Ecological constraints on the origin of neurones

Abstract

The basic functional characteristics of spiking neurones are remarkably similar throughout the animal kingdom. Their core design and function features were presumably established very early in their evolutionary history. Identifying the selection pressures that drove animals to evolve spiking neurones could help us interpret their design and function today. This paper provides a quantitative argument, based on ecology, that animals evolved neurones after they started eating each other, about 550 million years ago. We consider neurones as devices that aid an animal’s foraging performance, but incur an energetic cost. We introduce an idealised stochastic model ecosystem of animals and their food, and obtain an analytic expression for the probability that an animal with a neurone will fix in a neurone-less population. Analysis of the fixation probability reveals two key results. First, a neurone will never fix if an animal forages low-value food at high density, even if that neurone incurs no cost. Second, a neurone will fix with high probability if an animal is foraging high-value food at low density, even if that neurone is expensive. These observations indicate that the transition from neurone-less to neurone-armed animals can be facilitated by a transition from filter-feeding or substrate grazing to episodic feeding strategies such as animal-on-animal predation (macrophagy).

Appendices

Appendix A: Derivation of $p_{\text {fix}}$ for the idealised ecosystem

Our model ecosystem assumes that animals receive an independent and identically-distributed (i.i.d.) amount of energy on every timestep. Wald (1944) derived an identity that can be used to calculate not only absorption probabilities, but also the (conditional and marginal) distributions of the time to absorption, when the steps of a random walk are i.i.d.

Suppose that $S_t$ is a sum of $t$ i.i.d. random variables $V_i$, where $V_i \sim V$. Let the sum’s initial value $S_0$ fall between two constant absorbing barriers $b$ and $a;\, b < S_0 < a$. As long as the sum stays between $a$ and $b$, we continue adding observations of $V$ to the sum. When the sum hits either $a$ or $b$, it is said to ‘absorb.’ Let $T$ represent the (finite) random number of summations that have been performed when the sum hits one of the two absorbing barriers.

Let ${\mathbb {M}}_{V}(h)$ be the moment generating function of $V; \,{\mathbb {M}}_{V}(h) \equiv {\mathbb {E}}\left[ e^{Vh}\right]$, with $h$ as a free variable. Wald’s identity is (Miller 1961; Whittle 1964) :

$$\begin{aligned} {\mathbb {E}}\left[ e^{S_Th}({\mathbb {M}}_{V}(h))^{-T}\right] =e^{S_0 h}. \end{aligned}$$

(5)

Lemma 2 in Wald (1944) states that if ${\mathbb {E}}\left[ V\right] \ne 0$ and $\text {Var}[ V ] \ne 0$, then exactly one nonzero value $h_0 \ne 0$ makes ${\mathbb {M}}_{V}(h_0) = 1$. In other words, ${\mathbb {M}}_{V}(h) - 1$ is convex and has two real roots, one nonzero, under very general assumptions¹ (Barnett 1975). Evaluating Eq. 5 at $h_0$ gives:

$$\begin{aligned} {\mathbb {E}}\left[ e^{S_Th_0}\right] = e^{S_0 h_0}. \end{aligned}$$

(6)

If the absorbing barriers 0 and $V_{\text {rep}}$ are exactly reached upon absorption and not exceeded, then:

$$\begin{aligned} {\mathbb {E}}\left[ e^{S_T h_0}\right] = e^{a h_0} \Pr (S_T = a) + e^{b h_0} \Pr (S_T = b). \end{aligned}$$

Inserting $\Pr (S_T = b) = 1 - \Pr (S_T = a)$ and rearranging, Eq. 6 becomes:

$$\begin{aligned} \Pr (S_T = a) = \frac{e^{S_0 h_0} - e^{b h_0}}{e^{a h_0} - e^{b h_0}}. \end{aligned}$$

For the animals considered in our model, insert $S_0 = V_{\text {rep}}/2, a = V_{\text {rep}}$, and $b = 0$ to obtain our claimed expression for $p_{\text {rep}}$, Eq. 2.

Recall that $h_0$ is the unique nonzero value that makes ${\mathbb {M}}_{V}(h_0) = 1$. When ${\mathbb {E}}\left[ V\right] \approx 0, h_0 \approx 0$, and we can approximate $h_0$ very accurately by Taylor-expanding the moment generating function:

$$\begin{aligned} 1 = {\mathbb {M}}_{V}(h_0) \equiv {\mathbb {E}}\left[ e^{Vh_0}\right] = {\mathbb {E}}\left[ 1\right] + h_0 \cdot {\mathbb {E}}\left[ V\right] + h_0^2/2 \cdot {\mathbb {E}}\left[ V^2\right] + O(3). \end{aligned}$$

Neglecting higher-order terms and rearranging, we obtain Eq. 3:

$$\begin{aligned} h_0 \approx -2 {\mathbb {E}}\left[ V\right] / {\mathbb {E}}\left[ V^2\right] . \end{aligned}$$

We must stress that ${\mathbb {E}}\left[ V\right]$ must be very close to 0 for this approximation to be valid. A small error in our approximation of $h_0$ can result in a large error for our approximation of $p_{\text {rep}}$, so our approximation for $h_0$ must be very accurate. More accurate approximations, if needed, can be obtained by numerical evaluation of $h_0$ or by including more higher-order terms.

Finally we obtain the probability $p_{\text {fix}}$ that a single decider will fix in a population of idiots.

When ${\mathbb {E}}\left[ V\right] \approx 0$, the equilibrium total (decider + idiot) animal population $N$ will remain approximately constant throughout a decider’s invasion. Since $N$ is approximately constant, we can approximate our model ecosystem (see Fig. 1) as a Moran process.

The Moran process is a birth-death process; each change in the population structure involves one individual reproducing (R) and one individual perishing (P). We can describe our idealised ecosystem as a birth-death process as follows. During the simulation, deciders (D) and idiots (I) reproduce and perish. Define a birth-death event to be the instant when a reproduction–perish pair [R, P], composed of one animal from each clade is completed (i.e. [R, P] = either [D, I] or [I, D]). The approximately constant population size ensures that the number of births and deaths will be approximately equal, so the reproduction and perish components of the [R, P] pairs will occur reasonably close together in time.

In the original Moran process, $r$ is constant by definition, and $p_{\text {fix}}$ is (Ewens 2004; Monk et al. 2014):

$$\begin{aligned} p_{\text {fix}}= (1 - r ^ {-1}) / (1 - r ^ {-N}). \end{aligned}$$

This expression for $p_{\text {fix}}$ applies to our ecosystem if $r$ is constant throughout a decider’s invasion. We now show that $r$ indeed remains very nearly constant.

As in the Moran process, $r$ is the ratio of the probabilities of the decider population size $\delta _t$ increasing and decreasing by one:

$$\begin{aligned} r = \frac{\Pr (R = D, P = I| \delta _t)}{\Pr (R = I, P = D| \delta _t)} \approx \frac{\Pr (R = D| \delta _t) \Pr (P = I| \delta _t)}{\Pr (R = I| \delta _t) \Pr (P = D| \delta _t)}, \end{aligned}$$

since R and P are approximately conditionally independent, conditional on the decider population size. We may consider each of these probabilities as a ratio of reproduction or death rates $q$:

(7)

where $q_R^D$ is the rate that deciders reproduce, $q_R^I$ is the rate that idiots reproduce, $q_P^D$ is the rate that deciders perish, and $q_P^I$ is the rate that idiots perish:

Here, $p_{\text {rep}}^D$ is the probability that an individual decider will reproduce, and ${\mathbb {E}}[{T_D} | {R}]$ is the conditional expected time for an individual decider to absorb, given that it eventually reproduced. The other terms are defined analogously.

Since ${\mathbb {E}}\left[ V_t\right] \approx 0, {\mathbb {M}}_{V}(h)$ is well-approximated by a parabola in the neighbourhood $h \approx 0$. This symmetry of ${\mathbb {M}}_{V}(h)$ and the symmetry of the absorbing barriers for individual animals ($b = 0, S_0 = V_{\text {rep}}/2$, and $a = V_{\text {rep}}$) means that the conditional expected number of summations required to absorb, conditional on where absorption occurred, are equal. In other words, animals that start with energy $V_{\text {rep}}/2$ and eventually reproduce require the same expected number of summations as those that start with energy $V_{\text {rep}}/2$ and eventually starve.² This is shown in Fig. 5. Since ${\mathbb {E}}[{T_D} | {R}] = {\mathbb {E}}[{T_D} | {P}]$ and ${\mathbb {E}}[{T_I} | {R}] = {\mathbb {E}}[{T_I} | {P}]$, Eq. 7 simplifies to:

$$\begin{aligned} r = \frac{p_{\text {rep}}^D(1 - p_{\text {rep}}^I)}{p_{\text {rep}}^I(1 - p_{\text {rep}}^D)}. \end{aligned}$$

(8)

This expression for $r$ is virtually independent of the decider population size $\delta _t$, though not obviously so. $p_{\text {rep}}^I$ and $p_{\text {rep}}^D$ are dependent on $\delta _t$ because, as an invasion progresses, the food item density changes. Before a decider is introduced, the food item density is $\rho _I= (a+ m) / b$, where idiots break even on average. If a decider fixes, then the food item density has been driven to where deciders break even on average. This is shown in Fig. 5. Plotting $r$ over a range of decider population sizes, we see that $r$ remains approximately constant throughout a decider invasion (see Fig. 5).

Fig. 5

$r$ is virtually independent of the decider population size population size $\delta _t$. Upper plot Conditional (triangles) and marginal (lines) expected times to absorption for deciders (solid line and filled triangles) and idiots (dashed line and empty triangles) as a function of $\delta _t$. ${\mathbb {E}}[{T} | {R}] = {\mathbb {E}}[{T} | {P}]$, for both deciders and idiots, for all $\delta _t$; all the triangles perfectly overlap. When an invasion begins, it takes idiots longer to reach absorption than deciders, but when an invasion is nearly complete, it takes deciders longer to absorb than idiots. Lower left $p_{\text {rep}}$ as a function of $\delta _t$. As the population size of advantaged deciders grows, food density $\rho$ is driven down, so both deciders and idiots are less likely to reproduce as an invasion progresses. Lower right Plot of Eq. 8. $r$ is approximately constant for all $\delta _t$. The y-axis shows that the slope of the line is negligible. For all plots, at the beginning of a decider invasion, ${\mathbb {E}}\left[ V_t\right]$ = 3.5e-4, ${\mathbb {E}}\left[ V_t^2\right] = .2102, V_{\text {rep}}=500$, and $N = 80$

Since $r$ is nearly constant with respect to $\delta _t$, we can simplify Eq. 8 further by evaluating it at $\delta _t = 1$. At $\delta _t = 1, p_{\text {rep}}^I= 1/2$, so $r$ simplifies to:

$$\begin{aligned} r = \left. \frac{p_{\text {rep}}^D}{1 - p_{\text {rep}}^D} \right| _{\rho = \rho _I}, \end{aligned}$$

(9)

where the evaluation at $\rho =\rho _I$ means that we want to evaluate $p_{\text {rep}}^D$ at food item density $\rho _I= (a+ m) / b$.

Appendix B: Obtaining closed-form expressions for $p_i$

This Appendix defines models of the signals produced by food items and how a decider’s sensor spikes in response to those signals. It then shows how to calculate the $p_i$ given these models.

Let $r_k$ be the random distance between a decider’s sensor and the $k$th closest food item. Assume that food items produce signals $\sigma _k$ that fall as distance cubed:

$$\begin{aligned} \sigma _k = 1/r_k^3. \end{aligned}$$

Let the sensor spike with an intensity $I$ proportional to the additive strengths of the signals from the food items. The proportionality constant $g$ is the gain of the sensor, and let $n$ be the spontaneous firing rate:

$$\begin{aligned} I(\mathbf{{r}}) = g\sum _{k=1}^\infty \sigma _k + n. \end{aligned}$$

The sensor acts as a food item proximity detector; if food items are close to the sensor, then it experiences strong signals and spikes with a high intensity. The expected number of spikes $\lambda$ that the sensor produces in time $\varDelta t$ is:

$$\begin{aligned} \lambda (\mathbf{{r}}) = I(\mathbf{{r}}) \varDelta t. \end{aligned}$$

If the sensor spikes at least once in $\varDelta t$, the decider strikes:

$$\begin{aligned} \Pr (A=1|\mathbf{{r}}) = 1 - e^{-\lambda (\mathbf{{r}})}. \end{aligned}$$

With this model of $\Pr (A=1|\mathbf {r})$, we can calculate the probability that the decider strikes, given the random number of food items in the strike zone $Z$ and given that the number of food items that exist inside the decider $D$ is zero:³

$$\begin{aligned} \Pr (A=1|Z,D=0)&= \int _0^\infty \Pr (A=1,\mathbf{{r}}|Z,D=0)d\mathbf{{r}} \\&= \int _0^\infty \Pr (A=1|\mathbf{{r}}) \Pr (\mathbf{{r}}|Z,D=0) d\mathbf{{r}} \\&= {\mathbb {E}}[{\Pr (A=1|\mathbf{{r}})} | {Z,D=0}] = {\mathbb {E}}[{1-e^{-\lambda }} | {Z,D=0}]. \end{aligned}$$

If the decider almost always strikes (i.e. if $\Pr (A=1) \approx 1$), then $1-e^{-\lambda }$ is almost linear and Jensen’s inequality becomes an accurate approximation:

$$\begin{aligned} {\mathbb {E}}[{1-e^{-\lambda }} | {Z,D=0}] \approx 1 - \text {exp}\left( -{\mathbb {E}}[{\lambda } | {Z,D=0}] \right) . \end{aligned}$$

Inserting our expression for $\lambda$:

$$\begin{aligned} {\mathbb {E}}[{1-e^{-\lambda }} | {Z,D=0}] \approx 1 - \text {exp}\left( -g\varDelta t \sum _{k=0}^\infty {\mathbb {E}}[{\sigma _k} | {Z,D=0}] - n\varDelta t \right) . \end{aligned}$$

The conditional expected signal of the $k$th closest food item given the number of food items in the zone and given that no food items exist inside the decider is:

$$\begin{aligned} {\mathbb {E}}[{\sigma _k} | {Z,D=0}]&= \int _0^\infty \sigma _k \Pr (r_k|Z,D=0)dr_k \\&= \int _0^\infty \sigma _k \frac{\Pr (Z,D=0|r_k)\Pr (r_k)}{\Pr (Z,D=0)}dr_k \\&= \int _0^\infty \sigma _k \frac{\Pr (Z|r_k,D=0)\Pr (D=0|r_k)\Pr (r_k)}{\Pr (Z)\Pr (D=0)} dr_k. \end{aligned}$$

All of the above probability density functions are known. Let $\rho$ be the food item density, let the area of the decider be 1 (they are circular with radius $\sqrt{1/\pi }$) and let the area of the ‘strike zone’ be 1 (it is also circular with radius $\sqrt{2/\pi }$, see Fig. 1). Then:

$$\begin{aligned} \Pr (Z)&= \text {Poisson}(\rho ); \qquad \Pr (D=0) = e^{-\rho }; \\ \Pr (r_k)&= 2(\pi \rho )^k r_k^{2k-1} \text {exp}\left( -\pi \rho r_k^2 \right) / (k-1)!; \\ \Pr (D=0|r_k)&= \left\{ \begin{aligned}&0, \\&\text {Binom}(0;k-1,1/\pi r_k^2) \\ \end{aligned} \right. \qquad \qquad \begin{aligned}&r_k \le \sqrt{1/\pi } \\&r_k > \sqrt{1/\pi }; \\ \end{aligned} \\ \Pr (Z|r_k,D=0)&= \left\{ \begin{aligned}&\text {Poisson}(2\rho - \rho \pi r_k^2),\\&\text {Binom}(Z;k-1,1/(\pi r_k^2 - 1)),\\ \end{aligned} \right. \qquad \begin{aligned}&Z \ge k \\&Z < k. \end{aligned} \end{aligned}$$

$\Pr (A=1|Z,D=0)$ may now be numerically evaluated.