91Ó°ÊÓ

The metabolism process of a glucose molecule is given by

$38{\text{ADP}}^{+}{\text{+38PO}}_4\text{→38´¡°Õ±Ê}$

The ATP splits back into ADP and phosphate ions following the reaction

$38\text{´¡°Õ±Ê→}38{\text{ADP}}^{+}{\text{+38PO}}_4$

The force exerted by the muscle is 4 pN and the displacement is 11 nm.

Formula used:

Write the expression for the work done.

$\text{°Â=¹ó·³§Â·Â·Â·Â·Â·Â·Â·Â·Â·(1)}$

Here, F is the force, S is the displacement and W is the work done.

Write the expression for maximum work done per ATP.

${\text{W}}_{\text{max}}\text{=}\frac{\text{Energyreleased}}{\text{NumberofATPreleased}}\text{·········(2)}$

Here, ${\text{W}}_{max}$is the maximum work done.

Write the expression for the efficiency of the muscle.

$\text{E=}\frac{\text{W}}{{\text{W}}_{max}}\text{··········(3)}$

Here, E is the efficiency, W is the work done by a muscle.

Substitute $4\mathrm{pN}$for F and $11\mathrm{nm}$for S in expression (1).

$$\begin{gathered} W=\left(4\mathrm{pN}\right)(11\mathrm{nm}) \\ =\left(4\mathrm{pN}\left(\frac{{10}^{-12}\mathrm{N}}{1\mathrm{pN}}\right)\right)\left(11\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)\right) \\ =4.4×{10}^{-20}\mathrm{J} \end{gathered}$$

Substitute $2879\mathrm{kJ}/\mathrm{mol}$for energy released and 38 for the number of ATP released in expression (2).

$$\begin{gathered} {\text{W}}_{max}\text{=}\frac{2879\text{KJ/mol}}{38} \\ \text{=}\left(\frac{2879KJ/mol}{38}\right)\left(\frac{1mol}{6.023×{10}^{23}molecules}\right)\left(\frac{1000J}{1KJ}\right) \\ {\text{=1.26×10}}^{-19}\text{J} \end{gathered}$$

Substitute $4.4×{10}^{-20}\mathrm{J}$for W and $1.26×{10}^{-19}J$for ${\text{W}}_{\text{max}}$in expression (3).

$$\begin{gathered} \text{E=}\frac{4.4×{10}^{-20}J}{1.26×{10}^{-19}J} \\ \text{=0.349} \end{gathered}$$

The efficiency of the muscle is 34.9%.