Question

Question #16084

Answer 1

The turnover number for your enzyme is `"90.9 min"^(-1)`.

The turnover number, or `K_"cat"`, expresses how many substrate molecules are converted to product by the enzyme per unit time. Moreover, the turnover number is calculated when the enzyme has maximum efficiency `->` `V_"max"`.

`K_"cat" = V_"max"/([E])`, where

`[E]` - the total concentration of enzyme present.

Since `V_"max"` uses `mu"mol"` and the concentration of the enzyme is given in `"nmol"`, I'll convert the latter to `mu"mol"`

`121cancel("nmol") * (10^(-3)mu"mol")/(1cancel("nmol")) = 121 * 10^(-3)mu"mol"`

Once again, plug and play

`K_"cat" = (11.0cancel(mu"mol") * "min"^(-1))/(121 * 10^(-3)cancel(mu"mol")) = "90.9 min"^(-1)`