Question

The zero order rate is seen in an enzyme catalyzed reaction when:
A. There is too much enzyme and too little substrate
B. Enzyme reactions are always first order
C. The Michaelis constant is very high
D. The V_max of the reaction is twice that which is observed
E. The enzyme substrate comples is at its maximum concentration

What happens when you add more enzyme to a catalyzed reaction that already is proceeding at V_max?
A. The reaction is inhibited because of competition for substrate
B. The plot of velocity vs. [S] will show a sigmoidal (an S-curve) shape
C. The reaction goes faster
D. The K_m of the reaction will change
E. There is no change in the rate of the reaction

Given two enzymes with the same V_max but with very different K_m values
A. The enzyme with the higher K_m will approach V_max at smaller [S]
B. V_max values cannot be the same with different K_m values
C. The steady state concentration of the enzyme-substrate complex is the same for each enzyme at every substrate concentration
D. The enzyme with a lower K_m value approaches V_max at lower [S]
E. Both values change when more enzyme is sdded

I think first one is E, second one is E, and last one is C. What are the correct answers for above?

Answer 1

since V= Vmax*S/(KM+S)

where S= substrate concentration, Km is constant

when S>>KM, V= Vmax*S/S= Vmax

-dS/dt=Vmax= constant, zero order reaction.

Km= [S] [E]/[ES]

Km is less means , [ES] is high suggesting maximum concentration of ES since V= Vmax ( E is correct)

2. V= VmaxS/(KM+S)

when KM<<S, when substrate concentration is high only V approaches Vmax. So additional substrate does not effect the Vmax. So E is correct

3. V= Vmax*S/(KM+S)

since Vmax is same, they are competitive and hence they enzyme binds to substrate only, not the Enzyme-Substrate complex. So concentration of Enzyme-substrate concentration remain the same.( C is correct)