis of acetylcholine into choline and acetate ANTIcholinesterases, such as the insecticide Parathion, disrupt the activity of acetylcholinesterase, which means that acetylcholine can’t be degraded And this means that the concentration of acetylcholine in the synapse will INCREASE, especially since the presynaptic membrane will continue to secrete the same amount of acetylcholine that it normally secretes in response to incoming action potentials So, choice D must be wrong and choice C must be the correct answer Choice B is wrong because an anticholinesterase won’t affect the activity of other neurotransmitters, such as epinephrine and norepinephrine, so it certainly won’t bring ALL synaptic nervous transmission to a halt As for choice A, if acetylcholinesterase were blocked, then there would be an INCREASE, not a decrease in postsynaptic depolarization, because if acetylcholine is not degraded, it will continuously bind to its receptors on the postsynaptic membrane and depolarize it Hence, choice A is incorrect Again, choice C is the correct answer 22 The correct answer is choice B The number of genetically unique gametes produced by a given individual can be determined by using the formula “two to the nth power”, where “n” is the number of heterozygous gene pairs that the KAPLAN 11 MCAT individual has Heterozygous pairs are those in which the two alleles are not the same The number of homozygous gene pairs is not a factor in the formula because homozygous gene pairs not contribute to the total number of possible variations in a gamete’s chromosome content Why? Because all the gametes will have the same allele if the parent has two identical copies of the gene If an organism has only two traits, and its genotype is big A little a, big B big B, there are only different types of gametes that it can produce: big A big B, and little a big B Since the organism is homozygous for the B gene, there is no variation in the gametes for this allele every gamete will have the big B allele And if we used the formula, to the nth power we get different gametes for this organism How did we figure that out? Since there is only one heterozygous pair for this organism, which is big A little a, n is equal to So we get raised to the first power, which is Now let’s look at the question The individual in the question has the genotype big A little a, little b little b, big C little c, big D little d, big E big E Well, there are three heterozygous gene pairs out of a total of five pairs So, using the formula, we get two raised to the third power, which equals 8, choice B All eight of these gametes will have the little b gene and the big E gene; what these gametes differ in is which type of A gene, C gene, and D gene they wound up with as a result of independent assortment during gametogenesis, or meiosis Since you were only dealing with three gene pairs, you didn’t really need the formula to figure this one out you probably could have done it in your head The formula is much more useful as the number of heterozygous gene pairs increases Again, choice B is the right answer 23 The correct answer is choice D The important thing to remember about questions dealing with X-linked traits is that females have two X chromosomes one inherited from her mother, one inherited from her father, while males have one X chromosome inherited from his mother and one Y chromosome inherited from his father So if a male expresses an X-linked trait, that means he must have inherited it from his mother So if normal parents have a color-blind son, he MUST have inherited the color-blind gene, which is X-linked, from his mother His mother MUST be a carrier of the colorblindness allele So the probability that a color-blind son inherited the gene for color-blindness from his mother is 100%, choice D, which is the correct answer 24 The correct answer is choice A Recombinant frequencies of linked genes are used for mapping the relative locations of genes on a single chromosome Recombinant frequencies are determined by crossing individuals that differ in allele for all the genes in question, and then determining the genotypes of their offspring The recombinant frequencies are the frequencies at which nonparental genotypes appear, since these genotypes can arise only by genetic crossover The premise of mapping is based on the assumption that the PROBABILITY of a crossover occurring between two points INCREASES as the distance between these two points INCREASES Therefore, the FARTHER apart two genes are, the GREATER their recombinant frequency; that is, the chance that these two genes will be inherited together, or will be linked, is less the father apart two genes are on a given chromosome One map unit is defined as a 1% recombination frequency, and recombinant frequencies are roughly additive However, if the genes are very far away, then the recombination frequency will reach a maximum of 50%, at which point the genes are considered to be assorting independently As we go through the problem, it might help you write down your own gene map Okay, we’re given four genes D, E, F, and G and we’re given the recombinant frequencies between each possible pair In making your map, start with the pair that has the highest recombinant frequency In this case, the recombinant frequency is greatest between G and E; it’s 23%, which means that G and E are 23 map units apart, and they must be on the two ends of the genetic map we’re trying to construct From this one piece of information you can eliminate choices B and C because they not show G and E to be at the two ends So, your chances of getting this question correct have already improved to 50% Now fill in the intervening genes by looking for the genes that are closest to the two endpoints G and D are map units apart, and this is the shortest distance between G and anything, so D must be next to G So now you can eliminate choice B So by the process of elimination, the correct sequence of genes on this chromosome must be G, D, F, E, which makes choice A the correct answer By the way, EFDG is equally correct, which is the answer you would have gotten if you created your map in the opposite direction; but this isn’t one of the answer choices But let’s work through the rest of the problem anyway As a check, notice that D and E are 15 map units apart This makes sense, because the distance from G to D, which is 8, plus the distance from D to E, which is 15, equals the distance from G to E, which is 23 We also see that G and F are 15 map units apart, while F and E are units apart By the way, in this case the numbers add up exactly; that is, the distance from G to E equals the distance from G to D plus the distance from D to E This is not always the case, however Sometimes, the numbers will be off by one or two units, but this doesn’t necessarily mean that you’ve made a mistake As I said before, map distances are only roughly additive Again, choice A is correct 25 Choice D is the correct answer This is your basic cross using our friend, Drosophila melanogaster You’re told that the gene for wing type is located on an autosomal chromosome a non-sex chromosome, which means that it is NOT inherited as a sex-linked trait You’re also told that the dominant allele codes for wild-type wings; wild-type simply means that this is the phenotype that predominates in nature The recessive allele codes for the vestigial wing type, which is a stumpy wing The gender of the flies is of no relevance here; gender only comes into play for sex-linked traits So, our male fly is homozygous dominant for wing type and our female fly is homozygous recessive she has to be homozygous recessive, since she exhibits the recessive phenotype, vestigial wings A cross between a homozygous dominant and a homozygous recessive yields 100% heterozygous individual, which in this case, means that 100% of the progeny have wild-type wings-males and females Throwing in the condition about “female progeny” was just a red herring, since gender only comes into 12 as developed by Biology Discretes Test play with sex-linked inheritance patterns So, choice D is correct If you can’t figure out the results of this cross in your head, just assign letters to the alleles and work out the Punnett square; it’s common practice to assign the dominant allele a capital letter and assign the recessive allele the same lowercase letter Again, choice D is the correct answer 26 The correct answer is choice D DNA replication is semiconservative This means that the parental strands of the double helix unwind and each then acts as a template for synthesis of a complementary strand So, after the DNA has been replicated, two new daughter helices each consist of one parental strand base-paired to one newly synthesized strand Now that we understand the way in which DNA replicates, let’s get back to the question If a cell is grown in a medium containing radioactively labeled thymidine which is one of the nucleotides needed to synthesize DNA after the first round of replication, all of the DNA molecules will have one labeled strand and one unlabeled strand Why? Because the radioactively labeled thymidine will be incorporated into the newly synthesized strands If the cell undergoes ANOTHER round of replication in the same medium, the same sequence of events will occur The previously labeled parent strands, which are now the daughter strands, will base pair with free nucleotides, including more labeled thymidine, so these new daughter helices will have both of their strands labeled The unlabeled original parent strands will base pair with the labeled thymidine, so these new daughter helices will have only one labeled strand Hence, choice D is the correct answer; after the second round of replication, the label will appear in both strands of half the DNA molecules, and the other half will have only one labeled strand 27 The correct answer is choice D Because DNA is double-stranded and because of the rules of complementary base-pairing, the quantity of adenine in a DNA helix MUST EQUAL the quantity of thymine, and the quantity of cytosine MUST EQUAL the quantity of guanine Why? Because in DNA, adenine binds only with thymine, and cytosine binds only with guanine So you can eliminate choices A and B right off the bat because they don’t have either A and T or G and C in equal percentages Choice B can also be eliminated on account of it having the base uracil, U, which does not even appear in DNA So now you have to choose between C and D, which both have equal amounts of A and T, and equal amounts of C and G Well what about choice C? This choice has equal quantities of cytosine and guanine, and equal quantities of adenine and thymine However, if you’ll notice, adding all of the numbers together gives you 200%, which, of course, is impossible Therefore, choice C is incorrect and choice D must be the correct answer Choice D meets both requirements The base percentages are correct, and totaling these percentages gives you 100% So choice D is the right answer 28 Choice C is the correct answer Transcription is the synthesis of a strand of messenger RNA using a strand of DNA as a template The segment of DNA to be transcribed uncoils from its complementary strand, thereby exposing unpaired bases to the actions of RNA Polymerase the enzyme responsible for synthesizing mRNA Just like the DNA Polymerases, RNA Polymerase synthesizes only in the 5’ to 3’ direction What essentially happens is very similar to what happens during DNA synthesis: free RNA nucleotides temporarily bind to the DNA bases according to the rules of complementary base pairing, and the end result is a strand of mRNA with a base sequence complementary to the segment of DNA from which it was transcribed Another thing you need to keep in mind is that nucleic acids have a polarity there is a prime end and a three prime end to every strand of nucleic acid The two strands of DNA are situated antiparallel to each other; that is, if you imagine reading the strands from left to right, one strand is in the 5’ to 3’ orientation, while the other strand is in the 3’ to 5’ orientation The one other really important thing to remember is that RNA contains uracil instead of thymine So you’ll never find thymine in a segment of RNA any type of RNA Therefore, you should have immediately eliminated choice A because it contains thymine Okay, we want to know the base sequence of the strand of mRNA synthesized from a strand of DNA with the sequence ACGTCA in the 5’ to 3’ direction Since adenine pairs with uracil and guanine pairs with cytosine, the mRNA would have the sequence UGACGU in the 5’ to 3’ direction, which is choice C Choice B is wrong because it has the wrong polarity, and choice D is wrong because it has the wrong sequence Again, choice C is the correct answer 29 The correct answer is choice D To answer this question we must use the Hardy-Weinberg equation, p2 + 2pq + q2 = 1, where p equals the gene frequency of the dominant allele, and q equals the gene frequency of the recessive allele Hence, p2 is the frequency of homozygous dominants in the population, 2pq is the frequency of heterozygotes, and q2 is the frequency of homozygous recessives For a trait with only two alleles, p + q must equal 1, since the combined frequencies of the alleles must total 100% In this problem we are told that the frequency of the recessive allele for a particular trait is 0.6; hence, q = 0.6 Since p + q = 1, p = 0.4 You’re asked to determine the frequency of individuals expressing the dominant phenotype not the dominant genotype, so that’s equal to the number of individuals homozygous for the dominant trait, p2, added to the number of heterozygotes, 2pq, since they also express the dominant phenotype p2 = (0.4) x (0.4) = 0.16, and 2pq = x (0.6) x (0.4) = 0.48 So, p2 + 2pq = 0.16 + 0.48, which equals 0.64, and so the answer is choice D Choice A, 16, is the frequency of homozygous dominant individuals; choice B, 36, is the frequency of homozygous recessive individuals; and choice C, 48, is equal to the number of heterozygous individuals KAPLAN 13 MCAT 30 The correct answer is choice B Asexual reproduction is more efficient than sexual reproduction in terms of the number of offspring produced per reproduction, the amount of energy invested in this process, and the amount of time involved in the development of the young, both before and after birth Overall, sexual reproduction is a much more timeconsuming, energy-costly process So, choice A is incorrect However, asexual reproduction must rely heavily on mutation to introduce phenotypic variability in future generations, since asexual reproduction almost exclusively produces genetic clones of the parent Sexual reproduction, on the other hand, involves the process of meiosis, which consists of two rounds of cell division, which carries with it increased likelihood of cross-over events, chromosomal inversions, and nondisjunction events during the two anaphases Therefore, there is a much greater risk of mutation occurring with sexual reproduction, especially at the level of the entire chromosome, sometimes known as chromosomal aberrations or macromutations So, choice C is wrong, too The reason those species that reproduce sexually have a selective advantage is because gene recombination is a given with every fertilization two genetically unique nuclei the sperm nucleus and the egg nucleus fuse to form an equally genetically unique zygote This is the main means of introducing phenotypic variability into a population This variability does not always benefit or hinder the individual The new phenotype may be disadvantageous OR it may be advantageous; so choice D is also wrong If the new phenotype is advantageous, this individual is more likely to survive and pass his genes on to future generations; this is one of the main principles of natural selection Again, choice B is the correct answer 14 as developed by ... OF TEST KAPLAN MCAT ANSWER KEY: A 11 A 12 D 13 A 14 C 15 D C A B B 21 22 23 24 25 C B D A D 10 B C B A C 26 27 28 29 30 D D C D B D B B A C 16 17 18 19 20 as developed by Biology Discretes Test. .. increase lactic acid production will increase GO ON TO THE NEXT PAGE as developed by Biology Discretes Test Down’s syndrome is a human genetic disorder that can be caused by the nondisjunction... ED: 15% GF: 15% DF: 7% GDFE GFDE EFGD FGDE GO ON TO THE NEXT PAGE as developed by Biology Discretes Test In fruit flies, the gene for wing type is located on an autosomal chromosome The allele