一分子丙酰辅酶A彻底氧化生成多少ATP

2025-04-07 20:29:15
推荐回答(2个)
回答1:

先说结果,新算法为16.5ATP,老算法为20ATP,详细计算过程如下:
(1)首先,根据最新算法(电子传递链/呼吸链,以下‘=’为等同的意思,括号内的为老算法)
①1GTP=1ATP(1GTP=1ATP)
②1NADH=2.5ATP(1NADH=3ATP)
③1FADH2=1.5ATP(1FADH2=2ATP)
(2)丙酰-CoA变成琥珀酰-CoA,过程如下(根据为王镜岩生物化学第三版下册242页图28-14):
①丙酰-CoA+ATP+CO2+H2O→D-甲基丙二酰-CoA(此反应由丙酰-CoA羧化酶催化)
②D-甲基丙二酰-CoA→L-甲基丙二酰-CoA(此反应由甲基丙二酰-CoA消旋酶催化)
③L-甲基丙二酰-CoA→琥珀酰-CoA(此反应由甲基丙二酰-CoA变位酶催化)
(3)琥珀酰-CoA由部分TCA循环(柠檬酸循环/三羧酸循环)变成草酰乙酸,过程如下:
①琥珀酰-CoA→琥珀酸+1GTP(此反应由琥珀酰-CoA合成酶催化)
②琥珀酸→延胡索酸+1FADH2(此反应由琥珀酸脱氢酶催化)
③延胡索酸+H2O→苹果酸(此反应由延胡索酸酶催化)
④苹果酸→草酰乙酸+1NADH(此反应由苹果酸脱氢酶催化)
(4)草酰乙酸由部分葡糖异生变成磷酸烯醇式丙酮酸进而由部分EMP循环(糖酵解)变成乙酰-CoA,过程如下:
①草酰乙酸+GTP→磷酸烯醇式丙酮酸(此反应由磷酸烯醇式丙酮酸羧激酶催化)
②磷酸烯醇式丙酮酸→烯醇式丙酮酸+ATP→丙酮酸(前一反应由丙酮酸激酶催化,后一反应为非酶促反应)
③丙酮酸+CoA-SH+NAD+→乙酰-CoA+NADH(此反应由丙酮酸脱氢酶复合体催化)
(5)乙酰-CoA经完整TCA循环进行氧化
这里反应就略了,按新算法为等同产生10ATP(老算法为12ATP)
综上所述,丙酰-CoA完全氧化产生的ATP有:
(-1ATP)+(1GTP+1FADH2+1NADH)+(-1GTP+1ATP+1NADH)+10ATP=(-1+1+1.5+2.5-1+1+2.5+10)ATP=16.5ATP
老算法过程就略了,结果为20ATP(只要把上面NADH、FADH2等同的ATP数全都替换成老算法就行了)

回答2:

丙酰辅酶A先转变为琥珀酰CoA(要消耗一个碳酸氢根离子,没有能量的产生和消耗),再进入三羧酸循环彻底氧化分解。在三羧酸循环,产生1
GTP,1
FADH2,1
NADH,一共是1+2+3=6
ATP
对,我忘了,变成草酰乙酸还要→PEP(消耗1GTP)→丙酮酸(产生1ATP)→乙酰CoA(产生1NADH),再进入TCA循环彻底氧化分解。
所以能量一共是6-1+1+3+12=21ATP

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