Question

20. You run a reaction with 1 µg of enzyme QB26, buffer and 25 millimolar (mM) substrate and obtain an initial reaction velocity (vo) of 100micromolar (µM) product per second. You repeat the reaction the same as before but with 50 mM substrate and observe vo = 101µM/sec, repeat at 100mM substrate, get vo =102µM/sec. a. The substrate concentrations in these reactions are very close the Km b. The velocities in these reactions are very close to Vmax c. The reaction will go much faster if the substrate concentration were 500µM d. This is an example of “substrate inhibition”

Answer 1

At [S] = 25 mM ; Vo = 100 M/s

At [S] = 50 mM ; Vo = 101 M/s

At [S] = 100 mM ; Vo = 102 M/s

We know that,

Vo = Vmax [S] / (Km + [S])

100x10^-6 = Vmax (0.025) / (Km + 0.025) ...........(1)

102x10^-6 = Vmax (0.1) / (Km + 0.1) ...........(2)

Solving the two equations:

Km = 0.148 M = 148 mM

Vmax = 0.00025 M/s = 250 M/s

When [S] = 500 M = 0.0005 M

Vo = 0.00025*0.0005 / (0.00025 + 0.0005)

Vo = 0.000167 M/s = 167 M/s

Hence following are the correct statements:

c. The reaction will go much faster if the substrate concentration were 500µM