The Systems Biology of Spatial Organization

1. The Systems Biology of Spatial Organization Lecture I How is energy used: From equilibrium to steady states to transient state

2. General Rationale of Lecture • Systems Biology is conerned a lot about information use and information transmission • These involve non-equilibrium systems kept dynamic by unusual use of energy • We consider here how ATP energy is used in the cell and how it is physically linked to morphogenesis • In this lecture, as an introduction, we deal with more familiar uses of ATP to drive transmembrane transport and to power muscle contraction

3. In biochemistry we usually refer to how chemical energy is converted to the synthesis of biomaterials C6H12O6 hv 3O2 6CO2

4. Energy can be used for several purposes • Mechanical work-muscle contraction • Electrochemical work-polarization of membranes; electric fish • Spatial organization-cell shape and polarity • Processing information and relaying information-hormone action • Detecting, controlling and correcting errors • Chemical work-making cell walls, bones, etc

5. Energy can be used for several purposes • Mechanical work-muscle contraction • Electrochemical work-polarization of membranes; electric fish • Spatial organization-cell shape and polarity • Processing information and relaying information-hormone action • Detecting, controlling and correcting errors • Chemical work-making cell walls, bones, etc

6. Coupling ATP energy to achieve unfavorable synthetic steps Even to generate energy you need to overcome unfavorable steps.

7. Fructose-6-phosphate+ Pi Fructose 1,6-diphosphate DG0 = 16.8kjoules/mole DG0 = -RTln [B]/[A] So [B]eq/[A]eq = e- DG0/RT = e-16,800/8.3(300) = 1.2 x10-4

8. But coupling the reaction to ATP hydrolysis makes the overall process favorable Fructose-6-phosphate+ Pi 1 Fructose 1,6-diphosphate K1 = 10-4 K2 = 5x106 2 ATP ADP + Pi 3 Fructose-6-phosphate+ Pi + ATP Fructose-6-phosphate+ Pi + ADP + Pi K3= K1K2 = 500 Here the coupling is direct and covalent.

9. Passive Transport

10. But to drive molecules against a concentration gradient requires the coupling of transport to energetically favorable reactions, such as ATP hydrolysis

11. Coupled transport: how does one ion drive another molecule? What is the nature of the physical linkage?

12. Coupled Transport Na+/Ca++ 3Na+/ 1Ca++ This can only drive Ca++ to uM level; to get to submicromolar requires the Ca++ ATPase. But the Na+/Ca++ exchange is fast--2000 Ca++ per second, compared to 30 per second for ATPase. These are slow compared to a channel which lets about 107s-1

13. The process by which this works does not involve covalent chemistry but instead involves state selection, a very important principle in biology. We have already seen state selection in allosteric transitions. We shall see that state selection obviates the need for detailed chemistry and is very evolvable.In state selection, there is no direct linkage between molecules and effectors.

14. We will consider in detail the thermodynamics of one of the best understood transporter, lac permease, which uses co-transport of H+ and lactose in E. coli. H+ Lactose H+ Lactose

15. The structure and structural transitions of lac permease Outside (periplasm) Inside (cytoplasm) 12 helices De Felice Trends in Neurosci. 2004

16. S H+ H+ S

17. To understand the energetics and ultimately the kinetics of this process, consider an allosteric model, where E and E* are in a pre-equilibrium: E is the facing out conformation; E* is the facing in conformation E* E

18. H + E E•H Assume that S binds to E, only after H has bound. S + E•H S•E•H Also we have: H + E* E*•H And S + E* S•E*•H We also assume that the binding of S induces a conformational change or stabilizes a conformation so that: S + E•H S•E*•H Since S and H bind less strongly to E* than they do to E, both are transported down a concentration gradient. Finally, net transport is assured because CH+out/CH+IN >>CS out/CS in

19. E E* H H E•H E•H S S S•E•H S•E*•H

20. E E* H H Now if EH is converted to E*H E•H E•H S S S•E•H S•E*•H

21. E E* H H Now if EH is converted to E*H E•H E*•H S S S•E•H S•E*•H

22. E E* H H Now if EH is converted to E*H There will be a futile cycle of H+ from the outside to the inside of the membrane. This explains the often non-stoichiometric relationship or the unusual stoichiometries, such as 2 H+ for 1 substrate. E•H E•H S S S•E•H S•E*•H

23. We can use these cycle diagrams to understand how antiporters, synporters, and pumps work. Consider the Na+/K+ ATPase

24. The invention of the Na+/K+ ATPase was essential to maintaining the internal milieu of multicellular organisms NaKATPase Basal-lateral domain Modified from

25. How can an enzyme cause the vectorial flow of sodium ions out of the cell and potassium ions into the cell? • This was initially a mystery when it was demonstrated in the 19th century that frog skin could generate an electrical potential across two saline solutions and later that there was vectorial transport of Na+ across the skin. • There were a number of chemical mysteries here, such as how vectorial process could be generate by a scalar process. Also once again the non-integer stoichiometries were a problem. Furthermore. the chemical linkage of ATP to Na+ transport was hard to explain. • How can we account for this thermodynamically and can anything be said about the rates?

26. 1 E* (one conformation) 2 E (another conformation) 3 E*•2K 4 E•2K 5 3Na•E*•2K 6 3Na•E•2K 7 ATP•3Na•E*•2K 8 ATP•3Na•E•2K 1 2 K+ 4 3 Na+ 6 5 A complete cycle is 1 7 8 2 1 In reality the number of potential cycles is much larger. Also note that reactions 3 4 and 5 6 cause slippage and spoil the 1:2:3 stoichiometry ATP 7 8 From Terrell Hill, Free Energy Transduction and Biochemical Cycle Kinetics; Dover 2005

27. Subcycles 1 2 1 2 K+ K+ 4 3 4 3 Na+ 6 5 1 E* 2 E 3 E*•2K 4 E•2K 5 3Na•E*•2K 6 3Na•E•2K 7 ATP•3Na•E*•2K 8 ATP•3Na•E•2K ATP 7 8 From Terrell Hill, Free Energy Transduction and Biochemical Cycle Kinetics; Dover 2005

28. Subcycles 1 2 2K+ in 1 2 2K+out K+ K+ 4 3 4 3 But thermodynamically CK in> CK out And the process is unfavorable and will not occur. Na+ 6 5 1 E* 2 E 3 E*•2K 4 E•2K 5 3Na•E*•2K 6 3Na•E•2K 7 ATP•3Na•E*•2K 8 ATP•3Na•E•2K ATP 7 8 From Terrell Hill, Free Energy Transduction and Biochemical Cycle Kinetics; Dover 2005

29. Subcycles 1 2 K+ 3Na+ in 4 3Na+ out 3 4 3 Na+ Na+ 6 5 6 5 1 E* 2 E 3 E*•2K 4 E•2K 5 3Na•E*•2K 6 3Na•E•2K 7 ATP•3Na•E*•2K 8 ATP•3Na•E•2K Also unfavored as against a concentration gradient ATP 7 8 From Terrell Hill, Free Energy Transduction and Biochemical Cycle Kinetics; Dover 2005

30. Subcycles 1 2 K+ 4 3 Spontaneous! Na+ ATP ADP + Pi 6 5 6 5 1 E* 2 E 3 E*•2K 4 E•2K 5 3Na•E*•2K 6 3Na•E•2K 7 ATP•3Na•E*•2K 8 ATP•3Na•E•2K ATP ATP 7 8 7 8 From Terrell Hill, Free Energy Transduction and Biochemical Cycle Kinetics; Dover 2005

31. We can write the thermodynamic forces Xi associated with each reaction, where the thermodynamic force is equal to the difference in the chemical potential. The thermodynamic force is positive for a spontaneous process. For example, considering the K+ ion: XK = mK(outside) - mK(inside) = kT ln (CK out/CK in) For this reaction (CK out/CK in) < 1 and X has a negative value, reflecting the fact that K+ would spontaneously move in the opposite direction (from inside out).

32. The convention we used for sodium was that XNa = mNa in- mNa out XNa is also negative, since it is unfavorable These forces were defined for single ions so that that for the overall cycle the thermodynamic force, Xtotalis: Xtotal = 2XK +3XNa+XATP Even thought the first two terms are negative the positive force of ATP generates a large positive force.

33. In general the thermodynamics tells us nothing about the rates but that is not completely true. Since these are cycles, we can look at the constraints at equilibrium, where the individual rate constants have some relationship to each other. In a cyclic reaction A k3 k1 k2 C B The individual rates are not restricted

34. But at equilibrium there are limits on the rates: A k3 k1 k2 C B Not only must the overall flux go to zero, so that the concentrations of A, B and C are fixed but

35. But at equilibrium there are limits on the rates: A k3/k-3 k1/k-1 k2/k-2 C B Not only must the overall flux go to zero, so that the concentrations of A, B and C are fixed but the microscopic reactions must also be at equilibrium: [A]k1=[B]k-1 [B]k2= [C]k-2 [C]k3 =[A]k-3 This is called the principle of detailed balance. This puts restrictions on the rate constants because at equilibrium, the forward and reverse fluxes around any small cycle must also be balanced. For example in the K+ cycle: k13[K+]k34k42k21 = k31k43k24k12 At a non-equilibrium steady state: [K+ in]/[K+ out] = eX/kT

36. The point of this last exercise is to show that simple graphs, plus thermodynamics, plus the principle of detailed balance can begin get at a kinetic understanding of chemical cycles and lead to a clearer understanding of energy transduction. • We can also consider the efficiency of transport processes. The thermodynamic work term we described Xi is simply the change of free energy in one cycle (using kT or one mole of cycles using RT). Therefore we can describe the rate of free energy dissipation as the product of the flux and Xi for each process. If we denote that rate of dissipation byF • = JX. Also, F >0 • = JKXK+ JNaXNa +JATPxATP Now to measure the efficiency of the process one needs to sum over all the possible reactions. If the slippage reactions are zero. The expression is simple and the efficiency = JK((-XK)+ JNa((-Xna)/ JATPxATP

37. What have we learned from this thermodynamics exercise • Chemical energy can be linked directly (covalently) to generate chemical work, but this must be channeled by enzymes that determine the pathway of a complex reaction scheme • Chemical energy can do non-chemical work by state selection. In this case there is no direct connection between the source of energy (concentration gradient or ATP) and the substrate. • In order for the process to occur there must be at least one cycle in which the energy donor and the substrate are present. • The rate is limited by reactions arising from detailed balance.

38. Can we apply these concepts to even more complex systems, like muscle contraction?

39. The sliding filament model of muscle contraction

40. Myosin*ATP ATP 1 ADP + 45o 2 3 90o Actin•Myosin•ADP Actin•Myosin•ADP•Pi Pi Can one understand the mechanical force of muscle, the speed of contraction, the efficiency of contraction in terms of this proven model? This is a simplified model because it emphasizes the attached states and compresses the cycle but it is a simplification that emphasizes the work states.

43. #3 Dx F All of the work is exerted when the myosin is attached. We will initially assume only a single myosin head. dG= Fdx (T,P constant) Fi(x) = (¶G/¶x)T,P States 2 and 3 are the only attached states but every rate constant in our formulation involves a transition to an attached state. Applying detailed balance around this cyclic process provides some limit. We can write the ratio of the rate constants for each process in terms of the free energies.

44. (x)

45. By detailed balance, there is a restriction that going around the cycle must produce a free energy change independent of x, because the beginning and endpoint are independent of x. Hence the product: K12(x)K23(x)K31(x) = constant independent of X, XT

46. Consider one pass through the myosin cycle from position 1 to 1. G1 G/kT XT o x (A) G1-XT

47. G2 G1 G XT o x (A) G1-XT

48. G3 + mP G2 G1 G XT o x (A) G1-XT

49. Let us follow a path. In general the paths will move to lower free energy but uphill transitions can occur G3 + mP G2 G1 G XT o x (A) G1-XT

50. Let us follow a path. In general the paths will move to lower free energy but uphill transitions can occur G3 + mP G2 G1 G XT o x (A) G1-XT