Question #1
In an attempt to develop a vaccine for pneumonia, Fred Griffith performed a series of experiments in 1928 using mice and two strains of the pneumococcus bacteria: a virulent encapsulated strain and a nonvirulent unencapsulated strain. The encapsulated strain was called the "smooth strain" because the colonies looked smooth on a culture plate due to their polysaccharide capsules, whereas the unencapsulated strain was denoted as the "rough strain" due to the irregularity of its surface.
Four different groups of mice were injected with different combinations of the bacterial strains. The experimental results are shown in Table 1.
Table 1: Results of Griffith's experiment
Mice | Injected Bacteria | Result |
---|---|---|
1 | Live smooth | Died |
2 | Live rough | Survived |
3 | Heat-killed smooth | Survived |
4 | Heat-killed smooth + live rough | Died |
Autopsies performed on Group 4 mice revealed blood samples filled with live smooth-strain bacteria. Some component had been transferred from the heat-killed smooth-strain bacteria to the live rough-strain bacteria, transforming them into the virulent smooth-strain bacteria. This process later became known as transformation. Griffith believed the protein from the dead smooth-strain bacteria was the active transforming agent.
A colony of smooth strain bacteria is grown on a culture containing an experimental drug that cleaves nucleic acid base sequences wherever adenine is paired with uracil. Which of the following processes will be directly affected?
A. I only
B. I and II only
C. I and III only
D. I, II and III
Answer:
(B) is correct.
(B) I and II only is the correct answer
One of the ways that RNA differs from DNA is in the use of the nucleotide uracil. In RNA, uracil is substituted for thymine. In other words, RNA has uracil and DNA doesn't. During DNA replication, adenine pairs with thymine. During transcription (mRNA synthesis from a DNA template) and translation (translation of mRNA into a peptide chain), adenine pairs with uracil. Thus, if bacteria were exposed to a drug that cleaves adenine from uracil, transcription and translation would be disrupted, whereas transformation, which is the transfer of DNA from one bacteria to another, would not be directly affected by this drug. Therefore, statements I and II would be directly affected and statement III would not. Choice (B) is correct. Kaplan MCAT training will provide you with the content knowledge you need to quickly and confidently answer questions about molecular biology/DNA on test day.
Question #2
In an attempt to develop a vaccine for pneumonia, Fred Griffith performed a series of experiments in 1928 using mice and two strains of the pneumococcus bacteria: a virulent encapsulated strain and a nonvirulent unencapsulated strain. The encapsulated strain was called the "smooth strain" because the colonies looked smooth on a culture plate due to their polysaccharide capsules, whereas the unencapsulated strain was denoted as the "rough strain" due to the irregularity of its surface.
Four different groups of mice were injected with different combinations of the bacterial strains. The experimental results are shown in Table 1.
Table 1: Results of Griffith's experiment
Mice | Injected Bacteria | Result |
---|---|---|
1 | Live smooth | Died |
2 | Live rough | Survived |
3 | Heat-killed smooth | Survived |
4 | Heat-killed smooth + live rough | Died |
Autopsies performed on Group 4 mice revealed blood samples filled with live smooth-strain bacteria. Some component had been transferred from the heat-killed smooth-strain bacteria to the live rough-strain bacteria, transforming them into the virulent smooth-strain bacteria. This process later became known as transformation. Griffith believed the protein from the dead smooth-strain bacteria was the active transforming agent.
If Griffith had injected a fifth group of mice with a combination of heat-killed rough strain and heat-killed smooth strain, would the mice have died?
A. No, because the heat-killed rough strain cannot infect mice.
B. No, because the heat-killed smooth strain cannot transform mice cells.
C. Yes, because the heat-killed rough strain is transformed into the smooth strain.
D. Yes, because the mice do not have acquired immunity to pneumococcus bacteria.
Answer:
(A) is correct.
(A) is correct.
The only reason that the combination of heat-killed smooth strain plus live rough strain killed the Group 4 mice was because smooth-strain DNA coding for the polysaccharide capsule was incorporated into the genome of the rough strain. When the rough strain bacteria reproduced, all of the progeny inherited the ability to manufacture the capsule. In other words, the rough strain had been transformed into the smooth strain, and could now infect the mice and kill them. The key to the transformation was that the rough strain reproduced. Heat-killed bacteria, whether they are encapsulated or not, cannot reproduce and cause infection because their DNA has been denatured. Therefore, if Griffith had injected a fifth group of mice with a combination of heat-killed smooth strain and heat-killed rough strain the mice would have lived. So, choice (A) is correct.
Choice (B) is incorrect because even though the mice would live, it is not because the heat-killed smooth strain cannot transform mice cells. Bacterial cells can transform other bacterial cells; they cannot transform animal cells or plant cells. Choices (C) and (D) are incorrect because the mice would survive. Acquired immunity is the kind of immunity one gets from a vaccine; it is usually not permanent.
This is an example of a question that requires you to understand information given in a passage and, using critical thinking skills, apply this knowledge to a new situation. Kaplan-trained MCAT students know how to read a passage and annotate it, so that they can quickly find exactly the information they need to answer a particular question, without wasting valuable time.
Question #3
In which of the following cases will the normal force on a box be the greatest?
A. When the box is placed in a stationary elevator
B. When the box is placed in an elevator moving upwards at constant velocity
C. When the box is placed in an elevator accelerating downward
D. When the box is placed in an elevator accelerating upward
Answer:
(D) is correct.
(D) is correct.
Looking at the answer choices, we know that we want to find out how the normal force varies with the motion of the box. In all cases listed in the answer choices, there are two forces acting on the box: the normal force and the force of gravity. These two act in opposite directions: the normal force, N, in the upward direction and gravity, mg, in the downward direction. Taking the upward direction to be positive, we can express the net force on the box as N – mg.
From Newton's Second Law, this is also equal to ma, where a is the acceleration of the box (again with the upward direction being positive). For answer choices (A) and (B), the net acceleration of the box is zero, so N = mg. We can see how the acceleration of the elevator (and, hence, of the box) affects the normal force. The larger the acceleration (in the positive, i.e., upward, direction), the larger the normal force is to preserve the equality: N – mg = ma, N = ma+ mg. Answer choice (D), in which the elevator is accelerating upward, results in the greatest normal force, since in that case the magnitude of the normal force is greater than gravity by the amount ma.
Question #4
Which of the following addition reactions occurs the fastest?
A. I, because a primary cation is formed in the rate-determining step
B. II, because the cation formed in the rate-determining step is stabilized through an inductive effect
C. III, because a secondary cation which is resonance-stabilized is formed in the rate-determining step
D. II, because a secondary cation which is resonance-stabilized is formed in the rate-determining step
Answer:
(D) is correct.
(D) is correct.
Here's a question about reaction rates. The answer choices give some hints here, because they all mention "cation" formation in the "rate-determining step." The relative rates of three addition reactions are controlled by their rate-determining steps, all of which involve carbocation formation. Now remember some carbocation chemistry. The most stable carbocation is formed wmost quickly because its formation has a lower energy of activation. Primary carbocations are less stable than secondary carbocations so we can eliminate choice (A). All three reactions result in secondary carbocation formation in their rate-determining steps. Only the carbocation formed in reaction II has the extra stabilitly afforded by resonance. Choice (D) therefore is the correct answer.
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