Read the passage given below and answer the question:
Adenosine triphosphate (ATP) is the energy-carrying molecule found in the cells of all living things. ATP captures chemical energy obtained from the breakdown of food molecules and releases it to fuel other cellular processes. ATP is a nucleotide that consists of three main structures: the nitrogenous base, adenine, the sugar, ribose, and a chain of three phosphate groups bound to ribose. The phosphate tail of ATP is the actual power source which the cell taps. Available energy is contained in the bonds between the phosphates and is released when they are broken, which occurs through the addition of a water molecule (a process called hydrolysis). Usually only the outer phosphate is removed from ATP to yield energy, when this occurs ATP is converted to adenosine diphosphate (ADP), the form of the nucleotide having only two phosphates.
The importance of ATP (adenosine triphosphate) as the main source of chemical energy in living matter and its involvement in cellular processes has long been recognized. The primary mechanism whereby higher organisms, including humans, generate ATP is through mitochondrial oxidative phosphorylation. For the majority of organs, the main metabolic fuel is glucose, which in the presence of oxygen undergoes complete combustion to `CO_2` and `H_2O`:
`C_6H_(12)O_6 + 6O_2 rarr 6CO_2 + 6H_2O + "energy"`
The free energy (AG) liberated in this exergonic (AG is negative) reaction is partially trapped as ATP in two consecutive processes: glycolysis (cytosol) and oxidative phosphorylation (mitochondria). The first produces 2 mol of ATP per mol of glucose, and the second 36 mol of ATP per mol of glucose. Thus, oxidative phosphorylation yields 17-18 times as much useful energy in the form of ATP as can be obtained from the same amount of glucose by glycolysis alone.
The efficiency of glucose metabolism is the ratio of amount of energy produced when 1 mol of glucose oxidised in cell to the enthalpy of combustion of glucose. The energy lost in the process is in the form of heat. This heat is responsible for keeping us warm.
(source: Erecińska, M., & Silver, 1. A. (1989). ATP and Brain Function. Journal of Cerebral Blood Flow & Metabolism, 9(1), 2-19. `"https://doi.org/10.1038/jcbfm."` 1989.2 and `"https://www.britannica.com/science/adenosine-triphosphate"`)
What is the efficiency of glucose metabolism if 1 mole of glucose gives 38ATP energy?(Given: The enthalpy of combustion of glucose is 686 kcal, 1ATP= 7.3kcal)

Options

A. `100%`
B. `38%`
C. `62%`
D. `80%`

Explanation

To determine the efficiency of glucose metabolism when 1 mole of glucose yields 38 ATP, we can follow these steps:

### Step 1: Calculate the total energy produced from ATP
Each mole of ATP provides 7.3 kcal of energy. Therefore, if 1 mole of glucose produces 38 moles of ATP, we can calculate the total energy produced as follows:

\[ \text{Total energy from ATP} = \text{Number of ATP} \times \text{Energy per ATP} \]
\[ \text{Total energy from ATP} = 38 \, \text{ATP} \times 7.3 \, \text{kcal/ATP} = 277.4 \, \text{kcal} \]

### Step 2: Identify the enthalpy of combustion of glucose
The enthalpy of combustion of glucose is given as 686 kcal.

### Step 3: Calculate the efficiency of glucose metabolism
The efficiency of glucose metabolism can be calculated using the formula:

\[ \text{Efficiency} = \left( \frac{\text{Energy produced from ATP}}{\text{Enthalpy of combustion of glucose}} \right) \times 100 \]
Substituting the values we have:

\[ \text{Efficiency} = \left( \frac{277.4 \, \text{kcal}}{686 \, \text{kcal}} \right) \times 100 \]

### Step 4: Perform the calculation
Now, we can calculate the efficiency:

\[ \text{Efficiency} = \left( \frac{277.4}{686} \right) \times 100 \approx 40.5\% \]

### Step 5: Finalize the answer
Thus, the efficiency of glucose metabolism when 1 mole of glucose gives 38 ATP energy is approximately **40.5%**.

---