Medical Tests Medical Tests Certifications MCAT-TEST Questions & Answers

• Question 481:

  Before birth, the rodent brain is sexually undifferentiated. It is only in the first few days following birth, during a period referred to as the critical period, that the rodent brain differentiates along male or female lines. The hormone testosterone plays a critical role in this development. Specifically, sexual differentiation is determined by the presence of estradiol, an estrogen derivative of testosterone, in certain areas of the brain. Testosterone is converted to estradiol in critical brain cells that contain the enzyme aromatase. To study the effects of testosterone on the neonatal rodent brain, the following experiments were conducted: The above research, combined with additional studies, concluded that testosterone has two "organizational" effects on the male rodent brain: Defeminization Moderate levels of testosterone-derived estradiol during the critical period are sufficient for defeminization of the brain. Defeminization of the rodent brain results in loss of estrogen positive feedback on LH and FSH secretion and the ensuing loss of cyclicity, as well as loss of female sex behavior. Masculinization High levels of estradiol due to high levels of testosterone during the critical period results in masculinization of the brain. Masculinization leads to the induction of male sex behavior including antagonism towards other males and the mounting of females.

  

  In Experiment 2, researchers most likely waited 2-3 months before implanting ovaries:

  A. to allow the rat to recover from previous surgery.

  B. to allow testosterone levels to rise to necessary concentrations.

  C. to wait for the critical period to pass.

  D. to promote the defeminization of the rat brain.

  Correct Answer: C

  The experiments were designed to demonstrate the effects of testosterone on sexual differentiation. The ovaries were transplanted to allow the rodent to demonstrate whether or not it had the ability to cycle normally. Thus, it is important that

  transplantation of the ovaries did not influence the development of the rodent. To make sure that the transplantation of ovaries had no effect on the sexual differentiation of the rodent brains, the ovaries had to have been transplanted once the

  sexual differentiation of the rat brain had already occurred. According to the passage, this differentiation occurs during the critical period which begins a few days after birth. The researchers castrated the neonate and waited for the rodent to

  mature past the critical period before implanting the ovaries.

  Choice A is incorrect because it would not take 3 months for the rat to recover from surgery. Also, the rat is castrated at birth indicating that surgery is not an obstacle. Implanting the ovaries would require another surgery.

  Choice B is incorrect because testosterone levels would not rise in the castrated rat (no testes). Choice D is incorrect because defeminization would be promoted by estradiol (or testosterone converted by aromatase).

• Question 482:

  Before birth, the rodent brain is sexually undifferentiated. It is only in the first few days following birth, during a period referred to as the critical period, that the rodent brain differentiates along male or female lines. The hormone testosterone plays a critical role in this development. Specifically, sexual differentiation is determined by the presence of estradiol, an estrogen derivative of testosterone, in certain areas of the brain. Testosterone is converted to estradiol in critical brain cells that contain the enzyme aromatase. To study the effects of testosterone on the neonatal rodent brain, the following experiments were conducted:

  

  The above research, combined with additional studies, concluded that testosterone has two "organizational" effects on the male rodent brain: Defeminization Moderate levels of testosterone-derived estradiol during the critical period are sufficient for defeminization of the brain. Defeminization of the rodent brain results in loss of estrogen positive feedback on LH and FSH secretion and the ensuing loss of cyclicity, as well as loss of female sex behavior. Masculinization High levels of estradiol due to high levels of testosterone during the critical period results in masculinization of the brain. Masculinization leads to the induction of male sex behavior including antagonism towards other males and the mounting of females.

  A researcher proposes that very low doses of estradiol are required for induction of female sex behavior and cyclicity in the rodent brain. Which of the following observations would best support this hypothesis?

  A. Female neonates injected with the anti-estrogen tamoxifen are acyclic and show neither male nor female sex behavior.

  B. Female neonates lacking the aromatase enzyme develop normally.

  C. Female neonates injected with large doses of estradiol are acyclic and show male sex behavior.

  D. Male neonates injected with low dosages of estradiol develop normally.

  Correct Answer: A

  The passage indicates that estradiol is an estrogen. Choice A states that tamoxifen is an antiestrogen. Therefore, injecting tamoxifen blocks all estradiol activity. So, as predicted by the given experiments, masculinization would not occur in female neonates injected with tamoxifen and male sex behavior would not be induced. The lack of female sex behavior and cyclicity, and lack of male sex behavior, indicates that induction of female sex behavior and cyclicity also requires estradiol. Choice B is incorrect because aromatase is only responsible for converting testosterone to estradiol. So, lacking aromatase would have no effect on estradiol present in females that is not testosterone-derived. And, most of the estradiol in females is not derived from testosterone. This experiment, then, provides no information about the role of estradiol in inducing female sex behavior. Choice C is incorrect because large doses of estradiol lead to defeminization and masculinization as expected. This does not support the hypothesis. Choice D is incorrect because males convert testosterone to estradiol in the brain, leading to normal male development. This does not support the hypothesis. Note that while the incorrect choices did not oppose the hypothesis, they did not support the hypothesis either.

• Question 483:

  Before birth, the rodent brain is sexually undifferentiated. It is only in the first few days following birth, during a period referred to as the critical period, that the rodent brain differentiates along male or female lines. The hormone testosterone plays a critical role in this development. Specifically, sexual differentiation is determined by the presence of estradiol, an estrogen derivative of testosterone, in certain areas of the brain. Testosterone is converted to estradiol in critical brain cells that contain the enzyme aromatase. To study the effects of testosterone on the neonatal rodent brain, the following experiments were conducted:

  

  The above research, combined with additional studies, concluded that testosterone has two "organizational" effects on the male rodent brain: Defeminization Moderate levels of testosterone-derived estradiol during the critical period are sufficient for defeminization of the brain. Defeminization of the rodent brain results in loss of estrogen positive feedback on LH and FSH secretion and the ensuing loss of cyclicity, as well as loss of female sex behavior. Masculinization High levels of estradiol due to high levels of testosterone during the critical period results in masculinization of the brain. Masculinization leads to the induction of male sex behavior including antagonism towards other males and the mounting of females.

  A mutant male neonatal rodent with a defective aromatase enzyme is injected with large doses of testosterone during the critical period. Which of the following would occur?

  A. Induction of male sex behavior

  B. Development of female sex behavior

  C. Loss of both cyclicity and female sex behavior

  D. Loss of both cyclicity and female sex behavior and induction of male sex behavior

  Correct Answer: B

  The passage states that aromatase converts testosterone to estradiol. Estradiol that has been derived from testosterone leads to defeminization (loss of cyclicity and female sex behavior) and masculinization (induction of male sex behavior)

  of the rodent brain. If aromatase is defective, no estradiol will be produced even when large doses of testosterone are present. Thus, the brain will develop along female lines. Results of this situation are similar to those of experiment 1 in

  which the rodent is castrated and therefore has no source of testosterone.

  Choices A, C, and D are all incorrect because these effects are caused by estradiol in the brain. Without aromatase, there will be no estradiol in the male rodent brain.

• Question 484:

  Before birth, the rodent brain is sexually undifferentiated. It is only in the first few days following birth, during a period referred to as the critical period, that the rodent brain differentiates along male or female lines. The hormone testosterone plays a critical role in this development. Specifically, sexual differentiation is determined by the presence of estradiol, an estrogen derivative of testosterone, in certain areas of the brain. Testosterone is converted to estradiol in critical brain cells that contain the enzyme aromatase. To study the effects of testosterone on the neonatal rodent brain, the following experiments were conducted:

  

  The above research, combined with additional studies, concluded that testosterone has two "organizational" effects on the male rodent brain: Defeminization Moderate levels of testosterone-derived estradiol during the critical period are sufficient for defeminization of the brain. Defeminization of the rodent brain results in loss of estrogen positive feedback on LH and FSH secretion and the ensuing loss of cyclicity, as well as loss of female sex behavior. Masculinization High levels of estradiol due to high levels of testosterone during the critical period results in masculinization of the brain. Masculinization leads to the induction of male sex behavior including antagonism towards other males and the mounting of females.

  The conversion of testosterone to estradiol is what type of reaction?

  

  A. Reduction

  B. Aromatization

  C. Electrophilic aromatic substitution

  D. Methylation

  Correct Answer: B

  The passage states that the conversion of testosterone to estradiol is catalyzed by aromatase. In this reaction, an aromatic ring is formed. Thus, it is an aromatization reaction. Choice A is incorrect because a reduction involves addition of

  hydrogens to a molecule. In this case, hydrogens are removed. A pair of hydrogens is removed from the ring and the resulting cyclic dienone tautomerizes to its enol form. Choice C is incorrect because electrophilic aromatic substitution is

  substitution on an aromatic ring. This reaction did not begin with an aromatic ring.

  Choice D is incorrect because (Friedel-Crafts) alkylation requires an aromatic ring (it is an electrophilic substitution reaction).

• Question 485:

  Sugars are carbohydrates, that is, molecules usually with the empirical formula C(H2O), and structural formulas made up of polyhydroxy aldehydes or ketones. Because of their polyfunctional nature, sugars can undergo a wide variety of

  transformations upon treatment with acids, bases, or heat, and upon reaction with other simple reagents and enzymes. While many sugars occur in nature and are thus readily available, the synthesis and modification of simple sugars is a

  necessary step in studies of enzymatic processes.

  Higher sugars can be synthesized from the simple carbohydrate D-glyceraldehyde with the following procedure:

  D-glyceraldehyde (Compound A) is reacted with HCN to produce a cyanohydrin (Compound B). Compound B is then treated with hydrogen gas and a modified palladium catalyst (similar to the Lindlar reagent) to give Compound C.

  Compound C is hydrolyzed to give the higher sugars in Mixture D. This reaction is summarized in Figure 1. Mixture D contains two compounds, which can be separated by crystallization. Two doublets near 9.5 (, ppm) are observed in the 1H

  NMR spectrum of mixture D, with each doublet corresponding to one of the two products present in the mixture. IR spectroscopy shows broad absorptions for both products around 3300 cm?.

  

  The hydroxyl groups of carbohydrates can also participate in reactions. For example, D-glyceraldehyde can react with chloromethane under basic conditions to yield a completely methylated product. This SN2 reaction is shown in Figure 2.

  

  Figure 2 Methylation of D-glyceraldehyde

  In glucose, the carbonyl carbon can be attacked, intramolecularly, by the hydroxyl oxygen of carbon-5 to form:

  A. glucofuranose.

  B. a hemiacetal.

  C. a lactone.

  D. a glycoside.

  Correct Answer: B

  A hemiacetal is formed through reaction of a carbonyl group and a hydroxyl group. In this question, these two groups occur on the same molecule of glucose.

  

  Choice A is incorrect because glucofuranose is the five-membered ring that is formed through attack by the hydroxyl oxygen of carbon-4, not carbon-5 as the question stem states. Furanose refers to a five- membered ring while pyranose refers to the more energetically favorable six-membered ring. Choice C is incorrect because a lactone is a cyclic ester. No ester linkage is formed in this case. Choice D is incorrect because a glycoside is a sugar acetal. These acetals are formed during polysaccharide formation. They can also be formed by exposing a sugar like D-glucose to another alcohol in an acidic environment.

  

• Question 486:

  Sugars are carbohydrates, that is, molecules usually with the empirical formula C(H2O), and structural formulas made up of polyhydroxy aldehydes or ketones. Because of their polyfunctional nature, sugars can undergo a wide variety of

  transformations upon treatment with acids, bases, or heat, and upon reaction with other simple reagents and enzymes. While many sugars occur in nature and are thus readily available, the synthesis and modification of simple sugars is a

  necessary step in studies of enzymatic processes.

  Higher sugars can be synthesized from the simple carbohydrate D-glyceraldehyde with the following procedure:

  D-glyceraldehyde (Compound A) is reacted with HCN to produce a cyanohydrin (Compound B). Compound B is then treated with hydrogen gas and a modified palladium catalyst (similar to the Lindlar reagent) to give Compound C.

  Compound C is hydrolyzed to give the higher sugars in Mixture D. This reaction is summarized in Figure 1. Mixture D contains two compounds, which can be separated by crystallization. Two doublets near 9.5 (, ppm) are observed in the 1H

  NMR spectrum of mixture D, with each doublet corresponding to one of the two products present in the mixture. IR spectroscopy shows broad absorptions for both products around 3300 cm?.

  

  The hydroxyl groups of carbohydrates can also participate in reactions. For example, D-glyceraldehyde can react with chloromethane under basic conditions to yield a completely methylated product. This SN2 reaction is shown in Figure 2.

  

  Figure 2 Methylation of D-glyceraldehyde

  What are the missing compounds, respectively, in the following reaction?

  

  A. Option A

  B. Option B

  C. Option C

  D. Option D

  Correct Answer: D

  This reaction is called the Wolff-Kishner Reduction and reduces a carbonyl group to the corresponding alkane. The first step, reaction with hydrazine, leads to the formation of a hydrazone. The second step occurs in basic solution and leads

  to the evolution of nitrogen gas and the formation of the corresponding hydrocarbon.

  Choice A is incorrect because the indicated structures do not form. Choice B is incorrect because ethers do not form alkoxides under basic conditions. Ethers are relatively unreactive.

  Choice C is incorrect because hydrazine reacts with a carbonyl to form hydrazone. This is a condensation reaction, not an addition reaction, and the addition product shown would not form.

• Question 487:

  Sugars are carbohydrates, that is, molecules usually with the empirical formula C(H2O), and structural formulas made up of polyhydroxy aldehydes or ketones. Because of their polyfunctional nature, sugars can undergo a wide variety of

  transformations upon treatment with acids, bases, or heat, and upon reaction with other simple reagents and enzymes. While many sugars occur in nature and are thus readily available, the synthesis and modification of simple sugars is a

  necessary step in studies of enzymatic processes.

  Higher sugars can be synthesized from the simple carbohydrate D-glyceraldehyde with the following procedure:

  D-glyceraldehyde (Compound A) is reacted with HCN to produce a cyanohydrin (Compound B). Compound B is then treated with hydrogen gas and a modified palladium catalyst (similar to the Lindlar reagent) to give Compound C.

  Compound C is hydrolyzed to give the higher sugars in Mixture D. This reaction is summarized in Figure 1. Mixture D contains two compounds, which can be separated by crystallization. Two doublets near 9.5 (, ppm) are observed in the 1H

  NMR spectrum of mixture D, with each doublet corresponding to one of the two products present in the mixture. IR spectroscopy shows broad absorptions for both products around 3300 cm?.

  

  The hydroxyl groups of carbohydrates can also participate in reactions. For example, D-glyceraldehyde can react with chloromethane under basic conditions to yield a completely methylated product. This SN2 reaction is shown in Figure 2.

  

  Figure 2 Methylation of D-glyceraldehyde The two sugars in Mixture D are:

  A. enantiomers.

  B. anomers.

  C. epimers.

  D. disaccharides.

  Correct Answer: C

  Epimers are sugars that differ in the configuration about a single stereocenter. A new chiral center is formed in the first step of the sugar extension (cyanohydrin formation). This will lead to the formation of two different sugars in Mixture D.

  

  Choice A because enantiomers are mirror images that differ in configuration at every stereocenter. As shown above, these epimers differ at a single stereocenter. Choice B is incorrect because anomers aresugars that differ in their

  configuration about the new stereocenter formed when they adopt a ring configuration. The sugars shown above would generally not form rings because of the ring strain of 4- membered rings.

  Choice D is incorrect because dissaccharides are formed from the condensation of two monosaccharides (like the ones in mixture D).

• Question 488:

  Sugars are carbohydrates, that is, molecules usually with the empirical formula C(H2O), and structural formulas made up of polyhydroxy aldehydes or ketones. Because of their polyfunctional nature, sugars can undergo a wide variety of

  transformations upon treatment with acids, bases, or heat, and upon reaction with other simple reagents and enzymes. While many sugars occur in nature and are thus readily available, the synthesis and modification of simple sugars is a

  necessary step in studies of enzymatic processes.

  Higher sugars can be synthesized from the simple carbohydrate D-glyceraldehyde with the following procedure:

  D-glyceraldehyde (Compound A) is reacted with HCN to produce a cyanohydrin (Compound B). Compound B is then treated with hydrogen gas and a modified palladium catalyst (similar to the Lindlar reagent) to give Compound C.

  Compound C is hydrolyzed to give the higher sugars in Mixture D. This reaction is summarized in Figure 1. Mixture D contains two compounds, which can be separated by crystallization. Two doublets near 9.5 (, ppm) are observed in the 1H

  NMR spectrum of mixture D, with each doublet corresponding to one of the two products present in the mixture. IR spectroscopy shows broad absorptions for both products around 3300 cm?.

  

  The hydroxyl groups of carbohydrates can also participate in reactions. For example, D-glyceraldehyde can react with chloromethane under basic conditions to yield a completely methylated product. This SN2 reaction is shown in Figure 2.

  

  Figure 2 Methylation of D-glyceraldehyde

  The reaction shown in Figure 2 most likely proceeds via the formation of a(n):

  A. water leaving group.

  B. nucleophilic alkoxide ion.

  C. carbocation intermediate.

  D. tetrahedral intermediate.

  Correct Answer: B

  The passage states that an SN2 reaction occurs. This requires a good nucleophile. Alkoxides (RO-) are strong bases and thus good nucleophiles. Sodium hydroxide (NaOH) is used in this reaction to remove the hydrogen from the hydroxyl group to form the alkoxide. The SN2 reaction is shown below

  

  Choice A is incorrect because a hydroxyl group would be converted to water, a good leaving group, under acidic conditions. The reaction in Figure 2 occurs under basic conditions. Choice C is incorrect because carbocation intermediates occur in SN1 and E1 reactions, but not in SN2. Choice D is incorrect because tetrahedral intermediates are formed during addition reactions to a carbon/oxygen double bond. The reaction in Figure 2 does not involve attack on the carbonyl group.

• Question 489:

  Sugars are carbohydrates, that is, molecules usually with the empirical formula C(H2O), and structural formulas made up of polyhydroxy aldehydes or ketones. Because of their polyfunctional nature, sugars can undergo a wide variety of

  transformations upon treatment with acids, bases, or heat, and upon reaction with other simple reagents and enzymes. While many sugars occur in nature and are thus readily available, the synthesis and modification of simple sugars is a

  necessary step in studies of enzymatic processes.

  Higher sugars can be synthesized from the simple carbohydrate D-glyceraldehyde with the following procedure:

  D-glyceraldehyde (Compound A) is reacted with HCN to produce a cyanohydrin (Compound B). Compound B is then treated with hydrogen gas and a modified palladium catalyst (similar to the Lindlar reagent) to give Compound C.

  Compound C is hydrolyzed to give the higher sugars in Mixture D. This reaction is summarized in Figure 1. Mixture D contains two compounds, which can be separated by crystallization. Two doublets near 9.5 (, ppm) are observed in the 1H

  NMR spectrum of mixture D, with each doublet corresponding to one of the two products present in the mixture. IR spectroscopy shows broad absorptions for both products around 3300 cm?.

  

  The hydroxyl groups of carbohydrates can also participate in reactions. For example, D-glyceraldehyde can react with chloromethane under basic conditions to yield a completely methylated product. This SN2 reaction is shown in Figure 2.

  

  Figure 2 Methylation of D-glyceraldehyde

  The modified palladium catalyst is used instead of platinum to:

  A. avoid reduction of the cyanohydrin to the amine.

  B. make the reaction energetically favorable.

  C. act as a Lewis acid.

  D. reduce the likelihood of attack at the carbonyl carbon.

  Correct Answer: A

  The cyanohydrin (compound B) is reduced to an imine (compound C). If platinum were used, reduction would proceed all the way to the amine as shown below. This must be avoided in order to produce the carbonyl group on the extended sugar.

  

  Choice B is incorrect because a catalyst will not alter the thermodynamic favorability of a reaction. A catalyst will only increase the rate of reaction by lowering the activation energy. Choice C is incorrect because a Lewis acid is an electron

  pair acceptor. The palladium catalyst acts as an activator of hydrogen.

  Platinum would function in the same way.

  Choice D is incorrect because compound B does not even have a carbonyl group (carbon double bonded to oxygen).

• Question 490:

  Which of the following reactants should be used to form the product shown below?

  

  A. Sodium ethanoate

  B. 1-propanol

  C. Ammonia and ethane

  D. Dimethyl ether

  Correct Answer: B

  This question is primarily testing your knowledge of nomenclature. Reacting an alkanoyl halide with an alcohol will lead to formation of an ester, as shown. Note that we have added 3 carbons to our molecule, and so the unknown reactant must have three carbons. Propanol is the only three carbon choice, and the only alcohol.