CHAPTER 6 AN INTRODUCTION TO METBOLISM

1. CHAPTER 6 AN INTRODUCTION TO METBOLISM

3. Transformations between kinetic and potential energy

4. Transformations between kinetic and potential energy

5. Transformations between kinetic and potential energy There are also transformations between mechanical, chemical, and radiant energy

6. First law of thermodynamics Second law of thermodynamics Energy can be transformed but the entropy (disorder) in the never created or destroyed universe is increasing

7. Living organisms maintain order. Do we defy the second law?

12. ∆G = ∆HT∆S ∆G  change in free energy ∆H  change in heat or enthalpy; negative for an exothermic reaction T  absolute temperature will represent a constant of 1 ∆S  change in entropy; a larger number equals greater increase in disorder ∆G is negative for all reactions that occur spontaneously C6H12O6 + 6O2 6CO2 + 6H2O [greater potential energy][less potential energy] ∆H = 673 kcals/mole exothermic, ∆S is ∆G is negative and thus it is exergonic positive number + 13 kcal/mole -686

13. ENTHALPY; ENTROPY; AND SPONTANEITY [∆H +∆S] Spontaneous at all temperatures; wood burning [+∆H ∆S] Nonspontaneous regardless of temperatures; production of glucose during photosynthesis [+∆H +∆S] Spontaneous at high temperatures; ice to water, 2N2O54NO2 + O2 [∆H ∆S] Spontaneous only at low temperatures; steam to water;

14. -580 kcals/mole -620= ΔH – +40 ΔH= -620 + +40 ΔG=ΔH-TΔS ΔG= -620 kcals/mole ΔS=+40 kcals/mole ΔH= ΔG=-45 kcals/mole ΔH=+20 kcals/mole ΔS= ΔH=-222 kcals/mole ΔS=+30 kcls/mole ΔG= ΔG=-310 kcals/mole ΔS=-35 kcals/mole ΔH= ΔG=-350 kcals/mole ΔH=-295 kcals/mole ΔS= +65 kcals/mole -45=+20 – ΔS ΔS= -45 – +20 -252 kcals/mole ΔG= -222 kcals/mole - +30 kcals/mole -345 kcals/mole -310 = ΔH - -35 ΔH= -310 + -35 = -345 +55 kcals/mole

15. If you make a polysaccharide from monosaccharides would be the signs for the following be negative or positive? Δ G Δ H Δ S If you breakdown a polysaccharide into monosaccharides would be the signs for the following be negative or positive? Δ G Δ H Δ S + + - - - +

17. ΔG=-7.3 kcals/mole exergonic ATPase ATPsynthase endergonic ΔG= +7.3 kcals/mole

18. EQUILIBRIUMS • An equilibrium is reached when net changes cease but at any given time, some reactions are changing back and forth. 2. The concentrations of the reactants on each side do not have to be equal, only the rates of the forward and reverse reactions must be equal. 3. The relative proportions of AB and CD at equilibrium are determined by the free energy or the ∆G of the reaction. A + B⇌C + D 4. If ∆G for this reaction is 0 then there would be equal concentrations of A + B and C + D at equilibrium. 5. If the concentration of C + D is greater at equilibrium, then the potential energy of C + D must be less than the potential energy of A + B; then the ∆G for A + B  C + D is negative and this would be an exergonic reaction. The larger the ∆G value, the more exergonic the reaction is and the more the equilibrium will swing to the C + D side.

19. exergonic -ΔG A + B C + D +ΔG Less potential energy More potential energy endergonic Metabolic disequilibrium What if the cell wants A+B? It must add a coupled reaction. kinase D + ATP D P + ADP phosphorylated intermediate

20. endergonic +ΔG A + B C + D P -ΔG More potential energy Less potential energy exergonic

22. COUPLED REACTIONS An enzymatic pathway with all equilibriums exergonic in the direction of the desired products 40% ⇌ 60% 20% ⇌ 80% A + B⇌C + D⇌E + F⇌H +I 30% ⇌ 70% -ΔG -ΔG

23. COUPLED REACTIONS A pathway with an endergonic equilibrium in the direction of the desired products 40% ⇌ 60% 20% ⇌ 80% A + B ⇌ C + D ⇌ E + F ⇌ H + I 70% ⇌ 30% ∆G = +3kcals/mole -ΔG +ΔG ADD: a coupled reaction C + ATP ⇌ C P + ADPthe potential energy of C phosphorylatedhasbeen increased by intermediate7 kcals/mole kinase

24. COUPLED REACTIONS NOW: All the equilibriums are exergonic in the desired direction. 40% ⇌ 60% 20% ⇌ 80% A + B⇌CP + D⇌E + F+ P⇌H +I 35% ⇌ 65% ∆G = 4 kcals/mole

31. Induced Fit Hypothesis

32. THE INDUCED FIT HYPOTHESIS • The substrate is attracted to the partial  + and  charges on the enzyme’s active site and hydrogen bonds form between the enzyme’s active site and the substrate molecule but the fit is not perfect. 2. Formation of the initial H bonds causes the enzyme’s conformation to change slightly thus allowing it to form additional H bonds with the substrate. 3. Formation of these additional H bonds causes the enzyme to change its conformation again which then either pushes or pulls on the substrate through the hydrogen bonds.

33. THE INDUCED FIT HYPOTHESIS 4. This strain on the substrate results in the following: a. brings substrates into close proximity overcoming their mutual repulsion b. orients the substrate(s) so that they will fit together (or will come apart) c. puts stress on existing bonds in the substrate(s) resulting in those bonds being easier to break or new bonds easier to make 5. Once the substrate(s) are either put together or taken apart they change shape and charges so that they no longer fit or form H bonds with the active site. Thus the substrate is ejected from the active site.

34. Ca++ Fe++ Mg+ cofactors inorganic low molecular weight substances (usually ions) around which enzymes are wrapped; give the enzyme the proper shape and charges at the active site and are a permanent part of the enzyme;

35. Nicotinamide adenine dinucleotide NAD Nicotinamide adenine dinucleotide phosphate NADP coenzymeslarge nonprotein organic molecules which bind temporarily to enzymes and help them perform their function; usually help remove electrons and then transfer them to another molecule; made from vitamins

44. Cooperativity-when one substrate attaches to an active site it causes a conformational change the other subunits making them more attractive to their substrates

45. Enzymes are compartmentalized in a cell