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(D) Maize is a $C_4$ plant. In $C_4$ plants,the synthesis of one molecule of hexose (glucose) requires $30 	ext{ ATP}$ and $12 	ext{ NADPH}$. To cal

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$180 	ext{ ATP}, 72 	ext{ NADPH}$

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(D) Maize is a $C_4$ plant. In $C_4$ plants,the synthesis of one molecule of hexose (glucose) requires $30 	ext{ ATP}$ and $12 	ext{ NADPH}$. To calculate the requirement for $6$ hexose molecules,we multiply these values by $6$: $	ext{ATP requirement} = 6 	imes 30 = 180 	ext{ ATP}$. $	ext{NADPH requirement} = 6 	imes 12 = 72 	ext{ NADPH}$. Therefore,the total requirement is $180 	ext{ ATP}$ and $72 	ext{ NADPH}$.

The net requirement of assimilatory power for the formation of $6$ hexose molecules in a maize plant is:

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