C)[D]/[ED] = 5.20
D)[D]/[ED] = 5.20
E)[D']_T = 1.495* 10 ^-7 M
F)[D'] / [ED'] = 0.0579
Explanation:
E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K
Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)
E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)
ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)
C)
[E] =2.5*10^-7 mol/L
K_D = 1.3* 10^-6 M
K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]
= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7
= 13/25 * 10
=130/25 = 5.20
[D]/[ED] = 5.20
D)
ΔG =RTln Kd
ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6
ΔG_2 592.454 * [ln 1.3 +ln 10^-6]
ΔG_1 = 592.424 [0.2623 - 13.8155]
ΔG_2 = -592.424 * 13.553
ΔG_1 = -8184.633 cal/ mol
ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol
ΔG_1 = -34.245 KL/mol
so, ΔG_2 = ΔG_1 - 10.5 KJ/mol
ΔG_2 = -34.245 - 10.5
ΔG_2 = -44.745KJ / mol
ΔG_2 =RT ln K_D
-44.745 *10^3
=8.314 *298.15 lnK_D
lnK_D' = - 44745 / 2478.81 g
ln K_D' = -18.051
K_D' = -18.051
K_D' = e^-18.051
[D]/[ED] = 5.20
E)
[E] = 2.5* 10 ^-7 mol/ L = a
K_D' = [E][D] / [ED'] E +D' → ED'
K_D' = a/2(x-(a/2) / (a/2)
KD' = x - a/2
=2.447 *10^-8 = (2.5/2) * 10^-7
x=2.447 * 10^-8 + 1.25 * 10^-7
x = 2.447 *10^-8 + 1.25 * 10 ^-7
x= 10^-7 [1.25 + 0.2447]
x = 1.4947 * 10^-7
[D']_T = 1.495* 10 ^-7 M
F)
K_D' = [E][D'] / [ED']
[D'] / [ED'] = KD' / [E]
[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7
[D'] / [ED'] = 0.5788 * 10^-1
[D'] / [ED'] = 0.0579
