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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-------> [protein] --------kd--------->
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.
Now consider what would
happen if we change the rates of synthesis and/or degradation
in response to some external signal. The new rates of synthesis
and degradation are given by ks' and kd' respectively.
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
Thus, the rate of
approach to the new steady-state is determined solely by the degradation
rate of the protein. To respond quickly, one needs a rapid degradation
rate.
Refering to the figure above and assuming only the rate of synthesis changes:
| Case | ks | kd | ks' | kd' |
| a | 4µM/min | 4/min | 8µM/min | 4/min |
| b | 2µM/min | 2/min | 4µM/min | 2/min |
| c | 1µM/min | 1/min | 2µM/min | 1/min |
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)