※ 引述《Leni (喔…真不知該怎麼辦)》之銘言：
: 再過兩天就是期中考了，老師事先透露一題計算，
: 因為我本身是學生物的，對這個實在不了解，請大家救救我吧!
: The standard free energy difference between
: cis and trans proline amide bond conformations
: is about 5 kJ/mol (1.2 kcal/mol). For an
: isolated proline residue in a denatured protein
: at 300牵K, estimate the fraction of proteins
: that have proline in the cis conformation.
: p.s. proline為一胺基酸，較喜歡的是trans的狀態。
: 此題要求的是，算出proline是摺疊後，cis狀態的比例。
: ~
(可逆反應，非共振)
cis-proline ←→ trans-proline ΔG = -5 kJ/mol (向右為負)
C1 C2............ concentration at equilibrium
ΔG = -RT lnK ΔG: Gibbs' energy difference
R: gas constant
T: absolute temperature
K: equilibrium constant
K = C2/C1
-5000 = -8.314*298*lnK
計算得 K = C2/C1
X(C1) = C1/(C1+C2)
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※ 編輯: Keepgoing 來自: 140.112.243.210 (10/23 13:48)