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(B) In $C_3$ plants,the synthesis of one molecule of glucose $(C_6H_{12}O_6)$ requires $18$ $ATP$ and $12$ $NADPH$. In $C_4$ plants,an additional $ATP

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$6$

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(B) In $C_3$ plants,the synthesis of one molecule of glucose $(C_6H_{12}O_6)$ requires $18$ $ATP$ and $12$ $NADPH$. In $C_4$ plants,an additional $ATP$ is required per $CO_2$ molecule fixed to regenerate phosphoenolpyruvate $(PEP)$ from pyruvate. Since $6$ molecules of $CO_2$ are required to synthesize one molecule of hexose sugar,$6$ additional $ATP$ molecules are consumed in $C_4$ plants compared to $C_3$ plants. Therefore,the total $ATP$ required for $C_4$ plants is $18 + 6 = 24$ $ATP$.

How many additional $ATP$ molecules are required for the synthesis of one molecule of hexose sugar in $C_4$ plants compared to $C_3$ plants?

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