The following biochemical reaction occurs within a cell and is catalyzed by an imaginary enzyme called ABconvertase:
1) What is DG and
DG° for the reaction in the living cells ? In the
dead cells? In the test tube (at equilibrium) ?
2) If you add 1mM B to a test tube containing the
AB convertase, what will be the equilibrium concentrations of A and B?
3) If you induce the cells to produce more of the enzyme,
AB convertase, how would the concentrations of A and B change in the living
cells? In the dead cells? Explain why.
4) Give some possible reasons for the observation that the concentrations are different in living and dead cells.
Answer 1) In the test tube, I state specifically that the reaction has come to equilibrium. So the concentration ratio B/A (the equilibrium constant, K) is 1/9 or 0.111. We can calculate DG° = -RTln(K) = +5.45 kJ/mol (R=8.314 J/molK and T=298 K). Of course, if you chose a different temperature, or did the calculations in kilocalories (R=1.986 cal/molK) I accepted the answer.
In the living cells we don't know if the system is at equilibrium so we calculate DG (not DG°) as follows:
DG = DG° + RTln(Q) where Q is the concentration ratio in the live cells = 0.333
or DG = +5.45 + RTln(0.333) = +2.72 kJ/mol
This value tells you several important things:
a) The AB reaction is not at equilibrium (As someone nicely put it "The living cell is dynamic"). The enzyme can only move the reaction towards equilibrium, nothing more. This means that the concentrations of A and/or B are being changed by other reactions in the cell faster than the AB convertase can keep up.
b) The fact that DG is positive tells you that the reaction as I've written it (from A to B) is eager to procede from right to left. In other words the active metabolism is these dymanic cells is producing either a surplus of B or a deficit of A.
In the dead cells Q is 0.111 and DG
= 0. You should conclude from this that the AB reaction in
the dead cells has decayed to equilibrium state.
Answer 2) Simple. In the test tube the equilibrium concentration ratio will always be A:B = 9:1. In this case A=0.9 mM B=0.1 mM.
Answer 3) If you induce the living cells to make more enzyme the AB reaction will move closer toward equilibrium. The enzyme can't change the equilibrium concentrations, it can only speed up the rate at which the reaction occurs. If the cells produce more enzyme the concentration ratio of A:B will decrease, because DG is positive. In the dead cells, the reaction is already at equilibrium, so the amount of enzyme can have no effect. (I was assuming that you could induce the increase in enzyme before the cells died, but I will also accept the answer that you can't induce dead cells to make more enzyme!)
Answer 4) The live cells are dymanic. It could be that A is being depleted and/or B is being produced by other reactions faster than AB convertase can keep up. Several students also answered that the enzyme or substrates could be compartmentalized in the living cells. This is a good answer, too.
Note on grading this homework. About a quarter of the students gave me acceptable answers to questions 3 and 4 (good going!), there was clearly some ambiguity in the questions. As a result, I will not be counting these questions toward my part of your total grade.
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