I want to know the behavior of the solution without solving the ordinary differential equation for enzyme-substrate reactions.

Question

I want to know the behavior of the solution without solving the ordinary differential equation for the following enzyme-substrate reactions.

This chemical reaction represents the process by which substrate S is transformed into product P by the action of enzyme E. In detail, the reaction takes place as follows;

• First, the enzyme E and S become the enzyme-substrate complex ES
• Next, the enzyme-substrate complex ES splits into the enzyme E and the product P.　We also assume that the initial concentration of E is >0.

Here, $k_1,k_2,k_3>0$, and [S] is the concentration of S, and the others similarly represent the concentrations of substances sandwiched between brackets. ES represents a complex of E and S.

My question

Perhaps from the first equation, we expect to be able to say "that the concentration of substrate S become zero after a sufficient time interval , but can we prove this without solving the differential equation?

Answer 1

I'll use $T$ instead of $[ES]$ and drop all brackets. I'll also rescale $t$ so that $k_1=1$. The steady state equations read

$$\tag{1} 0=-SE+k_2T \\ 0=-SE +k_2 T +k_3T \\ 0=SE-k_2T-k_3T\\ 0=T $$

from which we learn that the product $SE\to 0$ as $t\to \infty$. For all $t$, a conserved quantity is

$$\tag{2} E(t)+T(t)=H $$

With an initial condition $E(0)\gt0$ it follows $H>0$ as all concentrations are non-negative. Evaluating (2) for large time demonstrates that

$$ E\to H >0 \qquad , \qquad t\to\infty $$

So the only way to satisfy the product $SE\to0$ is $S\to0$ as $t\to\infty$.

Answer 2

Although it is also a chemical reaction problem, this is a problem of qualitative consideration of solutions to differential equations, and is more of a mathematical problem than a chemistry problem. At one point, I obtained the following answers on my own. I think it is probably correct, but if it is wrong, I would appreciate comments.

From the equation ③ and ④, the material balance for substrate E can be obtained; adding ③ and ④, $d([E]+[ES])/dt = 0 $ regardless of time, and therefore, $-{k}_{1}[E]+{k}_{2}[ES] = constant$ is guaranteed on this system of differential equations alone, without making any a priori assumptions.

Taking into account the common knowledge of chemistry that at the t=∞, all velocities become zero (in other words, chemical equilibrium is reached),we get

• In equation ①, $d[S]/dt =0$ at t=∞, so $-{k}_{1}[S][E]+ {k}_{2}[ES]=0$ at t=∞.
• In equation ④, $d[P]/dt =0$ t=∞, so [ES]=0 at t=∞.
• Therefore, from equations ① and ④, we get that $[S][E]=0$ at t=∞.
• Furthermore, from equation ④, the behavior at time ∞, [ES]=0, and the material balance [E]+[ES]=constant, and initial consentration of E>0 we know that [E]>0 at t=∞. -Hence, in conjunction with the fact that [S][E]=0 at t=∞, we have [S]=0.