Simple model of mrna production

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1 Simple model of mrna production We assume that the number of mrna (m) of a gene can change either due to the production of a mrna by transcription of DNA (which occurs at a constant rate α) or due to degradation of the mrna (which occurs at a constant rate β). The equation of evolution for m is. = α βm At steady-state: = 0, so m = α/β. The general evolution for starting for molecules is m(t) m 0 m(t) = m (1 exp[ βt]) + m 0 exp[ βt] In [2]: import numpy as np import matplotlib.pyplot as plt %matplotlib inline In [18]: alpha=10 #mrna production rate (unit: molecules/time) beta=1 #mrna degradation rate (unit: 1/time) m0=0 #initial number of mrna minf=alpha/beta #steady-state number of mrna t = np.linspace(0, 10, 256, endpoint=true) #time m = minf*(1-np.exp(-beta*t))+m0*np.exp(-beta*t) #evolution of m plt.xlabel('time t') plt.ylabel('m(t)') plt.ylim((m0-1,minf+1)) plt.plot(t,m,'k') plt.plot(t,m0+0*t,'--r') plt.plot(t,minf+0*t,'--r') plt.plot(t,m0+(minf-m0)/2+0*t,'--b') plt.plot(np.log(2)+0*t,np.linspace(m0-1, minf+1, 256, endpoint=true),'--b') 1 sur 5 15/11/ :04

2 The convergence to steady-state is exponential with a half-life ( m( t 1/2 ) = m 0 + ( m m 0 )/2) of t 1/2 = log(2)/β. Thus, the time to approach steady-state is controlled by β the rate of decay/degradation. How can a organism increase the rate of response to a signal to turn on a gene? Increasing β alone increases the rate of response but also decreases steady state expression and results in more rapid turnover which is energitically costly. To increase the rate without affecting the magnitude requires α and β to be simultaneously tuned while preserving their ratio, ie α should be increased (more production) which is also costly. Evolution, find a compromise between energy and response rate? Including the protein level We add the protein level to the previous model by considering that protein (p) are produced from mrna at a constant rate γ and are degraded at a constant rate μ. The mass-action law gives: = α βm dp = γm μp At steady state: m = α/β and p = γ m /μ = (γα)/(βμ) (the product of production rates divided by the product of degradation rates). The general solution for p can be exactly derived μ p(t) = p [(1 exp[ μt]) + ( ) ( m 0 / m 1)(exp[ βt] exp[ μt])] + p 0 exp[ μt] μ β But in case where the dynamics of mrna is faster than the one of protein ( β >> μ), we can safely replace m(t) by its steady-state value m leading to dp = γm μp Hence p(t) p (1 exp[ μt]) + p 0 exp[ μt]. This is a simple form of an approximation known as quasisteady-state approximation, because to derive it we assume that some "fast" variable (here m) are quickly converging to their steady state ( / = 0). In general, we can make this approximation for any variable that has a relaxation time much shorter than the dynamics we are interested in. 2 sur 5 15/11/ :04

3 In [32]: alpha=10 #mrna production rate (unit: molecules/time) beta=1 #mrna degradation rate (unit: 1/time) gamma=4 #protein production rate (unit:1/time) mu=0.05 #protein degradation rate (unit: 1/time) t = np.linspace(0, 10/mu, 256, endpoint=true) #time m0=0 #initial number of mrna minf=alpha/beta #steady-state number of mrna m = minf*(1-np.exp(-beta*t))+m0*np.exp(-beta*t) #evolution of m p0=0 #initial number of protein pinf=alpha*gamma/(beta*mu) p=pinf*((1-np.exp(-mu*t))+mu/(mu-beta)*(m0/minf-1)*(np.exp(-beta*t)-np.exp(-mu* t)))+p0*np.exp(-mu*t) #exact evolution of p papprox=pinf*(1-np.exp(-mu*t))+p0*np.exp(-mu*t) #approximated evolution of p plt.xlabel('time t') plt.ylabel('m(t)/minf,p(t)/pinf') plt.plot(t,m/minf,'k') plt.plot(t,p/pinf,'r') plt.plot(t,papprox/pinf,'--r') Transcriptional regulation We consider that the transcription of the gene of interest is controlled by an «external» factor X (a protein, a chemicals, etc.) that can bind to a regulatory region present in the promoter of the gene. Therefore, the promoter can be in two states: (0) the promoter is free and the mrna production rate is α 0 ; (1) X is bound to DNA and the mrna production rate is α 1. The equation of evolution for m is then given by = α 0 Φ 0 (X) + α 1 Φ 1 (X) βm with Φ i (X) the probability that the promoter is in the (i) state. Here Φ 0 + Φ 1 = 1 (the promoter state is either 0 or 1). At the promoter, we have X + (0) (1). The law of mass-action gives dφ 1 = k 01 XΦ 0 k 10 Φ 1 Protein-DNA binding is often fast (see table) and reaches steady-state on a timescale (sec.) smaller than mrna transcription (min.). So we can use the quasi-steady-state approximation for the promoter state, this leads to Φ 0 = k 10 /( k 10 + k 01 X) = 1/(1 + (X/K)) and Φ 1 = k 01 X/( k 10 + k 01 X) = (X/K)/(1 + (X/K)) with K = k 10 / k 01 the (thermodynamical) equilibrium dissociation constant of the protein-dna association. 3 sur 5 15/11/ :04

4 In [34]: x = np.linspace(0, 10, 256, endpoint=true) #X/K plt.plot(x,1/(1+x),'k') #Phi_0 plt.plot(x,x/(1+x),'r') #Phi_1 plt.xlabel('x/k') plt.ylabel('phi_0 (black), Phi_1 (red)') Φ i is saturable and hyperbolic, converges slowly to 0 or 1 (not a good switch). Note that X in the Φ's formulas is the total number (or concentration) of free X molecules that may be different of the number of total molecules. Sometimes (when the number of molecules is very very low), it is necessary to account for such "depletion" effect, but usually (since the number of promoter is small compared to the total number of X), we can identified the number of free TF molecules to their total number. Going back to the equation for m, we can rewrite the mrna production rate as 1 (X/K) α 1 + ( α 0 α 1 ) = α 0 + ( α 1 α 0 ) 1 + (X/K) 1 + (X/K) If α 0 > α 1, X is a repressor of gene expression: transcription rate decreases with X ( α 1 is the leaky rate, amount not under control). On contrary, if α 1 > α 0, X is an activator of gene expression: transcription rate increases with X ( α 0 is the basal rate of expression). In [35]: alpha0=1 alpha1=5 x = np.linspace(0, 10, 256, endpoint=true) #X/K plt.plot(x,alpha0/(1+x)+alpha1*x/(1+x),'k') #transcription rate plt.xlabel('x/k') plt.ylabel('transcription rate') 4 sur 5 15/11/ :04

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