The steady-state kinetics of an enzyme are studied in the absence and presence o

Question

The steady-state kinetics of an enzyme are studied in the absence and presence of an inhibitor (inhibitor A). The initial rate is given as a function of substrate concentration in the following table: v[(mmol/L)min1]

Part A

What kind of inhibition (competitive, uncompetitive, or mixed) is involved?

Part B

Determine Vmax in the absence of inhibitor.

Part C

Determine Vmax in the presence of inhibitor.

part D

Determine KM in the absence of inhibitor.

Part E

Determine KM in the presence of inhibitor.

v[(mmol/L)min1] [S] (mmol/L) No inhibitor Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38

Explanation / Answer

Part D...Vo= Vmax [S}/ Km+S; 4.17 / 3.33 = 10 (Km+5) / 5 (Km+10) ;6.2 - 5 = Km- 0.62Km; 1.2 = 0.38Km

Km = 1.2/0.38 = 3.15 mmol (in absence of inhibitor)

Part E.. In presence of inhibitor Km

2.38/1.96 = 10(Km'+5) / 5 (Km'+10)

0.60Km' + 6 =Km'+5; 1= 0.4Km'

Km' = 2.5mmol.

Part B.

In absence of inhibitor, Vmax= Vo(Km+S)/ S ; Vmax = 4.17(13.15) / 10 = 5.48 mmol/lit/min

Part C.. In presence of inhibitor, Vmax = 2.38 (2.5+10) / 10 = 2.975

Part A.. Since Km and Vmax both are decreasing, it is uncompetitive inhibition.

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