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Question 1 of 5
1. Question
Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of the body. A patient was asked to perform cycles of deep inspiration and deep expiration. Fluoroscopy was used to measure the linear velocity of the movement of the diaphragm and the data was plotted against time. The origin in Figure 1 is the reference time 0 when the diaphragm was essentially in its equilibrium position.
At what time after t = 0 is the displacement of the diaphragm at a minimum?
Correct
Answer: C
Explanation: : To determine the displacement (NOT distance), take the area under the graph. The 2 similar sized trapezoids, one positive and the other negative, cancel each other. At time C, the diaphragm is back or near to its starting position; hence, its displacement has zero magnitude at that moment.Incorrect
Answer: C
Explanation: : To determine the displacement (NOT distance), take the area under the graph. The 2 similar sized trapezoids, one positive and the other negative, cancel each other. At time C, the diaphragm is back or near to its starting position; hence, its displacement has zero magnitude at that moment. -
Question 2 of 5
2. Question
The value of the water dissociation constant Kw varies with temperature. Its value is normally given as 1.00 x 10-14 mol2dm-6 at room temperature but 1.00 x 10-13 mol2dm-6 at 60 °C. What is the pH of pure water at 60 °C?
Correct
Answer: D
Explanation: : It is not necessary to calculate anything to solve this problem but we will go through the steps.
The dissociation of water (note carefully that the ratio of the products is 1:1; also keep in mind that this is the basis of the neutrality of pure water: acid units = base units):
H2O + H2O <-> H3O+ + OH-
The expression for the self-ionization of water or the water dissociation constant:
Kw = [H3O+][OH-] = [H+][OH-]
At room temperature, keeping in mind that the dissociation of the ions is 1:1, we get: 10-14 = [H+][OH-] = [H+]2
And so [H+] = 10-7 and then -log[H+] = 7 = pH
Doing the identical math but using Kw = 10-13 yields a pH of 6.5 BUT because the ratio of the ions is still 1:1 (if not, you would not have done the calculation) thus it must be neutral.Incorrect
Answer: D
Explanation: : It is not necessary to calculate anything to solve this problem but we will go through the steps.
The dissociation of water (note carefully that the ratio of the products is 1:1; also keep in mind that this is the basis of the neutrality of pure water: acid units = base units):
H2O + H2O <-> H3O+ + OH-
The expression for the self-ionization of water or the water dissociation constant:
Kw = [H3O+][OH-] = [H+][OH-]
At room temperature, keeping in mind that the dissociation of the ions is 1:1, we get: 10-14 = [H+][OH-] = [H+]2
And so [H+] = 10-7 and then -log[H+] = 7 = pH
Doing the identical math but using Kw = 10-13 yields a pH of 6.5 BUT because the ratio of the ions is still 1:1 (if not, you would not have done the calculation) thus it must be neutral. -
Question 3 of 5
3. Question
The isoelectric point of glycine is 6.0. When glycine is in a buffer with a pH of 6.0, which form predominates?
Correct
Answer: A
Explanation: : At pH 6.0, the isoelectric point, glycine is electrically neutral. This corresponds to the zwitterion form shown below. (Note that the form in answer B, though commonly written as shorthand, is impossible to find in nature.)
H3N+-CH2-COO-
Note the logic of the above zwitterion: the base component (the amino group) has accepted a proton from the acid component (the carboxylic acid COOH group). Answer choice C dominates below the isoelectric point (e.g. in a more protonated environment), and answer choice D dominates above the isoelectric point (e.g. in a less protonated environment).
Although it is possible to get a question like this on the real exam, it would be considered relatively easy! For the real exam, you will certainly have some questions involving amino acids that are dependent on your memory of the features of the side chains of the 20 common protein-building amino acids, how their charges change with pH, and their 3-letter and 1-letter representations. We will certainly be exploring all of these issues in the exams to follow. Hopefully, this will be a reminder following your MCAT content review:Ala A Alanine Leu L Leucine Arg R Arginine Lys K Lysine Asn N Asparagine Met M Methionine Asp D Aspartic acid Phe F Phenylalanine Cys C Cysteine Pro P Proline Gln Q Glutamine Ser S Serine Glu E Glutamic acid Thr T Threonine Gly G Glycine Trp W Tryptophan His H Histidine Tyr Y Tyrosine Ile I Isoleucine Val V Valine 
Incorrect
Answer: A
Explanation: : At pH 6.0, the isoelectric point, glycine is electrically neutral. This corresponds to the zwitterion form shown below. (Note that the form in answer B, though commonly written as shorthand, is impossible to find in nature.)
H3N+-CH2-COO-
Note the logic of the above zwitterion: the base component (the amino group) has accepted a proton from the acid component (the carboxylic acid COOH group). Answer choice C dominates below the isoelectric point (e.g. in a more protonated environment), and answer choice D dominates above the isoelectric point (e.g. in a less protonated environment).
Although it is possible to get a question like this on the real exam, it would be considered relatively easy! For the real exam, you will certainly have some questions involving amino acids that are dependent on your memory of the features of the side chains of the 20 common protein-building amino acids, how their charges change with pH, and their 3-letter and 1-letter representations. We will certainly be exploring all of these issues in the exams to follow. Hopefully, this will be a reminder following your MCAT content review:Ala A Alanine Leu L Leucine Arg R Arginine Lys K Lysine Asn N Asparagine Met M Methionine Asp D Aspartic acid Phe F Phenylalanine Cys C Cysteine Pro P Proline Gln Q Glutamine Ser S Serine Glu E Glutamic acid Thr T Threonine Gly G Glycine Trp W Tryptophan His H Histidine Tyr Y Tyrosine Ile I Isoleucine Val V Valine -
Question 4 of 5
4. Question
Consider the following structure.
Which of the following is the most accurate description of the structure provided?
Correct
Answer: C
Explanation: : 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).
Incorrect
Answer: C
Explanation: : 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).
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Question 5 of 5
5. Question
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. (Bioformatics Institute)
According to Figure 1, CDK2 activity would most reasonably increase due to all of the following EXCEPT:Correct
Answer: 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.Incorrect
Answer: 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.