# Change in synthesis rate of a molecule changes equilibrium concentration

I was reading the topic of 'The concentration of the molecule can be adjusted quickly only if the lifetime of a molecule is short' from Molecular Biology of the Cell by Alberts.

At the end of pg-837 (this part can also be found here), the author says -

"In fact, after a molecule's synthesis rate has been either increased or decreased abruptly, the time required for the molecule to shift halfway from its old to its new equilibrium concentration is equal to its normal half-life—that is, equal to the time that would be required for its concentration to fall by half if all synthesis were stopped."

What do they exactly mean by 'shift halfway'? I am not able to understand quite clearly what the above sentence means. I hope some of you may be able to help me. Thanks.

Note: The term equilibrium is different from steady-state w.r.t chemical reactions. Steady state is the right term for the above example. Equilibrium is used in the sense of forward and reverse reactions in a single reversible reaction. The above example considers two irreversible reactions — production and degradation.

... ... the time required for the molecule to shift halfway from its old to its new equilibrium concentration is equal to its normal half-life

This is not true if the degradation reaction is zeroth order or if these reactions are nonlinear. But for a linear production degradation set of reactions ($X$ is a biomolecule formed at a constant rate $\alpha$ and degraded at a first order rate $\beta$ ), the reaction rate can be represented by the following equation.

$$\frac{d[X]}{dt}\bigg|_1=\alpha _1 - \beta [X]$$

At steady state $\frac{d[X]}{dt}=0$, which means $[X_{ss_1}]=\Large\frac{\alpha _1}{\beta}$

Half life as defined, by setting $\alpha$=0 and solving the differential equation = ${\Large \frac{log(2)}{\beta}}$

When the formation rate is changed to a new value:

$$\frac{d[X]}{dt}\bigg|_2=\alpha _2 - \beta [X]$$

and $[X_{ss_2}]=\Large\frac{\alpha _2}{\beta}$

From the above equation following equation can be deduced (by integrating the differential equation):

$$-\frac{1}{\beta}.log(\alpha _2 - \beta [X]){\large|}_{X_{initial}} ^{X_{final}}= t_{\frac{1}{2}}$$

when you substitute $X_{initial}=X_{ss_1}=\frac{\alpha _1}{\beta}$ and $X_{final}=\Large\frac{\frac{\alpha _2}{\beta} - \frac{\alpha _1}{\beta}}{2}$ (halfway between new steady state and old steady state)

you would obtain: $$t_{\frac{1}{2}}= -\frac{1}{\beta}log\left(\frac{\alpha _2 -\beta\frac{\alpha _1}{\beta}}{\alpha _2 - \beta \frac{\frac{\alpha _2}{\beta} - \frac{\alpha _1}{\beta}}{2}}\right) = \frac{log(2)}{\beta}$$