Question

A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction HAPPY SAD. The researchers begin to characterize the enzyme In the first experiment, with [E_t] at 4 nM, they find that the V_max is 1.6 mu M s^-1. In another experiment, with [E]total at 1 nM and [HAPPY] at 30 µM, they find that v0 is 300 nM•s–1. Based on this second experiment, what is the Km for happyase?

Answer 1

a. kcat = Vmax/[Et] = (1.6 \muM/s) / 0.004 \muM = 400 s-1

b. Vmax = [Et] kcat = [1 nM] (400 s-1) = 400 nM/s which is 0.4 \muM/s.

Now we can use the Michaelis-Menten equation if all the units are similar (all molar conc in nM or in this case \muM):

V0 = Vmax*[S] / Km+[S] = 0.3 \muM/s = (0.4 \muM/s) (30 \muM) / (Km+30\muM)

Then solving for KM, we get:

(0.3 \muM/s) (Km + 30 \muM) = 0.4\muM/s (30 μM)

0.3 \muM/s(Km) + 9 \muM2/sec = 12 \muM2/s

0.3 \muM/s(Km) = 3 \muM2/s

Km = 10 \muM

Another way to do this would be to first rearrange the Michaelis-Menten equation to:

V0/Vmax = [S] / Km+[S]

(300 nM/s) / (400 nM/s) = 3/4 = [S] / (Km + [S])

4 [S] = 3 Km + 3[S]

Km = [S]/3 = 30 \muM / 3 = 10 \muM

c. After removal of ANGER, the Vmax increased to 4.8 \muM/s and the Km became 15 \muM. It is a mixed because it is affecting both Vmax and Km.

Because Vmax increased by a factor of 3, \alpha'=3.

Similarly, Km varies as a function of \alphaKm/\alpha'. Given that Km increased by a factor of 1.5 when ANGER was removed (that is, the inhibitor decreased the observed Km by 2/3 and \alpha'=3, then \alpha=2.

d. Because both \alpha and \alpha' are affected, ANGER is a mixed inhibitor.

Step-by-step explanation: