Enzyme Kinetics: Vmax and the Michaelis Constant · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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What Vmax and Km mean

Science · Organisms & Life Processes · ages 16-18
Name ______________________   Date ____________

  1. How do you find Km on a Michaelis-Menten rate curve?

    • Find half Vmax, meet the curve, then read substrate concentration
    • Find half the largest substrate value, then read the rate above it
    • Find the high-substrate plateau, then use its rate value as Km

  2. At fixed enzyme concentration, how do you estimate Vmax from a Michaelis-Menten rate curve?

    • Read the substrate concentration at the curve's halfway height
    • Read the limiting rate at high substrate concentration
    • Read the slope near zero substrate as the maximum rate

  3. For a Michaelis-Menten enzyme at fixed enzyme concentration, Vmax is the limiting initial rate as the active sites become saturated.

    Circle one:   True   False

  4. For the same substrate under matching conditions, enzyme A has Km = 2 mM and enzyme B has Km = 8 mM. What follows?

    • A reaches half its own Vmax at a lower substrate concentration
    • B reaches half its own Vmax at a lower substrate concentration
    • A must bind more tightly, regardless of all its later reaction steps

  5. Why does adding substrate raise the rate strongly in the low-substrate region of a Michaelis-Menten curve?

    • More substrate permanently changes the enzyme's folded structure
    • More substrate makes the enzyme concentration increase during each trial
    • Many sites are free, so substrate binding becomes more frequent

  6. An enzyme has Vmax = 80 rate units. What rate, in rate units, should you locate on the vertical axis before reading Km?

    Answer: ______________

  7. A student reports Km as 30 micromoles per minute because that is a Michaelis-Menten curve's limiting rate. What needs changing?

    • Keep 30 micromoles per minute as Km and label the horizontal axis time
    • Use the limiting rate as both constants because their units are identical
    • Call that rate Vmax; read Km as concentration at half that rate

  8. An enzyme is nearly saturated. A student predicts that doubling substrate will double its initial rate. Enzyme concentration and other conditions stay fixed. Which explanation challenges that prediction?

    • Doubling substrate must make twice as much enzyme available
    • Most sites are occupied, so turnover limits rate
    • Doubling substrate must permanently destroy the active sites
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Answer key

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What Vmax and Km mean W1-mt_ooF8eZ1Gzy-s1

  1. Find half Vmax, meet the curve, then read substrate concentration · Km is the substrate concentration at which the initial rate is Vmax/2.
  2. Read the limiting rate at high substrate concentration · Vmax is a rate: the vertical value approached as substrate concentration becomes high.
  3. True · As substrate saturates the enzyme, the initial rate approaches Vmax.
  4. A reaches half its own Vmax at a lower substrate concentration · A reaches half its own Vmax at 2 mM; B does so at 8 mM. These values alone do not prove which binds more tightly.
  5. Many sites are free, so substrate binding becomes more frequent · At low substrate concentration, many sites are free. Increasing substrate makes binding more frequent and raises the rate.
  6. 40 · Reading Km starts at Vmax/2. Here, 80 divided by 2 is 40 rate units.
  7. Call that rate Vmax; read Km as concentration at half that rate · 30 micromoles per minute is Vmax. Km is the substrate concentration where the rate is 15 micromoles per minute.
  8. Most sites are occupied, so turnover limits rate · Near saturation, most sites are occupied much of the time. More substrate gives only a small increase toward Vmax.
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