What is the effect of temperature on ligand integrin dimerization?

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The discussion focuses on the effect of temperature on the dimerization of integrin-ligand complexes using three RGD-containing ligands: cHArGd, cRGD, and Integrilin. At 20°C, Integrilin shows the highest degree of dimerization, while cRGD leads at 40°C. Participants express uncertainty about calculating the fraction of total integrin-ligand complexes and the Gibbs free energy, discussing the relationship between the association constant (Ka) and the dimer concentration. The equation ΔG = -RTlnKeq is proposed as a method to calculate Gibbs free energy, assuming Keq equals Ka. Overall, the thread emphasizes the complexities of calculating dimerization and thermodynamic properties in this context.
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Homework Statement



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



Homework Equations


none given.

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 \DeltaG=\DeltaH-T\DeltaS. Assuming H does not change, then would \DeltaS 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 \DeltaG=-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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Quickdry135 said:
would the equation \DeltaG=-RTlnKeq work for this equation assuming Keq=Ka in this case?
That is the way I read the problem. I don't know of any way to calculate entropy by inspection. It can be derived as:

\Delta S = \frac{\Delta H - \Delta G}{T}

where \Delta G is derived from \Delta G=-RTlnKeq
 
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thanks,

for part A, I was thinking that I could use Ka=[X]/[.5*initial complex-x]^2 to find X which would be the dimer concentration. Then I could use that over the initial complex concentration which is given to find the fraction. Would this work?
 
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