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(D) During glycolysis,each molecule of glucose $(C_6H_{12}O_6)$ is broken down into $2$ molecules of pyruvic acid $(CH_3COCOOH)$.Therefore,to calculat

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$26$

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(D) During glycolysis,each molecule of glucose $(C_6H_{12}O_6)$ is broken down into $2$ molecules of pyruvic acid $(CH_3COCOOH)$.Therefore,to calculate the number of glucose molecules required for the formation of $52$ pyruvic acid molecules,we use the following calculation:$	ext{Number of glucose molecules} = \frac{	ext{Total pyruvic acid molecules}}{	ext{Pyruvic acid molecules per glucose}}$$	ext{Number of glucose molecules} = \frac{52}{2} = 26$Thus,$26$ glucose molecules are required to produce $52$ pyruvic acid molecules.The correct option is $(D)$.

How many glucose molecules are required for the formation of $52$ pyruvic acid molecules at the end of glycolysis?

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