POPULATION GENETICS Predicting inheritance in a population 2016

POPULATION GENETICS Predicting inheritance in a population © 2016 Paul Billiet ODWS

Predictable patterns of inheritance in a population so long as… n n n there is no genetic drift The population is large enough not to show the effects of a random loss of alleles by chance events the mutation rate at the locus of the gene being studied is not significantly high mating between individuals is random no gene flow between neighbouring populations New individuals are not gained by immigration or lost by emigration the gene’s allele has no selective advantage or disadvantage (no natural selection). © 2016 Paul Billiet ODWS

SUMMARY n n n Genetic drift Mutation Mating choice Migration Natural selection All can affect the transmission of genes from generation to generation Genetic Equilibrium If none of these factors is operating then the relative proportions of the alleles (the ALLELE FREQUENCIES) will be constant. © 2016 Paul Billiet ODWS

THE HARDY WEINBERG PRINCIPLE Step 1 n Calculating the allele frequencies from the genotype frequencies n Easily done for codominant alleles (each genotype has a different phenotype). © 2016 Paul Billiet ODWS

Iceland Population n 313 337 (2007 est) Area n 103 000 km 2 Distance from mainland Europe n 970 km © 2016 Paul Billiet ODWS

Example Icelandic population: The MN blood group Sample Phenotypes Population Genotypes 747 Numbers Contribution to gene pool © 2016 Paul Billiet ODWS Type MN Type N Mm Mm Mm Mn Mn Mn 233 385 129 2 Mm alleles person 1 Mm allele person 1 Mn allele person 2 Mn alleles person

MN blood group in Iceland Total Mm alleles = (2 x 233) + (1 x 385) = 851 Total Mn alleles = (2 x 129) + (1 x 385) = 643 Total of both alleles =1494 = 2 x 747 (humans are diploid organisms) Frequency of the Mm allele = 851/1494 = 0. 57 or 57% Frequency of the Mn allele = 643/1494 = 0. 43 or 43% © 2016 Paul Billiet ODWS

In general for a diallellic gene A and a (or Ax and Ay) If the frequency of the A allele = and the frequency of the a allele = Then p+q = 1 © 2016 Paul Billiet ODWS p q

Step 2 Using the calculated allele frequency to predict the EXPECTED genotypic frequencies in the NEXT generation OR n to verify that the PRESENT population is in genetic equilibrium. n © 2016 Paul Billiet ODWS

Assuming all the individuals mate randomly NOTE the allele frequencies are the gamete frequencies too SPERMS EGGS © 2016 Paul Billiet ODWS Mm 0. 57 Mn 0. 43 Mm 0. 57 Mm Mm 0. 32 Mm Mn 0. 25 Mn 0. 43 Mm Mn 0. 25 Mn Mn 0. 18

Close enough for us to assume genetic equilibrium Genotypes Expected frequencies Observed frequencies Mm Mm 0. 32 233 747 = 0. 31 Mm Mn 0. 50 385 747 = 0. 52 Mn Mn 0. 18 129 747 = 0. 17 © 2016 Paul Billiet ODWS

In general for a diallellic gene A and a (or Ax and Ay) Where the allele frequencies are p and q Then p + q = 1 and SPERMS A p a q A p AA p 2 Aa pq a q Aa pq aa q 2 EGGS © 2016 Paul Billiet ODWS

THE HARDY WEINBERG EQUATION So the genotype frequencies are: AA Aa aa = = = p 2 2 pq q 2 or p 2 + 2 pq + q 2 = 1 © 2016 Paul Billiet ODWS

DEMONSTRATING GENETIC EQUILIBRIUM Using the Hardy Weinberg Equation to determine the genotype frequencies from the allele frequencies may seem a circular argument. © 2016 Paul Billiet ODWS

Only one of the populations below is in genetic equilibrium. Which one? Population sample Genotypes Allele frequencies AA Aa aa 100 20 80 0 100 36 48 16 100 50 20 30 100 60 0 40 © 2016 Paul Billiet ODWS A a

Only one of the populations below is in genetic equilibrium. Which one? Population sample Genotypes Allele frequencies AA Aa aa A a 100 20 80 0 0. 6 0. 4 100 36 48 16 0. 4 100 50 20 30 0. 6 0. 4 100 60 0 40 0. 6 0. 4 © 2016 Paul Billiet ODWS

Only one of the populations below is in genetic equilibrium. Which one? Population sample Genotypes Allele frequencies AA Aa aa A a 100 20 80 0 0. 6 0. 4 100 36 48 16 0. 4 100 50 20 30 0. 6 0. 4 100 60 0 40 0. 6 0. 4 © 2016 Paul Billiet ODWS

SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM haemoglobin gene n Normal allele Hb. N n Sickle allele Hb. S Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Genotypes Hb. N Hb. S Observed frequencies 0. 56 0. 4 0. 04 Expected frequencies © 2016 Paul Billiet ODWS Alleles Hb. N Hb. S

SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM haemoglobin gene n Normal allele Hb. N n Sickle allele Hb. S Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Genotypes Hb. N Hb. S Observed frequencies 0. 56 0. 4 0. 04 0. 76 0. 24 Expected frequencies 0. 58 0. 36 0. 06 © 2016 Paul Billiet ODWS Alleles

SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes Hb. N Hb. S Observed frequencies 0. 9075 0. 09 0. 0025 Expected frequencies © 2016 Paul Billiet ODWS

SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes Hb. N Hb. S Observed frequencies 0. 9075 0. 09 0. 0025 0. 95 0. 05 Expected frequencies 0. 9025 0. 095 0. 0025 © 2016 Paul Billiet ODWS

RECESSIVE ALLELES EXAMPLE ALBINISM IN THE BRITISH POPULATION Frequency of the albino phenotype = 1 in 20 000 or 0. 00005 © 2016 Paul Billiet ODWS

A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q Phenotypes Genotypes Hardy Weinberg frequencies Normal AA p 2 Normal Aa 2 pq Albino aa q 2 © 2016 Paul Billiet ODWS Observed frequencies

A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q Phenotypes Genotypes Hardy Weinberg frequencies Normal AA p 2 Normal Aa 2 pq Albino aa q 2 © 2016 Paul Billiet ODWS Observed frequencies 0. 99995 0. 00005

Albinism gene frequencies Normal allele Albino allele © 2016 Paul Billiet ODWS =A=p=? =q=?

Albinism gene frequencies Normal allele Albino allele © 2016 Paul Billiet ODWS =A=p=? = q = (0. 00005) = 0. 007 or 0. 7%

HOW MANY PEOPLE IN BRITAIN ARE CARRIERS FOR THE ALBINO ALLELE (Aa)? a allele = 0. 007 A allele But p+q Therefore p =q =p =1 = 1 - q = 1 – 0. 007 = 0. 993 or 99. 3% The frequency of heterozygotes (Aa) = 2 pq = 2 x 0. 993 x 0. 007 = 0. 014 or 1. 4% © 2016 Paul Billiet ODWS

Heterozygotes for rare recessive alleles can be quite common n Genetic inbreeding leads to rare recessive n n mutant alleles coming together more frequently Therefore outbreeding is better Outbreeding leads to hybrid vigour. © 2016 Paul Billiet ODWS

Example: Rhesus blood group in Europe What is the probability of a woman who knows she is rhesus negative (rhrh) marrying a man who may put her child at risk (rhesus incompatibility Rh– mother and a Rh+ fetus)? © 2016 Paul Billiet ODWS

Rhesus blood group A rhesus positive foetus is possible if the father is rhesus positive Rh. Rh x rhrh 100% chance Rhrh x rhrh 50% chance © 2016 Paul Billiet ODWS

Rhesus blood group Rhesus positive allele is dominant Rh Frequency = p Rhesus negative allele is recessive rh Frequency = q Frequency of rh allele = 0. 4 = q If p+q=1 Therefore Rh allele = p = 1 – q = 1 – 0. 4 = 0. 6 © 2016 Paul Billiet ODWS

Rhesus blood group n n Frequency of the rhesus positive phenotype = Rh. Rh + Rhrh = p 2 + 2 pq = (0. 6)2 + (2 x 0. 6 x 0. 4) = 0. 84 or 84% © 2016 Paul Billiet ODWS

Rhesus blood group Phenotypes Genotypes Hardy Weinberg frequencies Rhesus positive Rh. Rh p 2 Observed frequencies 0. 84 Rhesus positive Rhrh 2 pq Rhesus negative rhrh q 2 n n 0. 16 Therefore, a rhesus negative, European woman in Europe has an 84% chance of having husband who is rhesus positive… of which 36% will only produce rhesus positive children and 48% will produce rhesus positive child one birth in two. © 2016 Paul Billiet ODWS