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Biochemistry Module 6 Practice Exam

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About this Exam

Prepare with the Biochemistry Module 6 Practice Exam practice quiz. This question bank includes 10 questions covering mechanism, synthesis, michaelis-menten, enzymes, and describes. Use it to review important concepts, identify knowledge gaps, and build confidence for the related exam, course, or assessment.

Sample Questions

Question 1
What are aminotransferases and which cofactor do they require?
Aminotransferases catalyze oxidation of amino acids; require FAD
Aminotransferases catalyze transfer of amino groups between amino acids and α-keto acids; require pyridoxal phosphate (PLP)
Aminotransferases catalyze decarboxylation of amino acids
Aminotransferases transfer phosphate groups
Explanation:
Aminotransferases perform the transfer of amino groups between amino acids and α-keto acids in a reversible transamination reaction. They catalyze the movement of the amino group from a donor amino acid to an acceptor α-keto acid, producing a new amino acid and a new α-keto acid. The essential cofactor is pyridoxal phosphate (PLP), derived from vitamin B6. PLP forms a Schiff base with a lysine in the enzyme and then forms an external aldimine with the amino acid substrate, allowing the amino group to be removed and reattached through stabilized intermediates. This arrangement enables efficient shuttling of the amino group between molecules. The other options describe different enzyme activities: FAD is a redox cofactor used by dehydrogenases, decarboxylases remove carboxyl groups, and kinases transfer phosphate groups.
Question 2
What percentage of ribosomes' mass is composed of rRNA?
60%
65%
30%
50%
Explanation:
Ribosomes are ribonucleoprotein complexes, meaning RNA and protein both contribute to their structure and function, but the RNA portion forms the bulk of the mass. The rRNA molecules create the scaffold and the catalytic center, so they make up roughly two-thirds of the ribosome’s mass, with ribosomal proteins contributing the remaining one-third. In both bacteria and eukaryotes, this translates to about 60–65% of the mass being rRNA, which is why the commonly cited figure is around 65%.
Question 3
What mechanism transports acetyl-CoA from mitochondria to cytosol for lipid synthesis?
Citrate shuttle exporting citrate from mitochondria to cytosol, where it is cleaved to yield acetyl-CoA.
Direct transport of acetyl-CoA across inner mitochondrial membrane.
Pyruvate transporter.
Malate shuttle.
Explanation:
Acetyl-CoA itself can’t cross the inner mitochondrial membrane, so the cell moves the carbon for lipid synthesis as citrate. In the mitochondrion, acetyl-CoA condenses with oxaloacetate to form citrate. Citrate is then shuttled out of mitochondria into the cytosol, where ATP-citrate lyase cleaves it back to acetyl-CoA and oxaloacetate. The cytosolic acetyl-CoA is then used for fatty acid synthesis. The other routes don’t work for delivering acetyl units to the cytosol: there’s no direct acetyl-CoA transporter across the inner membrane, the malate shuttle mainly transfers reducing equivalents (NADH), and the pyruvate transporter moves pyruvate rather than acetyl groups.
Question 4
Which molecule serves as the two-carbon donor in the first committed step of fatty acid synthesis?
Acetyl-CoA
Acetoacetyl-CoA
Malonyl-CoA
Succinyl-CoA
Explanation:
In fatty acid synthesis, the chain is extended by a two-carbon unit donated from malonyl-CoA during the first condensation with the starter acetyl group. Malonyl-CoA is decarboxylated as it condenses with the growing chain on the fatty acid synthase complex, delivering two carbons and forming a four-carbon product. This decarboxylative condensation step is what makes malonyl-CoA the two-carbon donor in the initial committed step. Acetyl-CoA provides the initial primer, not the donor in this step. Succinyl-CoA isn’t part of this fatty acid synthesis pathway, and acetoacetyl-CoA is the product of the condensation, not the donor.
Question 5
What does Km represent in Michaelis-Menten kinetics?
Km is the substrate concentration at half-maximal velocity; lower Km indicates higher apparent affinity.
Km is the maximum rate.
Km is the turnover number.
Km is the substrate concentration at half-minimal velocity.
Explanation:
Km tells you the substrate amount needed to push the reaction to half of its maximum speed. In Michaelis-Menten terms, v = Vmax[S] / (Km + [S]); when [S] equals Km, the velocity is Vmax/2. This makes Km a practical measure of how tightly the enzyme binds the substrate: a smaller Km means the enzyme reaches half-max at a lower substrate concentration, i.e., higher apparent affinity. It is not the maximum rate (that’s Vmax) nor the turnover number (kcat). The standard description is the substrate concentration at half-maximal velocity.

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Additional Information

Biochemistry Module 6 Practice Exam

This practice set contains 10 questions from the matching question bank and focuses on mechanism, synthesis, michaelis-menten, enzymes, and describes. Work through each question carefully, review the provided solutions, and revisit topics that need more study before your next attempt.

This is an independent study resource intended for practice and review; it is not an official examination or an endorsement by any organization named in the title.

Frequently Asked Questions

This quiz contains a total of 10 practice questions carefully selected to test your knowledge on this subject.
Yes, you will have exactly 0 minutes to complete the exam. A countdown timer will be visible once you start.
Yes, you can retake this practice test as many times as you need. The questions and options may be randomized on subsequent attempts to ensure comprehensive learning.

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