Question

The metabolism of a glucose molecule (see previous problem) occurs in many steps, resulting in the synthesis of 38 molecules of ATP (adenosine triphosphate) out of ADP (adenosine diphosphate) and phosphate ions. When the ATP splits back into ADP and phosphate, it liberates energy that is used in a host of important processes including protein synthesis, active transport of molecules across cell membranes, and muscle contraction. In a muscle, the reaction ATP $\to$ ADP + phosphate is catalyzed by an enzyme called myosin that is attached to a muscle filament. As the reaction takes place, the myosin molecule pulls on an adjacent filament, causing the muscle to contract. The force it exerts averages about 4 piconewtons and acts over a distance of about $11\mathrm{nm}$. From this data and the results of the previous problem, compute the "efficiency" of a muscle, that is, the ratio of the actual work done to the maximum work that the laws of thermodynamics would allow.

Short answer

The efficiency of the muscle is $34.9\%$.

Step-by-step solution

Explanation

The metabolism process of a glucose molecule is given by

$38{\text{ADP}}^{+}{\text{+38PO}}_4\text{→38ATP}$

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

$38\text{ATP→}38{\text{ADP}}^{+}{\text{+38PO}}_4$

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

Formula used:

Write the expression for the work done.

$\text{W=F·S·········(1)}$

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.

Explanation

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

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

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

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

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

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

Step 3. Conclusion

The efficiency of the muscle is 34.9%.