Population Genetics Hardy Weinberg
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Population GeneticsHardy 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
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)
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
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!
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
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
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
After 5 generations (or any number):Allele frequencies do not change!Recessive alleles are maintained in the
population
*If some specific assumptions are made
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.
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!
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
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
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?
Answer to Fraggle Problems:Gene frequencies:
q2= .25, so:q= .5p= .5
Genotypic frequenciesFF = .25Ff = .5f f = .25
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
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…
Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells
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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
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.
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
Stem CellsNova