Question

help with b) please

3. Use we wyl Reaction rate (mol/min) (Substrate] Substrate Substrate + Unknown 1/[Substrate) 1N 1N (s. wow (UM) 0,055 0.1 I I 18.19 10 (2.5 10.115 0.183 0.4 8.70 5.46 5 10.17 0. 25 0 .2 5.884 0.250 .306 0. 1 4 3.29 20 0.31 0.35 0.05 3.23 2.86 SIOPE (r-oris)

Answer 1

20.00 y = 15.671x + 2.5121 15.00 10.00 . y = 7.4985x + 2.4935. 1/V 5.00 0.00 -0.6 -0.4. 0.2 0.2 0.4 0.6 0.8 1 1.2 -5.00 1/S • V . Unknown ....... Linear (V) ......... Linear (Unknown)

1/Vo = (Km/Vmax)1/S + 1/Vmax

Comparing this equation with Y=aX + c

where a is the slope and c is the intercept

1/Vmax = intercept

so V max = 1/intercept

For Uninhibited State.

y = 7.4985x + 2.4935

1/Vmax = 2.4935

Vmax = 1/2.4935 = 0.40 mol/min

Km/Vmax is equal to slope

Km/7.4985 = 0.4

Km = 7.4985 * 0.4

0.3 mol

For inhibited state

y = 15.671x + 2.5121

1/Vmax = 2.5121

Vmax = 1/2.5121 =0.398 or 0.40 mol/min

Km/Vmax is equal to slope

Km/15.671 = 0.40

Km =15.671*0.4 = 6.27 mol

Since there is no change in the Vmax bu Km is increased. So type of inhinition is competitive inhibition.

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