Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 561:

  Alleles are created when a single gene undergoes several distinct mutations. These alleles may have different dominance relationships with one another; for example, there are three alleles coding for the human blood groups, the IA, IB, and i alleles. Both the IA and IB alleles are dominant to the i allele, but IA and IB are codominant to each other.

  A multiple-allele system has recently been discovered in the determination of hair coloring in a species of wild rat. The rats are found to have one of three colors: brown, red, or white. Let B = the gene for brown hair; b = the gene for red hair; and w = the gene for white hair. The results from nine experimental crosses are shown below. The males and females in Crosses 1, 2, and 3 are all homozygous for hair color.

  

  Based on the experimental results, what is the genotype of the male in Cross 6?

  A. bw

  B. bb

  C. bw or bb

  D. Bb or bw

  Correct Answer: C

  Cross 6 is between a red male and a red female, and their offspring are 100% red. What we need to do is look at the offspring and work backwards to determine the genotypes of their parents. As we've just determined, there are two possible genotypes that correspond to a red phenotype -- little b, little b and little b, w, since the b allele is dominant to the w allele. With these two genotypes there are three possible types of crosses (you might want to write these down); little b, little b ?little b, little b; little b, little b ?little b, w; and little b, w ?little b, w. Note that we're not taking into account the gender of the parents. By this I mean we're not bothering with the fact that the little b, little b ?little b, w cross can occur in two different ways -- the male can be little b, little b and the female can be little b, w, and vice versa. All of the offspring of Cross 6 are red, therefore, we can eliminate the possibility of a little b, w ?little b, w cross since 25% of their offspring would be white. So now we're left with the other two crosses, both of which produce 10% red offspring. If the male is little b, little b, then the female can be either little b, little b or little b, w; and if the male is little b, w then the female must be little b, little b. Hence the male can either be little b, little b or little b, w and the correct answer is choice C.

• Question 562:

  A student was given a sample of an unknown liquid and asked to determine as much as possible about its structure. He was told that the compound contained only carbon, hydrogen, and oxygen, and had only one type of functional group. The student found its boiling point to be 206癈. Using mass spectroscopy, he determined its molecular weight to be 138 g/mol. Finally, he took the infrared spectrum of the compound, which is shown below.

  

  From this spectrum, the student quickly reached a conclusion about the functional group. He then turned his attention to the fingerprint region of the compound, which generally has a complicated pattern of peaks that are determined by the structure of the hydrocarbon portion of a molecule. The student decided that the large peak at 750 cm-1 must indicate that this was a disubstituted aromatic compound.

  Assuming that all of the student's deductions were correct, which of the following could be the structure of the unknown compound?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: B

  The student theorized that the peak at 750 cm-1 indicates a disubstituted aromatic compound, and since we're told to assume that all of his deductions were correct, this allows us to eliminate choice D, which has only one substituent on the ring. The passage also says that the compound contains only one type of functional group and so choice A, which contains two different functional groups, is also clearly wrong. By the way, neither choice A nor choice D corresponds to the correct molecular weight anyway You are now left with choices B and C, and to decipher between these, you have to look back at the spectrum. If this were an alcohol, as in choice C, the spectrum would contain a broad peak at about 3350 cm-1 to 3250 cm-1, which is characteristic of a hydroxyl group. This is much like the peak we mentioned for a phenol and for the O-H group of a carboxyl group; like those, it's broadened by the fact that it can form hydrogen bonds. This is a very characteristic feature of ANY type of hydroxyl-bearing group, and since it's not here, choice C must be incorrect.

• Question 563:

  A student was given a sample of an unknown liquid and asked to determine as much as possible about its structure. He was told that the compound contained only carbon, hydrogen, and oxygen, and had only one type of functional group. The student found its boiling point to be 206. Using mass spectroscopy, he determined its molecular weight to be 138 g/mol. Finally, he took the infrared spectrum of the compound, which is shown below.

  

  From this spectrum, the student quickly reached a conclusion about the functional group. He then turned his attention to the fingerprint region of the compound, which generally has a complicated pattern of peaks that are determined by the structure of the hydrocarbon portion of a molecule. The student decided that the large peak at 750 cm-1 must indicate that this was a disubstituted aromatic compound.

  The overlapping set of peaks near 3000 cm-1 includes one peak at 2850 cm-1. What type of functional group could this indicate?

  A. Methyl

  B. Phenol

  C. Carboxyl

  D. Aldehyde carbonyl

  Correct Answer: A

  The peak at 2850 cm-1 is characteristic of the C-H stretch of an alkyl group; in the case, it is attributed to the methyl group of the alkoxide. That's not a terribly exciting functional group, so you might not have known this right off the top of your head, but you could get it by a process of elimination. Choice B, a phenol group, would produce a broad peak somewhere between 3600 and 3200 cm-1, which represents the oxygen-hydrogen bond that is characteristic of all hydroxyl groups and is pretty unmistakable. Since this peak is not present in this spectrum, choice B must be incorrect. Choice C, a carboxyl group, would also produce a broad peak due to its hydroxyl group; this can be anywhere from 300 to 2400 cm-1, and tends to be even broader than a regular hydroxyl peak. The other peaks near the methoxy group peak are much too sharp to belong to a carbonyl group; they actually represent the carbon-hydrogen bonds of the aromatic ring. Whenever you look at the IR spectrum of an organic compound, remember that there will always be some kind of peak or peaks between 2800 and 3300 cm-1 that's simply due to the carbon backbone -- alkane, alkene, alkyne, aromatic, or mixtures of all of these. Anyway, getting back to the question, a carboxyl group would also produce another peak associated with the carbon-oxygen double bond, around 1700 cm-1, and that's also lacking in this spectrum, so choice C is definitely wrong. Finally, choice D, an aldehyde group, would have the same peak at about 1700; it would also produce two characteristic peaks in the range of 2720 and 2820 cm-1, called Fermi doublets. Neither of these are present, so D is also wrong.

• Question 564:

  Which of the following products may be formed in the reaction below?

  

  A. Option A

  B. Option B

  C. Option C

  D. option D

  Correct Answer: D

  This question asks you to predict the reaction products from the acid hydrolysis of an ester. When esters are hydrolyzed, they yield carboxylic acids and alcohols. From the hydrolysis of this ester, you would expect 2-butenoic acid and isobutanol. Choice D is isobutanol and is the correct answer. The other choices do not reflect possible products from this reaction. Choice A is an unsaturated aldehyde and choice C is an alkane; neither type of compound can be obtained by hydrolyzing an ester. Initially, choice B may look like the carboxylic acid that is obtained from hydrolysis of the ester, but in fact, the double bond is absent, so this answer choice is also incorrect.

• Question 565:

  Growth hormone decreases the sensitivity of cellular receptors to insulin. Therefore, a patient with acromegaly, which is caused by the oversecretion of growth hormone, would be expected to have:

  A. a low blood glucose concentration.

  B. a high blood glucose concentration.

  C. a decreased urine volume.

  D. a decreased cardiac output.

  Correct Answer: B

  According to the question stem, acromegaly is a condition that results from an oversecretion of growth hormone. In addition, you're told that growth hormone decreases the sensitivity of insulin receptors. Therefore, to answer this question you've got to determine the physiological effects of an insensitivity to insulin. You should know that insulin is secreted by the pancreas in response to high blood glucose concentration, and that it lowers blood glucose by stimulating the conversion of glucose into its storage form, glycogen. This means that the cell surface receptors of acromegaly patients do not bind insulin as much as normal cells. Therefore, the effects of insulin will be diminished in an acromegaly patient. What does this mean? If insulin cannot exert its effects on the cells, then excess glucose will not be converted into glycogen. Hence, the patient will have excess glucose -- in other words, a high blood glucose concentration. Therefore, choice B must be the correct answer and choice A must be the wrong answer. Choice C is wrong because a high blood glucose concentration leads to the excretion of glucose, along with the loss of water in the urine, meaning that patients with acromegaly will have an increased, NOT a decreased, urine volume. Choice D is wrong because cardiac output is defined as the volume of blood pumped by the heart per unit time, which is not even directly related to the sensitivity of insulin receptors.

• Question 566:

  A biochemist grows two cultures of yeast -- one aerobically and the other anaerobically -- and measures the amount of ATP produced by each culture. He finds that the aerobically-grown yeast produce about 18 times as much ATP as the anaerobically-grown yeast. These observations are consistent with the fact that in the aerobically grown yeast:

  A. oxygen is converted into ATP.

  B. oxygen is necessary to convert glucose into pyruvate.

  C. oxygen is the final electron acceptor of the respiratory chain.

  D. oxygen is necessary for the reduction of pyruvate into lactate.

  Correct Answer: C

  Aerobic respiration cannot occur without oxygen. Why? Because oxygen is necessary for the final step of aerobic respiration, which is the electron transport chain. Here's a brief summary of glycolysis, the tricarboxylic acid cycle, also known as the Krebs cycle of the citric acid cycle, and the electron transport chain -- which are the three stages of aerobic respiration. Glycolysis is a series of reactions that lead to the oxidative breakdown of glucose into pyruvate. These reactions occur in the cytoplasm and result in the production of two NADH and a net gain of two ATP. Oxygen is not required for glycolysis, therefore, choice B is wrong. In the absence of oxygen -- that is, under anaerobic conditions -- the pyruvate is reduced to lactate; it undergoes this fermentation step so that NAD+ can be generated. Hence, choice D is wrong. In the presence of oxygen -- that is, under aerobic conditions -- the pyruvate is oxidized to release the considerable energy still stored in its chemical bonds. The pyruvate is transported from the cytoplasm into the mitochondrial matrix. The oxidative decarboxylation of pyruvate into Acetyl CoA is catalyzed by pyruvate dehydrogenase complex; one NADH is generated during the formation of acetyl CoA. Next, the acetyl group from the acetyl CoA combines with oxaloacetate, forming citrate. Through a complicated series of reactions, the citrate is completely oxidized, two molecules of carbon dioxide are released, and NADH, FADH2, and ATP are generated. Next, all of the molecules of NADH and FADH2 generated during glycolysis, pyruvate decarboxylation, and the TCA cycle, transfer their high potential electrons to a series of carrier molecules located in the inner mitochondrial membrane. A series of redox reactions is coupled with the phosphorylation of ADP; this is known as oxidative phosphorylation. As the electrons are transferred from carrier to carrier, free energy is released, which is then used to produce ATP. So, choice A is also wrong. The final carrier of the electron transport chain, cytochrome a3, passes its electrons to the final electron acceptor, molecular oxygen. Oxygen picks up a pair of hydrogen ions, forming water. The final ATP tally for aerobic respiration is net gain of 36 ATP. This is eighteen times as much ATP as produced during anaerobic respiration. So, choices A, B, and D are all wrong, and choice C is true and consistent with the observations made by the biochemist in the question.

• Question 567:

  A certain chemical is found to inhibit the synthesis of all steroids. The synthesis of which of the following hormones would NOT be affected when a dose of this chemical is administered to a laboratory rat?

  A. Cortisol

  B. Aldosterone

  C. Epinephrine

  D. Testosterone

  Correct Answer: C

  We're told that a certain chemical inhibits the synthesis of steroids and we're asked to decide which one of the hormones listed in the answer choices would not be affected if a dose of this chemical was administered to a laboratory rat. In other words, we need to know which of these hormones is NOT a steroid hormone. Steroids are a type of lipid with a carbon skeleton of four fused rings. Different types of steroids differ in the functional group that is attached to the carbon skeleton. Cholesterol is a very important steroid, and is in fact the precursor steroid molecule from which most other steroids, including steroid hormones, are synthesized. For example, the sex hormones -- testosterone, estradiol, and progesterone -- which are synthesized in the gonads, are steroid hormones modified from cholesterol. This means that we can rule out choice D. The adrenal cortex is another gland that secretes a family of steroid hormones called the corticosteroids. The two main types of corticosteroids are the mineralocorticoids, such as aldosterone, and the glucocorticoids, such as cortisol. This means that we can also rule out choices A and B, which leaves us with choice C, epinephrine. The adrenal medulla secretes the hormone epinephrine in response to any type of stress -- good or bad. Epinephrine, along with norepinephrine, are compounds called catecholamines, and are synthesized from the amino acid tyrosine by chromaffin cells in the adrenal medulla. Hence, epinephrine synthesis would NOT be affected by this chemical, which makes choice C the correct answer.

• Question 568:

  Four major blood types exist in the human ABO blood system: types A, B, AB, and O; and there are three alleles that code for them. The A and B alleles are codominant, and the O allele is recessive. Blood types are derived from the presence of specific polysaccharide antigens that lie on the outer surface of the red blood cell membrane. The A allele codes for the production of the A antigen; the B allele codes for the production of the B antigen; the O allele does not code for any antigen. While there are many other antigens found on red blood cell membranes, the second most important antigen is the Rh antigen. Rh is an autosomally dominant trait coded for by 2 alleles. If this antigen is present, an individual is Rh+; if it is absent, an individual is Rh-. For example, a person with type AB blood with the Rh antigen is said to be AB+.

  These antigens become most important when an individual comes into contact with foreign blood. Because of the presence of naturally occurring substances that closely mimic the A and B antigens, individuals who do not have these antigens on their red blood cells will form antibodies against them. This is inconsequential until situations such as blood transfusion, organ transplant, or pregnancy occur.

  Erythroblastosis fetalis is a condition in which the red blood cells of an Rh+ fetus are attached by antibodies produced by its Rh- mother. Unlike ABO incompatibility, in which there are naturally occurring antibodies to foreign antigens, the Rh system requires prior sensitization to the Rh antigen before antibodies are produced. This sensitization usually occurs during the delivery of an Rh+ baby. So while the first baby will not be harmed, any further Rh+ fetuses are at risk. The Coombs tests provide a method for determining whether a mother has mounted an immune response again her baby's blood. The tests are based on whether or not agglutination occurs when Coombs reagent is added to a sample. Coombs reagent contains antibodies against the anti-Rh antibodies produced by the mother. The indirect Coombs test takes the mother's serum, which contains her antibodies but no red blood cells, and mixes it with Rh+ red blood cells. Coombs reagent is then added. If agglutination occurs, the test is positive, and the mother must be producing anti-Rh antibodies. The direct Coombs test mixes the baby's red blood cells with Coombs reagent. If agglutination occurs, the test is positive, and the baby's red blood cells must have been attacked by its mother's anti-Rh antibodies.

  A medical student suggested giving Rh- mothers of Rh+ fetuses a specific exogenous substance prior to delivery to prevent an immune response. Which of the following substances would likely be the safest and most effective?

  A. Rh antigen

  B. An immunosuppressive drug

  C. Anti-Rh antibody

  D. Iron pills

  Correct Answer: C

  It is, in fact, true that a medical student was responsible for the treatment currently used to prevent Rh incompatibility reactions. The exogenous substance is called RhoGAM, and is derived from Rh immunoglobulin. RhoGAM is anti-Rh antibody. The theory behind it is that if given prior to delivery -- the time at which exposure occurs -- rhoGAM will attack and coat the fetal Rh positive red blood cells. By coating the fetal cells, the Rh antigen is no longer accessible to the mother's system; it is effectively "hidden." The mother therefore does not produce anti-Rh antibody of her own, the baby's coated red blood cells are removed and destroyed by the mother's immune system, and sensitization has been prevented. So choice C is correct. In practice, however, rhoGAM is usually administered during the postpartum period, specifically within the first 72 hours following birth. Choice A, Rh antigen, would only enhance the mother's immune response if given prior to delivery. Exposing her to the antigen would lead to the production of anti-Rh antibody, which would increase the risk of erythroblastosis fetalis for future children. The whole idea of the treatment is to prevent the mother from producing her own anti-Rh antibodies, and thereby prevent sensitization, NOT enhance it. So, choice A is wrong. Choice B, an immunosuppressive drug, might be an effective way to suppress the immune response, but it would certainly not be safe. An immunosuppressive drug has a broad action, rather than a specific one, and would affect many aspects of the mother's immune system, not just her ability to produce anti-Rh antibodies in response to the Rh antigen on her baby's red blood cells. In fact, immunosuppressive drugs would make the mother more susceptible to infection and would have adverse side effects. So choice B is also incorrect. Choice D, iron pills, are often given to pregnant women as an iron supplement to combat the anemia associated with pregnancy. The anemia is due to the excessive circulatory demands exacted by the fetus and the placenta. Iron pills, however, in no way affect the immune system and would therefore not be effective in preventing an immune response of any sort; thus choice D is incorrect.

• Question 569:

  Four major blood types exist in the human ABO blood system: types A, B, AB, and O; and there are three alleles that code for them. The A and B alleles are codominant, and the O allele is recessive. Blood types are derived from the presence of specific polysaccharide antigens that lie on the outer surface of the red blood cell membrane. The A allele codes for the production of the A antigen; the B allele codes for the production of the B antigen; the O allele does not code for any antigen. While there are many other antigens found on red blood cell membranes, the second most important antigen is the Rh antigen. Rh is an autosomally dominant trait coded for by 2 alleles. If this antigen is present, an individual is Rh+; if it is absent, an individual is Rh-. For example, a person with type AB blood with the Rh antigen is said to be AB+. These antigens become most important when an individual comes into contact with foreign blood. Because of the presence of naturally occurring substances that closely mimic the A and B antigens, individuals who do not have these antigens on their red blood cells will form antibodies against them. This is inconsequential until situations such as blood transfusion, organ transplant, or pregnancy occur.

  Erythroblastosis fetalis is a condition in which the red blood cells of an Rh+ fetus are attached by antibodies produced by its Rh- mother. Unlike ABO incompatibility, in which there are naturally occurring antibodies to foreign antigens, the Rh system requires prior sensitization to the Rh antigen before antibodies are produced. This sensitization usually occurs during the delivery of an Rh+ baby. So while the first baby will not be harmed, any further Rh+ fetuses are at risk.

  The Coombs tests provide a method for determining whether a mother has mounted an immune response again her baby's blood. The tests are based on whether or not agglutination occurs when Coombs reagent is added to a sample. Coombs reagent contains antibodies against the anti-Rh antibodies produced by the mother. The indirect Coombs test takes the mother's serum, which contains her antibodies but no red blood cells, and mixes it with Rh+ red blood cells. Coombs reagent is then added. If agglutination occurs, the test is positive, and the mother must be producing anti-Rh antibodies. The direct Coombs test mixes the baby's red blood cells with Coombs reagent. If agglutination occurs, the test is positive, and the baby's red blood cells must have been attacked by its mother's anti-Rh antibodies.

  A woman who has never been pregnant has type B- blood. Which of the following antibodies would you expect to find in her serum?

  A. Anti-B antibody

  B. Anti-A antibody

  C. Anti-Rh antibody

  D. Both anti-A and anti-Rh antibodies

  Correct Answer: B

  To answer this question, you need to have an understanding of the ABO blood groups and what antigens its alleles code for. In addition, you need to have recalled from the passage that Rhimmunity requires prior sensitization, while ABO immunity does not, because of other naturally occurring antigens. Since the woman in the question has never been pregnant, it is highly unlikely that she has ever been exposed to the Rh antigen, which means that she would not pro- duce antibodies against it. So, choices C and D are wrong. Since her blood type is B, she would have anti-B antibodies, but would b expected to have anti-A antibodies because of a naturally occurring antigen that resembles the A antigen.

• Question 570:

  Four major blood types exist in the human ABO blood system: types A, B, AB, and O; and there are three alleles that code for them. The A and B alleles are codominant, and the O allele is recessive. Blood types are derived from the presence of specific polysaccharide antigens that lie on the outer surface of the red blood cell membrane. The A allele codes for the production of the A antigen; the B allele codes for the production of the B antigen; the O allele does not code for any antigen. While there are many other antigens found on red blood cell membranes, the second most important antigen is the Rh antigen. Rh is an autosomally dominant trait coded for by 2 alleles. If this antigen is present, an individual is Rh+; if it is absent, an individual is Rh-. For example, a person with type AB blood with the Rh antigen is said to be AB+.

  These antigens become most important when an individual comes into contact with foreign blood. Because of the presence of naturally occurring substances that closely mimic the A and B antigens, individuals who do not have these antigens on their red blood cells will form antibodies against them. This is inconsequential until situations such as blood transfusion, organ transplant, or pregnancy occur.

  Erythroblastosis fetalis is a condition in which the red blood cells of an Rh+ fetus are attached by antibodies produced by its Rh- mother. Unlike ABO incompatibility, in which there are naturally occurring antibodies to foreign antigens, the Rh system requires prior sensitization to the Rh antigen before antibodies are produced. This sensitization usually occurs during the delivery of an Rh+ baby. So while the first baby will not be harmed, any further Rh+ fetuses are at risk.

  The Coombs tests provide a method for determining whether a mother has mounted an immune response again her baby's blood. The tests are based on whether or not agglutination occurs when Coombs reagent is added to a sample. Coombs reagent contains antibodies against the anti-Rh antibodies produced by the mother. The indirect Coombs test takes the mother's serum, which contains her antibodies but no red blood cells, and mixes it with Rh+ red blood cells. Coombs reagent is then added. If agglutination occurs, the test is positive, and the mother must be producing anti-Rh antibodies. The direct Coombs test mixes the baby's red blood cells with Coombs reagent. If agglutination occurs, the test is positive, and the baby's red blood cells must have been attacked by its mother's anti-Rh antibodies.

  Based on information in the passage, what does the reaction below represent?

  

  A. Negative direct Coombs test

  B. Positive direct Coombs test

  C. Positive indirect Coombs test

  D. Negative indirect Coombs test

  Correct Answer: B

  As described in the passage, the Coombs tests are the screening procedures used to assess whether an Rh incompatibility reaction can or has occurred. The direct Coombs test will identify whether the baby's red blood cells have in fact been attacked by the mother's anti-Rh antibodies. The indirect Coombs test tests for the presence of anti-Rh antibodies in the mother's serum. The figure in the question shows red blood cells with anti-Rh antibodies already attached to them, being mixed with Coombs reagent. This is the test described in the fourth paragraph of the passage as being the direct Coombs tests. Therefore, choices C and D can be eliminated. The outcome shows that Coombs reagent reacted with the red blood cells; the results of this test are therefore positive; so choice A is wrong and choice B is right. In order for either a direct or indirect Coombs test to be negative, the anti-Rh antibody cannot be present. For this to have been a positive indirect Coombs test, there would have been an additional, previous step showing the mixing of the mother's serum with washed red blood cells.