BIO 209 Week 2 Thermodynamics for Enzymology

1. BIO 209 • Week 2 • Thermodynamics for Enzymology

2. Chemical Bond • An attraction between atoms that allows the formation of chemical substances that contain two or more atoms. • The strength of chemical bonds varies considerably; there are "strong bonds" such as covalent or ionic bonds and "weak bonds" such as dipole–dipole interactions, the London dispersion force and hydrogen bonding. • Strong chemical bonding is associated with the sharing or transfer of electrons between the participating atoms

3. Thermodynamics matters! Thermodynamics tells us which reactions will go forward and which ones won’t.

4. Energy in biological systems • Thermodynamicscharacterizes the energy associated with equilibrium conditions in reactions • Kineticsdescribes the rate at which a reaction moves toward equilibrium

5. Thermodynamics • Equilibrium constant is a measure of the ratio of product concentrations to reactant concentrations at equilibrium • Free energy is a measure of the available energy in the products and reactants • They’re related by DGo = -RT ln Keq

6. … but first: iClicker quiz! • 1. Which of the following statements is true? • (a) All enzymes are proteins. • (b) All proteins are enzymes. • (c) All viruses use RNA as their transmittable genetic material. • (d) None of the above.

7. Laws of Thermodynamics • First law: The energy of an isolated system is constant. • Second law: Entropy of an isolated system increases.

8. What do we mean by systems, closed, open, and isolated? • A system is the portion of the universe with which we’re concerned (e.g., an organism or a rock or an ecosystem) • If it doesn’t exchange energy or matter with the outside, it’s isolated. • If it exchanges energy but not matter, it’s closed • If it exchanges energy & matter, it’s open

9. That makes sense if… • It makes senseprovided that we understand the words! • Energy. Hmm. Capacity to do work. • Entropy: Disorder. (Boltzmann): S = klnW • Isolated system: one in which energy and matter don’t enter or leave • An organism is not an isolated system:so S can decrease within an organism! Boltzmann Gibbs

10. Enthalpy, H • Closely related to energy:H = E + PV • Therefore changes in H are:H = E + PV + VP • Most, but not all, biochemical systems have constant V, P:H = E • Related to amount of heat content in a system Kamerlingh Onnes

11. Units • Energy unit: Joule (kg m2 s-2) • 1 kJ/mol = 103J/(6.022*1023)= 1.661*10-21 J • 1 cal = 4.184 J:so 1 kcal/mol = 6.948 *10-21 J • 1 eV = 1 e * J/Coulomb =1.602*10-19 C * 1 J/C = 1.602*10-19 J= 96.4 kJ/mol = 23.1 kcal/mol James Prescott Joule

12. Quiz 2 • When molecules go into solution, their entropy -------------------- because they’re freer to move around

13. Free Energy • Gibbs: Free Energy EquationG = H - TS • So if isothermal, G = H - TS • Gibbs showed that a reaction will be spontaneous (proceed to right) if and only if G < 0

14. Free energy and equilibrium • Gibbs: Go = -RT ln Keq • Rewrite: Keq = exp(-Go/RT) • Keq is equilibrium constant;formula depends on reaction type • For aA + bB  cC + dD,Keq = ([C]c[D]d)/([A]a[B]b)

15. Spontaneity and free energy • Thus if reaction is just spontaneous, i.e. Go = 0, then Keq = 1 • If Go < 0, then Keq > 1: Exergonic • If Go > 0, then Keq < 1: Endergonic • You may catch me saying “exoergic” and “endoergic” from time to time:these mean the same things.

16. Free energy as a source of work • Change in free energy indicates that the reaction could be used to perform useful work • If Go < 0, we can do work • If Go > 0, we need to do work to make the reaction occur

17. What kind of work? • Movement (flagella, muscles) • Chemical work: • Transport molecules against concentration gradients • Transport ions against potential gradients • To drive otherwise endergonic reactions • by direct coupling of reactions • by depletion of products

18. Coupled reactions • Often a single enzyme catalyzes 2 reactions, shoving them together:reaction 1, A  B: Go1 < 0 reaction 2, C D: Go2 > 0 • Coupled reaction:A + C  B + D: GoC = Go1 + Go2 • If GoC < 0,then reaction 1 is driving reaction 2!

19. Revision Thermodynamics for Enzymology NEXT 5 Slides Go through them before next discussion session on Thursday

20. Gibbs Free Energy & Entropy Change in entropy (S) of surroundings, proportional to amount of heat (H) transferred from system,& inversely proportional to the temperature (T) of surroundings (heat content ‘H’ is enthalpy) Ssurroundings = - Hsystem/T (1) Total entropy change expression: Stotal = Ssystem + Ssurroundings (2) Substituting eq. 1 into eq. 2 yields Stotal = Ssystem - Hsystem/T (3) Multiplying by -T gives -TStotal = Hsystem - TSsystem (4) -TS has energy units, referred to as, Gibbs free energy G = Hsystem - TSsystem (5)

21. Gibbs Free Energy & Entropy, cont. -TS has energy units, referred to as, Gibbs free energy G = Hsystem - TSsystem (5) G used to describe energetics of biochemical reactions Equation (3) shows that total entropy will increase only if, Ssystem > Hsystem/T (6) {2nd Law} Multiplying by ‘T’ gives, TSsystem > Hsystem Therefore, entropy will increase only if, G = Hsystem - TSsystem < 0 (7) This means, free-energy change must be negative for a reaction to be spontaneous, with increase in overall entropy of the universe. Therefore, free-energy of the system is the only term we need consider, Any effects of changes within the system on the surroundings are automatically taken into account.

22. Free-energy change: Spontaneity not Rate G tells us if the reaction can occur spontaneously: If G is negative, reaction spontaneous, exergonic If G is zero, no net change, system at equilibrium If G is positive, free energy input required, endergonic G of a reaction depends only on free-energy of products minus free-energy of reactants. G of a reaction is independent of path (or molecular mechanism) of the transformation 2.G provides no information about the rate of a reaction

23. Go’ of a reaction is related to K’eq Go’ is standard free-energy change, K’eq is equilibrium constant To determine G, must consider nature of both reactants and products as well as their concentrations Consider this reaction A + B  C + D G is given by G = Go + RTln([C][D]/[A][B]) (1) Standard conditions: concentrations of reactants = 1.0 M (Go’) Convention for biochemical reactions: standard state has pH of 7, (if H+ is a reactant, its activity value = 1). H2O activity value = 1 (Go’) Relation between Go’ & K’eq expresses energetic relation between products and reactants in concentration terms. At equilibrium, G = 0. Equation 1 becomes, 0 = Go’ + RTln([C][D]/[A][B]) (2) Go’ = - RTln([C][D]/[A][B]) (3) and so

24. Go’ relation to K’eq, cont. Equilibrium constant under standard conditions, K’eq,is defined as K’eq =[C][D]/[A][B] (4) Substituting equation 4 into equation 3 gives Go’ = - RTlnK’eq (5) Go’ = - 2.303RTlog10K’eq (6) = - 2.303x1.987x10-3x298xlog10K’eq = - 1.36xlog10K’eq R = 1.987x10-3 kcal mol-1 deg-1 & T = 298K (=250C) For example, if K’eq = 10, Go’ = -1.36 kal mol-1 Note, for each 10-fold change in K’eq , Go’ changes by 1.36 kcal mol-1