Please Note! These problems are here to help you understand that the ΔG for any reaction can be changed by changing the concentrations of reactants and/or products. For example, a reaction that has a positive ΔG may be changed to a reaction having a negative ΔG by increasing the concentration of the substrates and/or decreasing the concentration of the products. Likewise, a reaction that has a negative ΔG may be changed to a reaction having a positive ΔG by decreasing the concentration of the substrates and/or increasing the concentration of the products. If you understand this, there is no reason to work out the problem described in this section of the text.

If you feel the need to work this out using natural logs, you will first have to work out Question 2 on page 345 and then go on to the REVERSIBILITY problem (second problem below). Natural logs will not be part of any objective for this course.

# The Second Q:/A: in Chapter 19

Question 2 on page 345 states: The ΔG0! for the conversion of glucose 6-P to glucose 1-P is +1.65 kcal/mole (Note!, the value given in the book of a negitive ΔG0! is incorrect). What is the ratio of [glucose-1-P] to [glucose-6-P] at equilibrium?

The answer for question 2 on page 346 states:  In equation 1 of Table 19.2, ΔG0!  = -RT ln Keq.  For this reaction, Keq = [glucose-1-phosphate]/[glucose-6-phosphate]. The constant R is 1.99 X 10-3 kcal/mole- oK, and T is (273 + 25) oK, so RT equals 0.593 kcal/mole. Substituting in equation 1 then gives 1.65 = -0.593 ln [glucose-1-P]/[glucose-6-P]. Thus, ln[glucose 1-P]/[glucose 6-P] = - 2.78, and [glucose-1-P]/[glucose-6-phosphate] = e-2.78, or 0.062.  So the ratio of [glucose-1-P] to [glucose-6-P] at equilibrium is 0.062.

You could answer this question by using log instead of ln.  Any examination question in this course will be stated in terms of log and not ln.

You can convert ln to log by multiplying by 2.3.  Thus, ln X = 2.3logX

You should remember that at equilibrium, the value for ΔG is zero.

ΔG = ΔG0! + RT ln[glucose-1-P]/[glucose-6-phosphate]

Since ΔG = 0 and ln X = 2.3log X

ΔG0!  = -RT (2.3) log [glucose-1-P]/[glucose-6-phosphate]

Since the ΔG0! for the conversion of glucose 6-P to glucose 1-P is +1.65 kcal/mole,

1.65 =  - (2 X 10-3 kcal/mole)(273+25) 2.3log [glucose-1-P]/[glucose-6-phosphate]

or 1.65 =  - 1.36 log [glucose-1-P]/[glucose-6-phosphate]

or 1.65 = 1.36 log [glucose-6-phosphate]/ [glucose-1-P]

or 1.21 = log [glucose-6-phosphate]/ [glucose-1-P]

101.21 = [glucose-6-phosphate]/ [glucose-1-P]

16.2 = [glucose-6-phosphate]/ [glucose-1-P]

You would need to use your calculator to convert 101.21 to 16.2

We have shown that, at equilibrium, the ratio of substrate to product is 16.2.

In this class, I would remind you that RT ln Keq is equal to 1.36 log Keq so don't bother to memorize the conversion.

# The Reversibility of the Phosphoglucomutase Reaction in the Cell

Later in the chapter, the author brings up another problem on the bottom, "THE REVERSIBILITY OF THE PHOSPHOGLUCOMUTASE REACTION IN THE CELL." The author illustrates that the ΔG for any reaction can be changed by changing the concentration of either the substrates or the products.  The author states:
"The effect of substrate and product concentration on ΔG and the direction of a reaction in the cell can be illustrated with conversion of glucose-6-P to glucose-1-P, the reaction catalyzed by phosphoglucomutase in the pathway of glycogen synthesis (see Fig. 19.3). The reaction has a small positive ΔG0! for glucose 1-P synthesis (+1.65.kcal/mole) and at equilibrium, the ratio of [glucose 1-P]/[glucose 6-P] is about 6 to 94 (This ratio was calculated in Question 2 above).  However, if another reaction utilizes glucose 1-P such that this ratio suddenly becomes 3 to 94, there is now a driving force for converting more glucose 6-P to glucose 1-P and restoring the equilibrium ratio.  Substitution in Equation 19.2 gives ΔG, the driving force to equilibrium, as  +1.65 + RT ln [G1P]/[G6P] = 1.65 + (-2.06) = -0.41, which is a negative value.  Thus, a decrease in the ratio of product to substrate has converted the synthesis of glucose 1-P from a thermodynamically unfavorable to a thermodynamically favorable reaction that will proceed in the forward direction until equilibrium is reached."

Using the values 6 and 94 for the concentrations of glucose-1-P and glucose-6-P, you can calculate that this reaction is pretty close to equilibrium, i.e., ΔG = 0

ΔG =  ΔG0! + RT ln [P]/[S]

If the ratio of [glucose 1-P]/[glucose 6-P] is 6/94

ΔG =  1.65 + (2 kcal/mole)(273+25) 2.3log (6/94)

ΔG =  1.65 + 1.36 log 0.064

ΔG =  1.65 + 1.36 (-1.19)

ΔG =  1.65 -1.63

ΔG =  0.02

0.02 is about 0, so the reaction is at equilibrium and there is no tendency for the reaction to occur in the forward direction.

However if you lower the product from 6 to 3,

ΔG =  1.65 + (2 kcal/mole)(273+25) 2.3log (3/94)

ΔG =  1.65 + 1.36 log 0.032

ΔG =  1.65 + 1.36 (-1.5)

ΔG =  1.65 -2.03

ΔG =  -0.38

Since the ΔG is negative, the reaction is now going to occur in the direction of product.