For the synthesis of one mole of hexose in ph | vedclass

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(D) In the $C_3$ cycle (Calvin cycle),the synthesis of one molecule of glucose $(C_6H_{12}O_6)$ requires $6$ turns of the cycle.Each turn of the Calvi

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$54 \ ATP \ eq.$

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(D) In the $C_3$ cycle (Calvin cycle),the synthesis of one molecule of glucose $(C_6H_{12}O_6)$ requires $6$ turns of the cycle.Each turn of the Calvin cycle requires $3 \ ATP$ and $2 \ NADPH$.Therefore,for $6$ turns: $6 	imes 3 = 18 \ ATP$ and $6 	imes 2 = 12 \ NADPH$.Since $1 \ NADPH$ is equivalent to $3 \ ATP$ in terms of energy,$12 \ NADPH = 12 	imes 3 = 36 \ ATP$.Total energy required = $18 \ ATP + 36 \ ATP = 54 \ ATP$ equivalents.

For the synthesis of one mole of hexose in photosynthesis,the energy used is equivalent to how many $ATP$?

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