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?

**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).