Question

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) A. 100% B. 38% C. 62% D. 80%​

Answer 1

Answer:

On complete combustion of glucose (1 mole) to CO

2

& H

2

O, approx. 686 kcal of energy is released. When 1 g of glucose respires aerobically by ETS, Glycolysis and Krebs cycle, around 38 ATP molecules are generated. The terminal group of a mole of ATP has around 10 kcal. Thus, 38 ATP molecules represent a yield of 380 kcal of energy.

So the correct option is '380'

Answer 2

If $1$ mole of glucose gives $38ATP$ energy, the efficiency of glucose metabolism is (B) $38 \%$ .

Step by step explanation:

Given:

No . of moles: $1$

Energy: $38ATP$

$1ATP= 7.3kcal$

To find:

Efficiency of glucose

Solution:

Catabolism of glucose produces a total of $38ATP$

                $38ATP \times 7.3k.cal$

$ATP= 262k.cal$

Glucose has a caloric value of $686 k.cal$

As a result,

Glucose metabolism efficiency is  $= \frac{262}{686} \times100$

                                                        $=0.38\times100$

                                                        $=38\%$

Glucose metabolism efficiency is $38\%$