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Chemistry LibreTexts

Turnover Number

In enzyme kinetics, we are interested to know how many maximum molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit time.

\[ k_{cat} =\dfrac{ V_{max}}{E_t} \]

with

  • \(k_{cat}\) = Turnover number,
  • \(V_{max}\) = Maximum rate of reaction when all the enzyme catalytic sites are saturated with substrate and
  • \(E_t\) =Total enzyme concentration or concentration of total enzyme catalytic sites.

The units of Turn over number (kcat) are \(k_{cat}\) = (moles of product/sec)/ (moles of enzyme) or sec-1.

Catalytic efficiency of enzyme

We also want to know the efficiency of our enzyme to use it in laboratory experiment. It can be use to compare different enzymes or it same enzyme for different substrates in order to make sure that we use the best possible enzyme for our experiment in laboratory. Theoretically the maximum value for the catalytic efficiency can be determined by diffusion limit because as soon as enzyme and substrate can come together it can make the products. An enzyme with 109 units of catalytic efficiency is very efficient enzyme to be use in the laboratory.

Contributors

  • Kiranpreet Kaur