Question

21 and 22 please with work shown

21/ Show that for competitive inhibition of an enzymatic reaction, the intercepts on the horizontal axis of a plot of 1/v versus 1/[S) at different inhibitor concentrations are equal to 1/Km (1 + [/K). 48 81 73 0.4 145 (22 [S] (UM) y - no inhibitor (um/min) y - with inhibitor (UM/min) Ti g t be 0.1 0.2 123 wool 0.6 95 0.75 160 108 Renin acts as a specific protease that cleaves a decapeptide unit from a serum globulin. Renin is inhibited by the microbial pentapeptide Pepstatin. Kinetic data are shown above. When present the Pepstatin concentration was 2.36 x 10-10 M. a) Determine Km and Vmax for Renin. b) What type of inhibition is being observed? c) Determine the value of K; for Pepstatin.

Answer 1

21.

Competitive inhibitor - An inhibitor for that that substrate for the active site is called competes with competes with tre competitive whisitor ki (Enzyme Ett a substrate) complex & ts ki. Es ka E+P k, a k-, - forward & reverse rate constants for ES I formatas EI + s - No reaction kg- Rate constant for Es decomposition't Here: Ky - [E] [I] & Inhibitor concentration product (C) formation [EIJ < Enzyme inhibitor Complex KI- Inhibitor constant The expression obtained in term of total enzyme concentration: TEJ - [E] + (EI] + [ES] Constant [ES] = [E] [s Menten Under steady state condition, it is known that: Ks = 6 [s] . Since kg=km Kg - Dissociation (ES] KM = [E] [5] KM- Michalles TEs ] =D KM LEJ: Constant Ky LES (s] for the enzyme inhibitor complex, putting values from @ TE LJ - EJ CI] Ky LES CI) KI (S] kq Putting ② & ③ in [E], & Ky [ES] & K [ES] [I] [ES]

TEDF Les3 S King 2 + 1} (E)< LES] Station CI+ +1} [ES] = [E], [s] KM (1 + I) + [s] It is known that ve K₂ LES] 2 R₂ [E], [s] also k[E] = Umax & 2 Vmax [s KM (HI) + (S) Where I ou Umax [s] &km + [s] Michaelis Menten Equation for competitive inhibitor Ring reciprocal lê get the Line wearer Buck. Plot we get. y = mx + c Slope = 2 km

22. a. The Lineweaver Burk Plot for the given data is as the following:

0.05 0.04 - 1/v (1/M/min) without inhibitor 1/v (1/um/min) lv uM/min with inhibitor Linear (1/v(1/UM/min) without inhibitor) Linear (1v uM/min with inhibitor) -5 0 5 10 15 -0.01 1/[S] (1/UM)

Vmax is the maximum velocity or the rate of reaction achieved by an enzyme. At VMAX, the enzyme is completely saturated with the substrate molecules; any further increase in the substrate concentration will not increase the rate of reaction any further.

KM is the substrate concentration at which the maximum velocity (VMAX) is half of the original value. Lower the KM of the enzyme, higher is its affinity towards the substrate.

KM and VMAX for Renin are as the following:

1/Vmax	Vmax (uM/min)	1/Km	Km (uM)
0.004	250	2.5	0.4

b. From the Lineweaver Burk Plot, it is observed that Pepstatin does not affect the VMAX of the enzyme but increases the KM of the enzyme. Hence it is a competitive inhibitor, which competes with the substrate for the active site of the enzyme due to similar geometrical shape that binds to the active site.

c. Ki for pepstatin:

For the inhibitor,

-1/αKM= -1

αKM=1

Also, KM=0.4 uM

Inserting this value in αKM, we get, α=10/4

α=1+ [I]/Ki

[I]= 2.36 x 10-10 M= 2.36 x 10-4uM

10/4= 1+ 2.36 x 10-4/Ki

Ki= 1.573 x 10-4 uM

Ki is also expressed in the same unit as as the inhibitor