Question

You have just completed assaying the change in absorbance for an enzyme. The initial absorbance was 0.022; the absorbance after 5 minutes was 0.444. Calculate the rate of enzyme activity for this sample (as the increase in A per 10 minutes). Show your work. HTML EditorKeyboard Shortcuts

Answer 1

Answer:

The Rate of Enzyme activity for this sample after 10 minutes will be -3.028

Explanation:

  k = $\frac{1}{t}\times ln(C-C0)$

Where,     t= time , C= final concentration , C₀= initial concentration

        k = $\frac{1}{5}\times ln (0.44-0.22)$ per minute

           = $\frac{1}{5}\times ln (0.22)$  per min

           = $\frac{1}{5}\times (-1.514)$ per min

k   = -0.3028 per minute

k in A per minute is = -0.3028

k in A per 10 minutes will be = 10× (-0.3028)

                                                = - 3.028