Question: ATP + H<sub>2</sub>O ¾® ADP + $PO_{4}^{3 -}$ D<sub>r</sub>Gŗ = – 31 kJ/mol Glucose + Fructose ¾® ...

Question

ATP + H₂O ¾® ADP + $PO_{4}^{3 -}$ D_rGŗ = – 31 kJ/mol

Glucose + Fructose ¾® Sucrose D_rGŗ = 23 kJ/mol

A 100 ml solution is 0.5 (M) with respect to glucose and 0.8 (M) with respect to fructose. To this how many litres of gastric juice containing 10^–2 (M) ATP is to be added to obtain maximum molarity of sucrose ?

Answer 1

Solution

Glucose + Fructose ¾® Sucrose

5 × 10^–2 mole 8 × 10^–2 mole

Glucose is the L.R. and maximum amount of sucrose to be

obtanined = 5 × 10^–2 mol

\ Energy required = 23 × 5 × 10^–2 kJ

Let V ml of ATP solution is required

V × 10^–5 × 31 = 23 × 5 × 10^–2

\ V = $\frac{23 \times 5}{31}$ × 10³ = 3.71 × 10³ mL = 3.71 L

Question: ATP + H<sub>2</sub>O ¾® ADP + \(PO_{4}^{3 -}\) D<sub>r</sub>Gŗ = – 31 kJ/mol Glucose + Fructose ¾® ...

Solution