# Question #8227c

Mar 6, 2015

The specific activity of your sample will be $3.0 \cdot {10}^{2} {\text{ nmol" * "min"^(-1) * "mg}}^{- 1}$.

First, the enzyme reaction mixture's activity is a measure of how many moles of substrate are converted per unit of time. In your case, the enzyme mixture converts 0.0252 $\mu \text{mol}$ of substrate per minute.

Specific activty, on the other hand, is defined as the activity of an enzyme per mg of total protein.

$\text{SA" = "activity of an enzyme"/"mg of total protein}$

Plug your values into the eqaution above to solve for the sample's specific activity

${\text{SA" = ("0.0252" (mu"mol")/"min")/"0.085 mg" = 0.296mu"mol" * "min"^(-1) * "mg}}^{- 1}$

Now, since you give the answer in nmol per mg per minute, set up a simple conversion factor to get you from micromol to nanomol

${\text{0.296" (mu"mol")/("min" * "mg") * "1000 nmol"/(1mu"mol") = "296 nmol" * "min"^(-1) * "mg}}^{- 1}$

If you round this off to two sig figs, the number of sig figs in 0.085 mg, the answer will be

${\text{SA" = 3.0 * 10^(2)" nmol" * "min"^(-1) * "mg}}^{- 1}$