Question

I only need help with part D. The answer is 0.065 sec^-1, but I need help to understand why.

You are studying a dehydrogenase enzyme that catalyzes the formation of NADH from NAD^+. Subsequent studies determined that NADU will absorb at 340 nm while NAD^+ will not. The extinction coefficient for NADH is 6220 M^-1 cm^-1. The reaction volume is 750 mu L total and the substrate molecular weight is 223 g/mol. You used 20 mu L of enzyme from an initial concentration of 20 mg/mL for every assay. The molecular weight of the enzyme was found to be 26 kDa. a. What volume of a 3.4 mg/mL initial stock solution for the substrate would be needed to measure a final concentration of 200 mu M for one reaction? 3.4^mg/mL = 3.4^g/L = 3.4g/L/223 g/mol = 0.0152 M = 15.2 mM V_1 = C_2V_2/C_1 = (0.2 mM) (750 mu L)/15.2 mM = 9.9 mu L 10 mu L b. After the reaction was complete, the initial velocity that you measured (Delta A_340/min which is the change in absorbance at 340 nm per minute) is 0.5/min (Delta A_340/min = 0.5/min). Convert this rate in terms of activity units (mu moles/min). Remember that A = epsilon bc with b = 1.0 cm and epsilon = 6220 M^-1 cm^-1 c = A/epsilon b = 0.52 min^-1/(6220 M^-1 M cm^-1) (1.0 cm) = 8.04 times 10^-5 M/min = (80.4 mu M/min) (80.4 mu M/min) (750 times 10^-6L) = 0.06 mu moles/min c. Convert the result from part lb to specific activity units (mu moles/min/mg). 20 mu L = 0.02 mL (0.02 mL) (20 mg/mL) = 0.4 mg of protein/assay Specific Activity = 0.06 mu moles/min + 0.4 mg = 0.15 mu moles/min/mg d. Assume the velocity that you measured from part 1b and part 1c was the V_max of the reaction, what is the k_cat in units of s^-1?

Answer 1

It is because kcat is equal to the product of vmax and concentration of the enzyme. As the vmax = 0.06 micro moles /min kcat will be 0.06 / s.