Chapter 4, Problem 2Q

Summary Introduction

To analyze:

Walking consumes 100kcal/mi (kilocalorie/mile). The hydrolysis of ATP (adenosine triphosphate) gives –7.3 kcal/mole as ΔG^{o f} (Gibbs free energy), which drives muscle contraction. Calculate the grams of ATP needed for walking a mile. ATP synthesis is linked to oxidation of glucose (ΔG^{o f} = –686 kcal/mol); so, calculate the grams of glucose needed to metabolizethat amount of ATP.

Given:

Glucose oxidation is necessary for the generation of ATP and 40% of the energy generated is used for phosphorylation of ADP (adenosine diphosphate).The gram molecular weight of glucose is 180 g (gram) and that of ATP is 507 g.

Introduction:

The energy currency of a living cell is called ATP. It is then hydrolyzed into ADP or AMP (adenosine monophosphate). The standard free energy is the energy released or utilized at 25˚C(degree Celsius)or 298 K(Kelvin) and 1 atm (atmospheric) pressure.

Explanation of Solution

The ΔG^{o f} will be positive when it is utilized by the body.

Given,

$\begin{array}{l}{\text{Hydrolysis of ATP gives ΔG}}^{\text{of}}\text{= 7}\text{.3 kcal/mol}\hfill \\ \text{Energy needed for walking = 100 kcal/mile}\hfill \end{array}$

$\begin{array}{c}\text{Number }\text{of}\text{moles}\text{of}\text{ATP}\text{}\text{needed}\text{for}\text{walking}\text{one}\text{mile =}\frac{\text{Energy}\text{needed}\text{for}\text{walking}}{\Delta G{°}^f\text{of}\text{hydrolysed}\text{ATP}}\\ \text{Number}\text{of}\text{moles}\text{of}\text{ATP}\text{}\text{needed}\text{for}\text{walking}\text{one}\text{mile =}\frac{\text{100 kcal/mi}}{\text{7}\text{.3}\text{kcal/mol}}\\ \text{Number}\text{of}\text{moles}\text{of}\text{ATP}\text{}\text{needed}\text{for}\text{walking}\text{one}\text{mile =}\text{13}\text{.69}\text{mol/mi}\end{array}$

Given,

$\begin{array}{c}\text{Gram molecular weight of one ATP = 507 g}\\ \text{Gram molecular weight of 13}\text{.69 ATP = }507\times 13.69\text{g}\\ \text{Gram molecular weight of 13}\text{.69 ATP = 6940}\text{.83 g}\end{array}$

31 molecules of ATP are formed when 1 molecule of glucose is oxidized. Therefore,

$$

$\begin{array}{c}\text{1}\text{ATP is formation requires = }\frac{\text{1}}{\text{31}}\text{Glucose}\text{molecule}\\ \text{13}\text{.69}\text{ATPs}\text{}\text{will}\text{requir}e=\frac{\text{1}}{\text{31}}\times \text{13}\text{.69}\text{}\text{Glucose}\text{}\text{molecule}\\ \\ \text{13}\text{.69}\text{ATPs}\text{}\text{will require}=\text{0}\text{.44}\text{}\text{glucose}\text{molecule}\end{array}$

$\begin{array}{c}\text{Gram molecular weight of one glucose molecule = 180 g}\\ \text{Gram molecular weight of 0}\text{.44 glucose molecule = 0}\text{.44}\times 180\text{g}\\ \text{Gram molecular weight of 0}\text{.44 glucose molecule = 79}\text{.2 g}\end{array}$

During oxidation of glucose, 40% of the ATP production is used for phosphorylating the ADP. Thus, the remaining amount of glucose utilized for ATP production= $\text{100\%}-\text{40\%}=\text{60\%}$ .

$\begin{array}{c}\text{Amount}\text{of}\text{glucose}\text{that is metabolized to generate ATP}=\text{79}\text{.2}\text{×}\text{60\%}\\ \text{Amount}\text{of}\text{glucose}\text{that is metabolized to generate ATP}=\text{79}\text{.2×}\frac{\text{60}}{\text{100}}\\ \text{Amount}\text{of}\text{glucose}\text{that is metabolized to generate ATP}=\frac{\text{4752}}{\text{100}}\\ \text{Amount}\text{of}\text{glucose}\text{that is metabolized to generate ATP}=\text{47}\text{.52}\text{g}\end{array}$

Conclusion

Therefore, it can be concluded that 6940.83 g of ATP is produced for a mile-longwalk. The amount of glucose, which is metabolized to get this amount of ATP is 47.52 g.