Analysis of Gray Squirrel Population in Northern

West Virginia

BIO-411

Shreya Ray

20091069

1. Find empirical data to construct the projection (Leslie) matrix for a particular population.

The following data represents a cohort of 530 female gray squirrels (Sciurus carolinensis) from a population in Northern West Virginia that was the focus of a decade-long study.

x nx bx

0 530 0.0

1 159 2.0

2 80 3.0

3 48 3.0

4 21 2.0

5 5 0.0

• x = age of squirrel in years• nx= number of individuals from the original cohort that are alive at the specified age (x).• bx= age-specific birth rates (considering only the number of females produced)

We now construct the cohort life-table to get the values of sx= age-specific survival rate.

x nx lx = (nx / n0) dx = (nx - nx+1) qx = (dx / nx) sx = 1 - qx

0 530 1.0 371 0.7 0.3

1 159 0.3 79 0.5 0.5

2 80 0.15 32 0.4 0.6

3 48 0.09 27 0.55 0.45

4 21 0.04 16 0.75 0.25

5 5 0.01 5 1.0 0.0

•lx= probability at birth of surviving to any given age•dx= age-specific mortality•qx= age-specific mortality rate

2*0.3 3*0.5 3*0.6 2*0.45 0*0.25 0

0.3 0 0 0 0 0

0 0.5 0 0 0 0

0 0 0.6 0 0 0

0 0 0 0.45 0 0

0 0 0 0 0.25 0

Using the values of sx and bx we shall now construct a Leslie projection matrix for the female gray squirrel population:

0.6 1.5 1.8 0.9 0 0

0.3 0 0 0 0 0

0 0.5 0 0 0 0

0 0 0.6 0 0 0

0 0 0 0.45 0 0

0 0 0 0 0.25 0

Which equals:

2. From the matrix, infer if the population is growing or declining. What is the population growth rate?

Using our projection matrix, we may construct a projection table for female grey squirrels starting with anyarbitrary population distribution, say, 20 squirrels that are much less than a year old, and going on for, say, 9years.

Year 0 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 Year 7 Year 8 Year 9

0 year old 20 12 16.2 20.52 24.462 29.2572 35.41482 42.68149 51.42643 61.99436

1 year old 0 6 3.6 4.86 6.156 7.3386 8.77716 10.62445 12.80445 15.42793

2 year old 0 0 3 1.8 2.43 3.078 3.6693 4.38858 5.312223 6.402224

3 year old 0 0 0 1.8 1.08 1.458 1.8468 2.20158 2.633148 3.187334

4 year old 0 0 0 0 0.81 0.486 0.6561 0.83106 0.990711 1.184917

5 year old 0 0 0 0 0 0.2025 0.1215 0.164025 0.207765 0.247678

Total population 20 18 22.8 28.98 34.938 41.8203 50.48568 60.89118 73.37472 88.44445

λ 0.9 1.266667 1.271053 1.20559 1.196986 1.207205 1.206108 1.205014 1.20538

Where, lambda (λ) is the ‘finite multiplication rate’, given by the quotient of the population at year t and thepopulation at year (t-1). Thus, every year, the population multiplies itself with an amount λ. Thus, if λ > 1, thepopulation is growing since dead members are being replaced by those born recently and even more. If λ < 1,the population is declining, since the new-born are unable to replace the dead.From our table, we observe that λ approaches a constant value as more and more years pass by. We say, thatthe population growth rate has got ‘stabilised’ and now the population will grow geometrically from herewith a constant ratio λ, keeping the age structure constant from now on. We will eventually reach the sameresult no matter what arbitrary initial population we start with.

For the female gray squirrel population in Northern West Virginia, λ turns out to be ≈ 1.2Since λ > 1 , we may infer that the population is growing.

Thus, Nt+1 = λt Nt , or, Nt+1 = (1.2)t Nt

Comparing this to the exponential population growth equation: Nt+1 = ert Nt (r=population growth rate)We have λt = ert , or, λ = er , or, r = ln (λ)So,r = ln (λ) = ln (1.2) ≈ 0.18

Hence, the population growth rate is = 0.18

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 2 4 6 8 10

λ(l

amb

da)

Year (t)

y = 16.947e0.1817x

R² = 0.9829

0

10

20

30

40

50

60

70

80

90

100

0 2 4 6 8 10

po

pu

lati

on

(N

)

Year (t)

The population projection table in the previous page was made using MS Excel.I’ve made a simulation in python too and I’ve got similar results. Following is the program code:

def go():

L=[[0.6,1.5,1.8,0.9,0.0,0.0],[0.3,0.0,0.0,0.0,0.0,0.0],[0.0,0.5,0.0,0.0,0.0,0.0],[0.0,0.0,0.6,

0.0,0.0,0.0],[0.0,0.0,0.0,0.45,0.0,0.0],[0.0,0.0,0.0,0.0,0.25,0.0]]

t=1

N0=[0.0]*6

print "enter the elements of the matrix N0“

for i in range(0,6):

N0[i]=float(raw_input("N0 for x="+str(i)+":"))

while(t>0):

Ntemp=N0

n=int(raw_input("How many times do you want to do the iteration? (enter 0 to quit) "))

if(n==0):

break

else:

pop=[0,0]

dist=[0,0,0,0,0,0]

for k in range(0,n):

Nt=[0.0]*6

for i in range(0,6):

for j in range(0,6):

Nt[i]+=L[i][j]*Ntemp[j]

if(k==n-2):

pop[0]+=Nt[i]

if(k==n-1):

pop[1]+=Nt[i]

dist[i]=Nt[i]

Ntemp=Nt

if(n==1):

for i in range(0,6):

pop[0]+=N0[i]

print Nt

print "lambda= ",pop[1]/pop[0]

for i in range(0,6):

dist[i]=dist[i]/pop[1]

print "Stable age distribution: ",dist

The output in the terminal upon entering some random values for initial population distribution converges to the same values for λ.

3. Assuming that the projection matrix remains the same from one year to another, what is the stable age structure that the population is eventually expected to arrive at?

The function described in #2 also calculates the stable age structure, along with the lambda value. No matter what initial population distribution we start with, the ratios of population numbers in each age group to the total population for that generation approach constant values.

Year 0 Year 9 Distribution

0 year old 20 61.99436 0.700941297

1 year old 0 15.42793 0.174436382

2 year old 0 6.402224 0.072386952

3 year old 0 3.187334 0.036037693

4 year old 0 1.184917 0.013397298

5 year old 0 0.247678 0.002800377

Total population 20 88.44445

λ 1.20538

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

1

2

3

4

5

Proportion of population

Age

of

squ

irre

l

4. Look for empirical data for the age structure (at any given point of time) of your population. If this is not available, assume that the proportion of the population at different ages is given by the survivorship curve (why is this a good assumption?). Using this as the initial composition of the population, simulate the distribution of individuals in the different age classes for a number of years. Do they approach the stable age structure you calculated in #3?

A population where age distributions are given by the survivorship curve represents a population that has stabilised at its carrying capacity, hence such a population distribution is a reasonable assumption to begin with. This phenomenon is discussed in more detail in #6.

x nx log10nx

0 530 2.724

1 159 2.201

2 80 1.903

3 48 1.681

4 21 1.322

5 5 0.699

0

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5 6

log 1

0nx

Age (x)

Survivorship curve

From the survivorship curve the initial composition of the population is approximated by thematrix [2.723,2.201,1.903,1.681,1.322,0.699] (decimals values are allowed as this is just ahypothetical initial composition)

Running these initial conditions through our simulation, we realise that the population composition,nevertheless, approaches the stable age structure calculated in #3, in about 9 years….

5. Calculate the life expectancy at birth and the generation time for this population.

= (1.4/2.63) = 1.878571 years

The life expectancy at birth for a female gray squirrel is given by the formula:

1.59 years

And,

The generation time is given by the following formula:

6. Given what you know about the ecology of this population, do you expect the prediction from #2 to hold? i.e., why may the population continue to grow/decline as expected, or to not do so?

#2 predicts an exponential population growth for the gray squirrels in Northern West Virginia.

However, this does not happen in reality.

In the paper, ‘Regulation of a Northern Gray Squirrel Population’, Donald C. Thompson shares hisamazement at the fact that the gray squirrel population in the forests of Northern West Virginia haveremained almost constant over the past few years, with neither growth nor decline in the populationnumbers, although the number of births every year and the survival probabilities of the squirrels suggest anexponential growth in population.

Hence, in spite of using the correct life-history parameters like survival rate and fecundity to construct aLeslie matrix and project a population composition for some generations down the line, we are definitelymissing out on something crucial.

The hypothesis is that the difference between the theoretical and the observed population number is dueto the fact that rate of recruitment of the young into the breeding population has not been taken intoaccount, which is a direct manifestation of intra-specific competition among squirrels.

The number of young which recruit into the adult population is a function of direct mortality and emigrationlosses.

The social system of the gray squirrel is such that established individuals possess home ranges which arepartially shared with a limited number of recognized neighbors (Thompson, 1978). Hence, there is lessstrife between individuals who recognize each other and aggression is directed towards unknownindividuals- the young squirrels who are trying to establish themselves, and the immigrants . This results ina positive deviation in mortality from the expected values that were used to construct our Leslie matrix, aswell as an added factor : emigration of young squirrels who are unable to take the stress. Of course, there isimmigration too, but immigrants aren’t mostly successful (indirect mortality). These factors regulate thesquirrel population and maintain it at a constant carrying capacity.

This carrying capacity, however, is strongly affected by increased food availability, which reduces aggressivebehavior. Hence, in the presence of abundant resources, the carrying capacity increases so much that all weobserve is an exponential growth rate, much similar to our simulation.

0

200

400

600

800

1000

1200

Carrying Capacity

Carrying Capacity upon increasing food availabilityInitial

population

Exponential growth

time

population

7. Suppose a mutant genotype leads to twice the fecundity at the age of half the life-expectancy at birth, but half the survival probability to the next age class. Will this mutant be selected?

This means, at the age of half the life-expectancy at birth, i.e., at the age of 1.59/2 = 0.795 years, the fecundity gets doubled but the survival probability to the next age class gets halved. ( We are assuming all other factors remain the same.)

0.795 years comes within the age group of 0-1 years. At age 0, the fecundity of gray squirrel is zero, and at age 1, it is 2. We don’t have information about the fecundity and survival for all the intermediate ages, so we’ll approximate the age of half the life-expectancy at birth as 1 (0.795 ≈ 1).

Then, fecundity increases from 2 to 4 for this mutant squirrel.And survival probability to the next age class decreases from 0.5 to 0.25.

Then, the new population projection matrix will be:

Running the simulation with the new values, we have:λ ≈ 0.96 hence, growth rate of the mutant population r = ln (λ) = ln (0.96) = -0.04

Since growth rate is negative, these mutants will slowly disappear from the population and thus they will not be selected.

0.3 1.5 1.8 0.9 0 0

0.3 0 0 0 0 0

0 0.25 0 0 0 0

0 0 0.6 0 0 0

0 0 0 0.45 0 0

0 0 0 0 0.25 0

Bibliography

The life history values about the female grey squirrels of Northern West Virginia have been taken from the book: Elements of Ecology, 6th Edition, by Thomas M. Smith and Robert Leo Smith.

I have also cited the paper: Regulation of a Northern Grey Squirrel ( Sciurus carolinensis) Population by Donald C. Thompson, published in 1978 by The Ecological Society of America.

I have also referred to:

Ohio Division of Wildlife- Life History Notes- Gray Squirrel -Publication 95 (1099)

USDA Forest Service Research Note NE-174 – 1973 – Northeastern Forest Experiment Station – Gray Squirrels reproduce in a 2-acre enclosure

The Vital Statistics of an Unexploited Gray Squirrel Population, by F. S. Barkalow, R. B. Hamilton, R. F. Soots, The Journal of Wildlife Management, 1970

I am attaching all the above, except the book Elements of Ecology. I am also attaching the excel sheet where I have done most of the calculation, analysis, and graph-plotting, plus I am attaching a short Kit Kat add featuring two squirrels.

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