Answer:
Explanation:
a. kcat = Vmax/[Et] = (1.6 Mμ/s) / 0.004 ∪M = 400 s-1
b. Vmax = [Et] kcat = [1 nM] (400 s-1) = 400 nM/s which is 0.4μM/s.
Now we can use the Michaelis-Menten equation if all the units are similar (all molar conc in nM or in this case μM):
V0 = Vmax*[S] / Km+[S] = 0.3μM/s = (0.4μM/s) (30μM) / (Km+30μM)
Then solving for KM, we get:
(0.3 M/s) (Km + 30μM) = 0.4μM/s (30 μM)
0.3μM/s(Km) + 9μM2/sec = 12μM2/s
0.3μM/s(Km) = 3μM2/s
Km = 10μM
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 = 30μM / 3 = 10μM
c. After removal of ANGER, the Vmax increased to 4.8μM/s and the Km became 15μM. It is a mixed because it is affecting both Vmax and Km.
Because Vmax increased by a factor of 3, ∝'=3.
Similarly, Km varies as a function of ∝Km/\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 ∝'=3, then ∝=2.
d. Because both ∝ and ∝' are affected, ANGER is a mixed inhibitor.