Question

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Similar to the lecture, our example is a drug molecule D and a drug target enzyme E. The drug target
enzyme Eis commonly found in concentrations [Eltot of 250 nM, i.e. 2.5 x 10-7 mol/l. The drug D binds to
Eto form the complex ED with a dissociation constant of KD = 1.30 MM (i.e. 1.30 x 10-6). The dissociation
constant has been determined at room temperature, i.e. for T = 298.15 K.
Physical Chemistry question
C) In order to estimate the effectivity of the drug D, calculate the ratio of the concentration of the free,
unbound drug molecules (D) and the bound complex (ED), i.e. (D]/(ED), for the conditions described in part
B. Provide your answer with 3 significant figures. The ratio of concentration (D]/[ED]= _____?
D) Long hours of work in the lab have reveled an improved version of the drug, D, which forms an
additional hydrogen bond with the drug target enzyme E. This results in an overall stabilization of the
complex ED relative to ED and the binding free energy for ED is 10.5 kJ/mol more negative than for ED.
What is the resulting dissociation constant for the complex ED at room temperature? Provide your answer
in units of mol/l and with 3 significant figures. The dissociation constant ko for the complex ED is ____?
E) Which total concentration of the drug D: (DItot, is needed to bind 50% of the drug target enzyme Einto
the complex ED? Provide your answer in units of mol/l and with 3 significant figures. The total
concentration of the drug D' required to bind 50% of the drug target enzyme (E) is (D']tot = ____mol/l?
F) In order to estimate the effectivity of the drug D, calculate the ratio of the concentration of the free,
unbound drug molecules [D] and the bound complex (ED), i.e. [D]/[ED). Provide your answer with 3
significant figures. The ratio of concentrations (D)/(ED) = ___
I answered part A & B correctly already with A being -33.6 kJ/mol and B being 1.43x10^-6 mol/l

Answer 1

To find the ratio ([D]/[ED]) and the new dissociation constant for the improved drug (KD'), we use the original KD and adjust for the change in the binding free energy. For 50% binding, KD is equal to the concentration of free drug ([D]). Enhanced binding affinity due to an additional hydrogen bond will decrease KD'.

Step-by-step explanation:

To calculate the ratio of the concentration of free, unbound drug molecules (D) to the bound complex (ED), we use the dissociation constant (KD). The following formula represents this equilibrium: KD = [D][E]/[ED] . With a given KD of 1.30 x 10⁻⁶ mol/L and knowing the total enzyme concentration ([E]tot), we can rearrange the formula to solve for the ratio ([D]/[ED]).

Since the question in part E asks for the concentration of the drug needed to bind 50% of the enzyme, we need to consider that at 50% binding, the concentration of the complex ([ED]) is equal to half of the total enzyme concentration. Therefore, KD will be equal to the concentration of free drug ([D]), since [E] is equal to [ED] at this point. The dissociation constant can be used, KD = [D] , to find the required [D].

For part F, to evaluate the improved version of the drug (D'), the change in the binding free energy (ΔG), which has become more negative by 10.5 kJ/mol, suggests stronger binding and a reduced KD'. This can be calculated using the relationship between ΔG and KD: ΔG = -RT ln(KD) , where R is the gas constant (8.314 J/mol·K) and T is the temperature in kelvins.

Answer 2

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Step-by-step 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