Chapter

Total number of E alleles Total number of e alleles =

Frequency of E allele (p)=11112/16140=0.6885 Frequency of e allele (q)=1-p=1-0.6885 = 0.3115

(ii) If the population was in HWE, we would expect the genotype frequencies to be equal to: p2 + 2pq + q2 = (0.6885)2 + 2(0.6885) (0.3115) + (0.3115)2 = 0.474 + 0.429 + 0.0970.

(iii) There are a total of 8070 individuals in this population. If the population was in HWE we would expect to find 8070(0.474) = 3825 individuals with the genotype EE (p2), 8070(0.429)=3462 individuals with the genotype Ee (2pq) and 8070(0.097)=783 individuals with the genotype ee (q2).

= (3969 - 3825)2/3825 + (3174 - 3462)2/3462 + (927 - 783)2/783 = 5.42 + 23.96 + 26.48 = 55.86

With one degree of freedom, this is highly significant (P<0.001), which means that the observed distribution of genotypes in this population is significantly different from that expected if the population was in HWE.

3.2. Ne = 4(Nef)(Nem)/(Nef + Nem) For population 1:

For population 2:

3.3. (i) Long-term Ne = i/[(1/Ne0 + (1/Ne2) + (1/N*) + ■■■ (1/Net)]

= 6/[(1/104) + (1/104) + (1/104) + (1/103) + (1/104) + (1/104)] = 6/0.0015 = 4000

3.5. Possible explanations:

• Inbreeding

• Insufficient sample size (sample size influences Ho more than it influences He)

• Wahlund effect

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