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

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

Step-by-step explanation:

k =
`(1)/(t)* ln(C-C0)`

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

k =
`(1)/(5)* ln (0.44-0.22)` per minute

=
`(1)/(5)* ln (0.22)` per min

=
`(1)/(5)* (-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