Deriving the Michaelis-Menten Equation from the Steady State · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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Where the rate equation comes from

Science · Biochemistry & Molecular Biology · ages 18-19
Name ______________________   Date ____________

  1. The steady-state balance is k1[E][S] = (k-1 + k2)[ES]. Total enzyme is [E]T = [E] + [ES]. Which substitution starts the derivation?

    • Replace [E] with [E]T + [ES] in the formation term
    • Replace [E] with [ES] - [E]T in the formation term
    • Replace [E] with [E]T - [ES] in the formation term

  2. ES forms by substrate binding and is lost by dissociation or product formation. What does the steady-state approximation say?

    • ES formation stops while product formation continues unchanged
    • ES formation approximately balances both routes of ES loss
    • ES formation exceeds both losses throughout the measurement

  3. In the Michaelis-Menten model, increasing substrate at fixed active enzyme concentration and fixed rate constants can raise the initial rate above Vmax.

    Circle one:   True   False

  4. In the one-site, one-complex Michaelis-Menten model, which expression sets Vmax?

    • k2 times total active enzyme concentration
    • k1 times total substrate concentration
    • k-1 times accumulated product concentration

  5. Expand k1([E]T - [ES])[S] = (k-1 + k2)[ES], then collect every [ES] term on the right. Which equation results?

    • k1[E]T[S] = (k1[S] - k-1 - k2)[ES]
    • k1[E]T[S] = (k1[S] + k-1 + k2)[ES]
    • k1[E]T[S] = (k1[S] + k-1 - k2)[ES]

  6. Which measurement window usually supports the initial-rate steady-state derivation when substrate greatly exceeds enzyme?

    • During ES buildup immediately upon mixing the reactants
    • After ES buildup, with little substrate loss or product buildup
    • After substrate depletion, when product has accumulated substantially

  7. The collected steady-state balance is [E]T[S] = (Km + [S])[ES]. Using v = k2[ES], which rate equation follows?

    • v = k2[E]T[S]/(Km - [S])
    • v = k2[E]T[S]/(Km + [E]T)
    • v = k2[E]T[S]/(Km + [S])

  8. A mutation increases k2 while k1 and k-1 stay fixed. What happens to Km = (k-1 + k2)/k1?

    • It increases because the numerator increases
    • It decreases because the denominator increases
    • It is unchanged because only binding can affect Km
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Where the rate equation comes from W1-mt_BFp0fwZ8_i-s1

  1. Replace [E] with [E]T - [ES] in the formation term · Subtract [ES] from the total-enzyme equation to obtain [E] = [E]T - [ES].
  2. ES formation approximately balances both routes of ES loss · Formation balances total ES loss, so the complex concentration changes very little.
  3. False · For positive Km, [S]/(Km + [S]) is below 1 at finite substrate concentration, so v cannot exceed Vmax.
  4. k2 times total active enzyme concentration · Vmax = k2[E]T: the product-formation constant multiplied by total active enzyme concentration.
  5. k1[E]T[S] = (k1[S] + k-1 + k2)[ES] · Expansion creates -k1[ES][S] on the left. Add k1[ES][S] to both sides, then factor [ES] from all three terms on the right.
  6. After ES buildup, with little substrate loss or product buildup · Measure after the ES transient, while substrate depletion and product effects remain small.
  7. v = k2[E]T[S]/(Km + [S]) · Divide the balance by Km + [S] to find [ES], then multiply that expression by k2 to obtain the initial rate.
  8. It increases because the numerator increases · Increasing k2 increases the numerator while k1 stays fixed, so Km increases.
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