Question

Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction

A <===> B

Researchers find that the K for the substrate A is 4 uM, and the kcat is 20 min^-1 (a) In an experiment, [A] = 6 mM, and the initial velocity, Vo was 480 nM min^-1 What was the [Et] used in the experiment?

(b) In another experiment, [Et] 0.5uM, and the measured Vo=5uM min^-1 What was the [A] used in the experiment? (

c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an ɑ of 10. In an experiment with the same [Et] as in part (a), but a different [A], an amount of Z is added that reduces the rate Vo to 240 nM min^-1 What is the [A] in this experiment?

(d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.

Answer 1

Answer - Page No - 0 Given that, Applying the enzyme is found Michaelis - Menten laration 11 Another that catalyzes the reaction , ی کد A ا where t is the enzyme Catalysing the reaction. Given i km = 4 um = 4x10 Klat = 20 min = 0.33 sec (a) As per Michael's - Menten quation, the initial velocity of the reaction to is given as; Yo = Vrau (A] where mase - Kat (E]] - [A] + km 3 NOW We bmw that CAJOnM6X10 Mandl Cineux CA] since [4] >> Com; [A] +km = (4] Vo = Vinau (A] ie, Voa Imaxe = 480 Moment => [@] = Vrau = 480 nm.my kat 20 min - (b) It fez] 2 0.5 cm i vince = 20X0.5 - = gunm = 10um, min! since Vo= Vmaxe CA] . Ta] tkm

le, 54106 Page No - 10CA] [A] +4 Now, Vo=5 um mint nyo = El max . (C) Vyax = Vrax CA] CA] +km » [A] + km = 24 i km = CA] - UMM For the same [ft] = 2unm; mau= usonin.min' - and Yo= 240 nm.mint ie, again Vo="2Y mau & As shown in (6) CA] = km = 4um when L=10; I'm = 4km = 10x4 = 40 UM = CA] No. catalytic Efficiency = Keatliam ~ 108-10 mist Here klat = 0.33 > 8.25*10m's' which is below the limit. (d) UXIT