1. The use of Energy to control cell morphogenesis
Lecture II
Marc W. Kirschner
March 22, 2005

2. The phenotype depends on the environment
The phenotype depends on the environment. But in the normal development and function of an organism, the rules are established by the organism, since so much of the environment is created by the environment.
But though the environment might be uniform the responses are only general and not stereotyped.
Endothelial cells migrating toward the margin of a wound. The microtubules of the cytoskeleton are stained. The arrow is perpendicular to the wound. From Gotlieb, et al. ( 1986) J. of Cell Biology. Note the variety of structures.

3. The gene number problem rears its head again
There are 22,500 human genes and 1014 neurons, even more synapses
The definition of systems biology could be restated in these terms: To discover the rules by which a small number of genes can generate the much larger complexity of the organism

4. intermediate filaments
There are three major filament types
in eukaryotic cells:
microtubules
intermediate filaments
actin
Intermediate filaments are different from the other in that they are not polar, they assemble without nucleotide hydrolysis, and they have highly diversified in the recent vertebrate lineage. They also differ in tensile strength, which is related to their structure. Microtubules and actin show unusual properties, involving nucleotide hydrolysis.

5. Kinetically linear polymers grow with a two phase growth curve, a nucleation phase and a linear growth phase. The nucleation phase can be complex.
For sickle cell hemoglobin the delay can be long and the polymerization abrupt. The sickling occurs when the HbS is deoxygenated in the capillary. The concentration dependence of the lag phase goes as the concentration to the 80th power!

6. K = trimer/dimer•monomer K = c3/c2c1 c3 = Kc13
Simple aggregation theory, with no cooperativity in polymer formation defines the critical concentration at infinite concentration.
K = dimer/monomer = c2/c12
c2 = Kc12 L
L +L L L L + L LL L
K = trimer/dimer•monomer
K = c3/c2c1
c3 = Kc13
In general: ci = K-1(Kc1)i
The total concentration in monomer units is cT
cT = Sici = c1/(1-Kc1)2 {f(x)= Sixi,=x/(1-x)2}
As cT goes to infinity c1 approaches 1/K

7. Simple linear aggregation of a non-helical polymer
C1 =1/K c1
10/K
20/K cT
So if the polymer can form and fall apart a any point the free monomer concentration grows slowly with the total concentration until it reaches asymptotically c1.

8. And g is a factor <1 which accounts for the additional energy
In a helical polymer there is deformation of the subunits to fit into the polymer. Once a nucleus is formed, however, additional subunits make bonds with two neighbors and this is more favorable. Thus there is a factor that makes the initial nucleus unfavorable and a different factor that makes more favorable subunit addition.
c3h = gc3
And g is a factor <1 which accounts for the additional energy
By the time the fourth subunit adds helically, the K changes to account for the additional binding surface.

9. The concentration of a four subunit polymer is:
c4h = Khc3hc1 = gK-1K3Khc14
Where Kh is the binding constant for the binding to the end of the helical polymer. Kh >>K
The favoring of binding to the end is given by s
= g(K/Kh) It is very small, appx for 20 kjoules difference due to g and 20 kjoules to Kh
This leads to a simple equation for the total concentration:
cT = c1 + sc1/(1-Khc1)2
If we look at this equation, at cT< c1; the second term is negligible(s appx 10-8), so cT is approximately equal to c1 . When cT approaches c1, as c1 approaches 1/Kh, the second term cannot be ignored. Almost all the subunits go into polymer. The value of the concentration where this happens (1/Kh) is called the critical concentration. If this were crystallization, it would be the solubility.

10. For a helical polymer c1 ch cc cT

11. a a¢ Ja = ac - a¢ + end J Jb = bc - b¢ ce - end -b¢ b¢ b -a¢
From now on we will consider the addition of subunits from the ends only and ignore nucleation or the first steps. Actin and tubulin, as well as HbS, and flagellin behave as helical polymers. We will now consider the kinetics of assembly that holds the clue to the problem we raised initially, what determines their placement in the cell. a a¢
L(solution)
Ja = ac - a¢
+ end J
LLLLLLLLLLL
Jb = bc - b¢ ce
- end
-b¢ b¢ b
-a¢
L(solution)

12. By the principle of detailed balance at equilibrium the two ends of the polymer must have the same critical concentration, which is another way of saying that despite the differences in rate the equilibrium constant must be the same for the two ends. Thus
At equilibrium: Ja= Jb = 0 = ac - a¢ = bc - b¢
Therefore cc = a¢/a = b¢/b = Kh
There is no concentration where the subunits grow from one end and shrink from the other. If the ends are occluded they will still have the same equilibrium constant. There is no growth below cc.

13. In microtubules: LT a2
a-2
a-1 a1 LD
LDLDLD
Similar processes
At the minus end

14. In microtubules: LT Ja= a1c - a2 Jb= b1c - b2 a2 a-2 a-1 a1 GTP Pi GDP
At the plus (a) end the main process is the addition of the tubulin monomer in the GTP form and the release of the subunit in the GDP form. Exchange of GTP for GDP occurs relatively rapidly in solution. Therefore the rate equation on the plus end is:
Ja= a1c - a2
Jb= b1c - b2
As this is steady state and not equilibrium, it is not true that both ends have the same critical concentration. a2
a-2
a-1 a1
GTP Pi
GDP LD
LDLDLD
Similar processes
At the minus end

15. J
-b¢
-a¢ ce
Ja = ac - a¢
Jb = bc - b¢
Note the critical concentrations are not the same. J
cca
ccb

16. J
-b¢
-a¢ ce
Ja = ac - a¢
Jb = bc - b¢
Note the critical concentrations are not the same. At a new steady state concentration cc, there is a net flow of subunits on the a end and a loss of subunits from the b end. This is called treadmilling. J cc
cca
ccb

17. This is not the whole story
This is not the whole story. The coupling of GTP hydrolysis makes it possible for the off rates of GDP subunits to be very different from that extrapolated back from the GTP curve. Considering for simplicity the reactions only at the a end, we have the separate rates for GDP reactions and GTP reactions:
Scale change
a1c- a-1
a1c - a2 J
a-2c- a2

18. a1c - a2 J ce(1) ce(2) Adding the reactions at the b end, we have
Scale change
a1c- a-1
a1c - a2 J
a-2c- a2
ce(1)
ce(2)

19. From Mitchison and Kirschner, 1984

20. POPULATION BEHAVIOR OF MICROTUBULES