For all questions, please show all work and/or provide explanations. The K d of

Question

For all questions, please show all work and/or provide explanations. The Kd of the insulin receptor is 1.5 x 10-10 M. There are 32,000 insulin receptors per cell. Assume one insulin binding site per receptor molecule.

The Kd of the insulin receptor is 1.5 x 10-10 M. There are 32,000 insulin receptors per cell. Assume one insulin binding site per receptor molecule.

a) At what [insulin] are 50% of the total receptors bound by insulin?

b) At what [insulin] are 80% of the total receptors bound by insulin?

c) 80% is a 1.6-fold increase over 50%. What is the fold-increase of [insulin] for this 1.6-fold increase in receptor/ligand complex concentration, [RL]?

d) If [insulin] is 4.2 x 10-12 M, what fraction of the total receptors are bound by insulin?

Explanation / Answer

Consider the dissociation equilibrium of the complex RI (receptor-insulin) in its components, receptor R and I (ligand, insulin):

          RI   < -- > R + I

The dissociation constant expression is :

          Kd = ([R][I])/[RI] .

Define the fraction r of the receptors bound by insulin as:

          r = [RI]/([RI] + [R])

([RI]= concentration of bound receptors, [RI] + [R] = overall, bound and unbound receptor concentration)

Divide by [RI] (rearrange), then replace [RI] in r by [RI] = ([R][I])/ Kd :

          r = [RI]/([RI] + [R]) =

          = 1/(1 + [R]/[RI]) =

          = 1/(1 + [R]/ (([R][I])/ Kd)) =

          = 1/(1 + Kd /[I]) =

          = [I]/ ([I] + Kd)

Use this equation to solve the questions:

a.

r = 0.5, then

[I]/ ([I] + Kd) = 0.5

[I] = Kd = 1.5 x 10-10 M

Observe this results: if [I] = Kd (in general if ligand [L] = Kd ), then 50% of the total receptors are bound by the ligand (here, insulin). This is true also for other cases.    

b.

r = 0.8, then

[I]/ ([I] + Kd) = 0.8

[I] = 0.8[I] + 0.8Kd

0.2[I] = 0.8Kd

[I]= 4Kd

[I]= 4x 1.5 x 10-10 M = 6 x 10-10 M

c.

6 x 10-10 M/1.5 x 10-10 M = 4-fold increase

d.

[I] = 4.2 x 10-12 M

r = [I]/ ([I] + Kd) =

  = 4.2 x 10-12 M /( 4.2 x 10-12 M + 1.5 x 10-10 M) =

          = 4.2 x 10-12 M/ 1.5042 x 10-10 M = (4.2 /1.5)x10-2 =

            = 0.028      (i.e., 2.8 %)

r = 2.8 %

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