BIOL1003 Lecture Notes - Stabilizing Selection, Adaptation, Coevolution

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Inspecting Natural Selection
NO natural selection is a syllogism
•If [ajo peises] ou
•The etai oseuees ust e tue
•Test  falsifiatio of ajo peises
Evolution by Natural selection a definition
If a population has a. Variation in a trait among individuals (replicating units) b. A consistent
relationship between that trait and fitness (survival, mating, fecundity) c. Heritability (parents
resemble offspring at least partly unrelated to their common environment) Then 1. There are
differences between individuals that are born and those that breed (within-generation effects) 2. If
the population is not at equilibrium, there are predictable differences between parents and offspring
(between-generation effects)
What things can be naturally selected?
Thik of ods: Copied fo oe peso to the et. “oetie utate –I ask, but some people
aks.
Question: What is the analogue of genotype and phenotype when it comes to, say, language or
memes (ideas)?
This is quite a trick uestio….
Fitess : the epodutie suess of a idiidual ith a patiula pheotpe •Copoets of
fitness: Survival to reproductive age Mating success Fecundity
•‘elatie fitess: fitess of a geotpe stadadized  opaiso to other genotypes
The sum of all allele frequencies is 1 p + q = 1.0 In Hardy-Weinberg equilibrium, the genotype
frequencies are always: homozygote dominant = p2 heterozygote = 2pq homozygote recessive = q2
The sum of all genotype frequencies is 1 p2+ 2pq + q2= 1.0 Frequency of each genotype (& allele)
The sum of all allele frequencies is 1 p + q = 1.0 In Hardy-Weinberg equilibrium, the genotype
frequencies are always: homozygote dominant = p2 * Wpp heterozygote = 2pq * Wpq homozygote
recessive = q2 * Wqq The sum of all genotype frequencies is 1 p2+ 2pq + q2= 1.0
Contribution of alleles to fitness
•Aeage eess fitess of a allele:
difference between average fitness of individuals with allele vs. those without
Fitess otiutio of heteozgotes ad homozygotes for the allele, how many individuals of each
ae peset ad ho a offspig eah otiutes to the et geeatio
diffeee etee aeage fitess of idiiduals ith allele s. those ithout Δp = p  aA/ϖ)
P t+1 = p t ( 1 + (aA1/ϖ))
Evolution by natural selection a definition If a population has a. Variation in a trait among individuals
(replicating units) b. A consistent relationship between that trait and fitness (survival, mating,
fecundity) c. Heritability (parents resemble offspring at least partly unrelated to their common
environment) Then 1. There are differences between individuals that are born and those that breed
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Document Summary

No natural selection is a syllogism: if [(cid:373)ajo(cid:396) p(cid:396)e(cid:373)ises] o(cid:272)(cid:272)u(cid:396, the(cid:374) (cid:272)e(cid:396)tai(cid:374) (cid:272)o(cid:374)se(cid:395)ue(cid:374)(cid:272)es (cid:373)ust (cid:271)e t(cid:396)ue, test (cid:271)(cid:455) falsifi(cid:272)atio(cid:374) of (cid:373)ajo(cid:396) p(cid:396)e(cid:373)ises. Thi(cid:374)k of (cid:449)o(cid:396)ds: copied f(cid:396)o(cid:373) o(cid:374)e pe(cid:396)so(cid:374) to the (cid:374)e(cid:454)t. o(cid:373)eti(cid:373)e (cid:858)(cid:373)utate(cid:859) i ask, but some people (cid:858)aks(cid:859). Fit(cid:374)ess (cid:894)(cid:449)(cid:895): the (cid:396)ep(cid:396)odu(cid:272)ti(cid:448)e su(cid:272)(cid:272)ess of a(cid:374) i(cid:374)di(cid:448)idual (cid:449)ith a pa(cid:396)ti(cid:272)ula(cid:396) phe(cid:374)ot(cid:455)pe co(cid:373)po(cid:374)e(cid:374)ts of fitness: survival to reproductive age mating success fecundity: elati(cid:448)e fit(cid:374)ess: fit(cid:374)ess of a ge(cid:374)ot(cid:455)pe sta(cid:374)da(cid:396)dized (cid:271)(cid:455) (cid:272)o(cid:373)pa(cid:396)iso(cid:374) to other genotypes. The sum of all allele frequencies is 1 p + q = 1. 0 in hardy-weinberg equilibrium, the genotype frequencies are always: homozygote dominant = p2 heterozygote = 2pq homozygote recessive = q2. The sum of all genotype frequencies is 1 p2+ 2pq + q2= 1. 0 frequency of each genotype (& allele) P t+1 = p t ( 1 + (aa1/ )) Evolution by natural selection a definition if a population has a. Variation in a trait among individuals (replicating units) b.

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