Map Lehninger Principles of Biochemistry Nelson Cox MHE/Freeman by Sapling Learn

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Map Lehninger Principles of Biochemistry Nelson Cox MHE/Freeman by Sapling Learning SEVENTH EDITION For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A For substrate A, she determined that Km 2.5 M and kcat -35 min1. Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows Michaelis-Menten kinetics In one experiment where [A]-3 mM, she found that the initial velocity, Vo was 574 nM min1 What was the [Et] she used in this experiment? b In another experiment with [Et]- 0.2 HM, she found that Vo = 6 M min". What was the [A] she used in this experiment? Number Number nM [A] The compound Z was found to be a very strong competitive inhibitor of the enzyme, with an of 10 In an experiment with the same [Et] as in part (a), but Menten curve at low [S], the second order rate a different [A], an amount of Z is added that reduces the rate Vo to 287 nM min. What was the [A] in this experiment? (d) A measure of the catalytic efficiency of an enzyme is the initial slope of the Michaelis- , constant kcat/Km. A "perfect" enzyme, limited only by diffusion, would have a value of 108 or 109 Ms-1. Calculate kcatlKm for this enzyme Number Number cat K,

Explanation / Answer

1.Since V= VmaxS/(KM+S_

Given S=[A] = 3mM=3*10-3M

Vo= 574*10-9 M/min

Since Vo= VmaxS/(KM+S)

574*10-9= Vmax*3*10-3/( 2.5*10-6+3*10-3)

Vmax=574.5*10-9 M/min= 574.5 nM/min

Since KCat= Vmax/ET

ET= Vmax/Kcat= 574.5/35 =16.41 nM

2. When ET=0.2uM= 0.2*10-6M

Vmax=Kcat*ET= 35*0.2*10-6 M/ min=7*10-6 M/min=7uM/min

Since V= VmaxS/(KM+S), Vo =6uM/min

6= 7*S/(2.5+S)

=15+6S= 7S

S= 15 uM

For competitive inhibition, Vmax remains the same and the Michaelis- Menten kinetics become

V= VmaxS/(KMapp+S)

KMapp = 1+I/KI

I/Ki= alpha

Given Vmax= 574.5 nM/min, KMapp= 2.5*10-6*(1+10)= 27.5*10-6 M=27.5*1000nM= 27500 nM

287= 574.5*S/(27500+S)

0.5 = S/(27500+S)

0.5S+13750= S

S=27500 nM=(27500/1000)uM= 27.5 uM

4. KCat/KM= 35/min/2.5*10-6 M)=1.4*107 /minM

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