You purify Yet Another Enzyme (YAB) and perform still more Michaelis-Menten kine

Question

You purify Yet Another Enzyme (YAB) and perform still more Michaelis-Menten kinetics experiments. You first test your purified enzyme with its substrate (uninhibited), then with either 5 M inhibitor A or 5 PM inhibitor B. You obtain the following results: 3. Uninhibited: y=207 3.6773x 1.8406+x 2.8392r 1.4395+x 2.1992x 1.2548+x Inhibitor B: y a. What are Vaas Kw. Vne and Kure for each line shown? b. What type(s) of inhibition does each exhibit? How do you know (briefly)? Calculate and/or , and K, and/or K1. as appropriate for each inhibitor. Does each bind E, ES, or both? Briefly justify your choices c.

Explanation / Answer

Ans. #a. Note that y-axis represents Vo and X-axis represents corresponding [S]. Also, Vo approaches to Vmax when [S] is highest in the graph. So, calculate Vmax using [S] = 8 nm for each curve.

#I. No Inhibitor:

Vmax = (y) = (3.6773 x 8 nM) / (1.8406 + 8 nM) = 2.9895 mM/sec

Km is the [S] at which ½ Vmax is attained.

So, putting the y = (1/2) x 2.9895 mM s-1 = 1.4948 mM s-1 in trendline equation to calculate corresponding [S]-

            1.4948= (3.6773 X) / (1.8406 + X)

            Or, 1.8406 + X = 3.6773X / 1.4948 = 2.4601X

            Or, 2.4601X – X = 1.8406

            Or, X = 1.8406/ 1.4601

            Hence, X = 1.2606

Therefore, Km = 1.2606 mM

#II. Inhibitor A:

Vmax,app = (y) = (2.8396 x 8 nM) / (1.4395 + 8 nM) = 2.406 mM/sec

Km is the [S] at which ½ Vmax is attained.

So, putting the y = (1/2) x 2.406 mM s-1 = 1.203 mM s-1 in trendline equation to calculate corresponding [S]-

            1.203 = (2.8396 X) / (1.4395 + X)

            Or, 1.4395 + X = 2.8396X / 1.203 = 2.3064 X

            Or, 2.3064X – X = 1.4395

            Or, X = 1.4395 / 1.3064

            Hence, X = 1.1019

Therefore, Km,app = 1.1019 mM

#III. Inhibitor B:

Vmax,app = (y) = (2.1992 x 8 nM) / (1.2548 + 8 nM) = 1.9010 mM/sec

Km is the [S] at which ½ Vmax is attained.

So, putting the y = (1/2) x 1.9010 mM s-1 = 0.950 mM s-1 in trendline equation-

            0.950 = (2.1992X) / (1.2548 + X)

            Or, 1.2548 + X = 2.1992X / 0.950 = 2.3149 X

            Or, 1.2548 = 2.3149X - X

            Or, X = 1.2548 / 1.3149

            Hence, X = 0.9543

Therefore, Km,app = 0.9543 mM

#b. Type of inhibition: Note that both Vmax and Km decreases in presence of inhibitor A as well as B. It is the characteristic of mixed inhibition.

Therefore, both the inhibitors A and B are mixed inhibitor.

#c.

#I. Inhibitor A:

a' = Vmax / Vmax,app = (2.9895 mM/sec) / (2.406 mM/sec) = 1.2425

a = (a’ x Km,app) / Km = (1.2425 x 1.1019 mM) / 1.2606 mM = 1.0861

KI = [I] / (a - 1) = 5 uM / (1.0861 - 1) = 58.0720 uM

KI’= [I] / (a’ - 1) = 5 uM / (1.2425 - 1) = 20.6186 uM

#I. Inhibitor B:

a' = Vmax / Vmax,app = (2.9895 mM/sec) / (1.9010 mM/sec) = 1.5729

a = (a’ x Km,app) / Km = (1.5729 x 0.9543 mM) / 1.2606 mM = 1.1907

KI = [I] / (a - 1) = 5 uM / (1.1907 - 1) = 26.2192 uM

KI’= [I] / (a’ - 1) = 5 uM / (1.5729 - 1) = 8.7275 uM

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