Lab 3 : Exact tests and Measuring Genetic Variation. χ 2 - test. Where, . Chi-square Assumptions : Finite # of observations. Observations are independent. Samples collected randomly. Large sample size (>20; >50) .
Example: Suppose you caught 5 Bluegill fish and detected two alleles (A1 and A2) and observed that all 5 fish were A1A2 heterozygotes. Calculate allele frequencies and do a χ2 – test to determine whether the population is in HWE.
Reject H0at α = 0.05 , because calculated χ2-value (=5) is more than critical χ2- value with 1 d.f. (≈ 3.84) i.e. Bluegill population is not in HWE.
Why is the previous alleles (A1 and A2) and observed that all 5 fish were A1A2 heterozygotes. Calculate allele frequencies and do a conclusion not reliable?
Because it violates the assumption of large sample size.
As a rule of thumb, the Chi-square test should not be used when the expected number for any genotype class is less than 5.
Exact Test alleles (A1 and A2) and observed that all 5 fish were A1A2 heterozygotes. Calculate allele frequencies and do a
2. Generate all possible permutations of 5 A1 alleles and 5 A2 alleles into 3 genotypes i.e. 10! =3,628,800.
3. Calculate probability of observing each of these samples under HWE using multinomial probability equation.
4. Determine proportion of samples, whose probability is ≤ 0.0313.
5. If proportion (p-value) is less than 0.05, then reject Ho at α = 0.05.
p – value = alleles (A1 and A2) and observed that all 5 fish were A1A2 heterozygotes. Calculate allele frequencies and do a [Sample with probability ≤ 0.0313] / [Total # of sample]
= 3/30 = 0.10
The p-value is more than 0.05, therefore we fail to reject H0
i.e. The bluegill population is in HWE at α = 0.05
Generation of all possible samples and calculation of probability for each sample is computationally intensive. It will require too much timeand is practically impossible for large samples.
In practice, exact tests are done by sampling a distribution generated from a Markov Chain(beyond the scope of this course).
Measures of Genetic Variation probability for each sample is
1. probability for each sample is Heterozygosity (Gene diversity)
= Expected homozygosity under HWE = p12 + p22 + p32 + …….+ pn2
For small sample size(< 50), unbiased HEcan be calculated by :
2 probability for each sample is . Number of Alleles (Na):
- Number of alleles present at a locus in a population.
- Also called allele diversity.
- Strongly influenced by sample size.
3. Effective number of Alleles (Ne):
The number of alleles a population would have if all alleles were at equal frequency
4. Proportion of polymorphic loci (P):
- Not so useful for highly variable loci like Microsatellites.
- Locus selection bias
Problem 1. probability for each sample is Use GenAlEx to perform the following analyses based on the human SSR data:
Problem 2: probability for each sample is Perform an exact test of HWE for all loci and all populations using Arlequin. Include the results for your assigned population in the lab report.