biochemistry - Finding steady state concentration given half life and production rate

Intracellular molecule B is normally synthesized at a constant rate of 1000 molecules/second. The lifetime B = 200 s.

How do I find the concentration of molecule B when it is at a steady state?

I Tried:

For molecule B. P is concentration of molecule B.

2^-(dt/200) * P + 1000 * dt = P

"2^-(dt/200)" is the half life formula. I multiply it by P to give me current concentration of molecule B. Then I add 1000 * dt to give me the amount of molecule B being produced at an exact moment. Rearranging the variables:

P = (1000 * dt) / (1-2^(-dt/200))

Now I limit dt --> 0 to give me P:

lim t->0 (1000 * dt) / (1-2^(-dt/200)) =

l'hopital's rule: lim t->0 1000/ (0.005 * ln2 * 2^(-dt/200))

Plugging in dt = 0, what I get is a steady state concentration of molecule B = 200000 / ln(2).

However, the correct steady state concentration for molecule B is just 200000.

Suggestions?

Answer

tl;dr besides making this a bit harder than necessary, your main problem is that you have confused the lifetime (average time to removal of a molecule), not the half-life (time until half of the molecules are removed, which is $\ln(2)$ times the lifetime ...) Wikipedia has some formulas ...

You can translate the problem into a differential equation (not as scary as it sounds):

$$\frac{dB}{dt} = \underbrace{1000\vphantom{\frac{1}{200}}}_{\textrm{production}} - \underbrace{\frac{1}{200} B}_{\textrm{removal}}$$

The only tricky part here is recognizing that if the lifetime is 200 s, a fraction 1/200 of the existing molecules will be removed per second.

Now we have to solve this at equilibrium: $dB/dt =0$. We get

$$\begin{split} 1000 - \frac{B}{200} & = 0 \\ 1000 & = \frac{B}{200} \\ B = 200000 \end{split}$$

Or you could memorize Little's Law: steady state = arrival (1000/s) $\times$ lifetime (200 s).