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Isothermal titration calorimetry

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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


    3. 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
     
  2. jcsd
  3. Sep 25, 2007 #2
    never mind, I think i figured it out using the van't hoff equation to find delta H at 37C since I can find the different Kds and am given the different temperatures.
     
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