Question

I think the answer to this biochemistry question is E but is this correct? Im looking for a good explanation.

Suppose you have 1X10 ^-6 moles of radioactive ATP where only the alpha phosphate group is radioactive (i.e. P-P-*P-suger-base, where the * indicates the radioactive phosphorous atom). If you add all the radioactive ATP to a tube containing hexokinase and 1X10 ^-6 moles of glucose, what percentage of phosphorylated glucose molecules would you expect to be radioactive after an appropriate amount of incubation?

A.	100%
B.	0%
C.	50% because at equilibrium the concentration of glucose and glucose-6-phosphate will be equal
D.	Impossible to tell
E.	Somewhere between 50-100% because the equilibrium does favor glucose-6-phosphate, but some glucose and ATP will remain.

Answer 1

Ans. The dG0’ value for the reaction catalyzed by hexokinase (Glu + ATP <---> Glu-P + ADP) is -16.7 kJ/mol. The calculated Keq value for the reaction is around 841.76, indicating that the concentration of products is far greater than that of reactants – say ,on a relative scale 842 product molecules per 1 reactant at the equilibrium.

# In other words, the reaction cab be assumed to goes to completion (100 %).

Therefore, correct option is- A. 100.0 %. That is, all (almost) the reactant glucose will be converted to Glu-6-P*.

AG RT In Keq Equation 1 Where, AGO standard/theoretical Gibb's free energy change T temperaten kelvin c 273.15) K Keq equilibrium constant under standard condition R0.008314462 k mol K10.001987204 kcal mol 1 K1 Given, T= 25.000C= 298.15 K Putting the values in equation 1 dGo'--16.70 kl mol-1 -16.70 kl mol-1- 一0.008314462 kJ mol-1 K-1x 298.15 Kx In Keq Or, 2.303 log Keq-2479 kJ mol -16.70 k mo6.736704526 Or, log Keq 6./367045262925186507 = 2.303 Or, Keqantilog 2.925186507 10 2.925186507 Hence, Keq841.76