The standard mass action kinetics model of gene expression arises from the differential equation $$df/dt = s - d f(t)$$, with s being the constant synthesis rate, d the constant degradation rate and $$f0=f(0)$$ (the abundance at time 0).

## Usage

f.old.equi(t, s, d)

f.old.nonequi(t, f0, s, d)

f.new(t, s, d)

## Arguments

t

time in h

s

synthesis date in U/h (arbitrary unit U)

d

f0

the abundance at time t=0

## Value

the RNA abundance at time t

## Functions

• f.old.equi(): abundance of old RNA assuming steady state (i.e. f0=s/d)

• f.old.nonequi(): abundance of old RNA without assuming steady state

• f.new(): abundance of new RNA (steady state does not matter)

## Examples

 d=log(2)/2
s=10

f.new(2,s,d)  # Half-life 2, so after 2h the abundance should be half the steady state
#> [1] 14.42695
f.old.equi(2,s,d)
#> [1] 14.42695
s/d
#> [1] 28.8539

t<-seq(0,10,length.out=100)
plot(t,f.new(t,s,d),type='l',col='blue',ylim=c(0,s/d))
lines(t,f.old.equi(t,s,d),col='red')
abline(h=s/d,lty=2)
abline(v=2,lty=2)

# so old and new RNA are equal at t=HL (if it is at steady state at t=0)

plot(t,f.new(t,s,d),type='l',col='blue')
lines(t,f.old.nonequi(t,f0=15,s,d),col='red')
abline(h=s/d,lty=2)
abline(v=2,lty=2)

# so old and new RNA are not equal at t=HL (if it is not at steady state at t=0)