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

2. Predictable patterns of inheritance in a population so long as…
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

3. 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 ALLELE FREQUENCIES) will be constant.
© 2016 Paul Billiet ODWS

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

5. Iceland Population 313 337 (2007 est) Area 103 000 km2
Distance from mainland Europe
970 km
© 2016 Paul Billiet ODWS

6. Example Icelandic population: The MN blood group
129
385
233
Numbers
747
MnMn
MmMn
MmMm
Genotypes
Type N
Type MN
Type M
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
© 2016 Paul Billiet ODWS

7. 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 = or 57%
Frequency of the Mn allele = 643/1494 = or 43%
© 2016 Paul Billiet ODWS

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

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

10. Assuming all the individuals mate randomly
NOTE the allele frequencies are the gamete frequencies too
SPERMS Mn Mm
Mn 0.43
Mm 0.57
EGGS
MmMm
MmMn
MmMn
MnMn
© 2016 Paul Billiet ODWS

11. Close enough for us to assume genetic equilibrium
Genotypes
Expected frequencies
Observed frequencies
MmMm
0.32
233  747 = 0.31
MmMn
0.50
385  747 = 0.52
MnMn
0.18
129  747 = 0.17
© 2016 Paul Billiet ODWS

12. 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
EGGS
AA p2
Aa pq
aa q2
© 2016 Paul Billiet ODWS

13. THE HARDY WEINBERG EQUATION
So the genotype frequencies are:
AA = p2
Aa = 2pq
aa = q2
or p2 + 2pq + q2 = 1
© 2016 Paul Billiet ODWS

14. 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

15. 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
© 2016 Paul Billiet ODWS

16. 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
Allele frequencies
Genotypes
Population sample
0.4
0.6
0.4
0.6
0.4
0.6
© 2016 Paul Billiet ODWS

17. 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
0.6
0.4 36 48 16 50 30 60 40
© 2016 Paul Billiet ODWS

18. 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
© 2016 Paul Billiet ODWS

19. 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
© 2016 Paul Billiet ODWS

20. 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
© 2016 Paul Billiet ODWS

21. 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
0.05
0.95
0.0025
0.095
0.9025
© 2016 Paul Billiet ODWS

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

23. 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
© 2016 Paul Billiet ODWS

24. 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
© 2016 Paul Billiet ODWS

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

26. Albinism gene frequencies
Normal allele = A = p = ?
Albino allele = q =  ( ) = or 0.7%
© 2016 Paul Billiet ODWS

27. 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%
© 2016 Paul Billiet ODWS

28. 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.
© 2016 Paul Billiet ODWS

29. 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

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

31. 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

32. Rhesus blood group
Frequency of the rhesus positive phenotype = RhRh + Rhrh
= p2 + 2pq
= (0.6)2 + (2 x 0.6 x 0.4)
= 0.84 or 84%
© 2016 Paul Billiet ODWS

33. Hardy Weinberg frequencies
Rhesus blood group
Phenotypes
Genotypes
Hardy Weinberg frequencies
Observed frequencies
Rhesus positive
RhRh p2
0.84
Rhrh
2pq
Rhesus negative
rhrh 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.
© 2016 Paul Billiet ODWS