Which ligand binds the tightest?
Ligand A, with a Kd of 10^-9M
Ligand B, with a Kd of 10^-3M
Ligand C, with a percent occupancy of 30% at one micromolar
Ligand D, with a percent occupancy of 80% at 10 nm
The answer is A, however the explanation previously given is incorrect.
The correct explanation is that the association fraction (theta) is given by the expression:
theta = [L]/([L] +Kd)
where [L] = ligand native concentration and Kd = the dissociation constant.
Given this equation, you can then solve for Kd in (C) and (D):
0.3 = 1 x 10^-6/(1 x 10^-6 + Kd)
Kd = 2.3 x 10^-6
0.8 = 1 x 10^-9/(1 x 10^-9 + Kd)
Kd = 2.5 x 10^-9
This allows you to compare each Kd, and conclude that (A) has the lowest dissociation constant, and, therefore, the greatest association constant of the ligand (the ligand binds the tightest).
the relative molecular mass of lead 2 trioxonitrate V is (pb=180,N=14, O=16).
To determine which ligand binds the tightest, we need to compare their binding affinities. Binding affinity is typically quantified using the dissociation constant (Kd), which represents the concentration of ligand at which half of the binding sites are occupied.
In this case, we have the following information:
- Ligand A has a Kd of 10^-9M.
- Ligand B has a Kd of 10^-3M.
- Ligand C has a percent occupancy of 30% at one micromolar concentration.
- Ligand D has a percent occupancy of 80% at 10 nanomolar concentration.
To compare the binding affinities, we want to find the ligand with the lowest Kd or the highest percent occupancy at a given concentration.
Comparing Ligand A and Ligand B:
Ligand A has a lower Kd (10^-9M) compared to Ligand B (10^-3M). Therefore, Ligand A binds tighter than Ligand B.
Comparing Ligand B and Ligand C:
To compare Ligand B and Ligand C, we need to convert the percent occupancy given for Ligand C into an equivalent Kd value. Since Ligand C has 30% occupancy at one micromolar concentration, this means that 30% of the binding sites are occupied at one micromolar concentration. To find the Kd, we need to determine the ligand concentration at which 50% occupancy would occur.
Assuming ligand occupancy follows a typical sigmoidal binding curve, let's calculate the equivalent Kd for Ligand C:
- At 30% occupancy, the ligand concentration is sufficient enough to occupy 30% of the binding sites.
- At 50% occupancy, the ligand concentration would be equivalent to the Kd.
- At 70% occupancy, the ligand concentration has surpassed the Kd.
Based on this analysis, it seems that Ligand C likely has a higher Kd compared to Ligand B. Therefore, Ligand B binds tighter than Ligand C.
Comparing Ligand C and Ligand D:
Again, we need to convert the percent occupancy given for Ligand D into an equivalent Kd value. Ligand D has 80% occupancy at 10 nanomolar concentration.
Using the same approach as before, we find that the Kd for Ligand D is likely lower than Ligand C, as 80% occupancy has a higher ligand concentration compared to 30% occupancy.
In conclusion, based on the information provided:
- Ligand A has the tightest binding (lowest Kd).
- Ligand B has tighter binding than Ligand C.
- Ligand D has tighter binding than Ligand C.
- The order from tightest to weakest binding is Ligand A > Ligand D > Ligand B > Ligand C.
Since the dissociation constant is equal to the inverse of the association constant, it is possible to figure this out. It would be the one with the lowest dissociation contant or the one with the highest association constant. Ligand A has a lower dissociation constant then B, so B is eliminated. Since 30% occupancy requires the ligand must greater than 3 times the number of it's dissociation constant, and the units for the dissociation constant are given in M. After converting everything to M, the equations for ligand C and D will be the following:
1x10^-6=3(Kd)
10x10^-9=8(Kd)
Solving for Kd,
Ligand C's Kd=3.3 x10^-7
Ligand D's Kd=1.2 x10^-9
The answer is Ligand A