The Significance of Reduced Conductance in the Sino-Atrial Node for Cardiac Rythmicity

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1 The Significance of Reduced Conductance in the Sino-Atrial Node for Cardiac Rythmicity Will Nesse May 14, 2005

2 Sino-Atrial Node (SA node): The Pacemaker of the Heart 1

3 The SA node and the Atria: How the pacemaker works Both the SA node and the Atria are excitable, spiking cells The SA node is more depolarized and has autonomous oscillations The Atria is more hyperpolarized and needs a stimulus to drive an action potential An action potential wave emanates from the SA node out through the Atria The Vagus nerve and the sympathetic cardiac nerve modulates the frequency of oscillation 2

4 The SA node Figure from Dobrzynski et al

5 The SA node Non contractile cells Small compared to the Atria Figure from Dobrzynski et al

6 The Atria Contractile cells Large compared to the SA node Figure from Dobrzynski et al. 2005t 5

7 The SA node is distinct from the Atria Figure from Dobrzynski et al

8 Electrical Conductance in the SA node- Atria The SA node expresses only low conductance gap junctions: 30-pS in SA node versus 160-pS in Atria Guess: The SA node has a lower electrical conductance 7

9 Question What is the Functional Significance of reduced conductance in the SA node? Dr Keener s result: There is a critical SA node size to achieve an oscillatory solution with homogeneous conductance Numerical studies (Joyner 1986, Winslow et al. 1995) suggest reduced conductance could help insulate the SA node from the Atria, reducing electrotonic loading from the Atria, preserving SA node oscillations My work: Reduced SA node conductance lowers the critical SA node size for an oscillatory solution 8

10 Why is a smaller SA node an advantage? Leaves more room for Atrial muscle Smaller SA node will have a more coherent oscillation Fewer cells Cells are more connected More robust to noise and intrinsic variation Easier to modulate 9

11 The Plan Use an analytically tractable 1D continuum model of excitable cardiac media: Piecewise linear FitzHugh-Nagumo model Model SA node as a small, excited part of the Atrial media Important parameters Excitability of SA node, Atria Conductance of SA node, Atria Size of SA node Specific aim: Find Hopf bifurcation condition for SA node- Atrial media relating SA node size and conductance 10

12 The Model The Continuum FitzHugh-Nagumo Model x [0, ) v = t x (k(x) v x ) + f(v) w w = ɛ(v γw α(x)) t k(x) = varying conductance α(x) = varying excitability level 11

13 Piecewise linear nonlinearity f(v) = v, v 1 4 v 1 2, 1 4 < v < v, v The FitzHugh Nagumo Piecewise Nonlinearity f(v) v 12

14 Oscillatory solution to the uncoupled FitzHugh- Nagumo model 0.4 The FitzHugh Nagumo Equation Recovery (w) Voltage (V) 13

15 Voltage trace of uncoupled FitzHugh-Nagumo Model 1.2 The FitzHugh Nagumo Equation Voltage (V) Time 14

16 Piecewise constant excitability and conductance α(x) = { α s α a, x σ, x > σ k(x) = { k s k a, x σ, x > σ s = SA node a = Atria Where α s > α a and k s k a = 1 σ is the SA node boundary 15

17 Steady state: v t = w t = SA Node Steady State Solution Steady State Voltage Region I II Region III 0.1 ξ σ x ξ is the boundary between excited (v > 1 4 ) and unexcited (v 1 4 ) states 16

18 Steady state solution V(x), W(x) The boundary and continuity conditions v I (0) = 0 v I (ξ) = v II (ξ) v I (ξ) = v II (ξ) v II (σ) = v III (σ) k s v II (σ) = k av III (σ) reduce to the consistency relation F (ξ, k s, σ) = 0 ξ(σ, k s ) 17

19 Steady state solution as a function of k s Steady State Solution as a Function of k s k s x

20 There is a larger excited region ξ as k s is reduced The Boundary of the Excited state For Different Values of k s ξ k s 19

21 Stability of steady state V (x), W (x) Linearize: v = V (x) + φ(x)e λt, w = W (x) + ψ(x)e λt Obtain Schrödinger equation µφ = x (k φ x ) + f (V (x))φ µ = λ + ɛ λ + ɛγ µ R and λ is pure imaginary (Hopf bifurcation) when µ = ɛγ 20

22 Schrödinger potential function f (V (x)) = Parameterized by ξ 1 x ξ 1 x > ξ 0.4 The FitzHugh Nagumo Piecewise Nonlinearity f(v) v 21

23 Schrödinger equation solution Break solution up into the linear pieces, solve them and using the boundary data and continuity constraints φ I (0) = 0 φ I (ξ) = φ II (ξ) φ I (ξ) = φ II (ξ) φ II (σ) = φ III (σ) k s φ II (σ) = k aφ III (σ) lim III(x) x bounded generate transcendental consistency equation Φ(µ, ξ(k s, σ), k s ) = 0 22

24 Hopf bifurcation problem Set µ = ɛγ to find the bifurcation manifold for σ and k s : Φ(ɛγ, ξ(k s, σ), k s ) = σ k s Bifurcation Manifold Stable Steady State k s Oscillatory Solution σ 23

25 Future directions Investigate parameter effects on SA node oscillation frequency Investigate traveling wave solutions Will SA node emit a traveling pulse? How does it depend on parameters? Parameter-dependent dispersion relation of the Atrial media. Will it support high frequencies? Higher dimensions Explore numerically the existence of bifurcation manifold in more biologically realistic models 24