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Shortcut used: Identify the tested biochemical principle, track charge/carbon/electrons or pathway direction, use the passage data, and reject choices that contradict cellular location or chemical behavior.
Solve in 3 steps:
- Mark the given values and the exact target.
- Apply the relevant formula, rule, or concept.
- Check the final value against the options.
Common mistake: Do not choose an option before confirming the exact demand of the question.
Question 2
Multiple Choice Question
Question
PASSAGE 1 — ENZYME KINETICS
Researchers study an enzyme that converts substrate S to product P. Without inhibitor, the enzyme has Vmax = 100 micromoles per minute and Km = 2 mM. In the presence of compound X, Vmax remains 100 micromoles per minute, but the apparent Km rises to 6 mM. No irreversible loss of enzyme activity is detected after X is removed.
QUESTION 2: In the presence of X, what is the reaction velocity when [S] = 6 mM?
Explanation
Correct Answer: B
ANSWER
Correct Answer: B
Method / Topic: Biochemistry
The correct answer is B: 50 micromoles/min. Use Michaelis–Menten kinetics with the apparent Km: v = Vmax[S]/(Km + [S]) = 100 × 6/(6 + 6) = 50 micromoles per minute. A substrate concentration equal to Km always gives half of Vmax.
Why this answer is correct
The correct option matches the required rule and the given information.
Short Explanation
The correct answer is B: 50 micromoles/min. Use Michaelis–Menten kinetics with the apparent Km: v = Vmax[S]/(Km + [S]) = 100 × 6/(6 + 6) = 50 micromoles per minute. A substrate concentration equal to Km always gives half of Vmax.
Detailed Explanation
Question
PASSAGE 1 — ENZYME KINETICS
Researchers study an enzyme that converts substrate S to product P. Without inhibitor, the enzyme has Vmax = 100 micromoles per minute and Km = 2 mM. In the presence of compound X, Vmax remains 100 micromoles per minute, but the apparent Km rises to 6 mM. No irreversible loss of enzyme activity is detected after X is removed.
QUESTION 2: In the presence of X, what is the reaction velocity when [S] = 6 mM?
ANSWER
Correct Answer: B
Method / Topic: Biochemistry
Use Michaelis–Menten kinetics with the apparent Km: v = Vmax[S]/(Km + [S]) = 100 × 6/(6 + 6) = 50 micromoles per minute. A substrate concentration equal to Km always gives half of Vmax. To solve this kind of MCAT biochemistry
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Question 2
Learn + Solve: Shortcut, Trap & Fast Route Click to view shortcut, steps, and common mistake
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Shortcut used, common trap, and the fastest solving route
Shortcut used: Identify the tested biochemical principle, track charge/carbon/electrons or pathway direction, use the passage data, and reject choices that contradict cellular location or chemical behavior.
Solve in 3 steps:
- Mark the given values and the exact target.
- Apply the relevant formula, rule, or concept.
- Check the final value against the options.
Common mistake: Do not choose an option before confirming the exact demand of the question.
Question
PASSAGE 1 — ENZYME KINETICS Researchers study an enzyme that converts substrate S to product P. Without inhibitor, the enzyme has Vmax = 100 micromoles per minute and Km = 2 mM. In the presence of compound X, Vmax remains 100 micromoles per minute, but the apparent Km rises to 6 mM. No irreversible loss of enzyme activity is detected after X is removed. QUESTION 2: In the presence of X, what is the reaction velocity when [S] = 6 mM?
Explanation
Correct Answer: B
ANSWER
Correct Answer: B
Method / Topic: Biochemistry
The correct answer is B: 50 micromoles/min. Use Michaelis–Menten kinetics with the apparent Km: v = Vmax[S]/(Km + [S]) = 100 × 6/(6 + 6) = 50 micromoles per minute. A substrate concentration equal to Km always gives half of Vmax.
Why this answer is correct
The correct option matches the required rule and the given information.
Short Explanation
The correct answer is B: 50 micromoles/min. Use Michaelis–Menten kinetics with the apparent Km: v = Vmax[S]/(Km + [S]) = 100 × 6/(6 + 6) = 50 micromoles per minute. A substrate concentration equal to Km always gives half of Vmax.
Detailed Explanation
Question
PASSAGE 1 — ENZYME KINETICS
Researchers study an enzyme that converts substrate S to product P. Without inhibitor, the enzyme has Vmax = 100 micromoles per minute and Km = 2 mM. In the presence of compound X, Vmax remains 100 micromoles per minute, but the apparent Km rises to 6 mM. No irreversible loss of enzyme activity is detected after X is removed.
QUESTION 2: In the presence of X, what is the reaction velocity when [S] = 6 mM?
ANSWER
Correct Answer: B
Method / Topic: Biochemistry
Use Michaelis–Menten kinetics with the apparent Km: v = Vmax[S]/(Km + [S]) = 100 × 6/(6 + 6) = 50 micromoles per minute. A substrate concentration equal to Km always gives half of Vmax. To solve this kind of MCAT biochemistry
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