Question

the ratio between NADH+H generated and ATP molecules utilizes in glycolysis​

Answer 1

Answer:

with respect to ncert ,it is 2/2=1

Answer 2

Answer:

Ratio: 2/2 = 1

Explanation:

• Glycolysis means the splitting of sugar. This process shows a sequence of reactions in which the glucose molecule is partially oxidized into two molecules of pyruvates with the production of a small amount of energy.

• Glycolysis is conveniently considered in two groups of reactions. The first set of reactions by which glucose (and fructose) derived from storage carbohydrates are converted into triose phosphates or Glyceraldehyde 3-phosphate via fructose 1,6-bisphosphate.

• This first group of reactions is called the 'preparatory or investment phase'. Since the energy in the form of ATP is consumed during this period.

• In the second set of reactions, triose phosphates or G3P are converted to pyruvate, which is the end product of glycolysis. This second set of reactions is characterized by a net gain of energy-rich molecules, such as ATP and NADH. This is known as the pay-off or energy-conserving phase.

• Each G3P molecule, being converted to pyruvate by a series of reactions during the payoff phase, produces 2 molecules of ATP and 1 molecule of NADH+ H⁺. So 1 molecule of glucose when converted to pyruvate by glycolysis can produce a total of 4 molecules of ATP and 2 molecules of NADH+H⁺.

• During the preparatory phase, activation of glucose (or fructose) to form Fructose 1,6-bisphosphate utilizes 2 molecules of ATP.

Thus net gain: 2 molecules of ATP and 2 molecules of NADH + H⁺,

(when 1 molecule of glucose or fructose is converted to 2 molecules of pyruvate during glycolysis).

Now coming to question,

• The ratio between NADH+H⁺ generated and ATP molecules utilized in the glycolysis pathway can be calculated as follows:

$Ratio: \frac{NADH+H ^{+}generated }{ATPutilized} = \frac{2 molecules of NADH+H^{+} }{2 molecules of ATP}=\frac{2}{2} = 1$

Hence the ratio is found to be 1.