含14个碳原子的脂肪酸在体内彻底氧化分解共生成多少个分子的FADH2和NADH

净生成多少分子ATP?
2025-03-15 08:18:40
推荐回答(4个)
回答1:

含14个碳原子的脂肪酸在体内彻底氧化分解共生成27个分子的NADH,13个分子的FADH,净生成92分子ATP。

含14个碳原子的脂肪酸在体内首先活化脂肪酸,消耗两个ATP,彻底氧化分解经6次β-氧化可生成7个乙酰CoA,6个NADH和6个FADH2。每个乙酰CoA进入TCA循环生成3个NADH、1个FADH和1个GTP(再转化为ATP)。

现在一般认为一个FADH2产生1.5个ATP,一个NADH产生2.5个ATP。

所以,共生成6+7*3=27个分子的NADH;

6+7=13个分子的FADH;

(14/2-1)*4+(14/2)*10-2=92个ATP。

扩展资料

1、脂肪酸氧化的其他途径:奇数碳原子脂肪酸的氧化。

人体含微量奇数碳脂肪酸,许多植物、海洋生物和石油酵母等含一定量的奇数碳脂肪酸。其β-氧化除生成乙酰CoA外,还生成1分子丙酰CoA,后者在β-羧化酶及异构酶的作用下生成琥珀酰CoA,经TCA途径彻底氧化。

2、生物体内由乙酰CoA合成脂肪酸的有:

①非线粒体酶系合成途径:即胞浆酶系合成饱和脂肪酸途径。该途径的终产物是软脂酸,故又称为软脂酸合成途径,它是脂肪酸合成的主要途径。

②线粒体酶系合成途径:又称为饱和脂肪酸碳链延长途径。

参考资料来源

百度百科-脂肪酸

回答2:

6分子FADH2和NADH,净生成112个ATP吧……7个乙酰辅酶A每个进行三羧酸循环生成12个,6分子FADH2和NADH一共30个,活化脂酰辅酶A消耗2个

回答3:

6分子FADH2和NADH,净生成92ATP

回答4:

各六分子

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