On oxidation of how many grams of glucose are | vedclass

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(A) The complete oxidation of $1$ mole of glucose $(C_6H_{12}O_6)$ yields approximately $38$ moles of $ATP$ in aerobic respiration.To calculate the am

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$180 \, 	ext{grams}$

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(A) The complete oxidation of $1$ mole of glucose $(C_6H_{12}O_6)$ yields approximately $38$ moles of $ATP$ in aerobic respiration.To calculate the amount of glucose required for $40$ moles of $ATP$:$1 \, 	ext{mole of glucose} = 180 \, 	ext{grams}$.If $38 \, 	ext{moles of ATP}$ are produced from $180 \, 	ext{grams}$ of glucose,then $1 \, 	ext{mole of ATP}$ is produced from $180 / 38 \, 	ext{grams}$ of glucose.Therefore,$40 \, 	ext{moles of ATP}$ are produced from $(180 / 38) 	imes 40 \approx 189.47 \, 	ext{grams}$.Given the standard options provided in typical biological contexts,if we assume the theoretical yield of $38 \, 	ext{ATP}$ per mole of glucose,the closest standard value for $1$ mole is $180 \, 	ext{grams}$.

On oxidation of how many grams of glucose are $40$ moles of $ATP$ synthesized?

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