Chapter 3, Problem 14Q

A.

Summary Introduction

To calculate: The number of ATP molecules generated from one glucose molecule.

Introduction: During cellular respiration, nutrients converted into biochemical energy in the form of ATP.  This process leads to the breakdown of large molecules into smaller molecules. This process releases energy and leads to the formation of stronger bonds in the products. Glucose breaks down into simpler molecules and produces energy in the form of ATP.

Explanation of Solution

Given,

Free energy (ΔG) for 1 mole of glucose = 2867 kJ

Chemical energy for phosphate bond of 1 mole ATP = 50 kJ

$\begin{array}{c}\text{Total number of generated ATP from 1 mole glucose =}\frac{\text{Total}\text{free}\text{energy}\text{present}\text{in}\text{1}\text{mole}\text{of}\text{glucose}}{\text{Energy}\text{for}\text{phosphate}\text{bond}\text{of}\text{1}\text{mole}\text{of}\text{ATP}}\\ \text{=}\frac{\text{2867}}{\text{50}}\\ \text{=}\text{57}\text{.3 molecules}\\ \simeq \text{ 57 molecules}\end{array}$

Conclusion

A maximum of 57 molecules are generated from one molecule of glucose.

B.

Summary Introduction

To calculate: The overall efficiency of ATP production from glucose as compared to the answer in part (A).

Introduction: Energy is required by cells to carry out metabolic activities. Cells can do so by yielding the energy stored in glucose molecules. This process is called glucose metabolism as glucose is converted to forms which can be easily used by the cell such as ATP.

Explanation of Solution

Refer to the Table 14-1, “products yields from glucose oxidation”, in the text book.

Given,

Respiration produces 30 moles of ATP from 1 mole of glucose.

$\begin{array}{c}\text{Overall efficiency of ATP from glucose =}\frac{\text{Actually}\text{produced}\text{ATP}\text{molecules}}{\begin{array}{l}\text{Number}\text{of}\text{ATP}\text{molecules}\\ \text{harvested}\text{as}\text{chemical}\text{energy}\end{array}}\text{×}\text{100}\\ \text{=}\frac{\text{30}}{\text{57}}\text{×}\text{100}\\ \text{=52}\text{.63\%}\\ \text{=}\text{53}\text{\%}\end{array}$

Conclusion

An overall efficiency of ATP production would be about 53%.

C.

Summary Introduction

To calculate: The increase amount of energy that heats the body.

Introduction: During cellular respiration, nutrients converted into ATP. This process releases energy and leads to the formation of stronger bonds in the products. Glucose breaks down into simpler molecules and produces energy in the form of ATP.

Explanation of Solution

Given,

$\begin{array}{c}\text{Energy}\text{not}\text{stored}\text{as}\text{chemical}\text{energy in ATP = Remaining}\text{after 47 \%}\text{of total free energy}\\ \times \text{ 0}\text{.24}\\ \text{ =}\text{1347}\times \text{0}\text{.24}\\ \text{=}\text{323kcal}\end{array}$

Body weight = 75kg

$\begin{array}{l}\text{The increase amount of energy}\text{that}\text{heats}\text{the}\text{body}\text{=}\frac{\text{322}}{\text{75}}\\ \text{= 4}\text{.3°C}\end{array}$

Conclusion

The temperature of the body increases by 4.3^°C, if the heat is not dissipated into the environment.

D.

Summary Introduction

To explain: The effect on the body when body converts the energy in food substances with 20% efficiency.

Introduction: In cellular respiration, oxygen reacts intracellularly with organic compounds to form water, carbon dioxide, and ATP as a source of energy.

Explanation of Solution

If body converts the energy in food substances with 20% efficiency, then 80% of available energy would be released as heat. The heat production would be higher than the normal. Hence the body would certainly overheat.

E.

Summary Introduction

To calculate: The amount of glucose from oxidation to produce energy.

Introduction: Cells can do so by yielding the energy stored in glucose molecules. This process is called glucose metabolism as glucose is converted to forms which can be easily used by the cell such as ATP.

Explanation of Solution

Given,

A resting body hydrolyzes about 40 kg of ATP in every 24 hours.

Refer to the Fig. 2-24, “Adenosine triphosphate (ATP) is a crucially important energy carrier in cells”, in the text book. The molecular formula of ATP is C₁₀H₁₂O₁₃N₅P₃.

$\begin{array}{l}\text{The molecular weight of ATP =}\text{(12×10 )+ (1×}\text{12}\text{)+}\text{(16}\text{×13)}\text{+(}\text{14×}\text{5}\text{)+ (31×}\text{3)}\\ \text{ =}\text{503 g/mole}\end{array}$

$\begin{array}{l}\text{The}\text{amount}\text{of}\text{ATP}\text{hydrolyzes}\text{in}\text{24}\text{hours}\text{=}\frac{\text{40kg}}{\text{0}\text{.503}\text{kg/mole}}\\ \text{=}\text{80 moles}\\ \text{=1000}\text{kcal}\text{liberated}\text{chemical}\text{energy}\end{array}$

$\begin{array}{l}\text{30}\text{moles}\text{of}\text{ATP}\text{=}\text{1}\text{mole}\text{glucose}\\ \text{ =}\text{180}\text{g/moles}\\ \text{80}\text{moles}\text{of}\text{ATP}\text{=}\frac{\text{180}\text{g/moles}}{\text{30}}\text{×}\text{80}\text{moles}\\ \text{ =}\text{480}\text{g}\end{array}$

Conclusion

1000 kcal chemical energy could be produced by the oxidation of 480 g glucose.