Week 5

Chloroplasts See Figure Outer membrane Inner Membrane –contains stroma - contain dark reaction enzymes Thylakoid membrane –stacked into “granum” –contain light harvesting pigments, electron transport chain thylakoid space

Endosymbiont Theory Theorizes about the origin of the chloroplast –Cyanobacteria? Evidence: –DNA –RNA –antibiotic sensitivity –16s RNA

Overall reaction 6CO H 2 O + energy---> C 6 H 12 O 6 + 6O 2 + 6H 2 O Energy source : sunlight - radiant light energy Photosynthesis –Light reactions –Dark reactions (syn. Calvin cycle, light independent reactions)

Light reactions Energy capturing reactions –convert light energy into chemical energy ATP and NADPH Occur on the thylakoid membrane Chlorophyll, carotenoids capture light energy –see Figures and 13-31

Light reactions Electrons moving down the electron transport chain establishes proton motive force across the thylakoid membrane –see Figures and Water is the ultimate source of electrons Products of light reaction are: ATP and NADPH

Dark reactions Carbon fixation by ribulose bisphosphate carboxylase - probably the most abundant enzyme in the world See Figure for pathway

Energetics

Laws of thermodynamics 1st law: –energy cannot be created or destroyed, but can be transferred or transformed from one form to another

2nd law of thermodynamics In the universe, or any isolated system, the degree of disorder can only increase. –Total entropy of the universe is always increasing –Free energy of a system is always decreasing

Entropy Measure of disorder in a system  S –measure of change in S with time  S system +  S surrounding > 0 –spontaneous reaction Difficult to quantify S

Free energy -  G –spontaneous reaction exergonic or exothermic reaction +  G –non spontaneous reaction endogonic or endothermic

Free energy Williard Gibbs Measure of energy to do work G = H - TS  G =  H - T  S –H = enthalpy –T = absolute temperature (K)

Free energy  G =  H - T  S  H =  E +  PV  PV is small for biochemical reactions  G =  E - T  S

Free energy  G is proportional to equilibrium constant of reaction See Panel 3-1  G =  G o + RT ln K –R = cal/mole-K o –T = absolute temperature (K o ) –  G o = standard free energy at 1M products and reactants. This is constant - never changes in value.

Calculating  G o At equilibrium  G = 0 therefore  G o = -RT lnK eq Example: –glucose 6 P Fructose 6 P –K eq = 0.5  G o = - (1.987) (298K o ) ln0.5  G o = cal/mole

Calculating  G Calculate  G for the following reaction: –dihydroxyacetone phosphate glyceraldehyde 3 P –Concentrations of dhap = 2 x M –concentration of G3P = 3 x M  G o = - (1.987) (298) ln  G o = +1.8kcal/mole

Calculation continued  G =  G o + RT ln K –  G o = +1.8kcal/mole –K = 3 x /2 x  G = +1.8kcal/mole + (1.987)(298) ln 3 x /2 x  G = kcal/mole Why is this a negative number when  G o is a positive number?