Medical Tests MCAT-TEST Online Practice Questions and Exam Preparation

Medical Tests MCAT-TEST Online Questions & Answers

• Question 431:

  Family violence, such as domestic violence, child abuse, and elder abuse, are serious and pervasive problems in the United States. On an annual basis, the National Crime Survey has found domestic violence results in 21000 hospitalizations, 99800 days of hospitalization, 28700 emergency department visits, and 39900 visits to physicians.

  Currently there is little consensus about the definitions of intimate violence. Even the terms employed are varied; for example, domestic violence, conjugal violence, intimate abuse, and partner abuse. Similarly, there are a range of causal explanations, and these are contingent upon the theoretical perspective employed. There is also controversy whether the term "violence," "abuse," or "aggression" should be used. Finally, within the terms adopted, there is no consensus about the victim-perpetrator relationship. For example, do the terms refer to a married co-habiting couple? Two heterosexual individuals who do not reside together but are dating? All this has implications for research, practice, and policy.

  The National Violence Against Women Survey was one of the largest studies sponsored by the National Institute of Justice and the Centers for Disease Control and Prevention. It systematically analyzed crime against women in the United States. A total of 8000 men and 8000 women in the United States were interviewed on the phone using a closed-ended survey. Table 1 displays the breakdown of figures when examining life time victimization by racial groups.

  Table 1 Percentage of people victimized by an intimate partner in lifetime, by victim gender, type of victimization, and victim race Source: Adapted from P Tjaden and N. Thoennes, "Consequences of Intimate Partner Violence: Findings from the National Violence Against Women Survey."

  

  How would a scholar utilizing social control theory explain domestic violence?

  A. Domestic violence is modeled and learned.
  B. Domestic violence is triggered by the lack of regulations and cohesive networks in society.
  C. Labelling as deviant the reactions of both the perpetuator and the victim of intimate violence reinforces such behaviors.
  D. Domestic violence is caused by an interaction of factors from the micro, meso, and macro-levels causing a feedback loop that then results in a loss of systems control.
  B. Domestic violence is triggered by the lack of regulations and cohesive networks in society.

  ---

  Explanation/Reference:

  B is correct. Social control theory maintains that social networks and social control mechanisms must be in place in society to regulate its members. These connections and their control mechanisms promote pro- social behaviors and reduce anti-social behaviors. Someone adopting this perspective would claim that lack cohesive social bonds and associated social control mechanisms would be responsive for violence. Thus, an intensification of social bonds and associated control mechanisms would diminish aspects such as domestic violence.

  A. This is incorrect. Social learning theory posits that all behaviors are learned through modeling. In this case, violence is learned by observing violent acts such as on the media. This is unrelated to social control theory. C. This is incorrect. According to labeling theory, society defines and creates deviance. Social units define (i.e., label) what is and is not normative, and what it means to be deviant. This act provokes a reaction at different levels (individual, institutional, and macrosocial). Under this perspective, deviance is not an intrinsic aspect of an act or person, but the result of the application of a label by someone to someone else. Changing labels would change reactions and likely decrease the performance of such "deviant" acts. Although labeling is an act of control, they recommend against it, unlike those in favor of social control theory. D. This is incorrect. Systems theory examines a problem like violence as a social interaction stemming from the micro-level (i.e., individual level), the meso-level (e.g., family, neighborhoods, schools, etc.) and macro-level (e.g., government, culture, institutions, etc.). This is unrelated to social control theory.

• Question 432:

  Aclown stands with her toes touching a fun house mirror with a convex bottom and a concave top. Which of the following best describes the distortion of the clown's image as she walks away from the mirror?

  A. Her head will shrink; her feet will grow, then shrink.
  B. Her head will shrink then grow; her feet will shrink, then grow.
  C. Her head will grow; her feet will grow.
  D. Her head will grow, then shrink; her feet will shrink.
  D. Her head will grow, then shrink; her feet will shrink.

  ---

  Explanation/Reference:

  Note that we can determine either the convex or concave mirror distortion to answer the question. We do not need to determine both. Solving for the convex bottom, we first note that the focal length of a convex mirror must be some negative value (while for the convex mirror it will be negative). We will solve this problem by drawing a diagram.Let's draw a diagram and examine the distortion of the clown's feet in the convex bottom of the mirror as she walks away from it. Recall that we can draw three kinds of rays in a ray tracing diagram:1.Rays that pass through the focal point will be reflected parallel. For a convex mirror, rays directed towards the focal point will be reflected parallel (because the focal point is in the virtual space behind the mirror).2. Rays that hit the axis of the mirror will be reflected with the same angle on the other side of the axis (like a plane mirror).3. Parallel rays will be reflected through the focal point. For a convex mirror, parallels rays will appear to pass through the focal point (because the focal point is in the virtual space behind the mirror.

  

  The image shrinks as the object moves farther from a convex mirror.

  From the diagram above, it is clear that the image in a convex mirror shrinks as the object moves away. Thus, the image of the clown's feet will shrink in the convex bottom of the mirror as she walks away.

• Question 433:

  The process of depolarization triggers the cardiac cycle. The electronics of the cycle can be monitored by an electrocardiogram (EKG). The cycle is divided into two major phases, both named for events in the ventricle: the period of ventricular contraction and blood ejection, systole, followed by the period of ventricular relaxation and blood filling, diastole.

  During the very first part of systole, the ventricles are contracting but all valves in the heart are closed thus no blood can be ejected. Once the rising pressure in the ventricles becomes great enough to open the aortic and pulmonary valves, the ventricular ejection or systole occurs. Blood is forced into the aorta and pulmonary trunk as the contracting ventricular muscle fibers shorten. The volume of blood ejected from a ventricle during systole is termed stroke volume.

  During the very first part of diastole, the ventricles begin to relax, and the aortic and pulmonary valves close. No blood is entering or leaving the ventricles since once again all the valves are closed. Once ventricular pressure falls below atrial pressure, the atrioventricular (AV) valves open. Atrial contraction occurs towards the end of diastole, after most of the ventricular filling has taken place. The ventricle receives blood throughout most of diastole, not just when the atrium contracts.

  Figure 1: Electronic and pressure changes in the heart and aorta during the cardiac cycle.

  

  The wall of the left ventricle is at least three times as thick as that of the right ventricle. This feature aids circulation by assuring that:

  A. blood entering the pulmonary artery is at a much higher pressure than blood entering the aorta.
  B. blood entering the aorta is at a much higher pressure than blood entering the pulmonary artery.
  C. the left ventricle has a higher blood capacity than the right ventricle at all times.
  D. the right ventricle has a higher blood capacity than the left ventricle at all times.
  B. blood entering the aorta is at a much higher pressure than blood entering the pulmonary artery.

  ---

  Explanation/Reference:

  It is commonly known that as a rule, the size of a muscle is proportional to its strength. The heart, which is a muscle, contains a chamber which must pump blood into the aorta to perfuse the grand majority of the body's tissues. Clearly, this chamber (= the left ventricle) must contain thicker muscle (= stronger) than a chamber that pumps blood only to the lungs (= the right ventricle through the pulmonary artery). The stronger chamber pumps blood with a greater force which means a higher pressure (recall from physics: P = F/A).

• Question 434:

  Compounds containing a hydroxyl group attached to a benzene ring are called phenols. Derivatives of phenols, such as naphthols and phenanthrols, have chemical properties similar to those of phenols, as do most of the many naturally-occurring substituted phenols. Like other alcohols, phenols have higher boiling points than hydrocarbons of similar molecular weight. Like carboxylic acids, phenols are more acidic than their alcohol counterparts. Phenols undergo a number of different reactions; both their hydroxyl groups and their benzene rings are highly reactive. A number of chemical tests can be used to distinguish phenols from alcohols and carboxylic acids.

  

  Thymol, a naturally occurring phenol, is an effective disinfectant that is obtained from thyme oil. Thymol can also be synthesized from m-cresol, as shown in Reaction A below. Thymol can then be converted to menthol, another naturally-occurring organic compound; this conversion is shown in Reaction B.

  Reaction A

  

  Reaction B

  

  What simple chemical test could be used to distinguish between the following two compounds?

  

  A. Option A
  B. Option B
  C. Option C
  D. Option D
  B. Option B

  ---

  Explanation/Reference:

  Compound I is para-cresol or para-methylphenol, and Compound II is benzyl alcohol, which behaves more like an aliphatic alcohol than a phenol. We can take advantage of this fact in order to distinguish between these two compounds. One of the easiest ways to do this is by their solubility behavior, which is also an easy way of separating them out of a mixture. Phenols are appreciably acidic, so they are quite soluble in aqueous sodium hydroxide. Alcohols, including benzyl alcohol, are NOT acidic and won't be soluble in aqueous sodium hydroxide. Now there is one exception; very small alcohols, with fewer than five carbons, are water-soluble, so they would dissolve in ANY aqueous solution. However, benzyl alcohol is too big a molecule to be water-soluble, so it won't be soluble in sodium hydroxide solution. So, the solubility of paracresol, Compound I, in aqueous sodium hydroxide would provide an effective test to distinguish between the two compounds, and thus choice B is the correct answer. Notice that this means you could separate a mixture of these two compounds by dissolving it in an organic solvent, and then extracting the solution in a separatory funnel with aqueous layer. Choice C is wrong, because compound I won't react with a bromine solution and decolorize it. Both compounds will react under stronger conditions, namely, the presence of a Lewis acid, but bromine in solution is too mild a reagent to react with the stable aromatic ring. Therefore, compound I, and compound II for that matter, won't decolorize bromine solution and choice C is incorrect. Choice D is incorrect because neither of the two compounds will be soluble in sodium bicarbonate. Sodium bicarbonate is a weak base, but since phenols are fairly weak acids, it takes a fairly STRONG base to make a phenol give up its acidic proton. Thus, para-cresol won't dissolve in sodium bicarbonate. As we've said, benzyl alcohol is an even weaker acid than para-cresol, so it certainly won't dissolve in sodium bicarbonate. And since neither compound will be soluble in sodium bicarbonate, sodium bicarbonate can't be used to distinguish between them.

• Question 435:

  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 student decides to carry out some simple tests on the compound in order to confirm his identification. Which of the following statements is NOT true?

  A. He could distinguish between a phenol and a benzoic acid by seeing if the unknown can be extracted with a weak base.
  B. He could distinguish between a benzyl aldehyde and a benzyl ketone by seeing if the unknown will react with cold KMn.
  C. He could distinguish between a benzyl alcohol and a phenol by attempting to dissolve the unknown in an aqueous solution of HCl.
  D. He could distinguish between a benzyl alcohol and a benzoic ester by attempting to dissolve the unknown in an aqueous solution of NaOH.
  C. He could distinguish between a benzyl alcohol and a phenol by attempting to dissolve the unknown in an aqueous solution of HCl.

  ---

  Explanation/Reference:

  This answer choice is false because neither one will dissolve in aqueous hydrochloric acid. You could use a strong base to distinguish between these two and then the slightly acidic phenol would deprotonate and dissolve as an ion in the aqueous solution. The alcohol, on the other hand, remains uncharged and would not dissolve in the aqueous solution because of its large organic phenyl group. The other three choices are all true statements. For choice A, the benzoic acid will become deprotonated, while the phenol won't. For choice B, the aldehyde will be oxidized, yielding a carboxylic acid, while the ketone won't. Finally, for choice D, sodium hydroxide will hydrolyze the ester-so it will dissolve, while benzyl alcohol will remain insoluble.

• Question 436:

  Musical instruments generate vibrations in the air that are perceived as musical tones. In many kinds of drums, these vibrations are created by a standing waves in a vibrating membrane. In a timpani drum, membrane vibration is coupled to the vibration of an enclosed volume of air. There may also be a second membrane whose vibration is coupled to that of the first by the enclosed air space, as in a snare drum. An idealized circular membrane will vibrate at normal mode frequencies given by Equation 1 where T is the membrane tension, r is the membrane radius, is the mass per unit area of the membrane, and frel is the relative frequency shown under each mode in Figure 1. The pitch of drums can be tuned by adjusting the membrane tension.

  

  Equation 1

  The modes are designated by two numbers, m and n. m indicates the number of diameter nodes, and n indicates the number of circular nodes. Several modes of vibration are shown in Figure 1.

  

  Figure 1

  The sound waves generated by a drum are:

  A. longitudinal waves because the displacement is perpendicular to the direction of propagation.
  B. longitudinal waves because the displacement is parallel to the direction of propagation.
  C. transverse waves because the displacement is perpendicular to the direction of propagation.
  D. transverse waves because the displacement is parallel to the direction of propagation.
  B. longitudinal waves because the displacement is parallel to the direction of propagation.

  ---

  Explanation/Reference:

  Sound waves are longitudinal waves in which pressure displacement occurs parallel to the propagation of the wave. This is in contrast to transverse waves, like those in the ocean, in which the displacement (up and down motion of the water) is perpendicular to the direction of propagation of the wave.

• Question 437:

  An enzyme catalyst is kept in a solution buffered to the pH that maximizes activity for those enzymes found in the blood. The enzyme catalyzes a reaction that produces an acid whose pKa is 4.5. In this buffer solution, the acid product will be:

  A. primarily deprotonated.
  B. primarily protonated.
  C. protonated and deprotonated in roughly equal amounts.
  D. not formed.
  A. primarily deprotonated.

  ---

  Explanation/Reference:

  The blood has a pH around 7.4, making it somewhat basic. If this enzyme is in a solution buffered to a point to maximize enzymes found in the blood, then the buffer solution will have a pH of 7.35 to 7.45. When an acid is in a solution whose pH is higher than its pKa, the acid is deprotonated, making choice (A) the correct answer.

  B: The acid would be protonated if the buffer solution had a pH that was below 4.5.

  C: The acid would have equal concentrations of the acid and conjugate base form if the buffer had a pH of around 4.5.

  D: There's no reason for us to assume the product wouldn't form.

• Question 438:

  The process of depolarization triggers the cardiac cycle. The electronics of the cycle can be monitored by an electrocardiogram (EKG). The cycle is divided into two major phases, both named for events in the ventricle: the period of ventricular contraction and blood ejection, systole, followed by the period of ventricular relaxation and blood filling, diastole.

  During the very first part of systole, the ventricles are contracting but all valves in the heart are closed thus no blood can be ejected. Once the rising pressure in the ventricles becomes great enough to open the aortic and pulmonary valves, the ventricular ejection or systole occurs. Blood is forced into the aorta and pulmonary trunk as the contracting ventricular muscle fibers shorten. The volume of blood ejected from a ventricle during systole is termed stroke volume.

  During the very first part of diastole, the ventricles begin to relax, and the aortic and pulmonary valves close. No blood is entering or leaving the ventricles since once again all the valves are closed. Once ventricular pressure falls below atrial pressure, the atrioventricular (AV) valves open. Atrial contraction occurs towards the end of diastole, after most of the ventricular filling has taken place. The ventricle receives blood throughout most of diastole, not just when the atrium contracts.

  Figure 1: Electronic and pressure changes in the heart and aorta during the cardiac cycle.

  

  The graph below shows the effects on stroke volume of stimulating the sympathetic nerves to the heart.

  

  According to the graph, the net result of sympathetic stimulation on stroke volume is to:

  A. approximately double stroke volume at any given end diastolic volume.
  B. decrease stroke volume at any given end diastolic volume.
  C. increase stroke volume at any given end diastolic volume.
  D. leave stroke volume relatively unchanged.
  C. increase stroke volume at any given end diastolic volume.

  ---

  Explanation/Reference:

  This question is easily answered by reading the graph. Compared to the curve labeled 'control', the curve labeled 'sympathetic stimulation' is always at a higher stroke volume for a given end-diastolic volume. However, by looking at stroke volume values for both curves at any end-diastolic volume value (i.e. 200), it is clear that the stroke volume for sympathetic stimulation is always less than twice the control (i.e. 160 vs. 100), thus eliminating answer choice A. Thus, we are left with answer choice C.

• Question 439:

  In 1972, Georges Ungar reported the discovery of a peptide that appeared to transfer learning. Ungar's claim was based on experiments in which rats placed in a chamber with specially designed dark and light regions were trained to avoid

  the dark regions of the chamber. Following their training, the rats were killed and brain extracts were prepared. These brain extracts were injected into naive rats which were then observed to acquire the fear of darkness without training. Two

  hypotheses were proposed to explain these remarkable results:

  Hypothesis 1

  Ungar concluded that the extracts contained some chemical that transmitted the learned fear of darkness to the naive rats. A fifteen amino-acid polypeptide was isolated from the brain extracts and sequenced. Ungar claimed that this peptide,

  called scotophobin, was a chemical transmitter of learning. The peptide had the primary structure shown below:

  C-ser-asp-asn-arg-gln-gln-gly-lys-ser-ala-arg-gln-glygly-tyr-N Scotophobin

  Hypothesis 2

  Other researchers, who tested scotophobin but could not reproduce Ungar's results, argued that scotophobin did not transfer the learned fear of darkness. Instead, they suggested that scotophobin, which is structurally similar to ACTH and

  vasopressin, acted to increase stress in the rats. Since stress increases sympathetic nervous activity, rats injected with scotophobin would become hyperactive and tend to spend less time in the dark regions of the experimental chamber.

  They argued that such stress responses in the rats could be misinterpreted as a fear of darkness. Ungar's claim was further weakened by chemical analysis in which both the scotophobin extracts which Ungar had injected into the naive rats

  and a sample of synthesized scotophobin peptide were subjected to SDS polyacrylamide gel electrophoresis, as shown in Figure 1.

  

  Figure 1

  Researchers were interested in purifying a second protein (protein X) from Ungar's extract. The gene segment encoding protein X was believed to consist of thirty nucleotides. According to Figure 1, which band could represent protein X?

  A. Band A
  B. Band B
  C. Band C
  D. Band D
  D. Band D

  ---

  Explanation/Reference:

  If the amino acid sequence encoding for Protein X is thirty nucleotides long, then Protein X consists of 10 amino acids. Protein X is thus smaller than scotophobin. The technique of electrophoresis uses an electrical field to separate proteins

  based on size. Larger proteins subjected to the same electric field will move more slowly than smaller proteins. Band C on Figure 1 which is common to the extract and to the purified protein represents scotophobin. Bands A and B which are

  closer to the top of the gel would represent proteins larger than scotophobin. Band D, which is below band C, represents a protein that has migrated further and is thus smaller than scotophobin. Of the choices, only band D could potentially

  represent Protein X.

  Choices A and B are incorrect because bands A and B, which have migrated less than band C, represent proteins that are larger than scotophobin. Choice C is incorrect because band C is common to both the extract and the purified protein,

  and must therefore represent scotophobin itself.

• Question 440:

  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
  A. Methyl

  ---

  Explanation/Reference:

  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.