Page 278 - Thermodynamics of Biochemical Reactions

278 Mathematica Solutions to Problems

4.1 (a) Calculate af G'" for the species of ATP at 298.15 K, pH 7, and I = 0.25. (b) Calculate A,G'" for ATP at 298.15 K,
pH 7, and I = 0.25 M. (c) Plot AfGo for ATP at 298.15 K, and I = 0.25 M versus pH. (d) Plot N,for ATP at I = 0.25 M
versus pH.

(BasicBiochemData2 has to be loaded)

(a) Use equation 4.4-10

0.5 2
2.91482 is (-nH + zi ) -PH
dgsp - - 2.47897 nH Log[lO 1
0.5
1 + 1.6 is
The data on the three species of ATP are in \the package.

The basic data for species of ATP are given by

atRsR
{{-2768.1, -3619.21, -4, 12}, {-2811.48, -3612.91, -3, 13},
{-2838.18, -3627.91, -2, 14}}
TableFOrm[atpSp]
-2768.1 -3619.21 -4 12
-2811.48 -3612.91 -3 13
-2838.18 -3627.91 -2 14
The first row gives the standard Gibbs energy of formation, the standard enthalpy of formation, the charge number, and the
number of hydrogen atoms of the species with the fewest hydrogen atoms. The standard transformed Gibbs energy of this
species can be calculated by using the function in the first line of this problem.

atpsgll2,lll
-2811.48

g1=atgsp~~~,~ll-~~*8.3l45l*.298l5*Log[lOA-7l-Z.9l482*((-4)A2-l2)*(.25A.5)/(l+l.6*.25A.5
1

-2291.86

~~~~~pSg[[~,111-13*8.31451*.29815*L0g[10~-71-2.91482*((-3)~2-13)*(.25~.5)/(1+1.6*~.25~.
5))
-2288.81

g3=atgsg[[3,1]]-14*8.3l45l*.Z98l5*Log[lOA-7]-2.9l482*((-2)A2-l4)*(.25A.5)/(l+l.6*(.25A.
5))
-2270. I

(b) Use equation 4.5-1 to calculate A, G'" for the pseudoisomer group