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Quickdry135
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Homework Statement
Using isothermal titration calorimetry, you calculate [tex]\Delta[/tex]H[tex]^{o}_{bind}[/tex] (= -5000 cal/mol) for a protein-ligand binding reaction at 25°C. You then perform a separate assay in which you measure equilibrium ligand binding at two different temperatures:
L[tex]_{0}[/tex] (nM) Ceq (nM) at 25°C Ceq (nM) at 37°C
0.01 :: 0.007 :: 0.006
0.03 :: 0.021 :: 0.017
0.1 ;: 0.070 :: 0.058
0.3 :: 0.197 ;: 0.166
1 :: 0.537 :: 0.439
3 ;: 0.830 :: 0.778
10 :: 0.943 :: 0.930
30 :: 1.002 :: 0.964
100 :: 0.981 :: 1.009
What is [tex]\Delta[/tex]S[tex]^{0}_{bind}[/tex] at 37C
Homework Equations
[tex]\Delta[/tex]G[tex]^{0}[/tex]=RTlnK[tex]_{D}[/tex]
[tex]\Delta[/tex]G=[tex]\Delta[/tex]G[tex]^{0}[/tex] + RTln[L][tex]_{eq}[/tex]/[P][tex]_{eq}[/tex][L][tex]_{eq}[/tex]
[tex]\Delta[/tex]G=[tex]\Delta[/tex]H-T[tex]\Delta[/tex]S
The Attempt at a Solution
I can find K[tex]_{d}[/tex] graphically and therefore find [tex]\Delta[/tex]G[tex]^{0}[/tex] to be -12300.1 cal. At the same temperature, [tex]\Delta[/tex]G[tex]^{0}[/tex]=[tex]\Delta[/tex]G at 25C, so i can find \DeltaS at 25C. But I don't know how this helps me find [tex]\Delta[/tex]S[tex]^{0}_{bind}[/tex] at 37C or if this helps me at all.
thanks for any help or direction