8 5 Allelic Frequencies and population genetics Learning

8. 5 Allelic Frequencies and population genetics Learning objectives: 1. Define the terms gene pool and allele frequency 2. Explain the Hardy-Weinberg equation and use it to calculate allele frequencies and genotype frequencies from data concerning the phenotypes within a population.

STARTER 1. Define allele. Give an example of a gene with two alleles. 2. Describe dominant and recessive alleles. 3. Define gene flow. Give a specific example of gene flow and describe how it causes changes in allele frequencies over time. 4. List and briefly describe examples of two real barriers to gene flow. 5. Describe how natural selection affects the allele frequencies of a population.

Allele Frequency • Population genetics is the study of genetic changes in populations. • Gene Pool is the sum total of the genes in a population. This term usually refers to one phenotypic character at a time. • Allelic frequency is the number of times an allele occurs within the gene pool. E. g. Huntingdon’s disease is an autosomal dominant degenerative neurological disease. Usually, people who have Huntington's disease don't start showing any symptoms until they are in their 30 s or 40 s. Mood changes, depression, irritability, or paranoia are often the first signs of the disease and are followed by a decline in coordination and an unsteady gait. WRITE DOWN ALL THE POSSIBLE GENOTPYES FOR THIS DISEASE. When looking at the gene pool each person has two alleles so if there are 1000 individuals then there will be twice as many alleles of this gene. How many will there be?

Allele Frequency When one allele is dominant to another the symbols p and q are used to represent their frequency. H = dominant allele p = its frequency H = recessive allele q = its frequency Frequency of dominant allele p + + Frequency of recessive allele q = gene pool = 1 i) If the frequency of the dominant allele for Huntington’s is 0. 4, what is the frequency of the recessive allele? ii) If everyone in a population was heterozygous (Hh), what would the frequency of the dominant allele be? iii) …. And the recessive? BUT populations are not made up of just one genotype!!!

1. 2. 3. 4. 5. 6. 7. 8. Work in groups of three. Each group is to take 100 beads – 80 red (dominant allele), 20 white (recessive lethal allele). These represent the gene pool of a population. Mix them and spill them out on the table. Work out the allele frequencies (p and q). With your eyes, each take a pair of alleles. The recessive allele is lethal so if you picked two white beads then you must discard these. This represents selection. With the remaining gene pool work out the allele frequency. Assuming your population reproduces top up your alleles to the percentages you have just calculated. This will keep the population at 100. Repeat the process ten times (or less if all the white alleles are gone). Generation no Dominant Alleles Recessive Alleles No of lethal alleles Total Alleles in gene pool 1. Starting no 80 20 e. g. 2 e. g. 98 Ending % 80/98 2. Starting no

Discussion 1. Which genotypes were selected to ‘survive’ and reproduce? 2. What are the effects of natural selection on a population? 3. What else could affect the gene pool? 4. How could you modify this modelling activity to look at the effect of gene flow on populations?

Timed Essay You have thirty minutes to write an essay. Spend five minutes planning out the essay. “Discuss monohybrid inheritance in humans” (25 marks)

Hardy-Weinberg Equation A Mendelian population is a group of sexually reproducing organisms of the same specie residing within defined geographic boundaries wherein interbreeding occurs. Gene Pool is the sum total of the genes in a population. This term usually refers to one phenotypic character at a time. The gene pool changes over time as new alleles enter of exit through events such as natural selection, migration or mutation. Allele frequency is not something that can be measured directly in populations. What can be seen or measured is the phenotype of each individual of the population.

Possible chance combinations of gametes If we know the relative frequencies of dominant and recessive alleles in a population (p and q) then we can calculate the expected frequencies of progeny genotypes and phenotypes (assuming random fusion of gametes). Male/Female p (A) q (a) p (A) p² AA pq Aa q (a) pq Aa q² aa NB. p + q = 1 i. e. the percentage of A and a gametes must add up to 100% in order to account for all the gametes in the gene pool.

Formula for the expected genotypes in the next generation The rule is named after G. H. Hardy and W. Weinberg who independently formulated it in 1908.

Several assumptions are needed: 1. 2. 3. 4. The population is infinitely large and mates at random. No selection pressures The population is closed (no immigration or emigration) There is no mutation of alleles (unless mutation rates are equal) 5. Meiosis is normal so that chance is the only factor in producing gametes. No population is infinitely large, mutations cannot be prevented, selection and migration pressures usually exist in most natural populations etc. so it may be surprising to note that despite this many populations do conform to equilibrium conditions between two generations.

Worked Example For example, if the frequency of A = 0. 6 and the frequency of a = 0. 4 then the expected frequencies would be p² = (0. 6)² = 0. 36 2 pq = 2 x 0. 6 x 0. 4 = 0. 48 q² = (0. 4)² = 0. 16 You can use this equation to calculate allele frequencies if you know the frequency of a least one of the genotypes. You can identify homozygous recessive individuals to calculate allele frequencies.

Using the HW equation 150 plants produce long pollen grains and 50 plants produce round pollen grains in a population of peas. The allele frequencies (p and q) can be found by calculating the frequency of homozygous recessive plants (round pollen grains). Number of round pollen grains = 50 Total number of plants = 200 q² = 50/200 = 0. 25 Frequency of recessive allele (q) = √ 0. 25 = 0. 50 Since p + q = 1 then p must be 1 -0. 50 = 0. 50 The frequency of the pollen grain genotypes may also be calculated; p² = (0. 5)² = 0. 25 (AA) or 50 individuals in a population of 200 2 pq = 2 x 0. 5 = 0. 50 (Aa) or 100 individuals q² = (0. 5)² = 0. 25 (aa) or 50 individuals

Questions 1. A population of Drosophilia contains 64 long-winged flies and 36 vestigial winged flies. a) What is the frequency of vestigial winged flies (q²) in this population? b) What is the frequency of vestigial-wings allele (q)? c) What is the frequency of long-wings allele? d) What is the frequency of homozygous long-winged flies? e) What proportion of long-winged flies are homozygous and what proportion are heterozygous? 2. Humans suffering from insulin-dependent mellitus cannot secrete the hormone insulin. This disorder is inherited as a recessive allele at a single locus. If the frequency of allele (q) in a human population is 0. 07, calculate the frequency of a) The normal allele b) People who suffer from diabetes c) Heterozygous carriers of the diabetes allele.

Homework • Complete the exam questions on HW. Due next Thursday.

STARTER – 6 marks Huntington’s disease is a human inherited condition resulting in gradual degeneration of nerve cells in the brain. It is caused by a dominant allele, but usually no symptoms are evident until the person is at least 30 years old. It is very rare in most populations. However, in one isolated area in Venezuela, 48% of the population possess a genotype which gives rise to Huntington’s disease. Many of the inhabitants can trace their origins to a common ancestor 200 years ago. a) Use the HW equation to estimate the % of this Venezuelan population which is heterozygous for Huntington’s disease. Show your working. b) Suggest why i) there is such a high incidence of Huntington’s disease in this population. ii) Huntington’s disease has not been eliminated from this population by natural selection

Answers – Self Assess a) 48% = 0. 48 1 – 0. 48 = 0. 52 = frequency of homozygous recessives. Square root of 0. 52 = 0. 72 = frequency of recessive allele. Frequency of dominant allele = 1 -0. 72 = 0. 28 Frequency of heterozygotes = 2 pq = 2 x 0. 28 x 0. 72 = 0. 403 = 40% b) i) Common ancestor/in-breeding/no migration/genetic isolation/small gene pool ii) Reproduction occurs before the symptoms of disease are apparent/possessors of the dominant allele are not at a selective disadvantage/Hh x hh … 50% offspring affected/ Hh x Hh …. 75% of offspring affected.