Harvard Systems Biology 200 Handout

Scaffold assembly equilibria. J´erˆome Feret, Walter Fontana November 10, 2007

The simple“star” case: a central hub B with n binding sites for one type of ligand A 1. B is a scaffold with n sites, each of which can bind another protein - or ligand for short - of just one type A. Bi is the number of cases in which B has i molecules of A bound to it. 2. The sites of B can be bound in any order. Question: what is the equilibrium concentration of fully occupied B’s, Bn , as a function of B, the total concentration of B in the mixture? Let na be the number of binding sites available for binding an A molecule. nb is the number of available binding sites that can bind a B. nab is the number of existing bonds between A’s and B’s in the system. This is a local view in terms of sites, rather than agents bearing those sites. With regard to agents, let A and B denote the total number of agents of type A and B, respectively. We have: na nb = Kd nab

(1)

nB = na + nab

(2)

A = nb + nab

(3)

with Kd = k−1 /k1 . The first equation is the equilibrium condition, the other two equations express site conservation. This allows us to express nab as a function of Kd , A, B, n. In addition, from ligand binding equilibria calculations (see pertinent handout), we have (using current notation): n  Bn nb = . (4) B K d + nb nb – the number of free sites (in equilibrium) available for binding to B – is the same as S in the lecture slides or the ligand binding handout of lecture 9. Combining equations (3) and (4), we obtain:  n A − nab . Bn = B Kd + A − nab Now we need to get nab . Substituting nb from equation (3) into equation (2), we express na in

1

(5)

500

400

Bn

300

200

100

0

0

1000

2000

3000

4000

B

Figure 1: Star equilibrium. n = 2, A = 1000, Kd = 1. Red: analytical [equations (5) with (8)], black: simulation

terms of nab and constants: na = Kd

nab . A − nab

(6)

Substituting (6) into (3) yields a quadratic equation for nab : 0 = n2ab − nab (Kd + A + nB) + nAB,

(7)

with solution nab =

 p 1 (Kd + A + nB) − (Kd + A + nB)2 − 4nAB . 2

(8)

The other solution is physically meaningless, because it violates site conservation. Using (8) in (5) yields the solution, whose graph is the solid line in Figure 1. The wiggly line is made of equilibrium ”measurements” of Bn from stochastic simulations (using the KappaFactory).

2