PCB 3023C | PCB3023C Exam 2: Cell Biology
Updated and Latest Questions and Correct
Answers with Rationale - Florida Gulf Coast
University
1. Which kinetic parameter is affected when a competitive inhibitor binds to an enzyme?
A. The Vmax decreases while Km remains the same
B. The Vmax increases while Km decreases
C. Both Vmax and Km decrease significantly
D. The Km increases while Vmax remains the same
Correct Answer: D
Expert Explanation: Competitive inhibitors compete with the substrate for the active site
of the enzyme. This competition leads to an increase in the apparent Km because higher
substrate concentrations are needed to reach half the maximal velocity. The Vmax remains
unchanged because a sufficiently high concentration of substrate can outcompete the
inhibitor. Non-competitive inhibitors would instead lower the Vmax without changing the
Km. Understanding these kinetic shifts is essential for analyzing metabolic regulation and
drug interactions.
2. Which of the following describes the net energy yield from one molecule of glucose during
glycolysis?
A. 4 ATP and 2 NADH
B. 2 ATP and 4 NADH
C. 2 ATP and 2 NADH
D. 32 ATP and 10 NADH
Correct Answer: C
Expert Explanation: Glycolysis consists of an energy investment phase and an energy
payoff phase occurring in the cytosol. Although four ATP molecules are produced, two are
consumed during the early steps, resulting in a net gain of two ATP. Additionally, two
molecules of NAD+ are reduced to NADH during the oxidation of glyceraldehyde-3-
phosphate. Other pathways like the Citric Acid Cycle contribute more significantly to NADH
production later in respiration. This net yield represents the primary energy source for
anaerobic organisms.
3. What is the primary role of Phosphofructokinase-1 (PFK-1) in cellular metabolism?
A. To act as a rate-limiting enzyme in glycolysis
,B. To catalyze the final step of the Citric Acid Cycle
C. To facilitate the transport of glucose across the membrane
D. To produce ATP through oxidative phosphorylation
Correct Answer: A
Expert Explanation: PFK-1 is the key regulatory enzyme that catalyzes the conversion of
fructose-6-phosphate to fructose-1,6-bisphosphate. It serves as a major flux control point
because it is regulated allosterically by the cell’s energy status. High levels of ATP inhibit
the enzyme, while high levels of AMP activate it to increase energy production. Citrate also
acts as an inhibitor to signal that the Citric Acid Cycle is saturated. This regulation ensures
that glucose is not broken down unnecessarily when energy stores are high.
4. During aerobic respiration, where is the highest concentration of protons (H+) located?
A. The mitochondrial matrix
B. The cytosol
C. The mitochondrial intermembrane space
D. The inner mitochondrial membrane
Correct Answer: C
Expert Explanation: The electron transport chain pumps protons from the mitochondrial
matrix into the intermembrane space. This creates a steep electrochemical gradient that
stores potential energy for ATP synthesis. The matrix actually has a lower concentration of
protons, making it more basic than the intermembrane space. This gradient is the driving
force behind the chemiosmotic coupling of oxidative phosphorylation. Without this spatial
separation, the cell could not generate the proton motive force required by ATP synthase.
5. Which molecule acts as the final electron acceptor in the mitochondrial electron transport
chain?
A. Oxygen (O2)
B. Pyruvate
C. NAD+
D. Water (H2O)
Correct Answer: A
Expert Explanation: Oxygen has a high electronegativity, allowing it to pull electrons
through the transport chain. At Complex IV, oxygen accepts electrons and combines with
protons to form water as a byproduct. If oxygen is absent, the entire chain backs up
because electrons cannot be passed forward. This cessation stops the production of the
, proton gradient and halts ATP synthesis. Therefore, aerobic organisms depend on oxygen
specifically for its role at the end of this biochemical pathway.
6. What occurs immediately after a ligand binds to a Receptor Tyrosine Kinase (RTK)?
A. The receptor is degraded by the proteasome
B. The receptor subunits dimerize and autophosphorylate
C. Cyclic AMP is produced by adenylyl cyclase
D. G-proteins exchange GDP for GTP
Correct Answer: B
Expert Explanation: Ligand binding induces a conformational change that causes two RTK
monomers to come together. Once dimerized, the intracellular kinase domains
phosphorylate each other on specific tyrosine residues. These phosphorylated tyrosines
then serve as docking sites for downstream signaling proteins like Grb2 or phospholipase
C. This mechanism is distinct from G-protein coupled receptors which use a different
activation method. RTKs are critical for growth factor signaling and cell cycle regulation.
7. How does the toxin from Vibrio cholerae affect cell signaling in the intestines?
A. It prevents the binding of ligands to GPCRs
B. It inhibits the activity of adenylyl cyclase
C. It locks the G-alpha subunit in an active, GTP-bound state
D. It destroys the secondary messenger cAMP
Correct Answer: C
Expert Explanation: Cholera toxin modifies the G-alpha subunit so that it can no longer
hydrolyze GTP into GDP. This causes the G-protein to remain constitutively active,
continuously stimulating adenylyl cyclase. The resulting overproduction of cAMP leads to
the massive efflux of chloride ions and water into the intestinal lumen. This biochemical
disruption results in the severe dehydration characteristic of the disease. It illustrates how
disrupting the ‘off switch’ in a signaling pathway can have physiological consequences.
8. Which enzyme is responsible for converting cAMP into 5’-AMP to terminate a signal?
A. Adenylyl cyclase
B. Phosphodiesterase
C. Protein kinase A
D. Phosphorylase kinase
Correct Answer: B
Updated and Latest Questions and Correct
Answers with Rationale - Florida Gulf Coast
University
1. Which kinetic parameter is affected when a competitive inhibitor binds to an enzyme?
A. The Vmax decreases while Km remains the same
B. The Vmax increases while Km decreases
C. Both Vmax and Km decrease significantly
D. The Km increases while Vmax remains the same
Correct Answer: D
Expert Explanation: Competitive inhibitors compete with the substrate for the active site
of the enzyme. This competition leads to an increase in the apparent Km because higher
substrate concentrations are needed to reach half the maximal velocity. The Vmax remains
unchanged because a sufficiently high concentration of substrate can outcompete the
inhibitor. Non-competitive inhibitors would instead lower the Vmax without changing the
Km. Understanding these kinetic shifts is essential for analyzing metabolic regulation and
drug interactions.
2. Which of the following describes the net energy yield from one molecule of glucose during
glycolysis?
A. 4 ATP and 2 NADH
B. 2 ATP and 4 NADH
C. 2 ATP and 2 NADH
D. 32 ATP and 10 NADH
Correct Answer: C
Expert Explanation: Glycolysis consists of an energy investment phase and an energy
payoff phase occurring in the cytosol. Although four ATP molecules are produced, two are
consumed during the early steps, resulting in a net gain of two ATP. Additionally, two
molecules of NAD+ are reduced to NADH during the oxidation of glyceraldehyde-3-
phosphate. Other pathways like the Citric Acid Cycle contribute more significantly to NADH
production later in respiration. This net yield represents the primary energy source for
anaerobic organisms.
3. What is the primary role of Phosphofructokinase-1 (PFK-1) in cellular metabolism?
A. To act as a rate-limiting enzyme in glycolysis
,B. To catalyze the final step of the Citric Acid Cycle
C. To facilitate the transport of glucose across the membrane
D. To produce ATP through oxidative phosphorylation
Correct Answer: A
Expert Explanation: PFK-1 is the key regulatory enzyme that catalyzes the conversion of
fructose-6-phosphate to fructose-1,6-bisphosphate. It serves as a major flux control point
because it is regulated allosterically by the cell’s energy status. High levels of ATP inhibit
the enzyme, while high levels of AMP activate it to increase energy production. Citrate also
acts as an inhibitor to signal that the Citric Acid Cycle is saturated. This regulation ensures
that glucose is not broken down unnecessarily when energy stores are high.
4. During aerobic respiration, where is the highest concentration of protons (H+) located?
A. The mitochondrial matrix
B. The cytosol
C. The mitochondrial intermembrane space
D. The inner mitochondrial membrane
Correct Answer: C
Expert Explanation: The electron transport chain pumps protons from the mitochondrial
matrix into the intermembrane space. This creates a steep electrochemical gradient that
stores potential energy for ATP synthesis. The matrix actually has a lower concentration of
protons, making it more basic than the intermembrane space. This gradient is the driving
force behind the chemiosmotic coupling of oxidative phosphorylation. Without this spatial
separation, the cell could not generate the proton motive force required by ATP synthase.
5. Which molecule acts as the final electron acceptor in the mitochondrial electron transport
chain?
A. Oxygen (O2)
B. Pyruvate
C. NAD+
D. Water (H2O)
Correct Answer: A
Expert Explanation: Oxygen has a high electronegativity, allowing it to pull electrons
through the transport chain. At Complex IV, oxygen accepts electrons and combines with
protons to form water as a byproduct. If oxygen is absent, the entire chain backs up
because electrons cannot be passed forward. This cessation stops the production of the
, proton gradient and halts ATP synthesis. Therefore, aerobic organisms depend on oxygen
specifically for its role at the end of this biochemical pathway.
6. What occurs immediately after a ligand binds to a Receptor Tyrosine Kinase (RTK)?
A. The receptor is degraded by the proteasome
B. The receptor subunits dimerize and autophosphorylate
C. Cyclic AMP is produced by adenylyl cyclase
D. G-proteins exchange GDP for GTP
Correct Answer: B
Expert Explanation: Ligand binding induces a conformational change that causes two RTK
monomers to come together. Once dimerized, the intracellular kinase domains
phosphorylate each other on specific tyrosine residues. These phosphorylated tyrosines
then serve as docking sites for downstream signaling proteins like Grb2 or phospholipase
C. This mechanism is distinct from G-protein coupled receptors which use a different
activation method. RTKs are critical for growth factor signaling and cell cycle regulation.
7. How does the toxin from Vibrio cholerae affect cell signaling in the intestines?
A. It prevents the binding of ligands to GPCRs
B. It inhibits the activity of adenylyl cyclase
C. It locks the G-alpha subunit in an active, GTP-bound state
D. It destroys the secondary messenger cAMP
Correct Answer: C
Expert Explanation: Cholera toxin modifies the G-alpha subunit so that it can no longer
hydrolyze GTP into GDP. This causes the G-protein to remain constitutively active,
continuously stimulating adenylyl cyclase. The resulting overproduction of cAMP leads to
the massive efflux of chloride ions and water into the intestinal lumen. This biochemical
disruption results in the severe dehydration characteristic of the disease. It illustrates how
disrupting the ‘off switch’ in a signaling pathway can have physiological consequences.
8. Which enzyme is responsible for converting cAMP into 5’-AMP to terminate a signal?
A. Adenylyl cyclase
B. Phosphodiesterase
C. Protein kinase A
D. Phosphorylase kinase
Correct Answer: B
