In: Chemistry
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?
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 %