EcoEvo Assignment 1

Question 1: Human population

The human population size is currently $8.2 \times 10^{9}$ . It is growing at 1.1 % /year, meaning average individual fitness $w = 1.011$ . How many years will it take to have doubled, assuming constant fitness (exponential growth)?

n_{t} = w^{t} n_{0}

2 n_{0} = w^{t} n_{0}

l n (2) = l n (w^{t})

t = l n (2) / l n (w)

t = l n (2) / l n (1.011) = 63.359

It will take 63.36 years to double thepopulation.

Discussion

Total Fertility rate

$w \approx T F R / 2.1$ is the relation between TFR (average births per woman) and individual fitness. The reason why it is divided by 2.1 instead of 2 (the number we would expect given there are equal numbers of men and women) is to account for rates of infant mortality and other conditions that would prevent their offspring from reproducing, account for women which don't have children, as well as the fact that women slightly outnumber men in the global population.

If the total individual fitness is 1.0 (replacement level), TFR will be 2.1. This means that in a population which is stable, not growing or shrinking, each woman will have 2.1 successful children on average, enough to replace themself and the other parent and ensure that there are enough offspring to also reproduce themselves.

Global Variation in fitness

Country	TFR	$w$
Finland	1.32	0.629
South Korea	0.7	0.333
Niger	6.6	3.143
Assuming no migration, the populations of Finland and South Korea will decline and Niger will increase. One possible reason for the lower birth rates in these countries is access to contraceptives. Another reason is that in many countries in the Global South like Niger with small scale agriculture based economies, having children is imporant for increasing economic output of a household once the children can work, while in countries like Finland and South Korea having children is an economic drain on the household.

Question 2. E. coli division

E. coli divides into two every 20 minutes under ideal conditions. Each individual is 1 cubic micrometer in volume. If there are no limiting resources, how long will it take for a single E. coli individual fill up the volume of the observable universe, $4 \times 10^{80} m^{3}$ ?

$n$ = number of bacteria needed to fill the universe = $4 \times 10^{80} / 10^{- 18} = 4 \times 10^{98}$
$n_{0} = 1$
$w = 2$
$t = 3 x$ , where $x$ is the number of hours required, since we are looking at 20-minute (1/3 hour) periods.

n = w^{3 x} n_{0}

l n (n / n_{0}) = 3 x * l n (w)

x = \frac{l n (4 \times 10^{98})}{3 l n (2)} = 109.18 hours or 4.5 days

This surprising answer shows that exponential growth can cause massive amounts of growth over orders of magnitude.

In nature, we don't observe this due to limited resources which introduces negative density dependence, competition for territory reduces effective fecundity, and bateria produce waste chemicals which kill nearby bacteria, reducing survivability.

Question 3: Beverton-Holt model

Based on the individual fitness equation $w = s + f_{e}$ , if effective fecundity is density-dependent according to the Beverton-Holt model but survivability is not, we replace $f_{e}$ with the Beverton Holt formula to get the following expression:
(i)

w = s + \frac{f}{1 + γ n}

At the stable equilibrium $w = 1$ , we have the following population size $n^{*}$ : $$1=s+{f \over 1+\gamma n^*}$$

1 + γ n^{*} = \frac{f}{1 - s}

γ n^{*} = \frac{f}{1 - s} - 1

γ n^{*} = \frac{f - (1 - s)}{1 - s}

(ii) So, the stable equilibrium population size is:

n^{*} = \frac{f - 1 + s}{γ (1 - s)}

(iii) As survival probability increases, the equilibirum population size also increases (since it is negative in the denominator and positive in the numerator). Here is an example using optimal fecundity $f$ = 3 and density dependence $γ$ = 0.1 . Here, the x-axis represents survival $s$ and the y-axis represents optimal population size $n^{*}$ :

left = -0.2
right = 1
bottom = 19
top = 150
---
f = 3
c = 0.1
y= (f +x-1)/(c(1-x))

Why does equilibrium population size increase with survival?

In our scenario, survival is not density dependent, only fecundity. Therefore, we are assuming that the effects of competition only reduce the amount of offspring an individual produces, not its average lifespan. If the average lifespan increases, this then means that more individuals exist in the population, but they all will not be able to reproduce as much. Therefore, a stable population must be larger in order to produce enough offspring to replace itself.

What are the assumptions of the model?

Here we assume that:
- It takes time for a predator to search for and handle prey
- There is a limited amount of prey per individual, so the amount of prey is inversely proportional to population size $n$ .
- There is no density-dependence associated with survival.
- Competition for space is not an important factor.
Something that might fit this scenario well is a predator searching for a limited amount of prey, like a bird hunting for insects or a fox hunting for mice, or a fish.
Something that wouldn't fit this scenario is a plant which is not in competition for nutrients, but is more limited by space, or a large grazing mammal like a buffalo in a fertile plain where there is no limit to grasses.

What is the relationship between effective births and deaths at equilibrium?

At equilibrium, the amount of effective births is approximately equal to the amount of deaths, meaning that the population does not change.

In terms our fitness equation $w = s + f_{e}$ , effective fecundity $f_{e}$ is related to the number of effective births per indvidual ( $births / n$ ) while survivorship $s$ represents the reverse of the probability of death of the individual (or $1 - deaths / n$ ). We can thus reformat our equation as:

w = (1 - \frac{deaths}{n}) + (\frac{births}{n})

w = 1 + \frac{births - deaths}{n}

When deaths and births are equal, we thus end up with individual fitness $w = 1$ , our equilibrium condition.

Density-dependence parameter variation between environments.

In an environment with very limited food like a desert, we would expect higher density dependence $γ$ . In a species that takes a long time to handle food, or that is not very efficient at passing on the energy to its offspring, we would also see a higher density dependence.

Homework Exercise 4:

Write 100-150 words about this RadioLab podcast