14.0: POPULATION GENETICS

Transcription

1 POPULATION GENETICS

2 14.0: POPULATION GENETICS 14.1 : GENE POOL CONCEPT 14.2 : HARDY- WEINBERG LAW

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4 INTRODUCTION Population genetics: Population is: A group of individuals Of the same species Occupying a given area Can freely interbreed And, produce fertile offspring

5 Population Genetic The study in the change of allele frequencies in population.

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7 GENE POOL CONCEPT At the end of this topic, students should be able to: Explain gene pool and allele frequency Explain population genetics and its relation to allele frequency Explain allele frequency and genetic equilibrium

8 GENE POOL CONCEPT total number of genes (alleles) of all sexually reproducing individuals in a population Gene pool consists of : all alleles at all gene loci in all individuals of a population.

9 A - Dominant allele (blue) a - Recessive allele (red) AA, Aa blue aa red What is the total no of allele in this population? Total no of blue butterfly 20 Total no of red butterfly 10 2 x 30 = 60 alleles Total no of butterfly 30

10 GENE POOL CONCEPT Example Question There are 10,000 individuals in a population. Calculate the gene pool.

11 GENE POOL CONCEPT Answer: Gene pool is the total number of genes (alleles) of all individuals in a population. Each individual is a diploid organism So, each individual carry 2 alleles. Gene pool = 2 x 10,000 = 20,000 alleles

12 GENE POOL CONCEPT The gene composition in a gene pool is NOT constant; always change from generation to generation.

13 REMEMBER! In the basis of evolution, each individual inherits the genes not the phenotypes

14 Each individuals inherit genes not the phenotypes

15 GENE POOL CONCEPT Allele frequency : The ratio of total number of particular allele to the total number of all alleles in gene pool

16 Example: In a population of 500 birds, the following genotypes was investigated AA = 150 individuals Aa = 250 individuals aa = 100 individuals Calculate the frequency of allele A and a. Solution: Frequency of allele A= Total number of allele A in population Total number of all allele in gene pool Total number of allele in gene pool = 2(150) + 2(250)+2(100) = 1000 Total number of allele A in population = (2 x 150) = 550

17 So, frequency of allele A = = 0.55 Frequency of allele a: Frequency of allele a = Total number of allele a in populations Total number of all allele in gene pool = (2 x 100) = 0.45

18 For a gene locus where only 2 alleles occur in a population, allele frequencies can be calculated using this equation : p + q = 1 p = frequency of the dominant allele q = frequency of the recessive allele If q = 0.45, p= 1 q = = 0.55

19 Genetic equilibrium : Condition where gene frequencies stay constant from generation to generation in a population.

20 EXAMPLE Genotype frequency for generation 1 = 0.2 Genotype frequency for generation 2 = 0.2 Has the population reached genetic equilibrium? Why? Yes. The population reached genetic equilibrium because the genotype frequency stay constant from generation to generation in a population.

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22 HARDY-WEINBERG LAW At the end of this topic, students should be able to: State the Hardy-Weinberg Law Explain five assumptions of Hardy- Weinberg Law for genetic equilibrium Calculate allele and genotype frequencies

23 HARDY-WEINBERG LAW Hardy-Weinberg Law states that: frequencies of alleles and genotypes in a population remain constant from generation to generation

24 HARDY-WEINBERG LAW Condition of a population for Hardy-Weinberg equilibrium due to five assumptions: Very large population size For a small-sized population, genetic drift may easily change allele frequencies Random fertilization If individuals choose mates only with certain traits, frequencies of certain alleles may change No net mutation Mutation may change an allele into another & this changes allele frequencies

25 No migration or emigration Migration may cause gene flow which changes the gene pool composition No natural selection All individuals must be equally fertile & able to pass the alleles to the next generation This law describes a non-evolving population Naturally, the conditions for the law of Hardy- Weinberg are rarely met

26 HARDY-WEINBERG LAW A allele p Dominant allele frequency a allele q Recessive allele frequency AA genotype p 2 Homozygous dominant genotype aa genotype q 2 Homozygous recessive genotype Aa genotype 2pq Heterozygous genotype

27 HARDY-WEINBERG LAW Related to allele frequencies are the genotypic frequencies Genotypic frequency is the ratio of individuals with certain genotype in a population Hardy-Weinberg equations are used to estimate the frequencies of alleles & genotypes in a population which is in genetic equilibrium

28 HARDY-WEINBERG LAW Calculate allele and genotype frequencies Allele frequency p + q = 1 Genotype frequency : p 2 + 2pq + q 2 = 1

29 HARDY-WEINBERG LAW p + q = 1 p 2 + 2pq + q 2 = 1 p = dominant allele frequency (A) q = recessive allele frequency (a) p 2 = homozygous dominant genotype (AA) 2pq = heterozygous genotype (Aa) q 2 = homozygous recessive genotype (aa)

30 Hardy-Weinberg Law Homozygous dominant(pp) Heterozygous (Pp) Homozygous recessive (pp) p 2 + 2pq + q 2 = 1 Gene pool Dominant phenotype Recessive phenotype

31 as a reminder, only in the case of dominant and recessive alleles we can use Hardy- Weinberg calculation.

32 It is possible to calculate all allele, genotype, dominant and recessive phenotype frequencies using the expressions : allele frequency p + q = 1, genotype frequency p 2 + 2pq + q 2 = 1, dominant phenotype frequency = p 2 + 2pq, and recessive phenotype frequency = q 2

33 EXAMPLE Resistance toward a type of pesticide for a population of rats is controlled by dominant allele, R. 64% of the rat population show the resistance. a) Calculate the frequency for R allele. Assume that the population is in genetic equilibrium and p 2 + 2pq + q 2 = 1 while p + q = 1

34 EXAMPLE Resistance toward a type of pesticide for a population of rats is controlled by dominant allele, R. 64% of the rat population show the resistance. a) Calculate the frequency for R allele. Assume that the population is in genetic equilibrium and p 2 + 2pq + q 2 = 1 while p + q = 1 36% of rat population are homozygous recessive (rr). Genotypic frequency for homozygous recessive (rr), = q 2 = 0.36 Frequency for recessive allele (r), q 2 = 0.36 q = 0.36 = 0.6 Frequency for dominant allele (R), p= 1 - q = = 0.4

35 TRY THIS.. In a population, the frequency for allele A is 0.4. for the next generation, the frequency for genotype AA is 0.2. Has the population reached genetic equilibrium? Give reason for your answer.

36 TRY THIS.. In a population, the frequency for allele A is 0.4. for the next generation, the frequency for genotype AA is 0.2. Has the population reached genetic equilibrium? Give reason for your answer. Generation 1: p = 0.4 Generation 2: frequency for genotype AA = p 2 = (0.4) 2 = 0.16 If the population is in Hardy- Weinberg equilibrium, then the frequency of AA should be 0.16 and not 0.2. Since the actual genotype frequency and the calculated one differ, our conclusion is that the population does not follow Hardy- Weinberg law and has not reached genetic equilibrium

37 Let s calculate : 1) If the frequency of allele A is 0.8, the frequency of allele a must be P + q = 1 P = 0.8 So, q = q = 0.2 2) In a population that is in equilibrium, 64% of its individuals have genotype rr. What is the frequency of R allele in the population? q2 = 64/100 q2 = 0.64 q = 0.64 = 0.8 p= 1- q p= p = 0.2

38 3) The ability to roll tongue is determined by the dominant allele, T. The recessive allele, t determines the inability to roll tongue. A research is conducted on 1000 students and the results are as follows: students that can roll tongue = 640 students that can t roll tongue = 360 If the population is in equilibrium, what is the percentage of students with heterozygous genotype? q 2 = 360/1000 q 2 = 0.36 q = 0.36 q = 0.6 p = 1-q p = p = 0.4 Heterozygous 2pq = 2 x 0.4 x 0.6 2pq= 0.48

39 4) Calculate the number of rats with genotypes RR, Rr and rr for a population of 200 rats. It is already known that p = 0.4 and q = 0.6 Genotypic frequency for homozygous dominant (RR), p 2 = (0.4) 2 = 0.16 Number of rats with genotype RR = 0.16 x 200 = 32 Genotypic frequency for heterozygous (Rr), 2pq = 2(0.4)(0.6) = 0.48 Number of rats with genotype Rr = 0.48 x 200 = 96 Genotypic frequency for homozygous recessive (rr), q 2 = 0.36 Number of rats with genotype rr = 0.36 x 200 = 72

40 5) For a population of Shorthorns, the following data was obtained: Genotype Phenotype Number of individuals C M C M red 110 C M C P red & white 150 C P C P white 48 Calculate the frequencies for alleles C M and C P. Total number of individuals for the population = 308 Frequency for allele C M = 2(110) (308) = 0.6 Frequency for allele C P = 2(48) (308) = 0.4