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Enzyme rates and inhibitors

14 min

Skills you'll build2 skills
  • Predict how enzyme reaction rate changes as substrate concentration rises
  • Tell competitive from noncompetitive inhibition by their effects on Km and Vmax

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?

Keep going to see the reasoning.

Keep going to see the reasoning.

Right. Keep going to follow the reasoning.

Keep going to see the reasoning.

Get the idea

Use the explanation to check your first decision.

Michaelis-Menten kinetics and saturation

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 reference points on the curve show how quickly the rate approaches Vmax:

  • 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.

Substrate excess in enzyme activity assays

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 very little on the substrate concentration.2 When a specimen contains so much enzyme that it depletes the substrate during the read, the measured rate falls and the reported activity is falsely low. A validated dilution restores the excess.2

Patterns of enzyme inhibition

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 inactivates part of the enzyme at any substrate concentration.1

Follow the bench demonstration

Sources for this step
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

Read each decision and its explanation.

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 gives the rate at each 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 measure the amount of enzyme, so small changes in substrate should have little effect on 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 the amount of enzyme determines the rate.

Practice this skill

Your turn

Problem 1 of 3 · Hints available

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 · Hints available

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 · On your own

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 inactivated.

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

Review Idealized inhibition patterns

Use it

Read the case evidence, then make each decision. The step is complete without 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 show?

Once substrate is no longer in excess, the rate falls during the read. The reported activity is then below the true CK.

An inhibitor would slow the rate from the start. This curve was fast at first and then flattened, the pattern of substrate depletion.

The reading is within the range because the substrate was depleted. The analyzer's flag shows that the rate no longer reflected the amount of enzyme.

Review Reaction rates

Decision 2 of 3

What do you do with the CK?

A result known to be falsely low would still be reported. 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. Multiplying by the dilution factor then gives 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 Concentration and patient-result calculation

The reaction curve flattened during the read, which showed substrate depletion, so the first result understated the CK. A tenfold dilution kept the substrate in excess and gave 14,200 U/L.

For reference

Sources checked