BIOSYNTHESIS OF MACROMOLECULES 621 
In the above, the efficiency abbreviations are: 
(2) Hy = Hixoi| (Li + %oi) ; A; = kuxi] (Kui + #1); 
Hy = hiyoi](Li’ + yo) By = Riniyi] (Ki + aire + Divs + Cai); 
M; = miyi’ Ti] (Di a OTs oh Revi’ of Sy Hr w Wo 
(Symbols like /;, L;, etc., are constants). 
It should be noted that we are treating the 7; equations in such a way as to imply 
(from equation 2) that the 7; and y,’ both enter as substrates in a simple bimo- 
lecular enzyme-catalyzed reaction!. Since we know nothing of the actual kinetics of 
such template reactions, this is speculative. We might well argue instead that the 
(empty or partly filled) template should enter linearly in the same way as the en- 
zyme, in which case we would have My = mTj_,yi' /(Di + yi’). The linear approxi- 
mation to the efficiencies MM; of equation 2, with respect to the partial templates 
T;_,, will probably be good in any case, since the total level of template may be 
very small. The precise form of M; is not too important for most of our discussion, 
so we may regard the symbol M; as non-committal between the two possibilities 
(and others) if we like. 
THE STEADY-STATE APPROXIMATION 
To expect to solve the complete system (1) rigorously is quixotic. One might solve 
it, for a given set of assumed values of the constants, with the help of a computer; 
but such a solution would not be very instructive. What we want is to know how 
the parameters affect the general properties of the solution, in particular for those 
parameters that represent measurable or controllable factors. We can get at this 
problem by dumping overboard most of the mathematical complexities of the com- 
plete system. It seems reasonable to assume that, once we wind up asystem like this 
and start it going, the intermediates of synthesis may rapidly reach steady values, 
where they are formed and removed at equal rates, while the end-products (e.g., the 
proteins) progressively change. 
This means, mathematically speaking, that (if we are willing to get along without 
the solution for the rapidly changing transient state while the intermediates approach 
their steady state) we can say that the time derivatives on the left sides of the 
equations of (1) are equal to zero, except for the time derivative of protein concen- 
tration. This means that the right sides are equal to zero. And this means that the 
differential equations become simply algebraic equations, which of course are a good 
deal less troublesome to solve. 
The problem is still simpler when we decide what we want to solve for, which is 
the value of d(Protein) /dt, the rate of protein synthesis. We note that this equals aTn. 
Looking at the last of the 7; equations, we see that aT, = M,E’. Tracing our way 
back through the system of 7; equations, we wind up with the result aT, = M,E’, 
since it appears that M, = My_, = ... = M, = M,. This brings us to the equation 
for y,’, from which M,E' = B,E,; this in turn leads to the y, equation from which 
BLE, = A,E,, + H,'P;'. The x, equation gives A,E,, = H,P,, hence, B,E, = 
H,P, + H,'P,'. Thus we have finally: 


(3) d(Protein)/di = HP, -- Ay Py. 
References p. 632 
