What is the effect of temperature on ligand integrin dimerization?

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SUMMARY

The discussion focuses on the effect of temperature on ligand integrin dimerization using three RGD-containing ligands: cHArGd, cRGD, and Integrilin. The equilibrium constants (Ka) for each ligand were measured at temperatures ranging from 20°C to 40°C, revealing that Integrilin exhibits the highest degree of dimerization at 20°C, while cRGD shows the greatest dimerization at 40°C. The participants also explored the calculation of Gibbs free energy (ΔG) using the relationship ΔG = -RTlnKeq, confirming that Keq can be equated to Ka in this context.

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  • Understanding of ligand-receptor interactions and dimerization.
  • Familiarity with equilibrium constants, specifically association constants (Ka).
  • Knowledge of Gibbs free energy and its calculation.
  • Basic principles of thermodynamics as they apply to chemical reactions.
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  • Calculate Gibbs free energy (ΔG) for the ligands at various temperatures using ΔG = -RTlnKa.
  • Investigate the relationship between temperature and dimerization in other ligand-receptor systems.
  • Explore the implications of dimerization on biological functions and signaling pathways.
  • Review methods for measuring equilibrium constants in biochemical systems.
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Researchers in biochemistry, molecular biology, and pharmacology, particularly those studying ligand-receptor interactions and thermodynamic properties of biomolecular complexes.

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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
 
Last edited:
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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
 
Last edited:
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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