Medical Tests MCAT-TEST Online Practice Questions and Exam Preparation

Medical Tests MCAT-TEST Online Questions & Answers

• Question 681:

  White light does not converge on a single point after passing through convex because of:

  A. chromatic aberration.
  B. spherical aberration.
  C. distortion.
  D. both spherical and chromatic aberration.
  B. spherical aberration.

• Question 682:

  Aqua regia is a mixture of HNO3 and HCl mixed in ratio of:

  A. 3:1
  B. 1:3
  C. 2:3
  D. 3:4
  B. 1:3

• Question 683:

  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
  B. Option B

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  Explanation/Reference:

  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 684:

  Suppose an -particle starting from rest is accelerated through a 5 megavolt potential difference. What is the final kinetic energy of the -particle? (Note: Assume that e = 1.6x10-19 C.)

  A. 1.6 x 10-12 J
  B. 8.0 x 10-13 J
  C. 6.4 x 10-26 J
  D. 3.2 x 10-26 J
  A. 1.6 x 10-12 J

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  Explanation/Reference:

  This question is a straightforward application of the conservation of energy. The absolute value of the change in kinetic energy equals the absolute value of the change in potential energy. Since the particle starts from rest, the change in kinetic energy is just the final kinetic energy. An -particle is a helium-4 nucleus consisting of 2 protons and 2 neutrons. The change in potential energy is equal to the charge times the potential difference. The charge of an -particle is equal to the charge times the potential difference. The charge of an -particle is equal to two times the charge of a proton, or 2e. So the final kinetic energy is therefore equal to the potential difference of 5 x 106 volts times the alpha particle charge of 2 x 1.6 x 10-19 coulombs. This is 1.6 x 10-12 joules, choice A.

• Question 685:

  Which structure shows two amino acids linked by 1 peptide bond?

  

  A. option A
  B. option B
  C. option C
  D. option D
  D. option D

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  Explanation/Reference:

  Amino acids are joined by a peptide bond: 1) the leading amino acid has a primary amine group to the left, 2) the peptide bond has a carbonyl group immediately following the first amino acid side chain, 3) bonded to a secondary amine leading the second amino acid side chain, and 4) the second amino acid finishes with a carboxylic acid group. Choice D is correct.

• Question 686:

  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.
  B. a hemiacetal.

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  Explanation/Reference:

  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 687:

  Consider the following structure.

  

  Which of the following is the most accurate description of the structure provided?

  A. Dipeptide
  B. Tripeptide
  C. Tetrapeptide
  D. Pentapeptide
  C. Tetrapeptide

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  Explanation/Reference:

  First a note regarding nomenclature: The shortest peptides are dipeptides, consisting of 2 amino acids joined by a single peptide bond, followed by tripeptides (3 amino acids, 2 peptide bonds), tetrapeptides (4 amino acids, 3 peptide bonds), pentapeptides (5 amino acids, 4 peptide bonds), etc. A peptide (= amide) bond is a covalent bond formed when the carboxyl group of one amino acid reacts with the amino group of another. In other words, focus on the number of times that you see C=O connected to N, as long as that bond is WITHIN the molecule (i.e. between amino acids). Since there are 3 such instances in the molecular structure provided, given the information provided above regarding 3 peptide bonds, the molecule must be a tetrapeptide. Going from left to right in the structure in the image, here are the 4 amino acids: Tyr-Pro-Phe-Pro-NH2

  Going deeper, the rules above apply to linear oligo/polypeptides. However, cyclic oligo/polypeptides will always have 1 additional bond where the amino end and carboxyl end bond to complete the ring.

  Beyond Required Assumed Knowledge (but the following includes some commonly explored ideas in MCAT Biochemistry passages): The tetrapeptide in the image is `morphiceptin'. It is an opioid (= a selective opioid receptor agonist/ligand) so that it has analgesic effects like opium and those effects are reversed by naloxone (= opioid inhibitor; `blocker'; used to block the effects of opioids especially in overdose).

• Question 688:

  Bebop lives! cries the newest generation of jazz players. During the 1980s, musicians like Wynton Marsalis revived public interest in bebop, the speedy, angular music that first bubbled up out of Harlem in the early 1940s, changing the face of jazz. That Marsalis and others thought of themselves as celebrating and preserving a noble tradition is, in one sense, inevitable. After the excesses of experimental or "free" jazz in the 1960s and the electronic jazz-rock "fusion" of the 70s, it is hardly surprising that people should hearken back to a time when jazz was "purer", perhaps even at the apex of its development. But the recent enthusiasm for bebop is also ironic in light of the music's initial public reception.

  In its infancy, during the first two decades of the 20th century, jazz was played by small groups of musicians improvising variations on blues tunes and popular songs. Most of the musicians were unable to read music, and their improvisations were fairly rudimentary. Nevertheless, jazz attained international recognition in the 1920s. Two of the people most responsible for its rise in popularity were Louis Armstrong, the first great jazz soloist, and Fletcher Henderson, leader of the first great jazz band. Armstrong, with his buoyant personality and virtuosic technical skills, greatly expanded the creative range and importance of the soloist in jazz. Henderson, a pianist with extensive training in music theory, foresaw the orchestral possibilities of jazz played by a larger band. He wrote out arrangements of songs for his band members that preserved the spirit of jazz, while at the same time giving soloists a more structured musical background upon which to shape their solo improvisations. In the 1930s, jazz moved further into the mainstream with the advent of the Swing Era. Big bands in the Henderson mold, led by musicians like Benny Goodman, Count Basie and Duke Ellington, achieved unprecedented popularity with jazz-oriented "swing" music that was eminently danceable.

  Against this musical backdrop, bebop arrived on the scene. Like other modernist movements in art and literature, bebop music represented a departure from tradition in both form and content, and was met with initial hostility. Bebop tempos were unusually fast, with the soloist often playing at double time to the backing musicians. The rhythms were tricky and complex, the melodies intricate and frequently dissonant, involving chord changes and notes not previously heard in jazz. Before bebop, jazz players had improvised on popular songs such as those produced by Tin-Pan Alley, but bebop tunes were often originals with which jazz audiences were unfamiliar.

  Played mainly by small combos rather than big bands, bebop was not danceable; it demanded intellectual concentration. Soon, jazz began to lose its hold on the popular audience, which found the new music disconcerting. Compounding public alienation was the fact that bebop seemed to have arrived on the scene in a completely mature state of development, without that early phase of experimentation that typifies so many movements in the course of Western music. This was as much the result of an accident of history as anything else. The early development of bebop occurred during a three-year ban on recording in this country made necessary by the petrol and vinyl shortages of World War II. By the time the ban was lifted, and the first bebop records were made, the new music seemed to have sprung fully-formed like Athena from the forehead of Zeus. And though a small core of enthusiasts would continue to worship bebop pioneers like Charlie Parker and Dizzy Gillespie, many bebop musicians were never able to gain acceptance with any audience and went on to lead lives of obscurity and deprivation.

  It can be inferred from the passage that the innovations of Fletcher Henderson (lines 27-34) were inspired primarily by:

  A. his admiration for Louis Armstrong.
  B. a hunger for international recognition.
  C. the realization that the public favored large bands over small combos.
  D. a desire to go beyond the structural limitations of early jazz music.
  D. a desire to go beyond the structural limitations of early jazz music.

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  Explanation/Reference:

  This is an inference question about the innovations of Fletcher Henderson, a key figure in jazz during the 1920s. Henderson's innovations are discussed in the sixth and seventh sentences of the second paragraph. Remember that the question stem seeks something that inspired Henderson. Sentence six says that Henderson, unlike many jazz musicians before him, had extensive training in music theory, and that he saw the creative possibilities in jazz played by a larger band. Sentence 7 says that Henderson's song arrangements gave jazz soloists "a more structured musical background upon which to shape their solo improvisations". We know from the second sentence of that same paragraph that most jazz prior to the 1920s was played by musicians who couldn't read music, and who consequently had to keep their improvisations fairly simple. That is, the structure of early jazz was simple, but in Henderson's arrangements the structure was raised to a grander scale and more sophisticated levels of complexity and structure. It can therefore be inferred that Henderson was inspired by the desire to go beyond the simple structure of early jazz, to go beyond its structural limitations. Therefore, choice (D) is correct. Choice (A) is a strong distracter. Armstrong is mentioned in the passage as the other key figure in jazz of the 1920s. But whether Henderson admired Armstrong or not is never discussed, so you can't infer that Henderson's innovations were inspired by his admiration for Armstrong. Choice (B) reflects the author's remark, in the third sentence of paragraph 2, that jazz attained international recognition in the 1920s. This choice distorts that point in suggesting that Henderson hungered for international recognition. There's simply no evidence for this in the passage, so (B) is wrong. Choice (C) is correct in suggesting that one of Henderson's innovations was in the application of jazz to the big band setting. A problem with choice C is that the public's preference for big bands over small combos did not become apparent until a decade after Henderson's contribution -- in the 1930s, during the Swing Era. Another problem with this choice is the implication that Henderson merely exploited the realization that the public liked the big band sound. The author gives only one reason why Henderson was drawn to large bands -- this is in the sixth sentence of paragraph 2: Henderson was drawn to large bands because, with his training in music theory, he "foresaw the orchestral possibilities of jazz played by a larger band."

• Question 689:

  A source emits a sound from one medium with a certain velocity, intensity, frequency and wavelength. When the sound exits the first medium and enters a denser medium, all of the following changes EXCEPT:

  A. velocity
  B. intensity
  C. frequency
  D. wavelength
  C. frequency

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  Explanation/Reference:

  This question is testing your understanding of the properties of sound as waves travel from one medium to another. To answer this question, you must know that the frequency of sound and light always remains constant when changing mediums. The wavelength on the other hand does change, and a change in the wavelength causes a proportional change in the velocity since v = f. Intensity is also proportional to velocity and so A, B, and D are all incorrect. Thus, C is the correct answer.

• Question 690:

  Historically, two different methods have been used to estimate the fluid pressure in capillary beds.

  Method 1 A glass pipette is inserted into the capillary. The level of blood rising in the pipette is measured and used to calculate the pressure. Alternatively, an inert fluid of density can be placed in the pipette and its height h can be measured. The pressure in the capillary is given by gh, where g is the acceleration due to gravity.

  

  Figure 1 Method 2

  The pressure can be measured indirectly in the following way. A section of gut tissue is removed from a specimen and placed on a beam balance. Blood is circulated through the tissue by a pump. The arterial pressure is then decreased. This leads to a decrease in the capillary hydrostatic pressure in the gut capillaries. The constant osmotic pressure of plasma proteins in the capillary causes absorption of fluid from the gut section which will decrease its weight. To prevent a change in the weight of the gut section, the venous pressure is increased. This tends to increase the capillary pressure, reducing the flow of fluid from the gut tissue into the capillaries. The capillary pressure is thus held constant (and the balance kept level) as the arterial pressure is decreased and the venous pressure increased. The arterial and venous pressures meet at the capillary pressure being measured.

  ( = MRT, where is the osmotic pressure, M the molarity of the solutes, R the universal gas constant, and T the temperature in Kelvin.)

  

  Figure 2

  Method 2 relies on keeping the beam balance level. Which of the following must be true if the balance is level?

  A. The arterial pressure equals the osmotic pressure.
  B. The weight of the gut equals the weight of the mass (m).
  C. The net force on the beam is zero.
  D. The net torque on the beam is zero.
  D. The net torque on the beam is zero.

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  Explanation/Reference:

  The net torque determines the angular acceleration of the beam. If the net torque is zero then the beam will remain level. Recall the following equation for calculating the torque. torque = Frsin where F is the force appliedr is the distance between the force and the fulcrum is the angle between the force and the vector r Choice A is incorrect because the passage does not describe any specific relationship between the arterial pressure and the osmotic pressure. It does state that when the capillary pressure equals the osmotic pressure, there is no net fluid flow into or out of the capillary bed and the gut section will not change weight, keeping the balance level. Note that the capillary pressure is related to both the osmotic pressure and the hydrostatic pressure. Choice B is incorrect because the net torque may not be zero even if the weight of the gut equals the weight of the mass (m). If the distance from the fulcrum r is not equal, then the net torque will not be zero and the beam will not remain level. Choice C is incorrect because the net torque may not be zero even if the net force on the beam is equal to zero. Note that when the beam is level, two forces are applied downwards (by the gut section and by the mass m) and the beam does not rotate. It is true that when the beam balance is not translated (moved without rotation), the net force is zero. Under these conditions, the fulcrum will apply an upwards force to counteract the two downwards forces described above. However, the beam balance can be translated, indicating a net force, and the beam will still remain level if the net torque is zero?consider pushing the beam balance around the room while the weights keep it level.