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Haldane Substrate-Inhibition Enzyme & Bioreactor Kinetics Formula

Explore a substrate-inhibition rate curve, its maximizing concentration and its actual peak. Distinguish the fitted rate-scale parameter from the peak and from a reactor feed concentration.

A rate model with a declining high-concentration branch

This worksheet uses v=V S/(Km+S+S²/Ki), often called a Haldane or Andrews substrate-inhibition model. Its use for a particular enzyme or microbial culture requires suitable data; a fitted curve does not by itself establish a molecular mechanism.

B2 is a positive rate-scale parameter V, often labeled Vmax in this parameterization. With finite positive Km and Ki it is not the attained peak of this inhibited curve. B3, B4 and B5 must share concentration units; the example uses mM. The output has the same rate units as B2.

Evaluate the rate at the current substrate concentration

B10 is the rate at the entered B5 concentration. The illustrative parameters V=120, Km=5 and Ki=80 give v=80 at S=20, and v≈52.173913 at S=100. At S=0 the predicted rate is zero.

For enzyme kinetics the rate unit might be amount per time; for specific microbial growth it is inverse time. Do not interpret the example value 120 as a recommended microbial growth parameter or mix those meanings.

=IF(B7<>"OK",B7,IF(B8<>"OK",B8,B2*B5/(B3+B5+B5^2/B4)))

Find the maximum of this curve

Differentiating the rate expression gives dv/dS=V(Km−S²/Ki)/(Km+S+S²/Ki)². With positive parameters, its sign changes from positive to negative at S*=SQRT(Km*Ki). B11 gives this concentration.

B12 gives the corresponding peak V/(1+2*SQRT(Km/Ki)). For the example, S*=20 mM and the peak is 80 rate units, below the parameter V=120. Inhibition is already present at S*; it is incorrect to call this a non-inhibiting concentration.

=IF(B7<>"OK",B7,B2/(1+2*SQRT(B3/B4)))

Experiment without treating the maximum as a feed setpoint

Try S=10, 20 and 40: the rates are approximately 73.846154, 80 and 73.846154. Try doubling V: both the current rate and the peak double, but S* stays unchanged. Quadruple Ki: S* doubles and the peak rises to 96.

S is the concentration experienced by the enzyme or organisms, not automatically the inlet/feed concentration. Selecting a reactor operating point also requires mass balances, residence time, biomass, stability and the chosen performance objective. Maximizing this kinetic curve alone does not maximize reactor throughput.

Validation and precision

The worksheet requires positive numeric V, Km and Ki, and nonnegative numeric S. Zero V is excluded so that the maximizing concentration remains a unique peak rather than an everywhere-flat zero curve.

Invalid substrate input blocks B10 while the parameter-only outputs B11 and B12 remain available. Invalid kinetic parameters block all outputs. Formulas retain full precision; use number formatting for display rather than rounding intermediate values.

Try the example

Use the original values or change them to test your understanding.

Further reading