Some elements of population genetics

• Population genetics perspective of evolution = change in allele frequencies

• E.g., cystic fibrosis gene product a protein:

• Cystic fibrosis transmembrane conductance regulator

• Expressed in mucus membrane (lining) of lungs and intestine.

• Enables cells to ingest and destroy Pseudomonas aeruginosa bacteria.

• Function lost with homozygous loss of function alleles

>500 loss of functionmutations at this singlelocus

• Population genetics and evolution.• Traditional theoretical starting point = Hardy-

Weinberg equilibrium• Principal use: a mathematic model demonstrating

the stasis of genotypic and allele frequencies in populations in which evolution is not taking place.

• The inertia that must be overcome if evolution is to take place..

• Evolution can be verified by comparing genotype or allele frequencies of a real example to the H-W model.

• p2 + 2pq + q2

The gene pool concept

Assumptions for H-W equilibrium• 1. Mating is random• 2. Population is infinitely large • 3. Alleles are not added or lost from the population

– gene flow (emigration or immigration)– asymmetrical mutation

• 4. No selection• One gene, two allele system:• (p + q) x (p + q) = (p + q)2 = p2 + 2pq + q2 = 1.0• Generally, in biology courses, either genotypic frequencies or allele

frequencies are provided.• To be tested against what would be expected under H-W conditions.• Sometimes a rough estimate of genotype and allele frequencies can

be gotten from the phenotype.

• True when a phenotypic alternative is produced by homozygous

recessives

An example• An unusual population of deer mice (some are albino).• Random sample of 1,000 individuals• 4 individuals are albino• What is the frequency of A and a in this population? • It takes a/a to produce albino; therefore,• We are interested in this term: p2 + 2pq + q2

• q2 = 0.004.• q = √ 0.004 = 0.0013• p + q = 1; therefore p = 1 – 0.0013 = 0.9987• How many in this population are carriers of the• a allele?

• Another question with bearing on evolution:• Q: How much influence can natural selection have on

a phenotypic character?• A: Depends on how much of the variation is based on

genetic variation.• Phenotypic variation

– Genetic component

– Environmental component

• Broad sense heritability = proportion of phenotypic variation based on genetic variation.

• H = VG/(VG + VE )

Conchas Lake, San Miguel County, New Mexico. Shown are coefficients of variation (CV), variances (s2), means ± SE, ranges of variation, and (sample sizes). Means with * are not significantly different at α = 0.05. ____________________________________________________________________ Character tesselata tesselata neomexicana sexlineata

GPI genotyped___________________________________________________________________

GAB 90.2 ± 1.56* 91.1 ± 1.21* 75.5 ± 0.34 73.4 ± 0.72

82–104 (21) 81–104 (30) 73–80 (28) 65–83 (31)

s2 = 51.0 s2 = 44.1 s2 = 3.1 s2 = 15.8

CV = 7.9 CV = 7.3 CV = 2.3 CV = 5.4

GPI variationGPI ac and GPI abAffects GAB

CV = (s/mean) x 100 So we’ll use thisparthenogen

A. neomexicana A. sexlineataH of GAB?

A. tesselata2 hybrids

• x = a variate (a individual score for a character)

= the sample mean for the variable

• S2 = (x - )2 / N-1

• S = √s2

• VG = (VE + VG) – VE

• H = VG / (VE + VG)

• A. neotesselata : VE

• A. sexlineata : VE + VG

• GAB character

• A. neomexicana: N = 29; s2 = 3.744

• A. sexlineata: N = 31; s2 = 15.845

• H = 15.845 – 3.744 / 15.845 = 0.76