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Population Genetics Hardy Weinberg

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Presentation on theme: "Population Genetics Hardy Weinberg"— Presentation transcript:

1 Population Genetics Hardy Weinberg

2 Population Genetics Mendelian genetics predicts the outcome of specific matings between individuals What about the genetics of an entire population? Population = all individuals of one species living in a given area Population genetics works with the entire gene pool or all the alleles present in the whole population

3 How will alleles change in the population?
Among a population of 2000 people: 720 have blue eyes (recessive) 1280 have brown eyes (dominant) DNA testing reveals: 320 are homozygous for Brown Eyes(BB) 960 are heterozygous for Brown Eyes (Bb)

4 Allele frequency f(B)= 960 + 320 + 320/ 4000 = 0.4
Allele frequency is the fraction: no. of a particular allele no. of all alleles in population For these 2000 people, there are 4000 alleles in the gene pool: 720 bb 960 Bb 320 BB how many B alleles? how many b alleles? f(B)= / 4000 = 0.4 F(b) = /4000 = 0.6

5 What happens in the next generation?
In all the matings for this generation, what is the chance that an egg with the B allele will be fertilized by a sperm with the b allele and create a person with Bb genotype? Recall: 40% of all eggs will carry B 60% of all sperm will carry b Recall the Rule of Multiplication: (prob. of event a) (prob of event b)= probability of both events happening 0.4 x 0.6 = 0.24 24% of offspring Bb *Assuming no difference between sexes and no mating preferences!

6 What happens in the next generation?
In all the matings for this generation: 0.4 x 0.4 = 16% BB 0.4 x 0.6 x 2 = 48% Bb Bb and bB (rule of addition) 0.6 x 0.6 = 36% bb 40% 60% B b BB Bb bb

7 What happens in the next generation?
In all the matings for this generation: if 4000 offspring are born: 0.4 x 0.4 = 16% BB BB 0.4 x 0.6 x 2 = 48% Bb Bb 0.6 x 0.6 = 36% bb bb 2560 Brown 1440 Blue

8 What happens in the next generation?
New allele frequencies: If 4000 offspring 640 BB 1920 Bb 1440 bb 3200 B allele/8000 = 0.4 4800 b allele/8000 = 0.6 After 5 generations: 64,000 offspring 10,240 BB 30,720 Bb 23,040 bb 51,200 brown alleles / 128,000 = 0.4 76,800 blue alleles / 128,000 = 0.6

9 After 5 generations (or any number):
Allele frequencies do not change! Recessive alleles are maintained in the population *If some specific assumptions are made

10 Hardy-Weinberg equilibrium
Godfrey Hardy (mathematician) and Wilhelm Weinberg (physician) (early 1900s): Given some assumptions, allele frequencies won’t change: The population is large Mating is random No migration in or out No mutation No selection (no allele is advantageous) How often in nature are ALL of these assumptions met? Rarely, if ever. This is an “ideal” state.

11 Does Hardy-Weinberg work?
In large populations, the Hardy-Weinberg equations predict results quite well for many traits If a population is not in equilibrium: Allele frequencies are changing Evolution is occurring!

12 Hardy-Weinberg equations
Allele frequency: Let p = frequency of the dominant allele Let q = frequency of the recessive allele Then, p + q = 1 Genotype frequency: p2 = frequency of homozygous dominant genotype q2 = frequency of homozygous recessive genotype 2pq = frequency of heterozygous genotype p2 + 2pq + q2 = 1

13 4 Steps to solving H-W Problems
set recessives = q2 Take square root of q2 1-q = p Plug into expanded equation Example: 16% of the cat population is white: q2= 0.16 square root = 0.4 = p p = 0.6 (.6) (.4) = 1 so, 36% of population is TT 48% of population is Tt

14 Another Example: Fraggles are mythical, mouselike creatures that live beneath flower gardens. Of the 100 fraggles in a population, 75 have green hair (FF or Ff) and 25 have grey hair (ff). Assuming genetic equilibrium: What are the gene frequencies of F and f? What are the genotypic frequencies?

15 Answer to Fraggle Problems:
Gene frequencies: q2= .25, so: q= .5 p= .5 Genotypic frequencies FF = .25 Ff = .5 f f = .25

16 Application of H-W principle
Sickle cell anemia inherit a mutation in gene coding for hemoglobin oxygen-carrying blood protein recessive allele = s normal allele = S low oxygen levels causes RBC to sickle breakdown of RBC clogging small blood vessels damage to organs often lethal

17 Sickle cell frequency High frequency of heterozygotes
1 in 5 in Central Africans = Ss unusual for allele with severe detrimental effects in homozygotes 1 in 100 =ss usually die before reproductive age Why is the s allele maintained at such high levels in African populations? Suggests some selective advantage of being heterozygous…

18 Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells 1 2 3

19 Heterozygote Advantage
In tropical Africa, where malaria is common: homozygous dominant (normal) die or reduced reproduction from malaria: SS homozygous recessive die or reduced reproduction from sickle cell anemia: ss heterozygote carriers are relatively free of both: Ss survive & reproduce more, more common in population Hypothesis: In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. Frequency of sickle cell allele & distribution of malaria

20 Sickle Cell Example: If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria? f(ss)= .09 = q2 q= .3 P= .7 2pq= .42 so 42% of the population is resistant to malaria.

21 Using Hardy-Weinberg Cystic Fibrosis: 1 in 1700 US Caucasian newborns have cystic fibrosis. Use an F for the normal allele and f for recessive: What percent of the above population have cystic fibrosis? What percent are healthy, non carriers? What percent are carriers of cyctic fibrosis? In a population of 1700 people, how many would you expect to be homozygous normal? In a population of 1700 people, how many would you expect to be heterozygous? = q2 q= .024 p= .976 P2= .9524 2pq= .0468 1700 x .9524= 1619 1700 x .0468= 80

22 Stem Cells Nova

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