Question

Enzymatic Asssay of LDH [Lac] 150 mM; [NAD+] = 6 mM 1.6 Add Enzyme 0.8 0.6 0.4 0.2 Find the Slope k1 Abs a 340 nm Base Line Slope ko 0 100 200 300 400 Time in second Figure IV

Take Figure IV as the example, if 10 µg of LDH protein (MW = 150 kDa) is used in the reaction, how you may calculate the kcat (the turnover number)? The turnover number is the number of mole of product generated per minute, catalyzed by 1 mol of enzyme at Vmax

Answer 1

We can calulate the kcat using Vmax.

LDH Calculations

Vmax (Units/mg) = ( ΔAbs₃₄₀/min)/(6.22xmg enzyme/ml rxn mix)

From the above graph it is inferred there is change of 0.400 Abs per 100 s.

Thus in 1 min there is change of 0.240 Abs

10 µg of LDH has been used the reaction and using the above formula we will get

Vmax of 0.3858 U/mg

Now converting putting the values in mg to g, and micromoles to moles and min to sec and multiplying with the molecular weight of LDH we will get the kcat value.

Kcat= 0.3858x1000x150000/(60x10⁶) = 0.9625 s^-1