 Biology
6 September, 01:12

# 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?

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Answers (1)
1. 6 September, 03:09
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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.
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