Marc R. Roussel

(University of Lethbridge, Alberta, Canada)

https://s3-seminar.github.io/seminars/marc-r-roussel

Title — Un modèle stochastique de la transcription d’un gène

Abstract — Nous étudions depuis quelques années des modèles stochastiques de la transcription, c’est-à-dire de la synthèse de l’ARN à partir de la séquence de l’ADN par une machine moléculaire, l’ARN polymérase. Pour le cas d’une seule polymérase, il est possible de solutionner exactement nos modèles. Lorsque les interactions entre les polymérases sont importantes, il faut par contre utiliser (pour le moment) des méthodes numériques. En forme d’introduction au sujet, je présenterai un de nos modèles les plus simples, et je démontrerai comment on peut obtenir tous les moments voulus de la distribution du temps de transcription, c’est-à-dire comment on peut solutionner ce modèle. Cette distribution pourra être utilisée dans des modèles d’expression génétique, où elle apparaitra comme distribution de retards de la production de l’ARN.

Biography — Marc R. Roussel is Professor at Alberta RNA Research and Training Institute, Department of Chemistry and Biochemistry, University of Lethbridge.

Transcript

1. The distribution of transcription times/delays Marc R. Roussel Alberta RNA Research and Training Institute Department of Chemistry and Biochemistry
2. Gene expression: the central dogma (in prokaryotes) DNA ⇓transcription (m/r/t)RNA ⇓translation (mRNA) Protein
3. Gene transcription Source: Wikimedia commons public domain image (https://commons.wikimedia.org/wiki/File:RNAP_TEC_small.jpg)
4. So why would you want to model transcription? Central process in living organisms To test our understanding of the biochemistry Rapid exploration of parameter space Discovery of unintuitive/unexpected properties of the process To generate probability distributions that can be used in differential equation models with distributed delays, or in fast delay-stochastic simulations Roussel and Zhu, Phys. Biol. 3, 274 (2006)
5. An example of a simple gene expression model Monk, Curr. Biol. 13, 1409 (2003) dM dt = αmG (P(t − τm)) − µmM dP dt = αpM(t − τp) − µpP G(x) = [1 + (x/p0)n] P: Hes1 protein . . . but we probably want (e.g.) dM dt = αm ∞ 0 G(P(t − τ)) ρ(τ) dτ − µmM.
6. A prokaryotic transcription model 6 U7 U8 U9 U ... 1 U2 U3 U4 O5 U RNA polymerase binding Polymerase + U1 k0 − − → O1 Activation of initial complex by 2 NTPs O1 k1 − − → A1 Activation by NTP Oi k3 − − → Ai Translocation Ai + Ui+1 k2 − − → Oi+1 + Ui Termination An k4 − − → Polymerase + RNA + Un Roussel and Zhu, Bull. Math. Biol. 68, 1681 (2006).
7. Solvable cases vs simulation The model is a Markov chain described by a master equation. Variables: joint probabilities that each site is in a specified state, e.g. p(O1, U[2,72] , A73, U[74,n−1] , On). Massive number of variables so solution is not feasible, but statistics can be obtained from stochastic simulations. The single-polymerase case is solvable. For shorter genes, we can compute ρ(τ). In general, we can recover all the moments of the distribution of transcription times.
8. Distribution of transcription times In our prokaryotic model, transcription consists of a sequence of steps i → i + 1. Jump time distributions obtained by solving conditional master equations The jumps are independent events, so their joint probability distribution is ρ(τ1, τ2, . . . , τn) = i ρi (τi ) The distribution of total transcription times is a convolution: ρ(τ) = · · · τi =τ ρ(τ1, τ2, . . . , τn) dτ1 dτ2 . . . dτn
9. Distribution of transcription times (continued) Using the convolution theorem for Laplace transforms, we get ˜ ρ(s) = i ˜ ρi (s) where ˜ f ≡ L(f ). Computing the inverse Laplace transform is possible but requires extended precision.
10. Moments of the distribution From the definition of the Laplace transform, ˜ ρ(s) = ∞ 0 e−sτ ρ(τ) dτ we get τm = ∞ 0 τmρ(τ) dτ = (−1)m dm ˜ ρ dsm s=0
11. Kurtosis Excess kurtosis: Measure of “peakedness”/“tailedness” γ2 = µ4 σ4 − 3 Aside: In many cases, γ2 really tells us whether different processes control the shape of the peak and tail. Special values: γ2 = 0 for a Gaussian. γ2 = 3 for a double exponential distribution. Heavy-tailed distribution: tails decay more slowly than an exponential
12. Kurtosis of the prokaryotic transcription model 1 2 3 4 5 6 7 8 9 10 200 400 600 800 1000 0 1 2 3 4 5 6 γ 2 k 0 /s-1 n γ 2 Polymerase + U1 k0 − − → O1 k1 = 8 s−1, k2 = 100 s−1, k3 = 100 s−1, k4 = 10 s−1
13. Distribution of transcription times k0 = 0.04 s−1 k1 = 8 s−1 k2 = 100 s−1 k3 = 100 s−1 k4 = 0.2 s−1 n = 250 nt γ2 = 5.56
14. Many-polymerase simulations ∆ minimum spacing
15. Distribution of transcription times Analytic distribution vs many-polymerase simulation 0 0.005 0.01 0.015 0.02 0.025 0.03 0 50 100 150 200 l(o)/s-1 o/s Analytic distribution 6 = 20 nt 6 = 40 nt 6 = 60 nt
16. An approximation to the single-polymerase distribution If (e.g.) k0 and k4 are significantly smaller than the other rate constants, then perhaps we can consider only those processes? ρ2(τ) = k0k4 k4 − k0 e−k0τ − e−k4τ (?) However, the fast processes have the effect of a fixed delay: τmin = 1 k1 + n − 1 k2 + n − 1 k3 . [Cooke and Grossman, J. Math. Anal. Appl. 86, 592 (1982)] If H(·) is the Heaviside function, the correct approximation is ρ(τ) ≈ H(τ − τmin)ρ2(τ − τmin)
17. Comparison of the approximate and exact distributions 0 0.005 0.01 0.015 0.02 0.025 0.03 0 20 40 60 80 100 120 140 ρ(τ)/s-1 τ/s Exact Approximate k0 = 0.04 s−1, k4 = 0.2 s−1, k1 = 8 s−1, k2 = 100 s−1, k3 = 100 s−1, n = 200 nt =⇒ τmin = 4.105 s
18. Noise minimization: an evolutionary possibility? 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 10-4 10-3 10-2 10-1 100 101 102 103 104 CV k1 /s-1 CV = σ τ O1 k1 − − → A1 k0 = 0.01 s−1, k2 = 10 s−1, k3 = 100 s−1, k4 = 0.1 s−1, n = 200 nt, ∆ = 40 nt
19. Single-polymerase theory fails under high traffic conditions 0.4 0.5 0.6 0.7 0.8 0.9 1 10-4 10-3 10-2 10-1 100 101 102 103 104 CV k4 /s-1 An k4 − − → Polymerase + RNA + Un k0 = 0.01 s−1, k1 = 5 s−1, k2 = 10 s−1, k3 = 100 s−1, n = 200 nt, ∆ = 40 nt
20. Phase transition from low- to high-traffic regime 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 0.0001 0.001 0.01 0.1 1 10 100 1000 10000 ξ k 4 /s-1 ξ = bound RNAP max. bound RNAP k0 = 0.01 s−1, k1 = 5 s−1, k2 = 10 s−1, k3 = 100 s−1, n = 4614 nt, ∆ = 40 nt
21. Modularity Because of the product form of ˜ ρ(s), it is easy to replace parts of the model by modules expressing different biochemistry. A module could replace one or several steps. It is only necessary to be able to solve the corresponding survival time problem. Example: Pausing module to replace a simple elongation module Ok kp − − − − k−p Pk Ok k3 − − → Ak Ak + Uk k2 − − → Ok+1 + Uk
22. Excess kurtosis and pausing One pause-prone site, k0 = 0.04 s−1, k1 = 8 s−1, k2 = 100 s−1, k3 = 100 s−1, k4 = 0.2 s−1, k−p = 0.01 s−1, n = 250 nt
23. Future directions: eukaryotes Eukaryotic transcription model under active development Vashishtha, M.Sc. thesis, University of Lethbridge, 2011 Major complication in eukaryotes: co-transcriptional splicing Source: Wikimedia commons/OpenStax College (https://commons.wikimedia.org/wiki/File:0326_Splicing.jpg)
24. Acknowledgments Dr Rui Zhu Saurabh Vashishtha RJ Murphy Katherine Gzyl Eric Hill Dr Theodore Perkins (OHRI) Funding: Invitation: Silviu Niculescu Alban Quadrat
25. Lethbridge
26. References Prokaryotic model: Roussel and Zhu, Bull. Math. Biol. 68, 1681 (2006). Roussel, Biomath 2, 1307247 (2013). Eukaryotic model: Vashishtha, M.Sc. thesis, University of Lethbridge (2011) https://www.uleth.ca/dspace/handle/10133/3190