
In the glycolysis pathway which of the following steps show reduction of coenzymes?
A. \[1\],\[3\] -diphosphoglycerate to \[3\]-phosphoglycerate
B. Glucose \[6\]-phosphate to fructose \[6\]-phosphate
C. Glyceraldehyde \[3\]-phosphate to \[1\],\[3\]-diphosphoglycerate
D. \[3\]-phosphoglycerate to \[2\]-phosphoglycerate
Answer
499.2k+ views
Hint: The metabolic pathway that converts glucose into pyruvic acid is termed as glycolysis. Free energy that is released during this process is used to form high-energy molecules Adenosine Triphosphate or ATP and reduced Nicotinamide Adenine Dinucleotide or NADH. The process of glycolysis is a sequence of ten reactions that are catalyzed by enzymes. It is a metabolic pathway which does not require oxygen.
Complete answer:
Option A: \[1\],\[3\]-diphosphoglycerate to \[3\]-phosphoglycerate: \[1\],\[3\] diphosphoglycerate is converted into \[3\] phosphoglycerate in the \[{7^{th}}\] step of glycolysis. Here, the phosphate is transferred from \[1\], \[3\]-bisphosphoglycerate to ADP which forms ATP with the help of phosphoglycerokinase. Hence, two molecules of phosphoglycerate and ATP are obtained by the end of this step. Here, there is no reduction of coenzymes.
Hence option A is not correct.
Option B: Glucose \[6\]-phosphate to fructose \[6\]-phosphate: The \[{2^{nd}}\] stage of glycolysis process is the conversion of glucose \[6\]-phosphate to fructose \[6\]-phosphate. With the help of enzyme phosphoglucomutase, the glucose \[6\]-phosphate is isomerized into fructose \[6\]-phosphate. This step does not show any reduction of coenzymes.
Hence option B is not correct.
Option C: Glyceraldehyde \[3\]-phosphate to \[1\], \[3\]-diphosphoglycerate: The \[{6^{th}}\] step in glycolysis involves conversion of glyceraldehyde \[3\]-phosphate to \[1\], \[3\]-diphosphoglycerate. In this step, the enzyme glyceraldehyde \[3\]-phosphate dehydrogenase adds a phosphate to oxidized glyceraldehyde phosphate. Also, \[NA{D^ + }\] is reduced into . Nicotinamide adenine dinucleotide or NAD is a coenzyme that is found in all living cells.
Hence option C is the correct answer.
Option D: \[3\]-phosphoglycerate to \[2\]-phosphoglycerate: This conversion happens in the \[{8^{th}}\] step of glycolysis. Here, the phosphate of both phosphoglycerate molecules is relocated from the third to the second carbon to yield two molecules of \[2\]-phosphoglycerate with the help of enzyme phosphoglyceromutase. No co-enzyme reduction is observed in this step.
Hence option D is not correct.
Therefore, Option C. Glyceraldehyde \[3\]-phosphate to \[1\], \[3\]-diphosphoglycerate is the correct answer.
Note:
The glycolysis pathway took almost about \[100\] years to fully elucidate. In order to understand the pathway as a whole, the combined results of many smaller experiments were required. The non-cellular fermentation experiments of Eduard Buchner provided insight into the component steps of glycolysis, during the \[1890\]. He demonstrated that the glucose to ethanol conversion was possible using a non-living extract of yeast, which is due to the action of enzymes in the extract.
Complete answer:
Option A: \[1\],\[3\]-diphosphoglycerate to \[3\]-phosphoglycerate: \[1\],\[3\] diphosphoglycerate is converted into \[3\] phosphoglycerate in the \[{7^{th}}\] step of glycolysis. Here, the phosphate is transferred from \[1\], \[3\]-bisphosphoglycerate to ADP which forms ATP with the help of phosphoglycerokinase. Hence, two molecules of phosphoglycerate and ATP are obtained by the end of this step. Here, there is no reduction of coenzymes.
Hence option A is not correct.
Option B: Glucose \[6\]-phosphate to fructose \[6\]-phosphate: The \[{2^{nd}}\] stage of glycolysis process is the conversion of glucose \[6\]-phosphate to fructose \[6\]-phosphate. With the help of enzyme phosphoglucomutase, the glucose \[6\]-phosphate is isomerized into fructose \[6\]-phosphate. This step does not show any reduction of coenzymes.
Hence option B is not correct.
Option C: Glyceraldehyde \[3\]-phosphate to \[1\], \[3\]-diphosphoglycerate: The \[{6^{th}}\] step in glycolysis involves conversion of glyceraldehyde \[3\]-phosphate to \[1\], \[3\]-diphosphoglycerate. In this step, the enzyme glyceraldehyde \[3\]-phosphate dehydrogenase adds a phosphate to oxidized glyceraldehyde phosphate. Also, \[NA{D^ + }\] is reduced into . Nicotinamide adenine dinucleotide or NAD is a coenzyme that is found in all living cells.
Hence option C is the correct answer.
Option D: \[3\]-phosphoglycerate to \[2\]-phosphoglycerate: This conversion happens in the \[{8^{th}}\] step of glycolysis. Here, the phosphate of both phosphoglycerate molecules is relocated from the third to the second carbon to yield two molecules of \[2\]-phosphoglycerate with the help of enzyme phosphoglyceromutase. No co-enzyme reduction is observed in this step.
Hence option D is not correct.
Therefore, Option C. Glyceraldehyde \[3\]-phosphate to \[1\], \[3\]-diphosphoglycerate is the correct answer.
Note:
The glycolysis pathway took almost about \[100\] years to fully elucidate. In order to understand the pathway as a whole, the combined results of many smaller experiments were required. The non-cellular fermentation experiments of Eduard Buchner provided insight into the component steps of glycolysis, during the \[1890\]. He demonstrated that the glucose to ethanol conversion was possible using a non-living extract of yeast, which is due to the action of enzymes in the extract.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

