Why Do it So Fast?
How does the cell know which proteins to degrade, and when?
Okay, so we have to degrade proteins in order to lower the steady-state concentrations, but why should that be a rapid process? It seems wasteful to try to maintain the concentration of a protein while it is simultaneously being degraded.
Consider the following scheme:
ks is the rate of synthesis of the protein and is experimentally a zero-order rate, while kd is the rate of degradation of the protein and is experimentally a first-order rate. [E] is the concentration of E at any time and [Eo] is the steady-state concentration of protein.
Then: dE/dt = ks - kd [E] At steady state: dE/dt = 0 Therefore: ks = kd[Eo] and [Eo] = ks/kd
Thus, it is obvious that the steady-state concentration depends on how fast the protein is synthesized (ie., mRNA levels, the transcriptional activity, etc.) and the rate at which it is degraded.
A new steady state concentration is reached with E' = ks'/kd' The rate of approach to the new steady state can be determined by:
writing the eqn for E as f(t) dE/dt = ks' - kd'[E] substituting from steady state ks'=kd'[Eo'] integrating and evaluating [Eo'-E]/[Eo'-Eo]=e^-kd't
Refering to the figure above and assuming only the rate of synthesis changes:
The role of mRNA and protein stability in gene expression. Hargrove JL; Schmidt FH. FASEB J 3: 2360-70 (1989).
Microcomputer-assisted kinetic modeling of mammalian gene expression. Hargrove, JL. FASEB J 7,1163-70 (1993)