NADH+H++1/2O2? NAD++H2O a) use Table 4.1 to calculate the Gibbs free-energy chan

Question

NADH+H++1/2O2? NAD++H2O

a) use Table 4.1 to calculate the Gibbs free-energy change of the above reaction as it takes place in mitochondria at pH 7 and 25°C. The concentration of reactants and products are [NADH]=2 mM, [NAD+]=1 mM, and PO2= 0.05 atm.

Table 4.1 gives E'=0.320 for NADH to NAD and E'=0.408 for 0.5O2 to H2O

b) Given that 4 mol of ATP is made from Pi and ADP for every mole of NADH oxidized, what fraction of the Gibbs free- energy change of part a) is use in making ATP in mitochondria during oxidative phosphorylation? [ADP]= 2 mM, [ATP]= 4 mM, and Pi=7 mM.

Explanation / Answer

delta G is change in gibbs free energy. It is equal to (-2.303RT log K), where T is temperature in kelvin, R is gas constant and K is Rate constant.

K = [products]/[reactants] = [NADH]/ [NAD+]= 2/1 =2

T= 25+273=298K; and R = 1.987*10-3 cal K-1milimol-1

Putting these values in the equation:-

delta G = -2.303 * 1.987 *10-3*298 * log 2 = 1.36 * log2 = 0.409 cal/ mM

log 2 = 0.3010

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