reactive nucleophile: by chymotrypsin, which of the following amino acids is the

Question

reactive nucleophile: by chymotrypsin, which of the following amino acids is the source of the c) G e) A 42. The following is a list of enzymes and their KM for a particular substrate. Which enzyme would be expected a) Vmax the fastest? values b) Catalase 1.1 x 100 c) x 10 s vin ila v max Crotonase/2.3 x 10 d) Fumarase/4,3 x 103 ee) 5.6 x 10 7 43. The following is a list of enzymes and their koa KM values for a particular substrate. Which enzyme is the most efficient? a) Catalase/4.0 x 10 b) Acetylcholinesterase 1.6 x 106 c) Crotonasel 2.8 x 10 d) Fumarase/3,6 x 10 e) Lactamasel 1.0 x 10

Explanation / Answer

Ans. 42. Correct option: E. smallest Km

Km is the substrate concertation at which half of Vmax is attained.

Smaller Km means that the enzyme can reach substrate saturation relatively faster than the enzyme with higher Km. So, the enzyme also reaches Vmax faster.

Consider an example. Two enzymes A and B have same Vmax for same substrate and are nearly identical to each other in all aspects except they differ in Km. A has a smaller Km (say, 10 mM) than B (100 mM). They are further incubated in separate reaction tubes under identical conditions of substrate, temperature, etc. Catalysis occurs when the substrate binds to the active site of enzyme.

Now, A requires to bind only 10 mM substrate to reach half Vmax while B requires 100 mM for the same. Since substrate binding occurs due to random collision between enzyme and substrate molecules, and also a time-dependent feature, A requires less time to get saturated with substrate than B- because A need to bind 10 mM substrate while B need to bind 10 times more substrate to give same Vmax.

Therefore, an enzyme with smaller Km reaches Vmax faster.

Ans. 43. Kcat = Vmax / Eo           ; Eo= amount of enzyme in the reaction mixture

Kcat measure product formation per unit enzyme per unit time.

Greater the value of Kcat, greater is the rate of product formation because Kcat is directly proportional to Vmax.

From Ans 42, a smaller value of Kcat makes enzyme reach Vmax faster.

So, (Kcat/ Km) = X, where X is higher for a smaller value of Km. That Kcat coupled with smaller Km gives greater rate of product formation.

So, Kcat/Km is taken as a measure of enzyme efficiency. Greater is the value of (Kcat/Km), more efficient is the enzyme, i.e. produces more product per unit time per unit enzyme concentration.

So, correct option. C. highest values of (Kcat/ Km)

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