1. D4: The Hardy-Weinberg Principle
2 hours

2. D.4.1 Explain how the Hardy–Weinberg equation is derived.
TT = p2 = freq of homoz dom
Tt = pq = freq of heterozygote
tt = q2 = freq of homoz rec
Punnett Square... Tt x Tt
p2 + 2pq + q2 = 100% (1)
p = freq of dominant allele
q = freq of recessive allele

3. D.4.2 Calculate allele, genotype and phenotype frequencies for two alleles of a gene, using the Hardy–Weinberg equation.
Examples from class!

4. --a population is large --with random mating
D.4.3 State the assumptions made when the Hardy–Weinberg equation is used. It must be assumed that:
--a population is large
--with random mating
--constant allele frequency over time
This implies
--no allele-specific mortality
--no mutation
--no emigration
--no immigration

5. Predicting inheritance in a population
POPULATION GENETICS
Predicting inheritance in a population
© 2008 Paul Billiet ODWS

6. Predictable patterns of inheritance in a population so long as…
the population is large enough not to show the effects of a random loss of genes by chance events i.e. there is no genetic drift
the mutation rate at the locus of the gene being studied is not significantly high
mating between individuals is random (a panmictic population)
new individuals are not gained by immigration or lost be emigration i.e. there is no gene flow between neighbouring populations
the gene’s allele has no selective advantage or disadvantage
© 2008 Paul Billiet ODWS

7. SUMMARY 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 GENE FREQUENCIES) will be constant
© 2008 Paul Billiet ODWS

8. THE HARDY WEINBERG PRINCIPLE
Step 1
Calculating the gene frequencies from the genotype frequencies
Easily done for codominant alleles (each genotype has a different phenotype)
© 2008 Paul Billiet ODWS

9. Iceland Population 313 337 (2007 est) Area 103 000 km2
Distance from mainland Europe
970 km
Google Earth
© 2008 Paul Billiet ODWS

10. Example Icelandic population: The AB blood group
129
385
233
Numbers
747 BB AB AA
Genotypes
Type B
Type AB
Type A
Phenotypes
Sample
Population
2 Mn alleles per person
1 Mn allele per person
1 Mm allele per person
2 Mm alleles per person
Contribution to gene pool
© 2008 Paul Billiet ODWS

11. AB blood group in Iceland
Total A alleles = (2 x 233) + (1 x 385) = 851
Total B alleles = (2 x 129) + (1 x 385) = 643
Total of both alleles =1494
= 2 x 747
(humans are diploid organisms)
p=Frequency of the A allele = 851/1494 = or 57%
q=Frequency of the B allele = 643/1494 = or 43%
© 2008 Paul Billiet ODWS

12. In general for a diallellic gene A and a
If the frequency of the A allele = p
and the frequency of the a allele = q
Then p+q = 1
© 2008 Paul Billiet ODWS

13. Step 2
Using the calculated gene frequency to predict the EXPECTED genotypic frequencies in the NEXT generation OR
to verify that the PRESENT population is in genetic equilibrium
© 2008 Paul Billiet ODWS

14. Assuming all the individuals mate randomly
NOTE the ALLELE frequencies are the gamete frequencies too
SPERMS B A
B 0.43
A 0.57
EGGS AA AB
p*p= p2
p*q AB BB
q*q= q2
p*q
© 2008 Paul Billiet ODWS

15. Close enough for us to assume genetic equilibrium
Genotypes
Expected frequencies
Observed frequencies AA
p2 = 0.32
233  747 = 0.31 AB
2pq =0.50
385  747 = 0.52 BB
q2 =0.18
129  747 = 0.17
© 2008 Paul Billiet ODWS

16. In general for a diallellic gene A and a
Where the gene frequencies are p and q
Then p + q = 1
and
SPERMS
A p
a q
EGGS
AA p2
Aa pq
aa q2
© 2008 Paul Billiet ODWS

17. THE HARDY WEINBERG EQUATION
So the genotype frequencies are:
AA = p2
Aa = 2pq
aa = q2
or p2 + 2pq + q2 = 1
© 2008 Paul Billiet ODWS

18. DEMONSTRATING GENETIC EQUILIBRIUM
Using the Hardy Weinberg Equation to determine the genotype frequencies from the gene frequencies may seem a circular argument
© 2008 Paul Billiet ODWS

19. 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 36 48 16 50 30 60 40
© 2008 Paul Billiet ODWS

20. Only one of the populations below is in genetic equilibrium. Which one?40 60
100 30 20 50 16 48 36
0.4
0.6 80 a A aa Aa AA
Gene frequencies
Genotypes
Population sample
0.4
0.6
0.4
0.6
0.4
0.6
© 2008 Paul Billiet ODWS

21. Only one of the populations below is in genetic equilibrium. Which one?
Population sample
Genotypes
Gene frequencies AA Aa aa A a
100 20 80
0.6
0.4 36 48 16 50 30 60 40
© 2008 Paul Billiet ODWS

22. SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM
 haemoglobin gene
Normal allele HbN
Sickle allele HbS
Phenotypes
Normal
Sickle Cell Trait
Sickle Cell Anaemia
Alleles
Genotypes
HbNHbN
HbN HbS
HbS HbS
HbN
HbS
Observed frequencies
0.56
0.4
0.04
Expected frequencies
© 2008 Paul Billiet ODWS

23. SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM
 haemoglobin gene
Normal allele HbN
Sickle allele HbS
Expected frequencies
0.04
0.4
0.56
Observed frequencies
HbS
HbN
HbS HbS
HbN HbS
HbNHbN
Genotypes
Alleles
Sickle Cell Anaemia
Sickle Cell Trait
Normal
Phenotypes
0.24
0.76
0.06
0.36
0.58
© 2008 Paul Billiet ODWS

24. SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION
Phenotypes
Normal
Sickle Cell Trait
Sickle Cell Anaemia
Alleles
Genotypes
HbNHbN
HbN HbS
HbS HbS
HbN
HbS
Observed frequencies
0.9075
0.09
0.0025
Expected frequencies
© 2008 Paul Billiet ODWS

25. SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION
Expected frequencies
0.0025
0.09
0.9075
Observed frequencies
HbS
HbN
HbS HbS
HbN HbS
HbNHbN
Genotypes
Alleles
Sickle Cell Anaemia
Sickle Cell Trait
Normal
Phenotypes
0.09
0.91
0.0081
0.16
0.8281
© 2008 Paul Billiet ODWS

26. RECESSIVE ALLELES EXAMPLE ALBINISM IN THE BRITISH POPULATION
Frequency of the albino phenotype = 1 in or
© 2008 Paul Billiet ODWS

27. Hardy Weinberg frequencies
A = Normal skin pigmentation allele Frequency = p
a = Albino (no pigment) allele Frequency = q
Phenotypes
Genotypes
Hardy Weinberg frequencies
Observed frequencies
Normal AA p2 Aa
2pq
Albino aa q2
© 2008 Paul Billiet ODWS

28. Hardy Weinberg frequencies
A = Normal skin pigmentation allele Frequency = p
a = Albino (no pigment) allele Frequency = q
Phenotypes
Genotypes
Hardy Weinberg frequencies
Observed frequencies
Normal AA p2 Aa
2pq
Albino aa q2
Use p+q=1 to determine
© 2008 Paul Billiet ODWS

29. Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q = ?
© 2008 Paul Billiet ODWS

30. Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q = ?
If q2 = …then q = …
 ( ) = or 0.7%
© 2008 Paul Billiet ODWS

31. HOW MANY PEOPLE IN BRITAIN ARE CARRIERS FOR THE ALBINO ALLELE (Aa)?
a allele = = q
A allele = p
But p + q = 1
Therefore p = 1- q
= 1 – 0.007
= or 99.3%
The frequency of heterozygotes (Aa) = 2pq
= 2 x x 0.007
= or 1.4%
© 2008 Paul Billiet ODWS

32. Heterozygotes for rare recessive alleles can be quite common
Genetic inbreeding leads to rare recessive mutant alleles coming together more frequently
Therefore outbreeding is better
Outbreeding leads to hybrid vigour
© 2008 Paul Billiet ODWS

33. 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)?
© 2008 Paul Billiet ODWS

34. Rhesus blood group R = Rh+ r = Rh-
A rhesus positive fetus is possible if the father is rhesus positive
RR x rr  100% chance
Rr x rr  50% chance
© 2008 Paul Billiet ODWS

35. Rhesus blood group Rhesus positive allele is dominant R Frequency = p
Rhesus negative allele is recessive r Frequency = q
Frequency of r allele = 0.4 = q
If p + q = 1
Therefore R allele = p = 1 – q
= 1 – 0.4
= 0.6
© 2008 Paul Billiet ODWS

36. Rhesus blood group
Frequency of the rhesus positive phenotype = RR + Rr
= p2 + 2pq
= (0.6)2 + (2 x 0.6 x 0.4) =
= 0.84 or 84%
© 2008 Paul Billiet ODWS

37. Hardy Weinberg frequencies
Rhesus blood group
Phenotypes
Genotypes
Hardy Weinberg frequencies
Observed frequencies
Rhesus positive RR p2
0.36
(0.84 total)
0.48 Rr
2pq
Rhesus negative q2
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
© 2008 Paul Billiet ODWS