1. Population genetics
Pages 538

2. Test I am going to give you your test back with a mark scheme
Please go through your test using this
I will speak to you individually
Questions
We will cover some generic points together.
For those of you who will need less time..there is another exam question for you to do!!!
LUCKY THINGS!
Hand out LO sheet

3. WHITEBOARDS STARTER: So key terms: Define
Learning Outcome
6.1.2 Patterns of inheritance
Use the Hardy–Weinberg principle to calculate allele frequencies in populations.
The equations for the Hardy– Weinberg principle will be provided
WHITEBOARDS STARTER: So key terms: Define
Population
Allele
Gene pool

4. Population genetics What is a population?
All of the organisms of the same species living in the same place at the same time that can interbreed
The set of genetic information carried by a population is the gene pool.
The gene pool is the total genetic information (the alleles) possessed by the reproductive members within a population of organisms
Allele is different versions of a gene
The relative frequency of a particular allele in a population is the allele frequency

5. Selection of tasks over the topics today
Homework
Selection of tasks over the topics today
Complete the Hardy Weinberg practice booklet
Complete the chi squared practice booklets

6. Gene pools and Evolution
Gene pools re dynamic
Define?
(of a process or system) characterised by constant change, activity, or progress
In what ways are gene pools dynamic?
Mutations
Immigration
Emigration
Gene flow
Natural selection
Mate selection
Geographical barriers
Genetic drift
Look at the handout and ensure you can explain these events to each other: 5 minutes

7. Gene pools and Evolution:
The allele frequency can change over time in response to changing conditions
Evolution involves long term change in the allele frequencies in a population

8. Allele Frequency
We can now understand that a group of individuals can carry a larger number of alleles than an individual in its gene pool.
This genetic diversity can be measured using the
Hardy-Weinberg equation
The HW equation provides a simple mathematical model of genetic equilibrium in a gene pool
In population genetics it is used to calculate allele and genotype frequencies in a population
Scientists use the frequency of alleles within a population, to measure the changes in alleles and so in genotype frequency from generation to generation
To measure the frequency of an allele you need to know:
Mechanism of inheritance of that trait
How many different alleles of the gene for that trait are in the population

9. Hardy-Weinberg principle
The Hardy-Weinberg principle is a fundamental concept of population genetics: GENTIC EQUILIBRIUM
‘For a very large, randomly mating population, the proportion of dominant to recessive alleles remains constant from one generation to the next’
It therefore makes the following assumptions
Population is very large
Random mating
No selective advantage/Natural Selection (no significant selective pressure against one of the genotypes)
No mutation
No Immigration/Emigration/genetic drift.

10. Hardy-Weinberg principle
The Hardy-Weinberg equation is applied to: Populations with a simple genetic situation With dominant and recessive alleles controlling a single trait The frequency of the dominant allele and recessive allele equals the total genetic complement and adds up to 1 (100%) Equation not needed in codominance, why? Because both alleles contribute to the phenotype, the genotypes of all the phenotypes are known.

11. Hardy-Weinberg principle
The Hardy-Weinberg principle shows the basic rules determining the proportion of each genotype in a large, randomly mating population.
The frequency of a genotype is its proportion of the total population
The total is the whole population =1
Frequencies are decimals of the total e.g. 0.25
So lets consider…
(all calculations are carried out using proportions not %)

12. Equations 1: alleles LEARN IT! Whiteboards!
A population (our beetles) is made up of 3 Genotypes: AA Aa and aa
p =frequency of the dominant allele (A)
q =frequency of the recessive allele (a)
The frequency of the allele will be in the range 0 – 1.
0 – no one has the allele
0.5 – half the population has the allele
1 – only allele for that gene in the population
Equation 1: The frequency of allele types
p + q = 1 (The whole population)
LEARN IT! Whiteboards!

13. Equation 2:Genotypes (allele combinations)
In most populations the frequency of the 2 alleles of interest is calculated from the proportion of homozygous recessive. (aa)
WHY?
Because it is the only genotype identifiable directly from its phenotype!
Equation 2: the frequency of allele combinations
p2 + 2pq + q2 = 1
Where
p2 = frequency of genotype (AA) Homozygous dominant
2pq =frequency of genotype (Aa) Heterozygous dominant
q2 = frequency of genotype (aa) Homozygous recessive

14. p2 = frequency of genotype (AA) Homozygous dominant
P =frequency of the dominant allele (A)
q=frequency of the recessive allele (a)
p2 = frequency of genotype (AA) Homozygous dominant
2pq =frequency of genotype (Aa) Heterozygous dominant
q2 = frequency of genotype (aa) Homozygous recessive
LEARN!
Whiteboards

15. How to solve Hardy-Weinberg equations..
Examine question: What piece of information have you been given about the population?
Usually it is the frequency of the homozygous recessive(aa) phenotype q2
Or the dominant phenotype (AA + Aa) 2pq+ p2
So the first objective is to find the value of p or q as everything can be worked out from there!
Take the square root of q2 to find q
Determine p by subtracting from 1 (p=1-q)
Determine p2 by multiplying p by itself (p2 =pxp)
Determine 2pq by multiplying: pxq x2
Check your values by adding up to get 1: p2 + 2pq + q2 = 1
Hand out summary sheet

16. Let’s have a go…
Firstly from memory onto your sheet write equations and key!
In American white population 70% population can taste the chemical PTC (the dominant phenotype), while 30% are non tasters (the recessive phenotype)
Determine the frequency of
Homozygous recessive phenotype
The dominant allele
Homozygous tasters
Heterozygous tasters
First write the symbols representing these next to them
Then what we know

17. Let’s have a go…
In American white population 70% population can taste the chemical PTC (the dominant phenotype), while 30% are non tasters.
Determine the frequency:
Homozygous recessive phenotype q % provided use 0.30 in equation
The dominant allele p
Homozygous tasters p2
Heterozygous tasters 2pq
NOW COMPLETE IN PAIRS CHECK!

18. p2 + 2pq + q2 = 1
Homozygous recessive phenotype q % provided use 0.30 in equation
So q= (square root of 0.30)
So p=0.4523
1-q=p
=0.4523
So then use to find out
Homozygous dominant p2
Heterozygous dominant 2pq

19. p2 + 2pq + q2 = 1
Homozygous recessive phenotype q % provided use 0.30 in equation
So q= (square root of 0.30)
So p=0.4523
1-q=p
=0.4523
So then use to find out
(AA) Homozygous dominant p2 so pxp =0.2046
(Aa)Heterozygous dominant 2pq
2x pxq
0.4523x0.5477=0.2477x2=

20. Let’s have a go…
In American white population 70% population can taste the chemical PTC (the dominant phenotype), while 30% are non tasters.
Determine the frequency
Homozygous recessive phenotype q
The dominant allele p
Homozygous tasters p
Heterozygous tasters 2pq
HOW DO YOU CHECK?? =1

21. Next example
A population of Hamsters has a gene consisting of 90% M alleles (black) and 10% m alleles (grey). Mating is random
Data: Frequency of recessive allele (10% m) and dominant allele (90% M)
Determine the proportion of offspring that will be black and the proportion that will be grey
p2 + 2pq + q2 = 1
Show your workings
Recessive allele q=
Dominant allele p=
Recessive phenotype q2 =
Homozygous dominant phenotype p2 =
Heterozygous dominant phenotype 2pq =

22. Next example
A population of Hamsters has a gene consisting of 90% M alleles (black) and 10% m alleles (grey). Mating is random
Data: Frequency of recessive allele (10% m) and dominant allele (90% M)
p2 + 2pq + q2 = 1
Show your workings
Recessive allele q= 10% so 0.1
Dominant allele p= 90% so 0.9
Recessive phenotype q2 = 0.1x0.1=0.01
Homozygous dominant phenotype p2 = 0.9x0.9=0.81
Heterozygous dominant phenotype 2pq = 2x 0.9x 0.1=0.09= 0.18
Determine the proportion of offspring that will be black and the proportion that will be grey
Black (AA and Aa)=2pq+ p2 x100%= 99%
Grey (aa) = q x100% =1%

23. Summarise in pairs/whiteboards/ into your books: 8 minutes
What is the Hardy Weinberg principle used for?
What does it state?
What assumptions does it make?
What are the 2 equations?
What is the key for the equations?
RECAP TOGETHER

24. Now to practice: 30 minutes
Ensure you have a decent summary of the equations set out so you understand and can use! THEN:
ALL: Complete the Hardy Weinberg calculations sheet /30 marks: 15 minutes
All: Complete the examination questions…… 4 marks and 8 marks 10 minutes
SOME: EXTENSION: Additional Hardy Weinberg practice problems
SOME: Gene pools and evolution sheet beetles page 2
ALL: Homework Pack /60 marks
PHEW!!!!!
Really!!