Life tables Age-specific probability statistics Force of mortality q x Survivorship l x l y / l x =...

Life tables

Age-specific probability statistics

Force of mortality qx

Survivorship lx

ly / lx = probability of living from age x to age y

Fecundity mx

Realized fecundity at age x = lxmx

Net reproductive rate R0 = lxmx

Generation time T = xlxmx

Reproductive value vx = (lt / lx ) mt

Ex = Expectation of further life

T, Generation time = average time from one gener-

ation to the next (average time from egg to egg)

vx = Reproductive Value = Age-specific expectation of all future offspring

p.143, right hand equation“dx” should be “dt”

vx = mx + (lt / lx ) mt

Residual reproductive value = age-specific

expectation of offspring in distant future

vx* = ( lx+1 / lx ) vx+1

Illustration of Calculation of Ex, T, R0, and vx in a Stable Population with Discrete Age Classes_____________________________________________________________________

Age Expectation Reproductive Weighted of Life Value

Survivor- Realized by Realized Ex vx

Age (x) ship Fecundity Fecundity Fecundity lx mx lxmx x lxmx

_____________________________________________________________________0 1.0 0.0 0.00 0.00 3.40 1.001 0.8 0.2 0.16 0.16 3.00 1.252 0.6 0.3 0.18 0.36 2.67 1.403 0.4 1.0 0.40 1.20 2.50 1.654 0.4 0.6 0.24 0.96 1.50 0.655 0.2 0.1 0.02 0.10 1.00 0.106 0.0 0.0 0.00 0.00 0.00 0.00Sums 2.2 (GRR) 1.00 (R0) 2.78 (T) _____________________________________________________________________E0 = (l0 + l1 + l2 + l3 + l4 + l5)/l0 = (1.0 + 0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 1.0 = 3.4 / 1.0E1 = (l1 + l2 + l3 + l4 + l5)/l1 = (0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 0.8 = 2.4 / 0.8 = 3.0E2 = (l2 + l3 + l4 + l5)/l2 = (0.6 + 0.4 + 0.4 + 0.2) / 0.6 = 1.6 / 0.6 = 2.67E3 = (l3 + l4 + l5)/l3 = (0.4 + 0.4 + 0.2) /0.4 = 1.0 / 0.4 = 2.5E4 = (l4 + l5)/l4 = (0.4 + 0.2) /0.4 = 0.6 / 0.4 = 1.5E5 = (l5) /l5 = 0.2 /0.2 = 1.0v1 = (l1/l1)m1+(l2/l1)m2+(l3/l1)m3+(l4/l1)m4+(l5/l1)m5 = 0.2+0.225+0.50+0.3+0.025 = 1.25 v2 = (l2/l2)m2 + (l3/l2)m3 + (l4/l2)m4 + (l5/l2)m5 = 0.30+0.67+0.40+ 0.03 = 1.40 v3 = (l3/l3)m3 + (l4/l3)m4 + (l5/l3)m5 = 1.0 + 0.6 + 0.05 = 1.65 v4 = (l4/l4)m4 + (l5/l4)m5 = 0.60 + 0.05 = 0.65v5 = (l5/l5)m5 = 0.1___________________________________________________________________________

Illustration of Calculation of Ex, T, R0, and vx in a Stable Population with Discrete Age Classes_____________________________________________________________________

Age Expectation Reproductive Weighted of Life Value

Survivor- Realized by Realized Ex vx

Age (x) ship Fecundity Fecundity Fecundity lx mx lxmx x lxmx

_____________________________________________________________________0 1.0 0.0 0.00 0.00 3.40 1.001 0.8 0.2 0.16 0.16 3.00 1.252 0.6 0.3 0.18 0.36 2.67 1.403 0.4 1.0 0.40 1.20 2.50 1.654 0.4 0.6 0.24 0.96 1.50 0.655 0.2 0.1 0.02 0.10 1.00 0.106 0.0 0.0 0.00 0.00 0.00 0.00Sums 2.2 (GRR) 1.00 (R0) 2.78 (T) _____________________________________________________________________E0 = (l0 + l1 + l2 + l3 + l4 + l5)/l0 = (1.0 + 0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 1.0 = 3.4 / 1.0E1 = (l1 + l2 + l3 + l4 + l5)/l1 = (0.8 + 0.6 + 0.4 + 0.4 + 0.2) / 0.8 = 2.4 / 0.8 = 3.0E2 = (l2 + l3 + l4 + l5)/l2 = (0.6 + 0.4 + 0.4 + 0.2) / 0.6 = 1.6 / 0.6 = 2.67E3 = (l3 + l4 + l5)/l3 = (error: extra terms) 0.4 + 0.4 + 0.2) /0.4 = 1.0 / 0.4 = 2.5E4 = (l4 + l5)/l4 = (error: extra terms) 0.4 + 0.2) /0.4 = 0.6 / 0.4 = 1.5E5 = (l5) /l5 = 0.2 /0.2 = 1.0v1 = (l1/l1)m1+(l2/l1)m2+(l3/l1)m3+(l4/l1)m4+(l5/l1)m5 = 0.2+0.225+0.50+0.3+0.025 = 1.25 v2 = (l2/l2)m2 + (l3/l2)m3 + (l4/l2)m4 + (l5/l2)m5 = 0.30+0.67+0.40+ 0.03 = 1.40 v3 = (l3/l3)m3 + (l4/l3)m4 + (l5/l3)m5 = 1.0 + 0.6 + 0.05 = 1.65 v4 = (l4/l4)m4 + (l5/l4)m5 = 0.60 + 0.05 = 0.65v5 = (l5/l5)m5 = 0.1___________________________________________________________________________

Stable age distribution

Stationary age distribution

Intrinsic rate of natural increase (per capita)

r = b – d

when birth rate exceeds death rate (b > d), r is positive

when death rate exceeds birth rate (d > b), r is negative

Euler’s implicit equation:

e-rx lxmx = 1

(solved by iteration)

If the Net Reproductive Rate R0 is near one,

r ≈ loge R0 /T

J - shaped exponential population growth

http://www.zo.utexas.edu/courses/THOC/exponential.growth.html

When R0 equals one, r is zero

When R0 is greater than one, r is positive

When R0 is less than one, r is negative

Maximal rate of natural increase, rmax

Instantaneous rate of change of N at time t

is total births (bN) minus total deaths (dN)

dN/dt = bN – dN = (b – d )N = rN

Nt = N0 ert (integrated version of dN/dt = rN)

log Nt = log N0 + log ert = log N0 + rt

log R0 = log 1 + rt (make t = T)

r = log or = er (is the finite rate of

increase)

Estimated Maximal Instantaneous Rates of Increase (rmax, per capita per day) and Mean Generation Times ( in days) for a Variety of Organisms___________________________________________________________________Taxon Species rmax Generation Time (T)-----------------------------------------------------------------------------------------------------Bacterium Escherichia coli ca. 60.0 0.014Protozoa Paramecium aurelia 1.24 0.33–0.50Protozoa Paramecium caudatum 0.94 0.10–0.50Insect Tribolium confusum 0.120 ca. 80Insect Calandra oryzae 0.110(.08–.11) 58Insect Rhizopertha dominica 0.085(.07–.10) ca. 100Insect Ptinus tectus 0.057 102Insect Gibbum psylloides 0.034 129Insect Trigonogenius globulosus 0.032 119Insect Stethomezium squamosum 0.025 147Insect Mezium affine 0.022 183Insect Ptinus fur 0.014 179Insect Eurostus hilleri 0.010 110Insect Ptinus sexpunctatus 0.006 215Insect Niptus hololeucus 0.006 154Mammal Rattus norwegicus 0.015 150Mammal Microtus aggrestis 0.013 171Mammal Canis domesticus 0.009 ca. 1000Insect Magicicada septendecim 0.001 6050Mammal Homo “sapiens” (the sap) 0.0003 ca. 7000

__________________________________________________________________ _

Inverserelationshipbetween rmax

and generation time, T

Demographic and Environmental Stochasticity

random walks, especially important in small populations

Evolution of Reproductive Tactics

Semelparous versus Interoparous

Big Bang versus Repeated Reproduction

Reproductive Effort (parental investment)

Age of First Reproduction, alpha,

Age of Last Reproduction, omega,

Mola mola (“Ocean Sunfish”) 200 million eggs!

Poppy (Papaver rhoeas)produces only 4 seeds whenstressed, but as manyas 330,000 under idealconditions

Indeterminant Layers

How much should an organism invest in any given act of reproduction? R. A. Fisher (1930) anticipated this question long ago:

‘It would be instructive to know not only by what physiological mechanism a just apportionment is made between the nutriment devoted to the gonads and that devoted to the rest of the parental organism, but also what circumstances in the life history and environment would render profitable the diversion of a greater or lesser share of available resources towards reproduction.’ [Italics added for emphasis.]

Reproductive Effort

Ronald A. Fisher

Asplanchna (Rotifer)

Trade-offs between present progenyand expectation of future offspring

Iteroparous organism

Semelparous organism

Patterns in Avian Clutch SizesAltrical versus Precocial

Patterns in Avian Clutch SizesAltrical versus Precocial

Nidicolous vs. NidifugousDeterminant vs. Indeterminant Layers

N = 5290 Species

Patterns in Avian Clutch SizesOpen Ground Nesters Open Bush Nesters Open Tree Nesters

Hole Nesters

MALE (From: Martin and Ghalambor 1999)

Patterns in Avian Clutch SizesClassic Experiment: Flickers usually lay 7-8 eggs, but in an egg removal experiment, a female flicker

laid 61 eggs in 63 days

Great Tit Parus major

David Lack

Parus major

European Starling, Sturnus vulgaris

Chimney Swift, Apus apus

Seabirds (N. Philip Ashmole)

Boobies, Gannets, Gulls, Petrels, Skuas, Terns, Albatrosses

Delayed sexual maturity, Small clutch size, Parental care

Albatross Egg Addition Experiment

Diomedea immutabilis

An extra chick added to eachof 18 nests a few days afterhatching. These nests with twochicks were compared to 18 othernatural “control” nests with onlyone chick. Three months later, only 5 of the 36 experimental chicks survived from the nests with 2 chicks, whereas 12 of the 18 chicks from single chick nests were still alive. Parents could not find food enough to feed two chicks and most starved to death.

Latitudinal Gradients in Avian Clutch Size

Latitudinal Gradients in Avian Clutch Size

Daylength Hypothesis

Prey Diversity Hypothesis

Spring Bloom or Competition Hypothesis

Latitudinal Gradients in Avian Clutch Size

Nest Predation Hypothesis Alexander Skutch ––––––>

Latitudinal Gradients in Avian Clutch Size

Hazards of Migration Hypothesis

Falco eleonora

Evolution of Death Rates

Senescence, old age, genetic dustbin

Medawar’s Test Tube Model

p(surviving one month) = 0.9

p(surviving two months) = 0.92

p(surviving x months) = 0.9x

recession of time of expression of the overt effects of a

detrimental allele

precession of time of expression of the effects of a

beneficial allele

Peter Medawar

Age Distribution ofMedawar’s test tubes

Percentages of people with lactose intolerance

Joint Evolution of Rates of Reproduction and Mortality

Donald Tinkle

Sceloporus

J - shaped exponential population growth

http://www.zo.utexas.edu/courses/THOC/exponential.growth.html

Instantaneous rate of change of N at time t

is total births (bN) minus total deaths (dN)

dN/dt = bN – dN = (b – d )N = rN

Nt = N0 ert (integrated version of dN/dt = rN)

log Nt = log N0 + log ert = log N0 + rt

log R0 = log 1 + rt (make t = T)

r = log or = er (is the finite rate of

increase)

Once, we were surrounded by wilderness and wild animals,But now, we surround them.