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The messenger RNA has steady state $X_{R,\ast }={\beta }_R/{\alpha }_R$. Since biological translation produces proteins at rate $r{X}_R$, setting the protein degradation rate equal to this production rate givesThe translational burst size is the expected number of proteins made during one messenger RNA lifetime, whose expected value is $1/{\alpha }_R$. It is dimensionless, rather than a production rate. With independent Poisson process translation and an exponential distribution for the lifetime, the resulting count has a geometric distribution on nonnegative integers with expected value $b$ and variance $b(1+b)$.

With zero initial messenger RNA and protein concentration, solving the two linear ordinary differential equations in sequence givesThe last expression applies after the short messenger RNA transient when ${\alpha }_R\gg {\alpha }_P$. All three mean curves have this shape. At a fixed steady state, a larger translational burst size requires fewer transcription events: ${\beta }_R={\alpha }_P{X}_{P,\ast }/b$. Thus $b=400$ gives infrequent large upward jumps followed by protein degradation, whereas $b=3$ gives many smaller jumps and a smoother trace.

Noise depends on molecule counts. If the fixed cell volume is $V$, write $\bar{N}=V{X}_{P,\ast }$. In the instantaneous-burst limit, independent Poisson process arrivals of geometrically distributed bursts, with count arrival rate $\lambda =V{\beta }_R$, and unit protein degradation giveIndeed, the moment equations are $d\mathbb{E}N/dt=\lambda b-{\alpha }_P\mathbb{E}N$ and $d\mathrm{var}N/dt=\lambda b(1+2b)+{\alpha }_P\mathbb{E}N-2{\alpha }_P\mathrm{var}N$. With zero initial counts the variance becomes $\bar{N}[(1+b)-e^{-{\alpha }_{P}t}-b{e}^{-2{\alpha }_{P}t}]$. These equations explain both the approach to the steady state and the scatter about it.

For illustration only, $V=1\mu {\mathrm{m}}^3$ gives $\bar{N}=10^4$ at the higher concentration: the two values of the coefficient of variation are $0.020$ and $0.200$. At the lower concentration, $\bar{N}=100$ and $b=3$ gives $0.200$. Decreasing the concentration by a factor of 100 therefore increases relative fluctuations tenfold, despite keeping small bursts. Absolute fluctuation sizes cannot be inferred without a cell volume. The original sketches below stipulate that volume; they are model realizations, not source data.^[ref]

The protein production rate is proportional to the current messenger RNA concentration, not an independently constant ${\beta }_P$: ${\beta }_P(t)=r{X}_R(t)$. Setting the two coupled rate equations to zero givesWith first-order removal, a messenger RNA molecule has mean lifetime $1/{\alpha }_R$. At biological translation rate $r$ per molecule, it produces an expected $r/{\alpha }_R=b$ proteins before its removal. Thus the translational burst size is a dimensionless mean number of offspring proteins per messenger RNA, not a concentration and not a fixed number produced in every event. When production and removal are independent Poisson events, mixing a Poisson number of biological translation events over an exponentially distributed lifetime makes the burst count geometrically distributed on the nonnegative integers. Its mean is $b$ and its variance is $b(1+b)$.

The protein fluctuation equation gives $\widetilde{\delta P}=(k_P\widetilde{\delta R}+{\tilde{\eta }}_P)/({\gamma }_P-i\omega )$. With the stated independent noises,Partial fractions and the supplied integral give the general varianceThe specific count-noise formula in the question additionally uses the usual steady birth-and-death noise strengths $q_R=2{\gamma }_R{R}_{\ast }$ and $q_P=2{\gamma }_P{P}_{\ast }$. Then $P_{\ast }=k_P{R}_{\ast }/{\gamma }_P$ reduces it to $P_{\ast }[1+k_P/({\gamma }_R+{\gamma }_P)]$. Arbitrary unspecified $q_R,q_P$ do not by themselves imply this particular Fano factor.

The requested rewriting isHere $b$ is the translational burst size, the expected number of proteins produced during one RNA's mean lifetime. The ratio $\varphi$ compares removal rates, equivalently RNA to protein mean lifetimes. Poisson molecule-number statistics have variance equal to mean, hence $F=1$.

When RNA is much less stable than protein, $\varphi \ll 1$ and $F\simeq 1+b$. At fixed degradation rates it rises linearly with the rate of biological translation, with exact slope $1/({\gamma }_R+{\gamma }_P)$. The formal intercept at zero rate is the limit from positive protein mean; at exactly zero production both mean and variance vanish and their ratio is undefined. A larger Fano factor need not imply a larger relative variance $\mathrm{var}P/P_{\ast }^2$, because the mean also increases.