Molecular Biology MCAT Biology Diagnostic Exam 1B Flashcards
Proteins are polymers of amino acids that play diverse roles in the body. Which of the following statements about proteins is true?
A. Proteins cannot contain more than 1000 amino acids.
B. The side chains of individual amino acids can have a significant effect on protein structure.
C. Protein synthesis always occurs in the 5
’
to 3
’
direction.
D. Protein functions include structure, transport, enzymatic catalysis, and emulsification of fats.
B. The tertiary structure of a protein (i.e., its final three-dimensional shape) is determined by interactions between the side chains of individual amino acids (choice B is correct). Proteins can contain any number of amino acids; some very large proteins include several thousand amino acids (choice A is wrong). Protein synthesis occurs in the NRight arrow.C direction; nucleic acid synthesis proceeds 5’ to 3’ (choice C is wrong). Proteins play structural roles, transport molecules around the cell and the body in general, and run reactions (among other functions), but emulsification of fats is carried out by bile, which is not a protein (choice D is wrong).
Concepts tested
Biochemistry: Biomolecules: Amino Acids and Proteins
Transposons are mobile genetic elements that can “jump” around the genome, causing chromosomal aberrations. All of the following are true of transposons EXCEPT:
A. a chromosomal inversion can be caused by two transposons in the opposite orientations.
B. chromosomal deletions and translocations can be caused by two transposons in the same orientation.
C. if the protein coding region of a gene is disrupted by a single transposon, then proteins levels will likely decrease.
D. the insertion of a single transposon into the promoter or regulatory region of a protein coding gene will always decrease protein levels
D. Two transposons in opposite orientations could cause chromosomal inversion, where a section of a chromosome is flipped to the reverse orientation (choice A is true and can be eliminated). Two transposons in the same orientation cause chromosomal deletions and translocations (choice B is true and can be eliminated). If a single transposon inserts into a protein-coding region of the DNA, the protein coding region will be disrupted and proteins levels will likely decrease (choice C is true and can be eliminated). However, if a single transposon inserts into the promoter or regulatory region of a protein coding gene, protein levels can either increase or decrease (choice D is not true of transposons and is the correct answer choice).
Concepts tested
Molecular Biology: Mutations and DNA Repair
A biochemist uses a sensitive assay to quantify the amount of energy required to translate an unknown protein. Given that he found 404 ATP equivalents of energy to be consumed in the process, how long was the peptide of interest?
A. 99 amino acids
B. 100 amino acids
C. 101 amino acids Correct Answer (Blank)
D. 404 amino acids
C. Translation is one of the most energy-intensive processes in the cell. While initiation requires only 1 GTP, tRNA loading requires 2 high energy phosphate bond equivalents per amino acid/tRNA pair, and elongation for each subsequent amino acid requires 2 GTP. Termination requires 1 GTP. This can be summarized in the equation 4n, where n is the number of amino acids in a peptide chain, and the equation represents the number of high energy bonds required to make the peptide. In this question, you are given the total quantity of energy consumed and need to determine the length of the peptide. If 4n = 404, then n = 101 (choice C is correct).
Concepts tested
Molecular Biology: Translation
α-ketoglutarate dehydrogenase (α-KGDH) is an enzyme used in cellular respiration that catalyzes the decarboxylation of isocitrate (a 5-carbon compound) to succinate (a 4-carbon compound). Inherited deficiencies of α-KGDH result in permanent lactic acidosis, among other symptoms, and lead to death at an early age. α-KGDH is mostly likely found in the:
A. mitochondrial matrix.
B. mitochondrial inner membrane.
C. cytosol
D. mitochondrial intermembrane space.
A. α-KGDH, from the description in the question text, sounds like an enzyme in the Krebs cycle. This is the only place in the cellular respiration reactions where a 5-carbon compound is converted to a 4-carbon compound. Furthermore, the fact that individuals with α-KGDH deficiency suffer from permanent lactic acidosis suggests that the lack of this enzyme shuts down the Krebs cycle and puts the individual into permanent fermentation. Thus, this enzyme is most likely found in the mitochondrial matrix, where the Krebs cycle takes place (choice A is correct and choices B, C, and D are wrong).
Concepts tested
Biochemistry: Glycolysis/Cellular Respiration
Enzymes are proteins that catalyze specific chemical reactions within a cell. Which of the following statements about enzymes is/are NOT true?
A common means of regulating enzyme activity is through phosphorylation of the enzyme.
Enzymes increase the energy of activation for a reaction, thereby making it go faster.
Enzymes shift the equilibrium of a reaction towards products.
A. I only
B. II only
C. I and II only
D. II and III only
D. Statement I is true and therefore an incorrect choice: protein kinases and enzyme phosphorylases attach phosphate groups to enzymes and are often used to regulate enzyme activity (choices A and C can be eliminated). Both remaining answer choices include Statement II, so it must be a false statement: while enzymes do make reactions go faster, they do this by reducing the energy of activation, not increasing it. Statement III is also false and a correct choice: enzymes do not change the equilibrium of a reaction, they only help the reaction reach equilibrium more quickly (choice B can be eliminated and choice D is correct).
Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition
High levels of ATP would:
A. stimulate phosphofructokinase and inhibit pyruvate kinase, thus stimulating glycolysis.
B. stimulate phosphofructokinase and stimulate pyruvate kinase, thus stimulating glycolysis.
C. inhibit phosphofructokinase and stimulate pyruvate kinase, thus inhibiting glycolysis.
D. inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis.
D. High levels of ATP indicate that the cell does not need to run glycolysis, and instead can use glycolytic intermediate in gluconeogenesis (choices A and B are wrong). Phosphofructokinase produces fructose-1,6-bisphosphate, driving the cell toward glycolysis; high ATP levels would inhibit this enzyme to inhibit glycolysis. Pyruvate kinase catalyzes the final step in glycolysis, the conversion of phosphoenolpyruvate to pyruvate. This enzyme must be inhibited by ATP to inhibit glycolysis (choice C is wrong and choice D is correct).
Concepts tested
Biochemistry: Glycolysis/Cellular Respiration
Which of the following statements is NOT true about competitive inhibition?
A. The Km of the uninhibited reaction is lower than the Km of the inhibited reaction.
B. Competitive inhibitors bind at the active site of an enzyme.
C. Even if you greatly increase substrate concentration, the Vmax of the inhibited reaction will never reach the same Vmax of the uninhibited reaction.
D. Competitive inhibitors can resemble the transition state of a reaction.
C. Competitive inhibitors can resemble either the substrate of a reaction or the transition state of a reaction (choice D is a true statement and can be eliminated), and as such, bind at the active site of an enzyme (choice B is a true statement and can be eliminated). If the substrate concentration is significantly increased, then it becomes more likely to bind substrate at the active site than to bind inhibitor, and the inhibited reaction can reach the same Vmax as the uninhibited reaction (choice C is a false statement and the correct answer choice). Km is the substrate required to reach 1/2 Vmax. Because you need more substrate to run the inhibited reaction at the same rate (V) as the uninhibited reaction, Km increases (choice A is a true statement and can be eliminated). Note that this Km, the Km measured in the presence of an inhibitor, is called the “apparent Km.”
Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition
Phosphofructokinase (PFK) catalyzes the phosphorylation of fructose-6-P in the third step of glycolysis. If ATP levels in the cell are high, ATP can bind to PFK (at a site other than the active site) to inhibit the reaction. This is most accurately described as:
A. competitive inhibition.
B. negative feedback.
C. allosteric regulation.
D. positive feedback.
C. When a molecule binds to an enzyme at a site other than the active site and regulates the activity of that enzyme, it is best described as allosteric regulation (choice C is correct). This might be an example of negative feedback (an end-product inhibiting an earlier enzyme) but it is more accurately described as allosteric regulation (choice C is better than choice B). In competitive inhibition, a molecule similar to the substrate binds at the active site to prevent product formation (choice A is wrong). In positive feedback, the end product stimulates an enzyme earlier in the series of reactions, and this is not occurring here (choice D is wrong).
Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition
Which of the following statements about lipids is NOT true?
A. Phospholipids are amphipathic and are used in the formation of lipid bilayers.
B. Triglycerides are formed by joining three fatty acids to a glycerol molecule, and the three fatty acids are always the same.
C. Saturated fats are solids are room temperature because their hydrocarbon chains can pack together tightly.
D. Cholesterol is a lipid formed from several hydrocarbon rings; it is used in the derivation of steroids such as estrogen, cortisol, and vitamin D.
B. Triglycerides are formed by joining three fatty acids to a glycerol molecule, however the three fatty acids do not have to be the same, and frequently aren’t (choice B is not true about lipids and is the correct answer choice). All other answer choices are true.
Concepts tested
Biochemistry: Biomolecules: Lipids
Mutation of which of the following codons would be most disruptive to the initiation of translation?
A. AUG
B. UGA
C. UUU
D. UAA
A. During translation initiation, the start codon (AUG) is recognized by the tRNA bound to methionine. By causing a mutation in this codon, the ribosome will either fail to initiate translation, or begin translation at an unintended site resulting in a protein which will likely be nonfunctional (choice A is correct). Also note that the codons UGA and UAA are stop codons (along with UAG).
Concepts tested
Molecular Biology: Translation
The pentose phosphate pathway (PPP) generates NADPH to be used as a reducing agent in biosynthetic reactions such as fatty acid synthesis, and generates ribulose-5-phosphate which can either be converted into ribose-5-phosphate or into glycolytic intermediates. Ribose-5-phosphate can also be generated from glycolytic intermediates without the reduction of NADPH. Which of the following would be favored in a rapidly dividing cell?
A. The formation of ribose-5-phosphate
B. The reduction of NADP+ to NADPH in the oxidative phase of the PPP
C. The conversion of ribulose-5-phosphate into glycolytic intermediates
D. The oxidation of glucose-6-phosphate to generate NADH
A. Cells that are rapidly dividing need to synthesize nucleic acids, thus the formation of ribose-5-phosphate, a precursor to nucleotides, would be favored (choice A is correct). NADPH is not necessary for nucleotide synthesis so the oxidative phase of the PPP would not be favored (choice B is wrong). The conversion of ribulose-5-P into glycolytic intermediates would shuttle the five-carbon sugar back into glycolysis when it would be more needed as a nucleotide precursor (choice C is wrong). NADH is not generated in the pentose phosphate pathway (choice D is wrong).
Concepts tested
Biochemistry: Carbohydrate Metabolism
Which of the following conditions is LEAST likely to lead to a spontaneous reaction?
A. A reaction with a large negative ΔH, run at a low temperature, that has a large increase in entropy.
B. A reaction with a small positive ΔH, run at a low temperature, that has a large increase in entropy.
C. A reaction with a small positive ΔH, run at a high temperature, that has a large increase in entropy.
D. A reaction with a large positive ΔH, run at a low temperature, that has a large decrease in entropy.
D. Consider the equation ΔG = ΔH – TΔS; reactions with a negative ΔG are spontaneous. ΔG can be negative under a variety of conditions. Reactions with an increase in disorder (entropy) have a large positive ΔS; when this value is combined with a large negative ΔH, or a high temperature, or a small positive ΔH, ΔG is likely to be negative and the reaction spontaneous (choices A, B, and C can be eliminated). However, a reaction with a large decrease in entropy is becoming MORE orderly and has a large negative ΔS. Combined with a large positive ΔH, this situation is least likely to result in a negative ΔG (choice D is the correct answer choice).
Concepts tested
Biochemistry: Thermodynamics/Kinetics
Which of the following differences between RNA and DNA could help explain the differences in secondary structure observed between the two types of nucleic acids?
A. The 2’ hydroxyl group present on RNA
B. The presence of thymine in RNA
C. Differing RNA binding proteins present in the nucleus of prokaryotes
D. Presence of a single phosphate per nucleotide in the RNA phosphodiester backbone
A. This question is asking for a difference between RNA and DNA and only one of the answer choices provided is a valid difference between the two nucleic acids. RNA contains a 2’ hydroxyl group, possesses uracil in place of thymine, and generally only exists in a single-stranded form in the cell (choice A is correct). Thymine should not be present in RNA and prokaryotes do not possess a nucleus (choices B and C are wrong). Finally, both RNA and DNA possess a single phosphate per nucleotide in their phosphodiester backbones (choice D is wrong).
Concepts tested
Biochemistry: Biomolecules: Nucleic Acids
Which of the following is an example of reciprocal regulation of glycogen metabolism?
A. High levels of citrate stimulate phosphofructokinase and inhibit fructose-1,6-bisphosphatase.
B. High levels of AMP inhibit glycolysis and stimulate gluconeogenesis.
C. Glucagon inhibits glycogen phosphorylase and stimulates glycogen synthase.
D. Insulin stimulates glycogen synthase and inhibits glycogen phosphorylase.
D. Reciprocal regulation occurs when a single molecule stimulates a pathway in one direction while inhibiting the pathway in the opposite direction. Insulin is released when blood sugar is high; it stimulates glycogen synthase (to store glucose as glycogen) and inhibits glycogen phosphorylase (the first enzyme in glycogen breakdown, choice D is correct). Glucagon is antagonistic to insulin; it is released when blood sugar is low to stimulate glycogen phosphorylase and glycogen breakdown while inhibiting glycogen synthase (choice C is wrong). While citrate and AMP do reciprocally regulate the indicated enzymes and pathways, how they regulate those pathways is not described correctly. Citrate inhibits phosphofructokinase to slow down glycolysis and stimulates fructose-1,6-bisphosphatase to increase gluconeogenesis (choice A is wrong); AMP acts to stimulate glycolysis (because ATP is low) and inhibit gluconeogenesis (choice B is wrong). And in either case, this is not regulation of glycogen metabolism.
Concepts tested
Biochemistry: Carbohydrate Metabolism
Which of the following statements about Km is true?
A. Km is the substrate concentration required to reach 1/2 Vmax, and is a measure of affinity between an enzyme and its substrate.
B. Km is a measure of affinity between an enzyme and its substrate; a high Km indicates a strong affinity.
C. Km is the substrate concentration required to reach Vmax, and a low Km indicates a high enzyme-substrate affinity.
D. Km is affected by enzyme concentration; if the enzyme concentration is reduced, Km decreases.
A. Km is the substrate concentration required to push a reaction to half of its maximum rate of product formation (choice C is wrong), and is a measure of affinity between an enzyme and its substrate. However a low Km indicates a high enzyme-substrate affinity, not the other way around. In other words, a low Km means that only a small amount of substrate is required to get to 1/2 Vmax, thus there must be a high affinity between the enzyme and the substrate. If Km is high, a lot of substrate is needed to reach 1/2 Vmax, indicating a low enzyme-substrate affinity (choice B is wrong). Km is unaffected by enzyme concentration; just because the amount of enzyme changes, does not mean its affinity for its substrate changes (choice D is wrong).
Concepts tested
Biochemistry: Enzymes and Enzyme Inhibition
