Population Genetics Hardy Weinberg

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Population Genetics Hardy Weinberg

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Population Genetics Hardy Weinberg. 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 - PowerPoint PPT Presentation

Transcript of Population Genetics Hardy Weinberg

Page 1: Population Genetics Hardy Weinberg

Population GeneticsHardy Weinberg

Page 2: Population Genetics Hardy Weinberg

Population GeneticsMendelian genetics predicts the outcome of

specific matings between individualsWhat about the genetics of an entire

population?Population = all individuals of one species living in a

given areaPopulation genetics works with the entire

gene pool or all the alleles present in the whole

population

Page 3: Population Genetics Hardy Weinberg

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)

Page 4: Population Genetics Hardy Weinberg

Allele frequencyAllele frequency is the fraction:

no. of a particular alleleno. of all alleles in population

For these 2000 people, there are 4000 alleles in the gene pool:720 bb960 Bb320 BB

how many B alleles?how many b alleles?

f(B)= 960 + 320 + 320/ 4000 = 0.4F(b) = 960 + 720 +720/4000 = 0.6

Page 5: Population Genetics Hardy Weinberg

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 B60% 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 = 0.2424% of offspring Bb

*Assuming no difference between sexes

and no mating preferences!

Page 6: Population Genetics Hardy Weinberg

What happens in the next generation?

In all the matings for this generation:

40% 60%

B b

40% B BB Bb

60% b Bb bb

0.4 x 0.4 = 16% BB0.4 x 0.6 x 2 = 48% Bb

- Bb and bB - (rule of addition)

0.6 x 0.6 = 36% bb

Page 7: Population Genetics Hardy Weinberg

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 640 BB0.4 x 0.6 x 2 = 48% Bb 1920 Bb0.6 x 0.6 = 36% bb 1440 bb

2560 Brown

1440 Blue

Page 8: Population Genetics Hardy Weinberg

What happens in the next generation?New allele frequencies:

If 4000 offspring

640 BB1920 Bb1440 bb

3200 B allele/8000 = 0.4

4800 b allele/8000 = 0.6After 5 generations:

64,000 offspring10,240 BB30,720 Bb23,040 bb

51,200 brown alleles / 128,000 = 0.476,800 blue alleles / 128,000 = 0.6

Page 9: Population Genetics Hardy Weinberg

After 5 generations (or any number):Allele frequencies do not change!Recessive alleles are maintained in the

population

*If some specific assumptions are made

Page 10: Population Genetics Hardy Weinberg

Hardy-Weinberg equilibriumGodfrey Hardy (mathematician) and Wilhelm

Weinberg (physician) (early 1900s):Given some assumptions, allele frequencies

won’t change:The population is largeMating is randomNo migration in or outNo mutationNo selection (no allele is advantageous)

How often in nature are ALL of these assumptions met?Rarely, if ever. This is an “ideal” state.

Page 11: Population Genetics Hardy Weinberg

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 changingEvolution is occurring!

Page 12: Population Genetics Hardy Weinberg

Hardy-Weinberg equationsAllele frequency:

Let p = frequency of the dominant alleleLet q = frequency of the recessive alleleThen, p + q = 1

Genotype frequency:p2 = frequency of homozygous dominant

genotypeq2 = frequency of homozygous recessive

genotype2pq = frequency of heterozygous genotypep2 + 2pq + q2 = 1

Page 13: Population Genetics Hardy Weinberg

4 Steps to solving H-W Problems1. set recessives = q2 2. Take square root of q2 3. 1-q = p4. Plug into expanded equation

Example: 16% of the cat population is white: 5. q2= 0.166. square root = 0.47. 1- 0.4 = p p = 0.68. .16 + 2 (.6) (.4) + .36 = 1

so, 36% of population is TT48% of population is Tt

Page 14: Population Genetics Hardy Weinberg

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?

Page 15: Population Genetics Hardy Weinberg

Answer to Fraggle Problems:Gene frequencies:

q2= .25, so:q= .5p= .5

Genotypic frequenciesFF = .25Ff = .5f f = .25

Page 16: Population Genetics Hardy Weinberg

Sickle cell anemiainherit a mutation in gene coding for

hemoglobin oxygen-carrying blood protein recessive allele = s

normal allele = Slow oxygen levels causes

RBC to sickle breakdown of RBC clogging small blood vessels damage to organs often lethal

Application of H-W principle

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Sickle cell frequencyHigh frequency of heterozygotes

1 in 5 in Central Africans = Ssunusual 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…

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

1

2

3

Page 19: Population Genetics Hardy Weinberg

Heterozygote AdvantageIn 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

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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= .3P= .72pq= .42so 42% of the population is resistant to

malaria.

Page 21: Population Genetics Hardy Weinberg

Using Hardy-WeinbergCystic 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?

.00058 = q2

q= .024p= .976

P2= .95242pq= .0468

1700 x .9524= 16191700 x .0468= 80