Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 591:

  Hemophilia is a genetically inherited disease that causes the synthesis of an abnormal clotting factor. As a result, hemophiliacs bleed excessively from the slightest injury. The figure below is a partial pedigree for the hemophilia trait in Queen Victoria's descendants. The pedigree indicates no history of hemophilia for either parent prior to the F1 generation.

  

  If Beatrice had married a hemophiliac and had a son, what is the probability that the son would have been a hemophiliac?

  A. 0%

  B. 25%

  C. 50%

  D. 100%

  Correct Answer: C

  If you look at the pedigree, you'll see that Beatrice, a member of the F1 generation, was a carrier of the gene for hemophilia, which means that she had one copy of it on one of her X chromosomes. In reality, Beatrice married a normal male, whose name you're not given; but for the purpose of this question, you're asked to determine the probability that any of her sons would have been hemophiliacs if she had in fact married a hemophiliac. So this is basically a cross between a carrier and a hemophiliac. Therefore, Beatrice's genotype is XhX, and her theoretical husband's genotype is XhY. So, in a cross between these two people, 50% of all their children are expected to be hemophiliacs. But you need to look at the disease in terms of gender; 50% of the daughters are expected to be hemophiliacs; the other 50% will be normal. Likewise, 50% of the sons are expected to be hemophiliacs; the other 50% to be normal.

• Question 592:

  Hemophilia is a genetically inherited disease that causes the synthesis of an abnormal clotting factor. As a result, hemophiliacs bleed excessively from the slightest injury. The figure below is a partial pedigree for the hemophilia trait in Queen Victoria's descendants. The pedigree indicates no history of hemophilia for either parent prior to the F1 generation.

  

  According to Figure 1, which of the following assumptions about the P1 generation must be true?

  A. Albert did not have the gene for hemophilia.

  B. Queen Victoria had two X chromosomes, each with the gene for hemophilia.

  C. Neither Albert nor Queen Victorian had the gene for hemophilia.

  D. Albert was a carrier of the hemophilia gene.

  Correct Answer: A

  This question asks you to conclude something about the P1, or parental, generation. The easiest way to do this is to go through each statement and analyze it. Choice A states that Albert did not have the gene for hemophilia. If we look at the pedigree, we see that this in fact is true; he doesn't have the gene -- as indicated by his unshaded square. If he did have the gene then he would have been a hemophiliac, since the gene is X-linked. So choice A is the correct answer. But let's just take a look at the other choices to be sure. Choice B states that Queen Victoria had two X chromosomes, each with the hemophilia gene. Now it's true that Queen Victoria had two X chromosomes -- all normal females do. However, as indicated by the half shading on her circle in the pedigree, she was a carrier of the gene for hemophilia; that is, she only had one copy of the gene. So choice B is wrong. Choice C states that neither Albert nor Victoria had the gene for hemophilia. Well, Victoria was a carrier, so choice C is wrong too. Choice D states that Albert was a carrier. First of all, men cannot be carriers of an X-linked trait. If they have a copy of the gene in question, then they express the trait. Males cannot have two copies of a X-linked trait because they have only one X chromosome. So, choice D is also wrong.

• Question 593:

  The reaction R -- Br + Br*? R -- Br* + Br -is always accompanied by inversion. If this reaction is carried out on an optically pure sample of a chiral compound, which of the following statements will be true? [Note: Br* represents a radioactive isotope of bromine.]

  A. The rate of Br* incorporation is half the rate of racemization.

  B. The rate of Br* incorporation is equal to the rate of racemization.

  C. The rate of Br* incorporation is twice the rate of racemization.

  D. The relation between the rate of Br* incorporation and the rate of racemization cannot be determined.

  Correct Answer: A

  This is another chemistry question about SN2 reactions and their effect on optical activity. Here you have to think about exactly what's happening as the stated reaction proceeds. Each time an alkyl halide molecule reacts, incorporating a radioactive bromine atom, it will be optically inverted; that is, it will go from R to S or vice versa. Each molecule thus inverted "cancels out" one of the molecules that still has the old configuration; thus each inversion event leads to the racemization of two molecules of alkyl halide. So the rate of radioactive bromine incorporation is equal to half of the rate of racemization, and so the correct answer is A.

• Question 594:

  Which of the following cell types does NOT contain the diploid number of chromosomes?

  A. Spermatogonium

  B. Spermatid

  C. Zygote

  D. Primary oocyte

  Correct Answer: B

  In males, diploid cells called spermatogonia undergo mitosis to produce diploid cells called primary spermatocytes. The primary spermatocytes undergo the first round of meiosis to yield secondary spermatocytes, which are haploid. The secondary spermatocytes undergo the second round of meiosis, resulting in four haploid cells called spermatids. The spermatids then mature into sperm -- the male gametes. So, choice A is incorrect because a spermatogonium is a diploid cell. And we've already found the correct answer -- choice B, spermatid, is NOT a diploid cell, it's haploid. Let's look at what happens in females anyway. In females, a diploid cell called a primary oocyte undergoes the first meiotic division to yield two haploid cells -- a polar body and a secondary oocyte. The secondary oocyte undergoes the second meiotic division to produce two more haploid cells -- a mature oocyte, or ovum, and another polar body. So, choice D, primary oocyte, is also incorrect because these are diploid cells. During fertilization, an ovum and a sperm fuse; two haploid cells fuse to form a single diploid cell called a zygote. Thus, choice C is also incorrect.

• Question 595:

  Which of the following structures plays a role in both the male excretory and male reproductive systems, but in the female excretory system only?

  A. Epididymis

  B. Prostate

  C. Urethra

  D. Ureter

  Correct Answer: C

  This question tests your knowledge of the male and female reproductive and excretory systems, which, as you should know, are NOT identical in structure. The epididymis, choice A, is a group of coiled tubes sitting on top of the seminiferous tubules in the male reproductive tract. Sperm are produced in the seminiferous tubules and mature and acquire motility in the epididymis. Sperm are stored in the epididymis until ejaculation. Hence, the epididymis functions in the male reproductive system only; so choice A is wrong. Choice B, prostate, is one of the glands associated with the male reproductive tract; the prostate gland secretes and alkaline milky fluid that protects the sperm from the acidic conditions in the female reproductive tract. So, choice B also functions only in the male reproductive system, and is therefore incorrect. The urethra, choice C, is a structure found in both men and women. During ejaculation, sperm travels from the epididymis, through the vas deferens, and through the urethra, which opens to the outside from the tip of the penis. The urethra is also directly connected to the bladder. Hence, in males, the urethra functions in both the reproductive and excretory systems. In females, however, the reproductive and excretory systems do NOT share a common pathway. Sperm enter the vagina and travel up through the cervix, uterus, and fallopian tubes, and urine leaves the body through the urethra; the vagina and the urethra never meet -- they are SEPARATE openings. So, choice C is the right answer. Choice D, ureter, is the duct connecting the kidney to the bladder. Urine is formed in the kidneys, it travels down to the bladder by way of the ureters, and is stored there until it is excreted through the urethra. This process is the same in both sexes.

• Question 596:

  Which of the following compounds share the same absolute configuration?

  

  A. I and III

  B. II and IV

  C. I and II

  D. II, III, and IV

  Correct Answer: B

  This question asks you to compare the absolute configurations of four compounds which are in the form of Fischer projections. There are certain rules which you need to be familiar with in order to decipher the absolute configuration. First, the substituent with the lowest priority must be positioned vertically, that is, either up or down. Remember that in 3-D terms, vertical lines represent bonds going into the page. Second, to move your diagram around on the page you have to remember that interchanging any two pairs of substituents will give you the same compound -- that is, it will preserve the absolute configuration. Interchanging just ONE pair of substituents will always REVERSE the absolute configuration. Also, you can hold one group steady while rotating the other three clockwise or counterclockwise. The third rule is that once you have the lowest-priority group positioned vertically, you should determine the order of priority among the other three substituents. If that order increases clockwise, that means the chiral carbon has an R configuration, while if it increases counterclockwise, the chiral carbon has an S configuration. If you look at compound I, you can see that the order of increasing priority is hydrogen, methyl, carboxyl, and hydroxyl. In order to get the hydrogen into a vertical position, you can hold the methyl group steady and rotate the others by 90? The order of increasing priority is clockwise and so the configuration is R, for compound II, the hydrogen is already placed vertically, so you don't have to worry about moving the molecule around the page. The order of increasing priority is hydrogen, methyl, carboxyl and amine, so when you draw arrows between the last three, the direction is counterclockwise and so the configuration is S. Therefore, you can discard choice C since these two molecules have opposite configurations. Compound III may catch you off guard, because if you look at this molecule closely you can see that it is achiral. Although they are written differently, there are two ethyl substituents in this molecule, therefore this molecule is neither R nor S, and so choices A and D can be eliminated. This leaves choice B, but let's just look at compound IV anyway. Again, the hydrogen is positioned vertically and the order of priority is the same as in compound I. However, if you connect the arrows this time, you can see that the direction is counterclockwise and so the configuration is S. This makes compounds II and IV the same, and so choice B is correct.

• Question 597:

  An increase in heart rate, blood pressure, and blood glucose concentration are all associated with stimulation of the:

  A. parasympathetic nervous system.

  B. sympathetic nervous system.

  C. somatic nervous system.

  D. digestive system.

  Correct Answer: B

  The nervous system is divided into the peripheral nervous system and the central nervous system. The central nervous system consists of the brain and the spinal cord. The peripheral nervous system is divided into the sensory division and the motor division. The sensory division consists of those receptors and neurons that transmit signals to the central nervous system. The motor division transmits signals from the central nervous system to effectors, and is divided into the somatic nervous system and the autonomic nervous system. The somatic system, choice C, innervates skeletal muscle, and its nervous pathways are typically under voluntary control. The autonomic nervous system regulates the internal environment by way of involuntary nervous pathways. The autonomic nervous system innervates smooth muscle in blood vessels and the digestive tract, and innervates the heart, the respiratory system, the endocrine system, the excretory system, and the reproductive system. The autonomic system is further divided into the sympathetic division, choice B, and the parasympathetic division, choice A. The sympathetic division innervates those pathways that prepare the body for immediate action; this is known as the "fight-or-flight" response. Heart rate and blood pressure increase, blood vessels in the skin vasoconstrict and those in the heart vasodilate, pathways innervating the digestive tract are inhibited, and epinephrine, or adrenaline, is secreted by the adrenal medulla, thereby increasing the conversion of glycogen into glucose. Based on this discussion, it's obvious that the correct answer is choice B; activation of the sympathetic nervous system is associated with an increased heart rate, blood pressure, and blood glucose concentration. The parasympathetic system innervates nervous pathways that return the body to homeostatic conditions following exertion. Heart rate, blood pressure, and blood glucose concentration all decrease, blood vessels in the skin vasodilate and those in the heart vasoconstrict, and the digestive process is no longer inhibited.

• Question 598:

  Electromagnetic radiation from space constantly bombards the earth. Most wavelengths are absorbed by the atmosphere; however, there are two "windows" of nonabsorption through which significant amounts of radiation reach the ground. The first transmits ultraviolet and visible light, as well as infrared light or heat; the second transmits radio waves. As a result, terrestrial organisms have evolved a number of pigments that interact with light in various ways: some capture light

  energy, some provide protection from light- induced damage, and some serve camouflage or signaling purposes.

  Among these compounds are many conjugated polyenes, which play important roles as photoreceptors. For every chemical compound, there are certain wavelengths of light whose quanta possess exactly the correct amount of energy to raise electrons from their ground state to higher-energy orbitals. For most organic compounds, these wavelengths are in the UV range. However, conjugated double bond systems stabilize the electrons, so that they can be excited by lower-frequency photons with wavelengths in the visible spectrum. Such a pigment, known as a chromophore, will then transmit the "subtraction color," a color complementary to the one absorbed. For instance, carotene, a hydrocarbon compound with eleven conjugated double bonds, absorbs blue light and transmits orange. The wavelength that is absorbed generally increases with the number of conjugated bonds; rings and side-chains also affect wavelength.

  Wavelength Color Subtraction Color 480 nm blue orange 580 nm yellow violet 680 nm red green

  Among the many biological molecules that are affected by light is DNA, the genetic material of living organisms. DNA absorbs ultraviolet light, and may be damaged by UVC (< 280 nm) and UVB (280-315 nm). UVA (315-400 nm) and visible light can actually repair light-induced damage to DNA by a process called photorepair. For this reason, UVA, which also stimulates tanning, was once considered beneficial. However, there is now increasing evidence that UVA can damage skin.

  The color-producing quality of conjugated polyenes is attributable to:

  A. antibonding orbitals.

  B. resonance.

  C. polarity.

  D. optical activity.

  Correct Answer: B

  As we mentioned earlier, resonance is responsible for reducing the frequency of light needed to excite an electron -- that is, raise it to a higher-energy, antibonding orbital. Choice A is wrong because almost any molecule can absorb light and have an electron raised to an antibonding orbital. This process does not normally produce color because the frequency of light involved is usually outside the visible range. Resonance reduces the amount of energy needed, bringing the frequency into the visible spectrum. Optical activity, choice D, is the property of being able to rotate plane-polarized light; this doesn't have anything to do with color. Polarity, choice C, also has nothing to do with color. Therefore, both of these are wrong. The correct answer is B.

• Question 599:

  Electromagnetic radiation from space constantly bombards the earth. Most wavelengths are absorbed by the atmosphere; however, there are two "windows" of nonabsorption through which significant amounts of radiation reach the ground. The first transmits ultraviolet and visible light, as well as infrared light or heat; the second transmits radio waves. As a result, terrestrial organisms have evolved a number of pigments that interact with light in various ways: some capture light energy, some provide protection from light- induced damage, and some serve camouflage or signaling purposes.

  Among these compounds are many conjugated polyenes, which play important roles as photoreceptors. For every chemical compound, there are certain wavelengths of light whose quanta possess exactly the correct amount of energy to raise electrons from their ground state to higher-energy orbitals. For most organic compounds, these wavelengths are in the UV range. However, conjugated double bond systems stabilize the electrons, so that they can be excited by lower-frequency photons with wavelengths in the visible spectrum. Such a pigment, known as a chromophore, will then transmit the "subtraction color," a color complementary to the one absorbed. For instance, carotene, a hydrocarbon compound with eleven conjugated double bonds, absorbs blue light and transmits orange. The wavelength that is absorbed generally increases with the number of conjugated bonds; rings and side-chains also affect wavelength.

  Wavelength Color Subtraction Color 480 nm blue orange 580 nm yellow violet 680 nm red green

  Among the many biological molecules that are affected by light is DNA, the genetic material of living organisms. DNA absorbs ultraviolet light, and may be damaged by UVC (< 280 nm) and UVB (280-315 nm). UVA (315-400 nm) and visible light can actually repair light-induced damage to DNA by a process called photorepair. For this reason, UVA, which also stimulates tanning, was once considered beneficial. However, there is now increasing evidence that UVA can damage skin.

  Which of the following compounds would be most likely to produce color?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: C

  This molecule, which is called dopamine quinone, has four conjugated bonds and produces a brownish color. Choice B has a larger number of double bonds, but they are not conjugated, so its absorbance wavelength is below the visible range. Choice A has too few double bonds to produce color, and although choice D has six double bonds, they are conjugated in two sets of three, not one set of six, and it's also less likely to be colored.

• Question 600:

  Electromagnetic radiation from space constantly bombards the earth. Most wavelengths are absorbed by the atmosphere; however, there are two "windows" of nonabsorption through which significant amounts of radiation reach the ground. The first transmits ultraviolet and visible light, as well as infrared light or heat; the second transmits radio waves. As a result, terrestrial organisms have evolved a number of pigments that interact with light in various ways: some capture light energy, some provide protection from light- induced damage, and some serve camouflage or signaling purposes.

  Among these compounds are many conjugated polyenes, which play important roles as photoreceptors. For every chemical compound, there are certain wavelengths of light whose quanta possess exactly the correct amount of energy to raise electrons from their ground state to higher-energy orbitals. For most organic compounds, these wavelengths are in the UV range. However, conjugated double bond systems stabilize the electrons, so that they can be excited by lower-frequency photons with wavelengths in the visible spectrum. Such a pigment, known as a chromophore, will then transmit the "subtraction color," a color complementary to the one absorbed. For instance, carotene, a hydrocarbon compound with eleven conjugated double bonds, absorbs blue light and transmits orange. The wavelength that is absorbed generally increases with the number of conjugated bonds; rings and side-chains also affect wavelength.

  Wavelength Color Subtraction Color 480 nm blue orange 580 nm yellow violet 680 nm red green

  Among the many biological molecules that are affected by light is DNA, the genetic material of living organisms. DNA absorbs ultraviolet light, and may be damaged by UVC (< 280 nm) and UVB (280-315 nm). UVA (315-400 nm) and visible light can actually repair light-induced damage to DNA by a process called photorepair. For this reason, UVA, which also stimulates tanning, was once considered beneficial. However, there is now increasing evidence that UVA can damage skin.

  The four compounds represented by the electronic spectra below were evaluated as potential sunscreens. What is the correct sequence of sunscreen strength, from strongest to weakest, among these four?

  

  A. I, II, III, IV

  B. IV, III, II, I

  C. III, II, I, IV

  D. IV, I, II, III

  Correct Answer: D

  Substance IV would make the best sunscreen because it absorbs over a broad spectrum, including all three types of ultraviolet light. By the way, the absorption spectrum of melanin, the pigment which protects human skin from sunlight, is similar to this. The next best choice is I, which absorbs UVB and UVC, the most dangerous wavelengths of ultraviolet light. Choice II absorbs UVA, so it would prevent tanning but would let through the dangerous UVB and UVC, and therefore allow burning. Choice III would be of little value, since it absorbs almost exclusively in the visible range (greater than 400 nm).