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Enzyme rates and inhibitors

14 min

  • Predict how enzyme reaction rate changes as substrate concentration rises
  • Tell competitive from noncompetitive inhibition by their effects on Km and Vmax

Read the full reference

Try first

Try first

An enzyme has a Km of 2 mmol/L. In an assay, the substrate concentration is raised from 20 to 40 mmol/L. How does the reaction rate change?

The next section explains it.

The next section explains it.

Right. The next section explains why.

The next section explains it.

Get the idea

The rate levels off

An enzyme binds its substrate in a reversible complex before releasing product:1

E + S ES → E + P

At low substrate, the rate rises nearly in proportion to substrate. As the active sites fill, the rate levels off toward a maximum velocity, Vmax. The Michaelis-Menten equation describes the curve:1

v = (Vmax × [S]) ÷ (Km + [S])

Km is the substrate concentration at which the rate is half of Vmax.1 Three points on the curve are worth knowing:

  • At [S] = Km, v is 50% of Vmax.
  • At [S] = 3 × Km, v is 75% of Vmax.
  • At [S] = 10 × Km, v is about 91% of Vmax.

Why activity assays saturate the enzyme

An activity assay measures the initial rate with substrate and cofactors in excess. At saturating substrate, the rate depends on the amount of enzyme in the specimen and hardly at all on the substrate.2 When a specimen holds so much enzyme that it uses up the substrate during the read, the rate falls away and the activity reads falsely low. A validated dilution restores the excess.2

Inhibitors change Km or Vmax

The idealized inhibition patterns differ in where the inhibitor binds:1

PatternWhere the inhibitor bindsVmaxApparent KmAdded substrate
CompetitiveFree enzyme, at the active siteUnchangedIncreasedRestores the rate toward Vmax
Pure noncompetitiveFree enzyme and the ES complex equallyDecreasedUnchangedCannot restore Vmax
UncompetitiveThe ES complexDecreasedDecreasedCannot restore Vmax

A competitive inhibitor competes with substrate for the active site, so enough substrate outcompetes it. A pure noncompetitive inhibitor takes some enzyme out of action at any substrate concentration.1

References
  1. International Union of Biochemistry and Molecular Biology. Recommendations on biochemical and organic nomenclature, symbols and terminology: symbolism and terminology in enzyme kinetics. Accessed September 27, 2026. https://iubmb.qmul.ac.uk/kinetics/
  2. Bishop ML, Fody EP, Van Siclen C, Mistler JM, Moy M. Clinical Chemistry: Principles, Techniques, and Correlations. 9th ed. Jones & Bartlett Learning; 2023.

Watch one

An enzyme has a Vmax of 200 U/L and a Km of 4 mmol/L. An assay design team compares substrate concentrations of 2, 4, 20 and 40 mmol/L. What rate does the model give at 40 mmol/L, and why does the assay run there?

  1. Write the model: v = (200 U/L × [S]) ÷ (4 mmol/L + [S]).

    The equation sets the rate at every substrate concentration.

  2. At [S] = 4 mmol/L: v = 800 ÷ 8 = 100 U/L, half of Vmax.

    Km is defined by the half-maximal rate, so it is a quick check on the setup.

  3. At [S] = 2 mmol/L: v = 400 ÷ 6 = 66.7 U/L. Doubling substrate from 2 to 4 mmol/L raised the rate 1.5 times.

    Below Km, the rate depends strongly on substrate.

  4. At [S] = 20 mmol/L: v = 4,000 ÷ 24 = 166.7 U/L. At [S] = 40 mmol/L: v = 8,000 ÷ 44 = 181.8 U/L. Doubling substrate here raises the rate by 9%.

    Far above Km, the curve is flat.

  5. Choose the flat part of the curve: at 40 mmol/L, ten times Km, the rate depends on the enzyme in the specimen.

    An activity assay should report the amount of enzyme, so small changes in substrate must barely move the rate.

At 40 mmol/L the rate is 182 U/L, 91% of Vmax. The assay runs there because the enzyme is nearly saturated and its amount sets the rate.

Your turn

Problem 1 of 3

In a Michaelis-Menten model, Vmax is 120 U/L and Km is 4 mmol/L. What is the reaction velocity (v) at a substrate concentration of 8 mmol/L?

Correct. v = (120 × 8) ÷ (4 + 8) = 960 ÷ 12 = 80 U/L, two-thirds of Vmax.

Incorrect. (120 × 8) ÷ 4 = 240 U/L leaves substrate out of the denominator and gives a rate above Vmax, which the model never reaches.

Incorrect. That is half of Vmax, the velocity when substrate equals Km. Here substrate is twice Km.

Hint
  1. Put the numbers into v = (Vmax × [S]) ÷ (Km + [S]).
  2. The substrate appears in both the top and the bottom of the equation.
  3. No answer can be above Vmax.

Review Reaction rates

Problem 2 of 3

Which change describes ideal competitive inhibition?

Incorrect. That is the ideal pure noncompetitive pattern, in which the inhibitor binds free enzyme and the enzyme-substrate complex with equal affinity.

Correct. The inhibitor binds free enzyme at the active site, so added substrate can restore the rate toward the unchanged Vmax.

Incorrect. That is the ideal uncompetitive pattern, in which the inhibitor binds the enzyme-substrate complex.

Hint
  1. Ask where a competitive inhibitor binds.
  2. Ask what happens to that inhibitor when a large amount of substrate is added.

Review Idealized inhibition patterns

Problem 3 of 3

Without inhibitor, an enzyme shows a Vmax of 150 U/L and a Km of 3 mmol/L. With compound X present, the rate levels off at 75 U/L and reaches half of that at 3 mmol/L substrate. Adding more substrate does not raise the plateau. Which pattern fits?

A competitive inhibitor raises the apparent Km and leaves Vmax unchanged, and added substrate restores the rate. Here Vmax fell and Km stayed the same.

Swapped competitive and noncompetitive inhibition effects

Competitive inhibition raises apparent Km and leaves Vmax unchanged, so added substrate restores the rate. Pure noncompetitive inhibition lowers Vmax and leaves Km unchanged. Swapping them predicts the wrong response to added substrate.

Vmax fell from 150 to 75 U/L, and Km stayed at 3 mmol/L. Added substrate cannot restore enzyme that the inhibitor has taken out of action.

An uncompetitive inhibitor lowers both Vmax and the apparent Km. Here Km is unchanged.

Review Idealized inhibition patterns

Use it

  • A 24-year-old man, MRN 3805416, arrives in the emergency department after a long crush injury to his leg.
  • The Meniscus CX-800 reports creatine kinase (CK) at 1,850 U/L. The method's analytical measurement range is 10 to 2,000 U/L.
  • The analyzer flags substrate depletion: the absorbance changed fast at first and then flattened during the read window.
  • The serum is clear, and QC is acceptable.
Decision 1 of 3

What does the flattened reaction curve tell you?

Once substrate is no longer in excess, the rate falls away as the read goes on. The reported activity then sits below the true CK.

An inhibitor would slow the rate from the start. This curve ran fast and then flattened, the shape of substrate running out.

The reading falls inside the range because the substrate ran out. The analyzer's flag shows that the rate stopped following the amount of enzyme.

Review Reaction rates

Decision 2 of 3

What do you do with the CK?

A number known to be falsely low would still reach the chart. A dilution gives the true activity from the same tube.

The specimen is sound. The problem is too much enzyme for the substrate, which a dilution solves.

The substrate-depletion flag shows the value understates the CK, and the true activity may be many times higher.

Diluting the serum lowers the enzyme so the substrate stays in excess for the whole read. The dilution factor then returns the true activity.

Review Reaction rates

Decision 3 of 3

The 1:10 dilution reads 1,420 U/L with no flag. What CK do you report?

That result came from a reaction that ran out of substrate. It understates the CK.

1,420 U/L × 10 = 14,200 U/L. The diluted reaction kept its substrate in excess, so its rate reflects the enzyme.

The reading describes serum diluted tenfold. It is multiplied by the dilution factor before it is reported.

Review Reaction rates

The clue that settled this case is the reaction curve that flattened during the read. It showed the substrate ran out, so the first result understated the CK. A tenfold dilution kept the substrate in excess and gave 14,200 U/L.

Keep

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