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The protein calcineurin binds to the protein calmodulin with an association rate of 8.9 × 103 M-1s-1 and an overall dissociation constant, Kd, of 10 nM. The dissociation rate kd is:_____. Please explain step by step.
A. 8.9 × 10^3 M-1s-1
B. 8.9 × 10^2 s-1
C. 1.1 × 10-10 s-1
D. 8.9 × 10-5 s-1

Respuesta :

ajeigbeibraheem

Answer:

D

Explanation:

From the information given:

Association rate [tex]K_a[/tex] = [tex]8.9 \times 10^3 M^{-1}s^{-1}[/tex]

dissociation constant [tex]K_D[/tex] = 10 nM

dissociation rate [tex]K_d[/tex] = ???

Using the following relation from equilibrium dissociation constant to determine the dissociation rate, we have:

[tex]K_D =\dfrac{ K_d}{K_a}[/tex]

[tex]K_d = K_D \times K_a[/tex]

[tex]K_d =(10*10^{-9} \ M) \times (8.9*10^3 \ M^{-1}{s^{-1})[/tex]

[tex]\mathbf{K_d =8.9*10^{-5} \ {s^{-1}}}[/tex]

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