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Analysis of Gray SquirrelPopulation in Northern West Virginia BIO-411 Shreya Ray 20091069
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1. Find empirical data to construct the projection (Leslie) matrixfor a particular population.The following data represents a cohort of 530 female gray squirrels (Sciurus carolinensis) from a populationin 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
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Using the values of sx and bx we shall now construct a Leslie projection matrix for the female gray squirrelpopulation: 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 Which equals: 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
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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 any arbitrary population distribution, say, 20 squirrels that are much less than a year old, and going on for, say, 9 years. 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 the population at year (t-1). Thus, every year, the population multiplies itself with an amount λ. Thus, if λ > 1, the population 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, that the population growth rate has got ‘stabilised’ and now the population will grow geometrically from here with a constant ratio λ, keeping the age structure constant from now on. We will eventually reach the same result no matter what arbitrary initial population we start with. For the female gray squirrel population in Northern West Virginia, λ turns out to be ≈ 1.2 Since 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 = 0.18 100 90 y = 16.947e0.1817x 80 R² = 0.9829 1.4 70 population (N) 1.2 60 1 50λ (lambda) 0.8 40 0.6 30 0.4 20 0.2 10 0 0 0 2 4 6 8 10 0 2 4 6 8 10 Year (t) Year (t)
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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: ",distThe output in the terminal upon entering some random values for initial population distribution converges tothe same values for λ.
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3. Assuming that the projection matrix remains the same fromone year to another, what is the stable age structure that thepopulation is eventually expected to arrive at?The function described in #2 also calculates the stable age structure, along with the lambda value. Nomatter what initial population distribution we start with, the ratios of population numbers in each agegroup 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 5 Age of squirrel 4 3 2 1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 Proportion of population
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4. Look for empirical data for the age structure (at any givenpoint of time) of your population. If this is not available,assume that the proportion of the population at different agesis given by the survivorship curve (why is this a goodassumption?). Using this as the initial composition of thepopulation, simulate the distribution of individuals in thedifferent age classes for a number of years. Do they approachthe stable age structure you calculated in #3?A population where age distributions are given by the survivorship curve represents a population that hasstabilised at its carrying capacity, hence such a population distribution is a reasonable assumption to beginwith. This phenomenon is discussed in more detail in #6. x nx log10nx Survivorship curve 3 2.724 2.5 0 530 2 log10nx 1 159 2.201 1.5 2 80 1.903 1 0.5 3 48 1.681 0 4 21 1.322 0 1 2 3 4 5 6 5 5 0.699 Age (x) From the survivorship curve the initial composition of the population is approximated by the matrix [2.723,2.201,1.903,1.681,1.322,0.699] (decimals values are allowed as this is just a hypothetical initial composition)
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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….
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5. Calculate the life expectancy at birth and the generationtime for this population.The life expectancy at birth for a female gray squirrel is given by the formula: 1.59 yearsAnd,The generation time is given by the following formula: = (1.4/2.63) = 1.878571 years
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6. Given what you know about the ecology of this population,do you expect the prediction from #2 to hold? i.e., why may thepopulation continue to grow/decline as expected, or to not doso?#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. 1200 population 1000 Exponential time 800 growth Carrying Capacity upon increasing food 600 availability Initial population 400 200 Carrying Capacity 0
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7. Suppose a mutant genotype leads to twice the fecundity atthe age of half the life-expectancy at birth, but half the survivalprobability 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, thefecundity gets doubled but the survival probability to the next age class gets halved. ( We are assuming allother 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 atage 1, it is 2. We don’t have information about the fecundity and survival for all the intermediate ages, sowe’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: 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 0Running the simulation with the new values, we have:λ ≈ 0.96 hence, growth rate of the mutant population r = ln (λ) = ln (0.96) = -0.04Since growth rate is negative, these mutants will slowly disappear from the population and thus they willnot be selected.
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BibliographyThe life history values about the female grey squirrels of Northern West Virginia have been taken from thebook: 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 byDonald 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 – GraySquirrels reproduce in a 2-acre enclosureThe Vital Statistics of an Unexploited Gray Squirrel Population, by F. S. Barkalow, R. B. Hamilton, R. F. Soots,The Journal of Wildlife Management, 1970I am attaching all the above, except the book Elements of Ecology. I am also attaching the excel sheet whereI have done most of the calculation, analysis, and graph-plotting, plus I am attaching a short Kit Kat addfeaturing two squirrels. ~END~
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