MCAT Q's
Caduceus Facts
UT Facts
100
Apoptosis is the process of programmed cell death that can occur in multicellular organisms. The proteins involved in apoptosis are associated with pathways for cell cycle arrest and DNA repair. These processes are mostly regulated through the interplay of various proteins involved in feedback loops including some of the ones shown in Figure 1.
Figure 1: Feedback loops forming a regulatory network affecting apoptosis, cell cycle arrest and DNA repair.
Figure 1: Feedback loops forming a regulatory network affecting apoptosis, cell cycle arrest and DNA repair. (Bioformatics Institute)
According to Figure 1, CDK2 activity would most reasonably increase due to all of the following EXCEPT:
- A. degradation of p21.
- B. high cyclin G concentrations.
- C. a mutation in the gene that produces PTEN.
- D. high p53 concentrations.
D. High p53 concentrations
Notice the key in the figure which will allow us to follow each arrow that stimulates the next protein and each symbol for negative feedback which means there will be some downregulation (amount/concentration goes down). [Notice a key step in the diagram: p21 inhibits CDK2]
Degradation of p21 implies that the concentration of p21 in its active form goes down. The diagram shows that p21 has a negative influence on CDK2. In other words, when p21 is high, CDK2 goes low. But in our instance, p21 is low (degraded) so this allows CDK2 to rise unchecked.
High cyclin G concentrations: From the bottom of Figure 1, we can see that high cyclin G leads to high mdm2 and low p53 (notice carefully, when we leave mdm2, there is only one place to go in the diagram because all the other symbols are pointing to mdm2 and only one symbol is pointing away). Note that we used the most direct route to get to CDK2 as the question used the words “most reasonably”. Low p53 means low p21 which we established will lead to a rise in CDK2.
A mutation in the gene that produces PTEN: The great majority of mutations will result in an ineffective gene product or none at all. Thus we have a decrease in PTEN which will lead to a rise in PIP3 (if you are unsure, think of what happens if PTEN goes up, then PIP3 must go down because of the negative feedback symbol), rise in AKT, rise in mdm2, decrease in p53 which we already established means an eventual rise in CDK2.
High p53 concentrations: clearly we get the opposite of the above, meaning a decrease in CDK2. High p53 stimulates p21 which has a negative feedback on CDK2.
100
When was Caduceus Founded?
2013
100
How many arces is UT Austin's main campus?
431 arces
200
The red bread mold Neurospora crassa grows well on a cultural plate with "minimal" medium which is a fluid containing only a few simple sugars, inorganic salts, and vitamin. Neurospora that grows normally in nature (wild type) has enzymes that convert these simple substances into the amino acids necessary for growth. Mutating any one of the genes that makes an enzyme can produce a Neurospora strain that cannot grow on minimal medium. The mutant would only grow if the enzyme product were to be added as a supplement. On the other hand, if a "complete" medium is provided, containing all required amino acids, then Neurospora would grow, with or without mutation.
Figure 1: A synthesis pathway for the amino acid arginine. Each gene in italics in the diagram produces one enzyme necessary for the synthesis of this essential amino acid required for growth.
Table 1: Growth response of mutant strains in "minimal" media with supplements as indicated. Growth is indicated by (+), and no growth is indicated by (-).
Question 2
According to the information provided, a conclusion that can be made with certainty is that neither mutant strain P nor Q have the defective enzyme:
- A. carbamoyltransferase.
- B. argininosuccinate synthase.
- C. argininosuccinase.
- D. None of the above enzymes are defective in either mutant strain P nor Q
- C. argininosuccinase.
You should recognize that enzymes typically end with –ase and so you can see that the figure provided indicates the 3 enzymes that are catalysts for the three reactions as shown. Now let’s reinterpret the question: if neither P nor Q have a defective enzyme X, that suggests that both P and Q have a functioning enzyme X (the other interpretation is that they have no enzyme but that is not consistent with the data in the table provided).
The information in the table shows a (+) symbol in the last two columns indicating that both P and Q are able to convert argininosuccinate to arginine (see Table 1). This necessarily means that both P and Q have a functioning enzyme argininosuccinase (see Figure 1).
Going Deeper: More about the 1 gene, 1 enzyme hypothesis and the importance within a metabolic pathway. Let’s consider some hypothetic examples using Figure 1 but no longer considering Table 1 so we can explore other possibilities and their consequences.
For example, let’s say that Mutant# 1 couldn't make ornithine. So, the gene that makes the enzyme for ornithine synthesis must have been mutated. If ornithine is added to the media, citrulline and then arginine would be made and Mutant# 1 could grow.
Similarly, consider a genetic mutation in Mutant# 2 affecting the enzyme that makes the arginine precursor citrulline. Adding citrulline as a supplement complements the mutation and drives arginine synthesis to completion. And consider a genetic mutation #3 affecting the final step of arginine synthesis — the conversion of citrulline to arginine. By adding arginine as a supplement, the mutation is complemented (like a ‘work around’) and Mutant# 3 could grow. With each mutated gene, only one step of the metabolic pathway is affected. Therefore, one gene is responsible for one enzyme or protein (of course, it’s a little more complicated because some genes are responsible for polypeptides which combine with other polypeptides to form a functioning protein).
200
How many Officers are there?
9
200
What is the name of the UT President?
Jim Davis
300
Experiments using the two mutant strains P and Q, reveal that strain P accumulates citrulline, but strain Q does not. Which of the following statements is most consistent with the data provided?
- A. Strain Q has only one mutation.
- B. Strain P has a mutation in argF only.
- C. Strain P has mutations in argF, argG and argH.
- D. Strain P has a mutation in argG only.
- D. Strain P has a mutation in argG only
On the Surface: In the previous question, we established that the enzyme argininosuccinase must be functional for both strains and that means that argH is not defective (no mutation) for either P or Q. Thus any answer choice suggesting an argH mutation is incorrect.
Strain P accumulates citrulline. This means that the synthetic pathway was working up to citrulline but then was blocked from progressing. This means that the next step in the production line, the gene product from argG, is not functional (mutation). Thus we have our answer.
Going Deeper: Strain Q: no accumulation of citrulline means that there must be a blockage before that point, so given the information presented, argF must have a mutation meaning ornithine accumulates. Adding ornithine: no growth and it accumulates. Adding citrulline, no growth because there must ALSO be an argG mutation. This is consistent with adding something after the argG step/mutation, argininosuccinate, resulting in growth. Thus strain Q has must have 2 mutations in the synthesis pathway shown.
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300
What is our philanthropy?
March of Dimes :)
300
Name 5 UT Austin Science Buildings
Norman Hackerman
Welch
Painter
J.T Patterson Labs
Moffett Molecular Biology Building
Marine Science Building
400
Biology: Passage (Questions 1–2)
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 above.
1. 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?
I. Transcription
II. Translation
III. Transformation
I only
I and II only
I and III only
I, II, and III
(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.
400
How many speaker series meeting do you need to attend?
4/6
400
Who is the student representatives for the college of natural sciences? (3)
Joseph Hyun
Michael Jeffords
Devin Almaguer
500
Biology: Passage (Questions 1-2)
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 above.
2. 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?
No, because the heat-killed rough strain cannot infect mice.No, because the heat-killed smooth strain cannot transform mice cells.Yes, because the heat-killed rough strain is transformed into the smooth strain.Yes, because the mice do not have acquired immunity to pneumococcus bacteria.
(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.
500
What were the names and the occupations of the 3 doctors who have presented in the speaker series so far?
Dr. Raymond J. Harshbarger: Cranomaxillofacial Surgeon
Dr Stuart Wolf, MD, FACS, Urologist, Endourologist
Dr Glendaliz Bosques Physiatrist Chief of Pediatric Rehab
500
How many schools are there at UT Austin?
18