## Ligand integrin dimerization

1. The problem statement, all variables and given/known data

3 different RGD-containing ligands (including Integrilin) are used to form receptor-ligand complexes, and measured dimerization of ligand-bound receptors as a fuction of temperature in the range 20-40C using a variety of methods. The 3 ligands are cHArGd, cRGD, and Integrilin. The dimerization reactin of ligand-bound receptors can be written as

2RL-> dimer

The values of the equilibrius constant K as a function of temperature for each of the 3 ligands are given in the following table

T (C) Ka, cHArGD (M-1) Ka, cRGD (M-1) Ka, Integrilin (M-1)
20 1.74*10^4 1.79*10^4 5.0*10^4
25 5.9*10^4 7.08*10^4 8.6*10^4
30 1.95*10^5 2.73*10^5 1.47*10^4
35 5.97*10^5 9.85*10^5 2.45*10^5
40 1.79*10^6 3.47*10^6 4.00*10^5

a) which ligand leads to the greates degree of dimerization of the integrin-ligand complex at 20C? at 40C? Calculate the fraction of total integrin-ligand complexes is 1*10^-6 M.

b) for each of the ligands, calculate the Gibbs free energy at 20C

2. Relevant equations
none given.

3. The attempt at a solution

Ka is the association constant, so the first part is easy. Integrilin at 20C and cRGD at 40C have the greatest degree of dimerization.

I'm not sure how to find the tfraction of total integrin-ligand complexes. The definition of Ka is [complex]/[ligand][integrin] so would the Ka be the same for the [dimer]/[.5*complex][.5*complex] of the same ligand type? Would this a correct dimer concentration? I'm a bit lost here

i know change in Gibbs free energy is $$\Delta$$G=$$\Delta$$H-T$$\Delta$$S. Assuming H does not change, then would $$\Delta$$S just be the number of complexes that have changed to dimers since 2 molecules have formed 1 molecule, lowering the entropy. I don't know how to find this change though.

::EDIT:: would the equation $$\Delta$$G=-RTlnKeq work for this equation assuming Keq=Ka in this case? I found it under a description of micelle formation.

Any help would be appreciated
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 Quote by Quickdry135 would the equation $$\Delta$$G=-RTlnKeq work for this equation assuming Keq=Ka in this case?
$$\Delta S = \frac{\Delta H - \Delta G}{T}$$
where $$\Delta G$$ is derived from $$\Delta G=-RTlnKeq$$