Solution of some equations in Biochemistry J.P Bennett and J.H. Davenport H.M. Sauro School of Mathematical Sciences Institue of Grassland and Animal Production University of Bath and Poultry Division Claverton Down Roslin, Bath BA2 7AY and Midlothian England Scotland Abstract It is possible to write down the equations governing a one-stage enzyme-catalysed reaction (ac- cording to Michaelis-Menton kinetics) quite easily, and deduce information about the steady-state flow in such a system. The situation is somewhat more complicated if several such reactions form a linear chain. We have applied Gr¨ ober-basis techniques to solve such systems. Introduction. If we consider an enzyme-catalysed reaction in steady-state, such as X 0 v -→X 1 (with the underlying first-order mechanism X 0 + E k1   k2 X 0 E k3   k4 X 1 + E in which we have written K eq = k 1 k 3 k 2 k 4 (overall equilibrium constant) K m,f = k 2 + k 3 k 1 (forward Michaelis constant) K m,r = k 2 + k 3 k 4 (reverse Michaelis constant) V max =[E TOT ]k 3 (maximum flux) [E TOT ]=[E]+[X 0 E] (conservation of enzyme) as is standard) the rate of conversion of X 0 to X 1 (flux) can be written as v = V m Km ([X 0 ] - [X 1 ]K eq ) 1+ [X 0 ] K m,f + [X 1 ] K m,r . (1) When the reactions are more complex, e.g. X 0 v1 -→S 1 v2 -→X 2 , it becomes harder to analyse the situation, even under the steady-state assumption that [S 1 ] is constant. We would like to eliminate [S 1 ] from the two equations (analogous to (1)) which determine the fluxes in the two stages. This has been done [...], and the result is that [S 1 ] satisfies a quadratic equation. As far as the authors are aware, the case of three-stage linear chains has not been solved. 1