Title: Population Genetics 12 th February 2014

1. Title: Population Genetics12th February 2014 Learning question: How can you calculate the frequency of alleles in a population? Homework: Starter Watch this video!

2. Key words • Population – group of individuals of one species that actually or potentially interbreed • Genotype frequency – fraction of population with given genotype • Allele frequency – fraction of gametes with given allele • Gene pool – all gametes made by breeding members in a generation (i.e. all inherited genes)

3. Population genetics • A group of individuals of the same species that can interbreed – a population – can carry a large number of different genes • The range of genetic diversity can be measured with a mathematical equation known as the Hardy-Weinberg equation • For this equation to work, certain factors need to be taken into consideration

4. Key Considerations • There is a stable gene pool • Because eye colour is not a “key” adaptation in regards to reproducing, there are no selection pressures on either characteristic. You are not more/less likely to reproduce based on your eye colour. NO SELECTION TAKING PLACE! 2. There are no mutations taking place. • A recessive blue allele is not going to “mutate” and code for brown eyes, vice versa. 3. There is a large population. • Large population = no chance of brown or blue allele randomly disappearing due to chance.

5. Key Considerations 4. Mating within the population is random. • No selective breeding occurring. 5. There is no immigration/emigration. • That way no alleles leave or enter the population. • The allele ratio always stays the same!

6. Measurement of allele and genotype frequencies • Observable characteristics of a genotype are the outward, physical characteristics (phenotype) • To measure the allele frequency, we need to know: • The mechanism of inheritance of a particular trait • How many different alleles of the gene for that trait are in the population

7. Measurement of allele and genotype frequencies • Co-dominance • Heterozygous phenotype = heterozygous genotype • Co-dominant alleles are easy to determine, therefore the frequency is easy to determine • Monohybrid inheritance • If an allele is recessive, it is impossible to determine the genotype from the phenotype (i.e. Dominant phenotype HH/Hh). • The frequency is therefore not as easy to determine

8. The Hardy-Weinberg principle • Godfrey Hardy and Wilhelm Weinberg asked same questionsand worked independently and at same time • Same conclusion about relation between allele and genotype frequencies: • “Hardy-Weinberg law” = mathematical model • 1. Allele frequency equation: p + q = 1 • 2. Genotype frequency equation: p2 + 2pq + q2 = 1

9. The Hardy-Weinberg principle • The principle makes the following assumptions: • The population is very large (eliminates sampling error) • The mating within the population is random • There is no selective advantage for any genotype • There is no mutation, migration or genetic drift

10. Worked example, cystic fibrosis • Cystic fibrosis, an inherited channel protein disorder of lungs. Heterozygotes CFcfare symptomless carriers. • Sufferers are cfcf. • 1/2000 in a population will suffer from CF. We wish to know how many of the population are carriers. • p represents the frequency of the dominant allele CF • q represents the frequency of the recessive allele cf • q2 is the frequency of the genotype cfcf • p2 is the frequency of the genotype CFCF • 2pq is the freqency of the genotype CFcf assuming random mating within the population, where any two individuals of CFcf (pq) mate with each other, the resulting genotypes of the offspring could be CFCF, CFcf, CFcf, cfcf. • These translate as p2 + 2pq + q2 • Within a population, p2 + 2pq + q2 adds up to 1 or 100%

11. Worked example, cystic fibrosis • We know that q2 is 1/2000, so q2 is 0.0005 • So q is the square root of 0.0005, which is 0.022 • If p + q = 1, then p = 1 – 0.022, which is 0.978 • The frequency of carriers is given by 2pq • So 2pq = 2 x 0.978 x 0.022 • So 2pq = 0.043 • This means that 4.3 people in 100 are carriers. • To find out how many people are carriers in our population of 2000, 2000 x 4.3/100 = 86

12. Genotype frequency • Frequency of recessive genetic disorder (ie, sickle-cell anaemia) = q2 frequency • q = square root of q2 frequency • p + q = 1 SO p = 1 – q • With p and q frequencies you can calculate • frequencies of three genotypes (p2 + 2pq + q2 = 1)

13. Predict allele frequency • You can find the frequency of one allele if you know the frequency of the other: p + q = 1 Total frequency of all possible alleles q = frequency of the recessive allele p = frequency of the dominant allele

14. Predict gene frequency • You can find the frequency of one genotype if you know the frequencies of the others: p2 + 2pq + q2= 1 Total frequency of all possible genotypes p2 = frequency of the homozygous dominant genotype 2pq= frequency of the heterozygous genotype q2 = frequency of the homozygous recessive genotype

15. The Hardy-Weinberg principle Imagine we live on a planet where there are only 2 possible eye colours. Blue: b • Brown: B recessive Dominant The only way someone can have blue eyes is if they inherit two recessive alleles (bb) There is a stable gene pool in regards to eye colour. E.G. Because eye colour is not a “key” adaptation in regards to reproducing, there are no selection pressures on either characteristic. You are not more/less likely to reproduce based on your eye colour. NO SELECTION TAKING PLACE!

16. Lets say we had a population of 2 people. Bb bb Allele frequency Phenotype frequency 75% (B = 1) (b=3) b b 50% B B b B b b b b b b

17. If we know that the following assumptions are true • No selection • No mutations • Large population • Allele frequency is constant Hardy weinberg equilibrium

18. Now the math part! p= frequency of b (blue eyes) q= frequency of B (brown eyes) p + q = 100% So, if 40% of the alleles are b The other 60% must be B ALLELE FREQUENCY

19. So we’ve covered allele frequency, lets now look at genotype frequency. p+q= 1 To get genotype frequency we now need to “square” the allele frequency (cover all possible genotypes) p2 + 2pq + q2 = 1 bbbB/BbBB

20. Say we have a population of 100,000 people. 9% of that population have blue eyes (phenotype) of the genotype (bb). (frequency of bb = 9%) But is this all the b alleles present in the population? No! There must be some other b alleles paired with the dominant B allele. Back to our equation - p2 + 2pq + q2 = 1 So we know the frequency of having two bb alleles, how do we work out the total number of b alleles in a population? p2 (bb) = 9% so to work out the frequency of the b allele in the population we must do another sum.

21. p2 + 2pq + q2 = 1 bbbB/BbBB bb (p2)= 9% - to get this you multiply the p by itself. p = √.09 = .3 or 30% - so if you counted all of the alleles in the population, you would find that 30% of them would be b. To put this another way, there is a 30% chance you would inherit one b allele from your mother, and a 30% you would inherit one b allele from your father, making the chances of having blue eyes 9%.

22. p2 + 2pq + q2 = 1 This means that there is a 70% chance you will inherit a B allele from either your mother or father, making the odds of having brown eyes 91%. 91% That your phenotype would be brown eyes 9% That your phenotype will be blue eyes 70% Chance of inheriting B allele 30% Chance of inheriting b allele

23. Questions For which phenotype can you always tell the genotype? Why? Homozygous recessive genotype, because the trait needs two recessive alleles, therefore, can only contain such. 2pq + q2 = 1 Finish this equation: p2 + Ok, now answer some of your own . . .

24. p2 + 2pq + q2 = 1 Ok, so how do we work out the percentage of people with homozygous dominant alleles? We know that the brown phenotype is 70%, so we reverse the equation and square instead of square root. BB = 72 = .49 or 49% bb = 9% BB = 49% Bb/bB = 42%