Download Bio 211 Genetics Laboratory Experiment 5: Human Population

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Molecular ecology wikipedia , lookup

Transcript
 Population Genetics
Bio 211 Genetics Laboratory
Experiment 5: Human Population Genetics
Hardy-Weinberg Equilibrium
Consider 2 alleles at a locus in a monoecious organism, with the following assumptions:
‐
‐
‐
‐
‐
Large population size
Random mating
No selection
No mutation (eg., T -> t, t -> T)
No migration (of genotypes into or out of the population)
Let p = frequency of one allele (such as T) and q = 1 – p = frequency of the other allele (such as
t). In a population under Hardy-Weinberg equilibrium, allele and genotypic frequencies do not
change and the genotypic frequencies are p2 : 2pq : q2 for the genotypes TT : Tt : tt.
2
χ Test for Hardy-Weinberg Equilibrium
There are 2 alleles at the TAS2R38 gene (T: taster, t: non-taster), and genotype of an individual
can be identified by the DNA banding patterns of HaeIII digest of PCR products. To determine
whether the class, treated as a population, exhibits Hardy-Weinberg equilibrium genotypic
frequencies for the TAS2R38 gene, a χ2 test for Hardy-Weinberg equilibrium can be conducted.
The null hypothesis for this test is:
H0: The population is in Hardy-Weinberg equilibrium.
In order to calculate the χ2 value, observed numbers of the different genotypes are compared to
the expected numbers under the assumption of Hardy-Weinberg equilibrium.
Genotype: TT Tt tt Observed number: NTT NTt Ntt Genotypic frequency: NTT/N NTt/N Ntt/N Let N = NTT + NTt + Ntt
If p = frequency of T and q = 1 – p = frequency of t, then allelic frequencies are:
p = (NTT/N) + (1/2)(NTt/N) and q = (Ntt/N) + (1/2)(NTt/N)
Hardy-Weinberg equilibrium genotypic frequencies and expected numbers for genotypes are:
Genotype: TT Tt tt Genotypic frequency: p2 2pq q2 Expected number: Np2 N2pq Nq2 χ2calc = Σ [(observed number – expected number)2/expected number]
χ2calc = (NTT – Np2)2 + (NTt – N2pq)2 + (Ntt – Nq2)2 Np2 N2pq Nq2 In this test, df = (number of classes or genotypes) – (number of alleles) = 3 – 2 = 1.
If χ2calc > χ2crit then H0 can be rejected, and the observed numbers of genotypes are not consistent with
a population in Hardy-Weinberg equilibrium.