Multiple substrate reactions An enzyme has a ping-pong (substituted) kinetics; S

Question

Multiple substrate reactions

An enzyme has a ping-pong (substituted) kinetics; Substrate A binds first producing E* and product X, Substrate B binds next returning E* to E and releasing product Y. At what concentration of substrate b would the Kmapp = 40microM ? If the KmA = 50microM and the KmB = 25microM ?

Allosteric control (Hill Coefficient) An Enzyme has a velocity (v) of 50 M/min at a substrate concentration of 15 M, a K0.5 of 40 HM, and a Hill coefficient (n) of 2; what is the max velocity of the enzyme at a saturating substrate concentration? What is the fold change in v if the substrate concentration changes from 15 JuM to 30 HM? Multiple substrate reactions An enzyme has ping-pong (substituted) kinetics; Substrate A binds first producing E* and product X, substrate B binds next returning E* to E and releasing product Y. At what concentration of substrate b would the kMapp 40 M? If the KMA-5011 and the kuB = 25 If the maximum velocity (V) when both substrates are in excess is 200 M/min, and the initial velocity is dependent upon the kMA calculate the Vapp under these conditions. Enzyme Inhibitiion Malonate increases the kM of succinate for succinate dehydrogenase (SDH) without affecting the Vmax. What type of inhibition is this? You can determine the kw pp (substrate kM in the presence of the inhibitor) using the Michaelis- Menten equation by substiting the term ku for kw If the intracellular concentration of succinate is 10 IN, and the V, ax is 40 M /min at a concentration of 10 M malonate the observed velocity v is 8 M/min; Calculate the kMapp. If the observed velocity (v) was 13.3 M/min in the absence of malonate, what is the true kM of SDH for succinate? How the ratio Vmax/km compare with and without inhibitor?

Explanation / Answer

Given

E + A ---> E*A----> E* + X

E* + B ---> E*A----> E + Y

KMA =50µM, KMB= 25µM

KM of a substance is the amount of substrate which allows the enzyme to achieve half of its Vmax

Apparent Km is the average of the two Km

or, Km apparent= (50+25)/2 = 37.5

concentration of B when Km apparent is 40 is

40= (37.5 * [B] )/ 25

[B]= (40*25)/ 37.5

[B]= 26.6 µM

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