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Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 31 )

Updated On: 30-Jan-2026
Page 31 of 164
PDF with 811 Questions

The polymerase chain reaction (PCR) is a powerful biological tool that allows the rapid amplification of any fragment of DNA without purification. In PCR, DNA primers are made to flank the specific DNA sequence to be amplified. These primers are then extended to the end of the DNA molecule with the use of a heat-resistant DNA polymerase. The newly synthesized DNA strand is then used as the template to undergo another round of replication.
The 1st step in PCR is the melting of the target DNA into 2 single strands by heating the reaction mixture to approximately 94° C, and then rapidly cooling the mixture to allow annealing of the DNA primers to their specific locations. Once the primer has annealed, the temperature is elevated to 72° C to allow optimal activity of the DNA polymerase. The polymerase will continue to add nucleotides until the entire complimentary strand of the template is completed at which point the cycle is repeated (Figure 1)


Figure 1
One of the uses of PCR is sex determination, which requires amplification of intron 1 of the amelogenin gene. This gene found on the X-Y homologous chromosomes has a 184 base pair deletion on the Y homologue. Therefore, by amplifying intron 1 females can be distinguished from males by the fact that males will have 2 different sizes of the amplified DNA while females will only have 1 unique fragment size.
The use of PCR for sex determination relies on the fact that:

  1. the amelogenin gene is responsible for an autosomal recessive trait.
  2. the X and Y homologous chromosomes have different sizes of intron 1 of the amelogenin gene.
  3. females have an X and Y chromosome and males have two X chromosomes.
  4. intron 1 has a different nucleotide length than intron 2.

Answer(s): B

Explanation:

According to the passage, sex determination can be determined by amplifying intron 1 of the amelogenin gene which is found on the sex chromosomes. Due to a deletion, intron 1 on the Y sex homologue is shorter than intron 1 on the X sex homolog. This difference in size can be used to distinguish between males and females because males have one X and one Y chromosome while females have two X chromosomes. Therefore, females will have only one uniform size of intron 1 which does not bear the deletion. In contrast, males will have 2 different sizes of intron 1 following its amplification. Hence, B. is the correct answer choice.



The polymerase chain reaction (PCR) is a powerful biological tool that allows the rapid amplification of any fragment of DNA without purification. In PCR, DNA primers are made to flank the specific DNA sequence to be amplified. These primers are then extended to the end of the DNA molecule with the use of a heat-resistant DNA polymerase. The newly synthesized DNA strand is then used as the template to undergo another round of replication.
The 1st step in PCR is the melting of the target DNA into 2 single strands by heating the reaction mixture to approximately 94° C, and then rapidly cooling the mixture to allow annealing of the DNA primers to their specific locations. Once the primer has annealed, the temperature is elevated to 72° C to allow optimal activity of the DNA polymerase. The polymerase will continue to add nucleotides until the entire complimentary strand of the template is completed at which point the cycle is repeated (Figure 1)

Figure 1
One of the uses of PCR is sex determination, which requires amplification of intron 1 of the amelogenin gene. This gene found on the X-Y homologous chromosomes has a 184 base pair deletion on the Y homologue. Therefore, by amplifying intron 1 females can be distinguished from males by the fact that males will have 2 different sizes of the amplified DNA while females will only have 1 unique fragment size.
What would PCR amplification of an individual's intron 1 of the amelogenin gene reveal if the individual were male?

  1. One type of intron 1 since the individual has one X chromosome and one Y chromosome.
  2. Two types of intron 1 since the individual has only one X chromosome.
  3. One type of intron 1 since the individual has only one X chromosome.
  4. Two types of intron 1 since the individual has one X chromosome and one Y chromosome.

Answer(s): D

Explanation:

A male individual, which contains one X and one Y chromosome, should have 2 types of intron 1. The passage states that intron 1 on the Y chromosome has a deletion which renders it smaller in length than the corresponding allele on the X chromosome, thereby providing males with 2 different size fragments. Females which have 2 X chromosomes will then have only 1 type of intron 1 of uniform size.



The polymerase chain reaction (PCR) is a powerful biological tool that allows the rapid amplification of any fragment of DNA without purification. In PCR, DNA primers are made to flank the specific DNA sequence to be amplified. These primers are then extended to the end of the DNA molecule with the use of a heat-resistant DNA polymerase. The newly synthesized DNA strand is then used as the template to undergo another round of replication.
The 1st step in PCR is the melting of the target DNA into 2 single strands by heating the reaction mixture to approximately 94° C, and then rapidly cooling the mixture to allow annealing of the DNA primers to their specific locations. Once the primer has annealed, the temperature is elevated to 72° C to allow optimal activity of the DNA polymerase. The polymerase will continue to add nucleotides until the entire complimentary strand of the template is completed at which point the cycle is repeated (Figure 1)

Figure 1
One of the uses of PCR is sex determination, which requires amplification of intron 1 of the amelogenin gene. This gene found on the X-Y homologous chromosomes has a 184 base pair deletion on the Y homologue. Therefore, by amplifying intron 1 females can be distinguished from males by the fact that males will have 2 different sizes of the amplified DNA while females will only have 1 unique fragment size.
Both copies of intron 1 of the amelogenin gene on the 2 sex chromosomes can be referred to as:

  1. homologous chromosomes.
  2. recessive traits.
  3. alleles.
  4. spliced RNA.

Answer(s): C

Explanation:

Alleles are two or more alternative forms of a gene that are found at the same place on homologous chromosomes. Since there is a 184 base pair deletion on the Y homologue and there is no deletion at the same position in the X homologue, intron 1 represents two different alleles of the same amelogenin gene.



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.

Feedback loops forming a regulatory network affecting apoptosis, cell cycle arrest and DNA repair.
Figure 1:
(Bioformatics Institute)
According to Figure 1, CDK2 activity would most reasonably increase due to all of the following EXCEPT:

  1. degradation of p21.
  2. high cyclin G concentrations.
  3. a mutation in the gene that produces PTEN.
  4. high p53 concentrations.

Answer(s): D

Explanation:

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.



Several models have been developed for relating changes in dissociation constants to changes in the tertiary and quaternary structures of oligomeric proteins. One model suggests that the protein's subunits can exist in either of two distinct conformations, R and T. At equilibrium, there are few R conformation molecules: 10 000 T to 1 R and it is an important feature of the enzyme that this ratio does not change. The substrate is assumed to bind more tightly to the R form than to the T form, which means that binding of the substrate favors the transition from the T conformation to R.
The conformational transitions of the individual subunits are assumed to be tightly linked, so that if one subunit flips from T to R the others must do the same. The binding of the first molecule of substrate thus promotes the binding of the second and if substrate is added continuously, all of the enzyme will be in the R form and act on the substrate. Because the concerted transition of all of the subunits from T to R or back, preserves the overall symmetry of the protein, this model is called the symmetry model. The model further predicts that allosteric activating enzymes make the R conformation even more reactive with the substrate while allosteric inhibitors react with the T conformation so that most of the enzyme is held back in the T shape.
Experiment Evaluating Non-Symmetry Model Enzymes
Experiments were performed with enzyme conformers that did not obey the symmetry model. The data is summarized in Figure 1.

Equilibrium distribution of two conformers at different temperatures given the free energy of their Figure 1:
interconversion. (modified from Mr.Holmium).
The symmetry model describes a form of cooperative binding. Most enzymes do not engage in cooperative binding. The predicted shape of a graph representing reaction rate versus the addition of substrate to most enzymes would be expected to be:

  1. a hyperbola.
  2. a straight line with a positive slope.
  3. a straight line with a negative slope.
  4. sigmoidal.

Answer(s): A

Explanation:

The amount of substrate-enzyme complex would increase steadily as more substrate is added until a point at which all enzymes are involved in a substrate-enzyme complex, and any more substrate added will have no effect (saturation kinetics). The graph would show a steadily slowing curve of positive slope which reaches a point at which it levels off into a horizontal line. This curve is called a hyperbola (see image below). A sigmoidal shape would be expected in cooperative binding (i.e. the symmetry model as described in the passage or hemoglobin).
Note: This was a classic question in the old MCAT and, not surprisingly, the same concept comes up in the AAMC's new MCAT practice materials: the difference between the simple (rectangular) hyperbola and the sigmoidal curve suggesting cooperative binding (and also, the ability to recognize the shapes of these 2 curves independently). Also note that the myoglobin saturation curve is a hyperbola, but hemoglobin has a sigmoid shape due to the cooperative binding of oxygen molecules.
And finally, note the positions of Vmax (= maximum velocity/reaction rate) and Km (substrate concentration at 1/2 Vmax) displaying Michaelis-Menten kinetics associated with the hyperbolic curve on the left, as opposed to the sigmoidal curve on the right (image from the GS BIO book or ebook, BCM 2.9):



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