I had studied population genetic via SSR marker in 5 different wildlife habitats. Three methods of I.A.M, T.P.M and S.S.M were studied by BOTTLENECK software. This result related to one of the mentioned population, which had tested as below. I would greatly appreciate it if you kindly give me some opinion on test methods and discussion:
SIGN TEST
Assumptions: all loci fit I.A.M., mutation-drift equilibrium.
Expected number of loci with heterozygosity excess: 3.33
4 loci with heterozygosity deficiency and 1 loci with heterozygosity excess.
Probability: 0.04330
Assumptions: all loci fit T.P.M., mutation-drift equilibrium.
Expected number of loci with heterozygosity excess: 3.14
4 loci with heterozygosity deficiency and 1 loci with heterozygosity excess.
Probability: 0.05649
Assumptions: all loci fit S.M.M., mutation-drift equilibrium.
Expected number of loci with heterozygosity excess: 2.72
4 loci with heterozygosity deficiency and 1 loci with heterozygosity excess.
Probability: 0.12497
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STANDARDIZED DIFFERENCES TEST
Caution: only 5 polymorphic loci (minimum 20).
Assumptions: all loci fit I.A.M., mutation-drift equilibrium.
T2: -4.229 Probability: 0.00001
Assumptions: all loci fit T.P.M., mutation-drift equilibrium.
T2: -5.744 Probability: 0.00000
Assumptions: all loci fit S.M.M., mutation-drift equilibrium.
T2: -6.130 Probability: 0.00000
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WILCOXON TEST
Assumptions: all loci fit I.A.M., mutation-drift equilibrium.
Probability (one tail for H deficiency): 0.04688
Probability (one tail for H excess): 0.96875
Probability (two tails for H excess and deficiency): 0.09375
Assumptions: all loci fit T.P.M., mutation-drift equilibrium.
Probability (one tail for H deficiency): 0.04688
Probability (one tail for H excess): 0.96875
Probability (two tails for H excess or deficiency): 0.09375
Assumptions: all loci fit S.M.M., mutation-drift equilibrium.
Probability (one tail for H deficiency): 0.04688
Probability (one tail for H excess): 0.96875
Probability (two tails for H excess or deficiency): 0.09375
Regards,
Mahmoud