Retinal oxygen distribution and the role of neuroglobin

Abstract

The retina is the tissue layer at the back of the eye that is responsible for light detection. Whilst equipped with a rich supply of oxygen, it has one of the highest oxygen demands of any tissue in the body and, as such, supply and demand are finely balanced. It has been suggested that the protein neuroglobin (Ngb), which is found in high concentrations within the retina, may help to maintain an adequate supply of oxygen via the processes of transport and storage. We construct mathematical models, formulated as systems of reaction–diffusion equations in one-dimension, to test this hypothesis. Numerical simulations show that Ngb may play an important role in oxygen transport, but not in storage. Our models predict that the retina is most susceptible to hypoxia in the regions of the photoreceptor inner segment and inner plexiform layers, where Ngb has the potential to prevent hypoxia and increase oxygen uptake by 30–40 %. Analysis of a simplified model confirms the utility of Ngb in transport and shows that its oxygen affinity ($P_{50}$ value) is near optimal for this process. Lastly, asymptotic analysis enables us to identify conditions under which the piecewise linear and quadratic approximations to the retinal oxygen profile, used in the literature, are valid.

Appendices

Appendix 1: Placing a bound on the concentration of pentacoordinate neuroglobin

It was found in all single and eight layer model simulations that the concentration of pentacoordinate Ngb, n, at any given point in space is significantly lower than that of hexacoordinate Ngb, $n_h$, or oxygen bound Ngb, $n_o$, at that point. In this appendix we derive a bound on n to explain why this is the case.

Beginning with the time-dependent equation for n, Eq. (10), we can re-write it in the form

$$\begin{aligned} \frac{\partial n}{\partial t} = D\frac{\partial ^2 n}{\partial x^2} + f(x,t) - \left( \frac{1}{\kappa } + g(x,t)\right) n, \end{aligned}$$

(42)

where $1/\kappa = k_3$, $f(x,t) = k_4n_h + \alpha k_2n_o \ge 0$ and $g(x,t) = \alpha k_1 c \ge 0$. Whether we are using the single or the eight layer model, this equation will be defined on a finite domain, which we can denote as $x \in [0,L]$ without loss of generality. As usual, we impose zero-flux boundary conditions at either end of the domain.

Consider the function $M = M(t)$ which satisfies

$$\begin{aligned} \frac{{\mathrm {d}} M}{{\mathrm {d}} t} = \left( \max _{x,t}(f(x,t)) + \beta \right) - \frac{1}{\kappa }M, \end{aligned}$$

(43)

where the constant $\beta > 0$ is arbitrarily small. This can be written as

$$\begin{aligned} \frac{{\mathrm {d}} M}{{\mathrm {d}} t} = F - \frac{1}{\kappa }M, \end{aligned}$$

(44)

where $F = (\max _{x,t}(f(x,t)) + \beta ) > f(x,t) \ge 0$ is a constant. We close the system by imposing the initial condition

$$\begin{aligned} M(0) = \max _{x}(n(x,0)). \end{aligned}$$

(45)

Solving this initial-value problem using the integrating factor method we obtain

$$\begin{aligned} M(t) = e^{-\frac{t}{\kappa }}M(0) + \kappa F(1 - e^{-\frac{t}{\kappa }}). \end{aligned}$$

(46)

Taking the limit as t tends to infinity we find that

$$\begin{aligned} \lim _{t \rightarrow \infty }M(t) = \kappa F. \end{aligned}$$

(47)

Let $u(x,t) := n(x,t) - M(t)$, then

$$\begin{aligned} \frac{\partial u}{\partial t} - D\frac{\partial ^2 u}{\partial x^2} = (f - F) - \frac{u}{\kappa } - gn, \end{aligned}$$

(48)

from Eqs. (42) and (44). Since $(f - F) < 0$ and $gn \ge 0$ we have that

$$\begin{aligned} \frac{\partial u}{\partial t} - D\frac{\partial ^2 u}{\partial x^2} + \frac{u}{\kappa } < 0, \end{aligned}$$

(49)

and therefore

$$\begin{aligned} \frac{\partial }{\partial t}\left( e^{\frac{t}{\kappa }}u\right) - D\frac{\partial ^2}{\partial x^2}\left( e^{\frac{t}{\kappa }}u\right) < 0. \end{aligned}$$

(50)

Defining $v(x,t) := e^{\frac{t}{\kappa }}u(x,t)$, we have that

$$\begin{aligned} \frac{\partial v}{\partial t} - D\frac{\partial ^2 v}{\partial x^2}< 0, \end{aligned}$$

(51)

with initial condition

$$\begin{aligned} v(x,0) = n(x,0) - \max _{x}(n(x,0)), \end{aligned}$$

(52)

and zero-flux boundary conditions.

By the maximum principle for parabolic PDEs we have that v must achieve its maximum on one of the boundaries $x = 0$, $x = L$ or $t = 0$ (Ockendon et al. 2003). Suppose that v takes its maximum value on $x = 0$ for some $t = t_1 > 0$. Since $n_x(0,t_1) = 0$ by the zero-flux boundary condition and since $M_x(t_1) = 0$, as M is independent of x, this implies that $u_x(0,t_1) = 0$ and so $v_x(0,t_1) = 0$. Therefore, in order for this point to be a maximum, we require both that $v_t(0,t_1) = 0$ and $v_{xx}(0,t_1) \le 0$. This implies that $v_t(0,t_1) - Dv_{xx}(0,t_1) \ge 0$, which contradicts (51). Therefore, the maximum cannot lie on the boundary $x = 0$. Using a similar argument we can show that the maximum cannot lie on the boundary $x = L$. Therefore, the maximum must lie on the boundary $t = 0$. As a check we can see that if the maximum lies on the boundary $t = 0$ for some $x = x_1$, where $0 < x_1 < L$, then we must have that $v_t(x_1,0) \le 0$, $v_x(x_1,0) = 0$ and $v_{xx}(x_1,0) \le 0$. Thus (51) may be satisfied, provided $|v_t(x_1,0)| > D|v_{xx}(x_1,0)|$.

Since v(x, t) takes its maximum value on $t = 0$ this means that $v(x,t) \le \max _{x}(v(x,0)) = \max _{x}(n(x,0) - \max _{x}(n(x,0))) = \max _{x}(n(x,0)) - \max _{x}(n(x,0)) = 0$. Therefore $u \le 0$, which implies that $n \le M$. Therefore, by (47), we have that at steady-state $n \le \kappa F$. Taking the limit as $\beta$ tends to zero, we find $n \le \kappa \max _{x,t}(f(x,t)) = \kappa \max _{x,t}(k_4n_h + \alpha k_2n_o) \le \kappa \alpha k_2 n_T = \frac{\alpha k_2 n_T}{k_3} \approx 4n_T\times 10^{-4}$. This bound was found to be satisfied in all simulations.

Appendix 2: Asymptotic analysis

In the absence of Ngb, Eqs. (21)–(23) reduce to a single equation for oxygen:

$$\begin{aligned} 0 = \frac{{\mathrm {d}}^2c}{{\mathrm {d}}x^2} - \frac{Q_ic}{\gamma + c}, \end{aligned}$$

(53)

for $Q_i$ $(1,\ldots ,7^*).$

In layers 2 and 4, $Q_2 = Q_4 = 0$, so that an exact analytical solution to Eq. (53) may be obtained. In all other layers, $Q_i$ is strictly positive, so that an exact analytical solution cannot be derived. In these layers we look instead for a leading order solution. Since the value of $Q_i$ is not, in general, continuous between layers, we cannot use standard matching techniques to construct a solution which is valid across all layers. Instead we use patching, ensuring that the solution in each layer satisfies the boundary conditions between it and the adjoining layers (see Bender and Orszag 1999, pgs. 335–336 for a discussion of patching).

We construct an asymptotic expansion for c(x):

$$\begin{aligned} c(x) \sim c_0(x) + \varepsilon c_1(x) + O(\varepsilon ^2) , \end{aligned}$$

(54)

choosing $\varepsilon = 0.1$, so as to provide a clear separation between the various scales, and introduce $\gamma ^* = \varepsilon ^{-2} \gamma$ so that $\gamma ^* = O(1)$. We also note that $Q_i = O(1)$ for $i \in \{1,3,5,6^*,7^*\}$ and that $Q_2 = Q_4 = 0$. Applying the scaling on $\gamma$, Eq. (53) becomes

$$\begin{aligned} 0 = \frac{{\mathrm {d}}^2c}{{\mathrm {d}}x^2} - \frac{Q_ic}{\varepsilon ^2 \gamma ^* + c}. \end{aligned}$$

(55)

We now consider the leading order solution to Eq. (55) within each layer, grouping layers that have the same scaling.

1.1 Layer 1

In layer 1, $c = O(1)$, as can be seen in Fig. 9, and Eq. (55) supplies, at leading order,

$$\begin{aligned} c(x) \sim c_0(x) = \frac{Q_1x^2}{2} + A_1x + B_1, \end{aligned}$$

(56)

where $A_1$ and $B_1$ are constants.

1.2 Layers 2 and 4

In layers 2 and 4, $Q_2 = Q_4 = 0$, so that Eq. (55) can be solved exactly to yield

$$\begin{aligned} c(x) = A_ix + B_i, \end{aligned}$$

(57)

for $i = 2, 4,$ where $A_i$ and $B_i$ are constants.

1.3 Layers 3, 5 and 7$^*$

In layers 3, 5 and 7$^*$, $c = O(\varepsilon )$, as can be seen in Fig. 9. Rescaling oxygen as $c = \varepsilon c^*$, we then rescale x as $x = \varepsilon ^{1/2}x_i^* + L_{i-1}$ in order to achieve a dominant balance, so that, after returning to our original scaling on c and x we have, at leading order,

$$\begin{aligned} c(x) \sim \frac{Q_ix^2}{2} + A_ix + B_i, \end{aligned}$$

(58)

for $i \in \{3,5,7^*\}$, where $A_i$ and $B_i$ are constants.

1.4 Layer 6$^*$

The situation in layer 6$^*$ is less straightforward. On the left- and right-hand sides of layer 6$^*$, $c = O(\varepsilon )$, whereas, toward the centre of the layer, in a neighbourhood around the local minimum, $c = O(\varepsilon ^2)$. Where $c = O(\varepsilon )$, we could scale c and x as in layers 3, 5 and 7$^*$; however, in the central region, where $c = O(\varepsilon ^2)$, we would regain Eq. (53) in the dominant balance after dropping the stars. Since seeking leading order solutions does not allow us to avoid dealing with Eq. (53), we instead retain Eq. (53) across the whole of layer 6$^*$ and use quadrature methods to derive approximate analytical solutions for the oxygen profile in this layer.

We will derive separate approximations to the oxygen profile for the left-hand side, centre and right-hand side of layer 6$^*$, in order to account for the variation in c between $O(\varepsilon )$ and $O(\varepsilon ^2)$. We will also derive approximations to the minimum oxygen concentration in layer 6$^*$, $c_{min}$, and its position, $x_{min}$.

Multiplying Eq. (53) by ${\mathrm {d}}c/{\mathrm {d}}x$ and integrating between $x_{min}$ and x, we find that

$$\begin{aligned} \frac{{\mathrm {d}}c}{{\mathrm {d}}x} = \pm \sqrt{2Q_{6^*}}\sqrt{c - c_{min} - \gamma \log {\left( \frac{\gamma + c}{\gamma + c_{min}}\right) }}, \end{aligned}$$

(59)

where we take the positive (negative) root to the right (left) of $x_{min}$, since the gradient of the oxygen profile is positive (negative) there. We note that the values of $c_{min}$ and $x_{min}$ are unknown at this stage.

We begin by seeking the left-hand and right-hand approximations near $L_5$ and $L_{6^*}$ respectively. For the left-hand approximation, we integrate Eq. (59) between x and $L_5$ to obtain

$$\begin{aligned} \int _{c(x)}^{c_L}\left( s - c_{min} - \gamma \log \left( \frac{\gamma + s}{\gamma + c_{min}}\right) \right) ^{-\frac{1}{2}}{\mathrm {d}}s = \sqrt{2Q_{6^*}}(x - L_5), \end{aligned}$$

(60)

where $c_L = c(x=L_5)$ is unknown at this stage. Since $c \approx c_L$ in the left-hand region, where $c_L = O(\varepsilon )$, and since $c_{min} = O(\varepsilon ^2)$ and $\gamma = O(\varepsilon ^2)$ (see Fig. 9 for values of $c_L$ and $c_{min}$), we may approximate the integrand by $s^{-\frac{1}{2}}$ to obtain the left-hand approximation as

$$\begin{aligned} c(x) \approx \frac{Q_{6^*}}{2}(L_5 - x)^2 + \sqrt{2Q_{6^*}c_L}(L_5 - x) + c_L. \end{aligned}$$

(61)

In a similar way, we obtain the right-hand approximation:

$$\begin{aligned} c(x) \approx \frac{Q_{6^*}}{2}(x - L_{6^*})^2 + \sqrt{2Q_{6^*}c_R}(x - L_{6^*}) + c_R, \end{aligned}$$

(62)

where $c_R = c(x=L_{6^*})$ is unknown at this stage. Both $c_L$ and $c_R$ may be found by applying the boundary conditions as described at the end of this section.

To derive the central approximation, valid in the neighbourhood of $x_{min}$, we integrate Eq. (59) between $x_{min}$ and x. Writing $\log ((\gamma + s)/(\gamma + c_{min})) = \log (1 + (s - c_{min})/(\gamma + c_{min}))$, we expand the integrand (which is the same as in Eq. (60)) about $s = c_{min}$, in powers of $(s - c_{min})/(\gamma + c_{min})$, where s is the variable of integration and $|(s - c_{min})/(\gamma + c_{min})| \ll 1$. Retaining only the first term and neglecting higher order terms, we obtain the central solution:

$$\begin{aligned} c(x) \approx \left( 1 + \frac{Q_{6^*}(x_{min} - x)^2}{2(\gamma + c_{min})}\right) c_{min}, \end{aligned}$$

(63)

in which $x_{min}$ and $c_{min}$ are presently unknown.

To determine $x_{min}$, we integrate Eq. (59) between the limits $x_{min}$ and $L_5$, and $x_{min}$ and $L_{6^*}$, to obtain the following pair of equations:

$$\begin{aligned} \int _{c_{min}}^{c_L}\left( s - c_{min} - \gamma \log \left( \frac{\gamma + s}{\gamma + c_{min}}\right) \right) ^{-\frac{1}{2}}{\mathrm {d}}s= & {} \sqrt{2Q_{6^*}}(x_{min} - L_5), \end{aligned}$$

(64)

$$\begin{aligned} \int _{c_{min}}^{c_R}\left( s - c_{min} - \gamma \log \left( \frac{\gamma + s}{\gamma + c_{min}}\right) \right) ^{-\frac{1}{2}}{\mathrm {d}}s= & {} \sqrt{2Q_{6^*}}(L_{6^*} - x_{min}). \end{aligned}$$

(65)

Subtracting Eq. (64) from Eq. (65), and noting that $\min (c_L,c_R) < s < \max (c_L,c_R)$ in the integrand, where $c_L = O(\varepsilon )$, $c_R = O(\varepsilon )$, $c_{min} = O(\varepsilon ^2)$ and $\gamma = O(\varepsilon ^2)$ (see Fig. 9 for values of $c_L$, $c_R$ and $c_{min}$), we may approximate the integrand by $s^{-\frac{1}{2}}$ to obtain

$$\begin{aligned} x_{min} \approx \frac{L_5 + L_{6^*}}{2} - \left( \frac{\sqrt{c_R} - \sqrt{c_L}}{\sqrt{2Q_{6^*}}}\right) . \end{aligned}$$

(66)

We obtain an implicit expression for $c_{min}$, by substituting for $x_{min}$ from Eq. (66) into Eq. (64):

$$\begin{aligned} \int _{c_{min}}^{c_L}\left( s - c_{min} - \gamma \log \left( \frac{\gamma + s}{\gamma + c_{min}}\right) \right) ^{-\frac{1}{2}}{\mathrm {d}}s \nonumber \\ \quad \approx \sqrt{\frac{Q_{6^*}}{2}}(L_{6^*} - L_5) - (\sqrt{c_R} - \sqrt{c_L}), \end{aligned}$$

(67)

where the integral must be calculated numerically. With $x_{min}$ and $c_{min}$ specified by (66) and (67) respectively, we can use Eq. (63) to calculate the central solution.

1.5 Using iteration to improve accuracy

We can iteratively improve the accuracy of our approximation using the exact derivative, given by Eq. (59), in the oxygen-flux boundary conditions at $x=L_{6^*}$ and $x=L_{7^*}$, given by Eq. (20), to yield:

$$\begin{aligned} 0= & {} \sqrt{2Q_{6^*}}\sqrt{c_L - c_{min} - \gamma \log \left( \frac{\gamma + c_L}{\gamma + c_{min}}\right) } + \nonumber \\&\displaystyle \quad (Q_5L_5 + A_5(c_L)) - {\hat{h}}_5(c_v - c_L),\nonumber \\ \end{aligned}$$

(68)

$$\begin{aligned} 0= & {} \sqrt{2Q_{6^*}}\sqrt{c_R - c_{min} - \gamma \log \left( \frac{\gamma + c_R}{\gamma + c_{min}}\right) } + \nonumber \\&\displaystyle \quad Q_{7^*}(L_{7^*} - L_{6^*}) - {\hat{h}}_{6^*}(c_v - c_R).\nonumber \\ \end{aligned}$$

(69)

The constant $c_{min}$ takes the value calculated by solving (67), using the original values of $c_L$ and $c_R$, whilst the constant $A_5$ is a function of $c_L$, given by:

$$\begin{aligned} A_5(c_L) = \frac{1}{L_5}\left[ c_L - c_c - \frac{1}{2}\left( Q_5(L_5^2 + L_4^2) + Q_3(L_2^2-L_3^2) - Q_1L_1^2\right) \right] . \end{aligned}$$

(70)

Equations (68)–(69) can be solved numerically, using the Matlab routine fsolve, to find updated values for $c_L$ and $c_R$. The updated $c_L$ and $c_R$ can then be used to calculate an updated value for $c_{min}$, using Eq. (67). We may then repeat the iteration, using the updated value of $c_{min}$ in Eqs. (68)–(70), to find new values for $c_L$ and $c_R$. Once the solution has converged, the final values of $c_L$ and $c_R$ can then be used to calculate improved values for $A_1,\ldots ,A_5,B_{7^*},x_{min}$ and $c_{min}$, by applying the boundary conditions to Eq. (41) as described in Sect. 4.3 and using Eqs. (66) and (67).

It was found that iteration results in a small improvement in the accuracy of the approximation and that the solution does not change significantly (by $O(10^{-4})$ or greater) after the third iteration. Therefore, we use the parameters generated by the third iteration for the approximate solution in Sect. 4.3.