Question

) Oxygen gas is reduced to water: O2 +4H+4e2H20. If each gram of 20. 06 pt \ rluscle tissue consumes O2 at a rate of 12.0 μmol/min, what rate does it consume glucose (glucose being its only fuel), and what rate does it produce ATP?

Answer 1

Glucose + 6O₂ + 6P_i^2- + 36 ADP^3- + 36H⁺ $\rightarrow$ 6CO₂ + 42H₂O + 36ATP^4-

Rate of O₂ consumption = 12.0×10^-6mol/min.

-d[glucose]/dt = -(1/6)d[O₂]/dt

Rate of glucose consumption = (1/6)×12.0×10^-6mol/min

= 2.0×10^-6mol/min. (Answer)

Rate of ATP generation is 6 times more than that of rate of consumption of oxygen.

Rate of ATP generation = 6×12.0×10^-7mol/min

= 72.0×10^-5mol/min

= 7.20×10^-4mol/min . (Answer)