1摩尔丙酮酸彻底氧化生成二氧化碳和水时,净生成多少摩尔ATP

2025-04-07 06:46:12
推荐回答(2个)
回答(1):

12.5分子atp
解析:
分两个阶段:
【1】丙酮酸氧化脱羧形成乙酰辅酶a:
该过程发生在线粒体的基质中,释放出1分子co2,生成一分子nadh+h+.
【2】乙酰辅酶a参与三羧酸循环,产生二氧化碳:
主要事件顺序为:
(1)乙酰coa与草酰乙酸结合,生成六碳的柠檬酸,放出coa.柠檬酸合成酶.
(2)柠檬酸先失去一个h2o而成顺乌头酸,再结合一个h2o转化为异柠檬酸.顺乌头酸酶
(3)异柠檬酸发生脱氢、脱羧反应,生成5碳的a-酮戊二酸,放出一个co2,生成一个nadh+h+.异柠檬酸脱氢酶
(4)
a-酮戊二酸发生脱氢、脱羧反应,并和coa结合,生成含高能硫键的4碳琥珀酰coa,放出一个co2,生成一个nadh+h+.酮戊二酸脱氢酶
(5)碳琥珀酰coa脱去coa和高能硫键,放出的能量用于驱动gtp(哺乳动物中)或atp(植物和一些细菌中)的合成.琥珀酰辅酶a合成酶
(6)琥珀酸脱氢生成延胡索酸,生成1分子fadh2,琥珀酸脱氢酶
(7)延胡索酸和水化合而成苹果酸.延胡索酸酶
(8)苹果酸氧化脱氢,生成草酸乙酸,生成1分子nadh+h+.苹果酸脱氢酶
小结:
一次循环,消耗一个2碳的乙酰coa,共释放2分子co2,8个h,其中四个来自乙酰coa,另四个来自h2o,3个nadh+h+,1fadh2.此外,还生成一分子atp.
三羧酸循环总反应:
乙酰coa+3nad++fad+gdp+pi—→2co2+3nadh+fadh2+gtp(atp)+2h+
+coa-sh
再加上丙酮酸氧化脱羧形成一分子nadh,所以共产生:4个nadh、1个fadh2和1个gtp(atp)
一分子nadh通过电子传递链的氧化,形成2.5分子atp;一分子fadh2通过电子传递链的氧化,形成1.5分子atp.【《生物化学》王镜岩
第三版
下册
107页】
一分子丙酮酸在线粒体内氧化成二氧化碳和水可生成atp分子的数目为:
2.5×4
+
1.5
+
1
=
12.5
即,可以生成12.5分子的atp

回答(2):

12.5分子ATP解析:分两个阶段:【1】丙酮酸氧化脱羧形成乙酰辅酶A:该过程发生在线粒体的基质中,释放出1分子CO2,生成一分子NADH+H+.【2】乙酰辅酶A参与三羧酸循环,产生二氧化碳:主要事件顺序为:(1)乙酰CoA与草酰乙酸结合,生成六碳的

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