The net gain of glycolysis of one molecule of glucose is the formation of

Question

The net gain of glycolysis of one molecule of glucose is the formation of (A) 2NADH+H++4ATP+1 Pyruvic acid (B) 2NADH+H++2ATP+2 Pyruvic acid (C) 8ATP+2NADH+H++2 Pyruvic acid (D) 8ATP+2NADH+H++2 Pyruvic acid +CO2

Tardigrade Admin · Accepted Answer

Aerobic respiration has three major steps:
1. Glycolysis
2. TCA
3. ETS
Glycolysis occurs in the cytoplasm and results in the formation of two molecules of pyruvate, and is followed by a transition reaction, occurring in the mitochondrial matrix and is then followed by Krebs or TCA cycle which occurs inside the mitochondria.This leads to the conversion of pyruvate (3C molecule) into Acetyl CoA (2C molecule).
During glycolysis, one mole of glucose is broken down to 2 moles of Pyruvic acid, in many steps. In these steps, 4 moles of ATP are directly formed, 2 ATP molecules are utilized, and 2 moles of NADH+H⁺ are formed. The net reaction of this can be written as:

C_{6} H_{12} O_{6} (Gl u cose) + 2 N A D^{+} + 2 A D P + 2 H_{3} P O_{4} \to 2 C H_{3} COCOO H (P yr uv i c a c i d) + 2 N A D P H + 2 H^{+} + 2 A TP

Q. The net gain of glycolysis of one molecule of glucose is the formation of

$2 N A DH + H^{+} + 4 A TP + 1$ Pyruvic acid

$2 N A DH + H^{+} + 2 A TP + 2$ Pyruvic acid

$8 A TP + 2 N A DH + H^{+} + 2$ Pyruvic acid

$8 A TP + 2 N A DH + H^{+} + 2$ Pyruvic acid $+ C O_{2}$

Q. The net gain of glycolysis of one molecule of glucose is the formation of

2NADH+H++4ATP+1 Pyruvic acid

2NADH+H++2ATP+2 Pyruvic acid

8ATP+2NADH+H++2 Pyruvic acid

8ATP+2NADH+H++2 Pyruvic acid +CO2​

$2 N A DH + H^{+} + 4 A TP + 1$ Pyruvic acid

$2 N A DH + H^{+} + 2 A TP + 2$ Pyruvic acid

$8 A TP + 2 N A DH + H^{+} + 2$ Pyruvic acid

$8 A TP + 2 N A DH + H^{+} + 2$ Pyruvic acid $+ C O_{2}$