Question

3. Calculation, fill in the blank: Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5pL of a 2mg/mL enzyme stock solution to an assay mixture with a 1mL total volume, and taking into account that XYZase is a monomeric enzyme with a molecular mass of 45,000 Daltons, what is the keat of XYZase given in s1 (give the answer in to two significant figures)? s1 Vo (Murder Weapon- Cmpd. X) Vo(No inhibitor) [S] 50 0.14 0.040 75 0.050 0.19 150 0.32 0.080 400 0.47 0.14

Answer 1

Chart Title 30 y= 1023.3x5.2884 25 20 15 10 y 291.46x +1.3179. 0 0.005 0.02 0.01 0.015 0.025 1/[S] No inhibitor Compound X . Linear (No inhibitor) .. Linear (Compound X) 1/[Vo] un o

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.

y = 291.46x + 1.3179

Vmax = 1/1.3179

Vmax = 0.76

Vmax = Kcat [Etotal]

[Etotal] = Enzyme concentration.

2gm/ml = 2ug/ul.

The volume of enzyme used = 5 ul.

So the amount of enzyme = 5*2 = 10ug.

Molarity = mass/molecular weight * 1/ volume in litre

mass = 10ug

moleculer weight = 45000 dalton

volume = 1ml. = 0.001litre

Molarity = 10*10-6 /45000 *1/0.001

M = 222 *10-9 M or 222 nM.

Kcat = Vmax. / Etotal

= 0.79 /222 * 10^-9

= 0.00355 * 10 ⁹

= 3.55 *10⁶ s^-1

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