A pesticide inhibits the activity of a particular enzyme whi

A pesticide inhibits the activity of a particular enzyme, which can therefore be used to assay for the presence of the pesticide in an unknown sample. a. In the laboratory, the initial rate data shown in the table below were obtained. What type of inhibitor (competitive, uncompetitive, non-competitive) is the pesticide? Determine the values of V_max, K_M, and K_1. b. After 50 ml of the same enzyme solution in part (a) is mixed with 50 ml of 8 times 10^-4 M substrate and 25 ml of a test sample, the initial rate observed is 18 mu M/min. What is the pesticide concentration in the test sample assuming no other inhibitors or substrates are present?

Solution

Vo = Vmax [S]/Km + [S]

Without inhibitor,

149 = Vmax * 0.00221/Km + 0.00221

149 (Km + 0.00221) = Vmax * 0.00221

149/0.00221 (Km + 0.00221) = Vmax

67420.81 Km + 149 = Vmax ------------------------------ (1)

129 = Vmax * 0.00165/Km + 0.00165

129/0.00165 (Km + 0.00165) = Vmax ------------------------ (2)

Applying (1) into (2)

78181.81 Km + 129 = 67420.81 Km + 149

10761 Km = 20

Km = 20/10761 = 1.8* 10^-3 M = 1.8 mM

Vmax = 67420.81 Km + 149 = 274.3 * 10⁶ M/min

With inhibitor,

125 = Vmax’ * 0.00221/Km’ + 0.00221

125 (Km’ + 0.00221) = Vmax’ * 0.00221

125/0.00221 (Km’ + 0.00221) = Vmax’

56561.1 Km’ + 125 = Vmax’ ------------------------------ (1)

103 = Vmax’ * 0.00165/Km’ + 0.00165

103/0.00165 (Km’ + 0.00165) = Vmax’

62424.2 Km’ + 103 = Vmax’ ------------------------ (2)

Applying (1) into (2)

62424.2 Km’ + 103 = 56561.1 Km’ + 125

5863.1 Km’ = 22

Km’ = 22/5863.1 = 3.75mM

Vmax’ = 62424.2 Km’ + 103 = 337.2 * 10⁶M/min

In the above case Km has increased. Km increases only in Competitive inhibition. In the other two types of inhibition, Km decreases. So, you can call this as \'competitive\'.

Note: Logically, if it was an inhibitor, then Vmax shouldn’t have increased. May be some misprint is there in the readings.