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An enzyme catalyzes the reaction A --> B. The enzyme is present at a concentration of...

An enzyme catalyzes the reaction A --> B. The enzyme is present at a concentration of 2 nM, and the Vmax is 1.2 mm/s. The Km for substrate A is 10 mM. Calculate the initial velocity of the reaction, Vo, when the substrate concentration is a) 4 mM, b) 15 mM, c) 45 mM.

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Answer #1

Et= concentration of enzyme= 2nM =2*10-9 M   , Vmax= 1.2 mM/s= 1.2*10-3 M/s, KM= 10mM= 10*10-3 M

From Michaelis-Menten Kinetics, V= initial velocity = VmaxS/(KM +S)

a) for S= 4mM= 4*10-3M, V= 1.2*10-3* 4*10-3/( 10*10-3 +4*10-3)= 4.8*10-6/ 14*10-3= 0.0003429 M/s

b) for S= 15mM= 15*10-3M, V= 1.2*10-315*10-3/ (10*10-3 +15*10-3)= 1.2*15/*10-3/25= 7.2*10-4 M/s

c) for S= 45mM =45*10-3M, V= 1.2*10-3*45*10-3/ (10*10-3+45*10-3)= 1.2*45*10-3/58= 9.3*10-4 M/s

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