Question

The data for an Enzyme with a Kcat is 5,000m-1. This experiment was performed by adding a given amount of substrate of 100 micrograms of enzyme in buffer solution. What is the molecular weight of the enzyme?

Substrate Concentration mM	rate (Vo) micromol/min
1	167
2	250
4	334
6	376
100	498
1000	499

Answer 1

In the given table if we look at the last two values, we see that when substrate concentration increases by 10 times (from 100 mM to 1000 mM) then rate increases by only one unit. We can safely assume that maximum rate will be constant at 500 umol/min even if we increase the substrate concentration further.

maximum rate = 500 umol/min

moles of enzyme = (maximum rate) / (k_cat)

moles of enzyme = (500 umol/min) / (5000 min^-1)

moles of enzyme = 0.1 umol

moles of enzyme = 0.1 x 10^-6 mol

mass of enzyme = 100 ug = 100 x 10^-6 g

molar mass of enzyme = (mass of enzyme) / (moles of enzyme)

molar mass of enzyme = (100 x 10^-6 g) / (0.1 x 10^-6 mol)

molar mass of enzyme = (100 g) / (0.1 mol)

molar mass of enzyme = 1000 g/mol