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Advanced Biochemistry Board Review: Master Complex Metabolic Pathways & Molecular Mechanism Practice Questions & Detailed Explanations

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Advanced Biochemistry Board Review: Master Complex Metabolic Pathways & Molecular Mechanism Practice Questions & Detailed Explanations

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Advanced Biochemistry Board Review:
Master Complex Metabolic Pathways &
Molecular Mechanism Practice Questions &
Detailed Explanations
Subject: Advanced Biochemistry

Subtopic: Enzyme Kinetics, Metabolic Regulation, and Molecular Mechanisms
(Questions 1–30)

Question 1: An uncompetitive inhibitor is added to an enzyme-catalyzed reaction. Which of the
following correctly describes the effect of this inhibitor on the apparent kinetic parameters, and
the molecular rationale behind it?

A) $V_{max}$ decreases and $K_m$ increases because the inhibitor binds to both the free
enzyme and the enzyme-substrate complex.

B) $V_{max}$ decreases and $K_m$ decreases because the inhibitor binds exclusively to the
enzyme-substrate complex, shifting the equilibrium toward the complex.

C) $V_{max}$ remains unchanged and $K_m$ increases because the inhibitor competes directly
with the substrate for the active site.

D) $V_{max}$ decreases and $K_m$ remains unchanged because the inhibitor alters the
turnover number without affecting substrate affinity.

Correct Answer: B) $V_{max}$ decreases and $K_m$ decreases because the inhibitor binds
exclusively to the enzyme-substrate complex, shifting the equilibrium toward the complex.

Explanation: Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex, not to the
free enzyme. Because it removes active ES complexes from the system, the effective $V_{max}$
decreases. Simultaneously, by depleting the ES complex, Le Chatelier's principle drives the
binding equilibrium toward more ES formation, which artificially increases the apparent affinity
of the enzyme for the substrate, thereby lowering the apparent $K_m$.

Question 2: During periods of intense starvation, the liver significantly upregulates
gluconeogenesis. A critical regulatory step involves the reciprocal control of
phosphofructokinase-1 (PFK-1) and fructose-1,6-bisphosphatase (FBPase-1). Under these
conditions, what is the phosphorylation state and activity of the bifunctional enzyme PFK -
2/FBPase-2?

A) Phosphorylated; PFK-2 domain is active, increasing fructose-2,6-bisphosphate levels.

,B) Dephosphorylated; FBPase-2 domain is active, decreasing fructose-2,6-bisphosphate levels.

C) Phosphorylated; FBPase-2 domain is active, decreasing fructose-2,6-bisphosphate levels.

D) Dephosphorylated; PFK-2 domain is active, increasing fructose-2,6-bisphosphate levels.

Correct Answer: C) Phosphorylated; FBPase-2 domain is active, decreasing fructose-2,6-
bisphosphate levels.

Explanation: Glucagon signals starvation via cyclic AMP and protein kinase A (PKA). PKA
phosphorylates the bifunctional enzyme PFK-2/FBPase-2. Phosphorylation inhibits the PFK-2
domain and activates the FBPase-2 domain. This leads to a drop in fructose-2,6-bisphosphate
levels, relieving the inhibition on FBPase-1 and shutting down PFK-1, thereby promoting
gluconeogenesis.

Question 3: A patient presents with a severe, hereditary metabolic defect resulting in a non-
functional pyruvate carboxylase enzyme. Which of the following intermediates would
experience the most immediate and profound depletion in the hepatic mitochondria during
fasting?

A) Acetyl-CoA

B) Oxaloacetate

C) Succinate

D) Malate

Correct Answer: B) Oxaloacetate

Explanation: Pyruvate carboxylase is a critical anaplerotic enzyme that converts pyruvate
directly into oxaloacetate (OAA) within the mitochondria, requiring ATP and biotin. During
fasting, OAA is heavily consumed for gluconeogenesis. Without pyruvate carboxylase, the liver
cannot replenish OAA pools from glycolytic products, leading to its immediate exhaustion. While
malate is also downstream, OAA is the direct product.

Question 4: Consider the electron transport chain (ETC) and oxidative phosphorylation. If a
novel xenobiotically engineered molecule acts as a pure protonophore that selectively increases
the permeability of the inner mitochondrial membrane to $H^+$ ions, what will be the observed
effect on oxygen consumption and ATP synthesis?

A) Oxygen consumption stops; ATP synthesis stops.

B) Oxygen consumption increases; ATP synthesis increases.

C) Oxygen consumption increases; ATP synthesis decreases or stops.

,D) Oxygen consumption decreases; ATP synthesis remains constant.

Correct Answer: C) Oxygen consumption increases; ATP synthesis decreases or stops.

Explanation: A protonophore acts as an uncoupler of oxidative phosphorylation. It dissipates the
proton motive force ($\Delta p$) across the inner mitochondrial membrane by allowing protons
to leak back into the matrix without passing through ATP synthase ($F_oF_1$ complex).
Because the feedback inhibition of a high proton gradient on the ETC is removed, electron
transport and oxygen consumption run at maximal rates, while ATP synthesis drops drastically
or halts entirely.

Question 5: A mutant strain of E. coli expresses a version of DNA Polymerase III that lacks $3'
\rightarrow 5'$ exonuclease activity. What is the most likely phenotypic consequence of this
mutation?

A) Complete cessation of DNA replication on the lagging strand.

B) Inability to remove RNA primers synthesized by primase.

C) A drastically elevated spontaneous mutation frequency.

D) Failure to separate parental DNA strands at the replication fork.

Correct Answer: C) A drastically elevated spontaneous mutation frequency.

Explanation: The $3' \rightarrow 5'$ exonuclease activity of DNA Polymerase III is its
proofreading mechanism, which recognizes and excises mismatched nucleotides immediately
after incorporation. Lacking this activity does not arrest replication elongation, but it eliminates
the primary error-correction machinery during synthesis, resulting in a dramatic increase in
replication errors and a high mutation rate.

Question 6: The regulation of glycogen phosphorylase in skeletal muscle involves fine-tuned
allosteric and covalent modifications. During a sudden, vigorous sprint, which cellular signal
initiates the fastest, immediate allosteric activation of glycogen phosphorylase $b$ before
hormonal signaling can take full effect?

A) High levels of Glucose-6-phosphate

B) High levels of AMP

C) High levels of ATP

D) Low levels of intracellular Calcium ($Ca^{2+}$)

Correct Answer: B) High levels of AMP

, Explanation: Glycogen phosphorylase exists in a less active '$b$' form and a more active '$a$'
form. Under sudden metabolic stress/muscle contraction, rapid ATP hydrolysis generates large
amounts of AMP. AMP acts as a powerful allosteric activator of glycogen phosphorylase b,
shifting it to the active R-state without requiring covalent phosphorylation by phosphorylase
kinase, providing an immediate source of glucose-1-phosphate.

Question 7: A specific tRNA contains the anticodon sequence $5'\text{-IAU-}3'$, where I
represents the modified base inosine. According to Wobble hypothesis rules, which of the
following codons in an mRNA transcript can this tRNA pair with during translation?

A) $5'\text{-AUA-}3'$ only

B) $5'\text{-AUG-}3'$, $5'\text{-AUC-}3'$, and $5'\text{-AUU-}3'$

C) $5'\text{-AUA-}3'$, $5'\text{-AUC-}3'$, and $5'\text{-AUU-}3'$

D) $5'\text{-UAU-}3'$ and $5'\text{-UAC-}3'$

Correct Answer: C) $5'\text{-AUA-}3'$, $5'\text{-AUC-}3'$, and $5'\text{-AUU-}3'$

Explanation: According to Crick's Wobble Hypothesis, a hypoxanthine/inosine (I) base at the
$5'$ position of the anticodon (the wobble position) can form unconventional hydrogen bonds
with adenine (A), uracil (U), or cytosine (C) at the $3'$ position of the mRNA codon. The mRNA
codon must match in the first two positions ($5'\text{-AU-}3'$), allowing pairing with $5'\text{-
AUA-}3'$, $5'\text{-AUC-}3'$, and $5'\text{-AUU-}3'$.

Question 8: The primary regulatory step of fatty acid de novo synthesis is catalyzed by Acetyl-
CoA Carboxylase (ACC). Which of the following accurately describes the short-term allosteric
and covalent regulation that maximizes ACC activity?

A) Activation by palmitoyl-CoA and phosphorylation by AMPK.

B) Activation by citrate and dephosphorylation by Protein Phosphatase 2A.

C) Inhibition by citrate and phosphorylation by Protein Kinase A.

D) Activation by AMP and dephosphorylation by Protein Phosphatase 1.

Correct Answer: B) Activation by citrate and dephosphorylation by Protein Phosphatase
2A.

Explanation: Acetyl-CoA Carboxylase (ACC) is active as a polymer and inactive as a monomer.
Citrate (signaling a high energy state and abundant carbon precursors) acts as an allosteric
activator that promotes polymerization. Covalently, ACC is inactivated by phosphorylation via
AMP-activated protein kinase (AMPK) during low energy states, and activated by
dephosphorylation via protein phosphatases stimulated by insulin signaling.

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