[s]/M: v 0 /10 -6 M . min -1 2.5 x10 -5 38.0 4.00 x 10 -5 53.4 6.00 x 10 -5 68.6

Question

[s]/M:                                v0/10-6 M . min -1

2.5 x10-5                                        38.0

4.00 x 10-5                                       53.4

6.00 x 10-5                                       68.6

8.00 x10-5                                         80.0

16.00 x10-5                                      106.8

20.00 x 10-5                                114.0

how would an incerase in the enzyme concentration by a factor of 2 affect of the following question KM; V max AND v0 AT [S] =5X10-5 M

Explanation / Answer

Vo = Vmax[s] / ([s]+Km) Vmax = Vo ([s]+Km) / [s]) Vmax - KmVo/[s]= Vo -----(main equation for solving) by subtituting any two values, we can get two equation in which Vmax and Km are the variable. by solving the two equations we can get the value of Vmax by substituting Vo=38 and substrate is doubled id [S] = 2[s]= 2* 2.5 x10-5 we get Vmax - 0.76 Km = 38*10^-6 ----------(1) Vo = 80 and [s] = 8.00 x10-5 Vmax - 0.5 Km = 80 * 10^-6 ------------(2) by solving (1) and (2) we get Vmax = 0.000160769230769231 = 16*10^-5 (ans) Km = 0.000161538461538462 = 16 * 10^-5 (ans) therefore with an increase in [s] by a factor by 2, there is an increase in Km value, but the Vmax is constant with the change in substrate concentration. !!! :)

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