Glycogen Phosphorylase Equilibrium Glycogen phosphorylase...

Glycogen Phosphorylase Equilibrium Glycogen phosphorylase catalyzes the removal of glucose from glycogen. The $\Delta G^{\prime \circ}$ for this reaction is $3.1 \mathrm{kJ} / \mathrm{mol}$ (a) Calculate the ratio of $[\mathrm{Pi}] \text { to [glucose } 1 \text { -phosphate }]$ when the reaction is at equilibrium. (Hint: The removal of glucose units from glycogen does not change the glycogen concentration.) (b) The measured ratio $\left[\mathrm{P}_{\mathrm{i}}\right] /[\text { glucose } 1 \text { -phosphate }]$ in myocytes under physiological conditions is more than $100: 1 .$ What does this indicate about the direction of metabolite flow through the glycogen phosphorylase reaction in muscle? (c) Why are the equilibrium and physiological ratios different? What is the possible significance of this difference?

Glycogen Phosphorylase Equilibrium Glycogen phosphorylase catalyzes the removal of glucose from glycogen. The $\Delta G^{\prime \circ}$ for this reaction is $3.1 \mathrm{kJ} / \mathrm{mol}$
(a) Calculate the ratio of $[\mathrm{Pi}] \text { to [glucose } 1 \text { -phosphate }]$ when the reaction is at equilibrium. (Hint: The removal of glucose units from glycogen does not change the glycogen concentration.)
(b) The measured ratio $\left[\mathrm{P}_{\mathrm{i}}\right] /[\text { glucose } 1 \text { -phosphate }]$ in myocytes under physiological conditions is more than $100: 1 .$ What does this indicate about the direction of metabolite flow through the glycogen phosphorylase reaction in muscle?
(c) Why are the equilibrium and physiological ratios different? What is the possible significance of this difference?

Key Concepts

Standard Free Energy Change and Equilibrium Constant

This concept relates the standard free energy change (?G°') of a reaction to its equilibrium constant (K) through the relationship ?G°' = -RT ln(K). It allows us to predict how far a reaction lies from equilibrium under standard conditions and to calculate the ratio of concentrations of products to reactants at equilibrium.

Reaction Equilibrium and the Mass Action Ratio

At equilibrium, the reaction's forward and reverse rates are equal, which means the ratio of the concentrations of products and reactants (the mass action ratio) remains constant and is equal to the equilibrium constant. Calculating such ratios can provide insight into the relative concentrations of involved metabolites when the reaction is balanced.

Physiological Steady-State Conditions

In living cells, metabolic reactions rarely achieve true chemical equilibrium; instead, they operate in a dynamic steady state. Physiological conditions such as substrate availability, enzyme activity, and compartmentation often create concentration ratios that differ markedly from those predicted by equilibrium thermodynamics, allowing cells to maintain necessary metabolic flux.

Metabolic Flux Direction and Regulation

The direction of metabolite flow through an enzyme-catalyzed reaction is influenced by the departure of the in vivo concentration ratios from their equilibrium values. A large disparity between the measured physiological ratio and the equilibrium ratio indicates that the reaction is driven in one preferential direction, which is a key aspect of metabolic regulation and adaptation to cellular demands.

Transcript

00:01 To the nth power, so the nth power is some sort of random number, plus a loose phosphate yielding glycogen, and the glycogen is n to the minus 1, so that's minus the one that we just used, and that'll equal, or plus, excuse me, glucose 1 phosphate.

00:20 Glucose 1 phosphate is the one that's tied to the original glycogen that we put together to begin with, with the phosphate.

00:28 Now, on the glycogen to the nth power and glycogen to the nth minus 1, we don't actually use those concentrations because they are so larged.

00:41 There's so much of both sets of those that it's basically equal to 1.

00:45 And so instead, what we're going to use is the loose phosphate and the glucose 1 phosphate concentrations to help us with understand the steps that's involved with our calculation.

00:57 We also are going to have this equation of delta g equals minus rt natural log of k sub eq.

01:06 The delta g is our energy unit, and then the rt we're going to use as 2 .48.

01:16 The ln is natural log, and the case of eq, what is this? it's basically variations depending on our equations of the equilibrium constant.

01:28 That we're going to have as our number for that particular equilibrium of that reaction.

01:35 So take that delta g equals negative r t natural log k sub eq and rearrange it.

01:41 And what we're going to end up happening is delta g negative negative delta g divided by capital r capital t equals natural log of k sub eq.

01:52 So plug in basically plug and play.

01:55 What we know is the delta g is 3 .1 kilojoules per mole.

02:00 So we have a negative, because of the negative delta g, negative 3 .1 kilojoules per mole divided by, and then rt is 2 .48 kilojoules per mole.

02:10 We can eliminate the kilojoules per mole.

02:12 What we end up with is 3 .1 divided by 2 .48, which is 1 .25, and the whole thing is negative since it was negative 3 .1 to start with.

02:22 So then we need to do the natural log case of eq.

02:25 Is equal to negative 1 .25.

02:28 In order for us to be able to find the concentration of, sorry, the calculation of k sub eq, what we need to do is plug into our calculator, hit the buttons for 1 .25, then put into the button of our calculator, the plus minus sign, and then last put the button on the calculator of the e superscript x...

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