7.If the equilibrium constant of binding of NADH to an enzyme [E] is2.7=2,7 vKo = [NADH][E]/[E-NADH] = 1.5 x 10-6 M, and solution concentrations of [E] and [NADH] are 10 nM and1.0 µM, respectively, what is the concentration of the [E-NADH] complex?%3D51210(a) 16 µM0.:0(b) 0.62 x 109 M1.5E-NADH(c) 62 nM .0000000(d) None of the above.5-2.51x

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Asked Dec 6, 2019
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7.
If the equilibrium constant of binding of NADH to an enzyme [E] is
2.7=2,7 v
Ko = [NADH][E]/[E-NADH] = 1.5 x 10-6 M, and solution concentrations of [E] and [NADH] are 10 nM and
1.0 µM, respectively, what is the concentration of the [E-NADH] complex?
%3D
51210
(a) 16 µM0.:0
(b) 0.62 x 109 M
1.5
E-NADH
(c) 62 nM .0000000
(d) None of the above.
5
-2.51x
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7. If the equilibrium constant of binding of NADH to an enzyme [E] is 2.7=2,7 v Ko = [NADH][E]/[E-NADH] = 1.5 x 10-6 M, and solution concentrations of [E] and [NADH] are 10 nM and 1.0 µM, respectively, what is the concentration of the [E-NADH] complex? %3D 51210 (a) 16 µM0.:0 (b) 0.62 x 109 M 1.5 E-NADH (c) 62 nM .0000000 (d) None of the above. 5 -2.51x

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Expert Answer

Step 1

The concentration of E is calculated in the unit of molar as follows:

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[E] =10 nM 10-° M = 10 nMx- 1 nm =1x10-8 M

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Step 2

The concentration of NADH is calculated ...

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[NADH] =1.0 µM 106 M =1.0 µM x 1 um =1.0×10 M

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