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from last time… Pressure = force/area Two major pressures in plants.

from last time… Pressure = force/area Two major pressures in plants. 1. The positive pressure (turgor) inside living cells and that’s required for cell and tissue growth. 2. The negative pressure (tension) that exists in the cells of the xylem of transpiring plants.

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from last time… Pressure = force/area Two major pressures in plants.

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  1. from last time… • Pressure = force/area • Two major pressures in plants. • 1. The positive pressure (turgor) inside living cells and that’s required for cell and tissue growth. • 2. The negative pressure (tension) that exists in the cells of the xylem of transpiring plants. In general we’ll use units of pressure to express the energy status of water, the “water potential” . How is pressure like energy/volume?

  2. force/area = energy/volume? N m2 multiply by unity as m/m N x m m2 m N . m = force x distance = energy m3 volume volume So, we can use units of pressure to express the energy status of water. We’ll see that water tends to move from areas of higher to lower energy/vol, or pressure.

  3. energy must equal volume x pressure a bit more review…. The gas constant, R Remember PV = nRT? R, the “gas constant” makes the relationship among P, V, n, and T work. R shows up in lots of energy equations. Values and units for R 8.314 J mol-1 K-1 8.314 m3 Pa mol-1 K-1 We’ll use R a lot!

  4. How to put some numbers to all the energy • expended in doing the work of life. • Chemical reactions - synthesizing compounds, • degrading others • Solute transport - maintaining concentration • differences across membranes • Maintaining electrical potentials and moving • ions across charged membranes. • How can we understand when these processes require energy and how much?

  5. Bioenergetics and Free Energy Free Energy, G, is the energy available to do work. DG is the change or difference (D) in G during a process or reaction. DG equations help us understand whether a reaction: 1) yields energy and can happen spontaneously (DG < 0), 2) requires energy input to occur (DG > 0), 3) or is at equilibrium (DG = 0). We will use DG equations for understanding bioenergetics of chemical reactions and the transport of charged and uncharged solutes.

  6. the free energy equations give us values of energy per mole, J mol-1

  7. Gof a chemical reaction General equations: • G= G0 + 2.3 RT log (K) • or • G= G0 + RT ln (K) G0 is the standard free energy change, defined for standardized conditions. It allows comparisons of G of different reactions. K is the equilibrium constant K = ([product]*[product]) ([reactant]*[reactant]) So, G= G0 + 2.3 RT log (K) can be written as: G= G0 + 2.3 RT log ([product]*[product]) ([reactant]*[reactant])

  8. Example using a very important reaction • G for ATP hydrolysis: ATP  ADP + Pi • G0 = - 33kJ mol-1 • In typical cellular conditions • G = –50 kJ mol-1 to –65 kJ mol-1, • The reaction releases energy and can happen spontaneously. • Compare with ATP synthesis: • ADP + PI ATP • G > 0, requires energy, is not spontaneous.

  9. 2. Solute transport Transport is from C1 to C2 [C1] -------------> [C2] The standard equation: G = 2.3 RT log [C2]/[C1] There are 3 possibilities: 1. [C1] < [C2], so log [C2]/[C1] > 0, so G > 0 2. [C1] > [C2], so log [C2]/[C1] < 0, so G < 0 3. [C1] = [C2]solog [C2]/[C1] = 0, so G = 0 Which happens spontaneously?

  10. 2. Solute transport G = 2.3 RT log [C2]/[C1] transport is from C1 to C2 [C1] > [C2]means that log [C2]/[C1] < 0 so G < 0 This can happen spontaneously, without energy input C1 C2 ------------->

  11. 2. Solute transport G = 2.3 RT log [C2]/[C1] transport is from C1 to C2 [C2]= [C1]meanslog[C2]/[C1]= 0 So, G = 0, equilibrium C1 C2 <---------->

  12. Numerical example • [C2] = 100mM, [C1] = 10mM • 37 0C • How much energy to transport a mole of C? C1 C2 ------------->

  13. G = 2.3 RT log[C2]/[C1] Dimensional analysis - do the units make sense? G R T log[C2]/[C1] Units: J mol-1 = J mol-1 K-1 Kmol l-1/mol l-1 J mol-1 = J mol-1

  14. Now fill in the numbers G = 2.3 RT log[C2]/[C1] G = (2.3)(8.314 J mol-1 K-1)(310 0K) log(100/10) = 2.3 x 8.314 x 310 x 1 J mol-1 = 5928 J mol-1 or 5.928 kJ mol-1 Energy is required to move a solute “up” a concentration gradient C1 C2 ------------->

  15. 3. DG for ion transport: ions are charged solutes Fig. 6.4

  16. Biological membranes are electrically polarized, like a battery. K+ NO3- Ca+2 SO4-2 Which ion requires the most energy to move across the membrane, assuming the same concentration gradient for all four?

  17. 3. G for ion transport G = zF Em z is charge on the ion: K+ = +1, NO3- = -1, Ca+2 = +2, SO4-2 = -2 other molecules have not net charge F is Faraday’s constant = 9.65 x 104 J vol-1 mol-1 Em is membrane potential, volts

  18. Example: uptake of NO3-against -0.15 volt potential G = zF Em G = (-1)x9.65x104 J Volt-1 mol-1 x (-0.15Volt) = 1.45 x 104 J mol-1 = 14.5 kJ mol-1 NO3-

  19. 4. Movement along electrical and concentration gradients • G = zF Em + 2.3 RT log(C2/C1) • note rearrangment as Em = -2.3 RT/zF log(C2/C1) • R = 8.314 J mol-1 K-1 • z is charge of solute • F is Faraday’s constant = 9.65 x 104 J vol-1 mol-1 • Em is membrane potential, volts

  20. Enzyme kinetics What are enzymes? What kinds of molecules are they made of? What do they do to reaction rates? How do they work? What’s the world’s most abundant enzyme? What conditions affect the rate of enzyme-catalyzed reactions?

  21. Vmax reaction rate, V (moles of product per second) Substrate concentration, S (moles/liter)

  22. Vmax x S Km + S V = Vmax V 1/2 Vmax Michaelis-Menten equation Km S (substrate concentration)

  23. Enzyme specificity is not perfect, other molecules can compete for the active site. “competitive inhibition” Enzyme structure can be modified by other molecules, reducing enzyme activity. “non competitive inhibition”

  24. Substrate concentration Competitive inhibitor increases Km and does not affect Vmax, but a higher [S] is required to reach Vmax. Mechanisms Competitive inhibitor binds at same active site as substrate, making less enzyme available to catalyze E+S reaction. Competitive inhibitor binds at another site on enzyme, causing a conformational change in active site that reduces affinity for the primary substrate. “allosteric inhibitor”. No inhibitor With competitive inhibitor

  25. Conditions affecting enzyme activity. 1. pH Enzymes have an optimum pH at which activity is maximum, with sharp declines in activity at lower and higher pH. pH affects enzyme activity by altering ionization state of active site or by affecting the 3-D conformation of the active site. 2. Temperature Enzyme activity has an optimum temperature, with sharp declines in activity at lower and higher pH. Reaction rates increase with temperature because enzymes and reactants are moving faster and have higher probability of encountering one another. Enzyme activity decreases at temperatures high enough to cause “denaturation”, the unfolding of protein structure and loss of proper conformation for catalysis.

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