Class 19.Steady State Case PDF

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BCMB3100 with Paula Lemons and Takahiro Ito...

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Lemons BCMB 3100 Spring 2019

Name: Paris Chey

Steady State Case Modified learning objectives     

Explain why biochemists are interested in different ways of looking at the energetics of a chemical reaction, i.e., the various Gs of a reaction. Distinguish between ΔG and ΔG°’. Understand the equation: ΔG = ΔG°’ + RT ln Q. Use ΔG°’ and the concentrations of reactants and products to calculate actual ΔG for a reaction. Define steady state (i.e., homeostatic condition) and its importance in biological systems. Distinguish between steady state, standard state, and equilibrium and use these terms appropriately when describing biochemical systems. Explain the mathematical relationship between Keq and ΔG°’.

In this case we will consider the reaction shown below, which is step 5 in glycolysis. We will consider why biochemists want to understand the energetics of this reaction in a variety of conditions. What unique things can they learn about this reaction if they consider it under standard conditions or steady state conditions? How might this information be useful in understanding the function of a cell or organism?

Part 1: Standard delta G and steady state delta G 1. First, let’s think about the standard delta G of this reaction: ΔG°’. What are the conditions used to determine ΔG°’ for any reaction? Be sure to include the concentrations of products and reactants. To determine the value of ΔG°’, you must consider the temperature and an equilibrium constant, as well as a constant, R. 2. The ΔG°’ of DHAP   GAP is 7.5 kJ/mol. Based on this value is the forward or reverse reaction favored under standard conditions?

Based on this value, the reverse reaction is favored. When you plug ΔG°’=7.5 kJ/mol into the ΔG equation, you will get a positive ΔG value. 3. Imagine glycolysis happening in your cells right now! It’s a 10-step pathway, and the end product is pyruvate. If the fifth step of glycolysis DHAP   GAP has a ΔG°’ of 7.5 kJ/mol, how is pyruvate ever formed? Hint: Look back to your reading under the heading “Change in Concentration and dG.”

It is in the cells, so it is under nonstandard conditions (not at 1M of all of our solutes). There is homeostatic activity Q = [products]/[reactants] (mass action ratio, concentration of products over reactants) K is a special case of Q when we are at equilibrium; [products]/[reactants] Standard conditions: Q = 1 Steady state conditions: Q = different number

4. Your assigned reading explained that ΔG°’ does not tell us about the actual ΔG in a cell. However, we can use ΔG°’ calculate the actual ΔG. To do so, we will use this equation: ΔG = ΔG°’ + RT ln Q A. What is ΔG? Free energy under non-standard conditions (if ΔG = 0  equilibrium conditions) B. What is ΔG°’? Biochemical standard free energy C. What is R? Gas constant D. What is T? Temperature E. What is Q? Value of mass action expression of a system that is not at equilibrium Equilibrium conditions are not standard, so we don’t have a different symbol Q = products/reactants (only use K if it is at equilibrium) 5. As it turns out, cells typically maintain the concentration of GAP at 1/25 of the concentration of DHAP. Given this information, what is the value of Q that should be used in the equation above? 0.04 (1/25 = Q) 6. For the sake of comparison, remind yourself what the value of Q is under standard conditions? 1

7. Now calculate ΔG of DHAP   GAP in the cell. Remember to use ΔG = ΔG°’ + RT ln Q (use 310 K for T) (7.5) + (0.008324*310*ln(0.04)) = -0.806 ΔG is negative. This means the forward reaction is favored under actual cellular conditions. The actual cellular ΔG is closer to equilibrium; this is important because at equilibrium the forward and backward reactions are happening at the same rate. If we are near equilibrium, we are close to the backward reaction occurring as well. 8. You have just calculated the “steady state” ΔG for DHAP   GAP. Define steady state (i.e., homeostatic condition). The condition of internal optimal functioning for organisms; when organisms are at homeostasis Steady parameters under which life occurs; what’s happening in the cell Conditions in life are not at Equilibrium because the body an open system, things come in and out and not everything is constant/steady ΔG = 0  equilibrium, useful in a test tube ΔG°’ is useful because we can then use it to find the actual ΔG; group of simple consistent conditions that allow you to compare any set of reactions (we standardize in order to compare)

9. Now that you have completed this calculation, fill in the following table: [DHAP]

Standard conditions Steady state conditions

[GAP]

1

1

Mass Action Ratio ([GAP]/ [DHAP]) 1/1

25

1

1/25 = 0.04

Free Energy

ΔG°’=7.5 kJ/mol ΔG = -0.806

10. Pull together all you’ve learned so far to explain why ΔG and ΔG°’ are important to biochemists?  Steady state allows you to modify the concentrations and see what ΔG is. If the concentrations are equal (under standard conditions), they are equal because they are all at 1M. ΔG can be used in vivo, you can use ΔG to decide what a reaction is going to do at certain conditions in vivo. ΔG°’ allows us to compare across reactions, as well as tell us information about reactions that are under standard conditions.  There are not really standard conditions inside of the cell normally.

ΔG tells us if the reaction is spontaneous or not, and which direction the reaction is favored (towards the products or reactants). ΔG°’ is the change of ΔG that accompanies the formation of 1 mole of that substance from its component elements at their standard states. This helps determine if the reaction is spontaneous or not under standard conditions while other factors are involved and things are being altered in a standard state, which alters ΔG.

Keq = ([products]/[reactants]) Part 2: Steady state, Standard State, and Equilibrium The following questions tie together material from the carbonic anhydrase case and the steady state case: 11. There is one other G condition that biochemists are interested in: the G at equilibrium. What is the value of G at equilibrium? Does this mean that the reaction is not occurring? Explain. ΔG at equilibrium is 0. This means that the reaction is not driven towards the products or the reactants; same rate of formation for both products and reactants, concentrations remain constant. 12. The equilibrium condition is useful for biochemists because it allows us to understand the nature of a reaction. That is, if we examine the equilibrium condition, we can understand the direction of the reaction that will be favored when the reaction is NOT at equilibrium. What term do we use to describe the mass action ratio under equilibrium conditions? K is used to describe the mass action ratio under equilibrium conditions. Q is used to describe the mass action ratio when it is not under equilibrium conditions. 13. Keq for DHAP   GAP reaction is 0.05447. Reference Question 10 of the Carbonic Anhydrase case and Table 6.3 of your textbook. Given this Keq value is the G°’ of the reaction positive or negative? [NOTE: Yes, you already know this – see question 2, but I’m wanting you to think about the relationship between Keq and G°’] Keq is less than 1, so the reaction favors the reactants (reverse reaction), therefore the ΔG°’ of the reaction should be positive. 14. Use the mathematics provided in Appling section 3.3, Equations 3.19 – 3.21 to explain the relationship between Keq and G°’. Can set ΔG equal to 0 (because we are at equilibrium) and then solve for ΔG°’ (standard); Q at equilibrium is equal to Keq It is not saying you will have equilibrium concentrations of your reactions and products; it is saying you manipulated the equations so that if you had ΔG°’ you could find Keq and if you had Keq you could find ΔG°’.

15. Below I’ve added a row to the table from Question 9. Copy the values you obtained from question 9. Then add information for equilibrium conditions.

[DHAP]

Standard conditions Steady state conditions Equilibrium conditions

[GAP]

Mass Action Ratio ([GAP]/ [DHAP])

Free Energy

1

1

1

ΔG°’= 7.5

25

1

0.04 (1/25)

ΔG = -0.806

18

1

0.05447

ΔG = 0

Keq1 ΔG°’ will be negative Keq doesn’t have to be 1...