Question

(3) With habitat loss increasing the overall population of African wild dogs (Lycaon Dictus) in Southern Africa has become a primary concern for wildlife management officials. To assess diversity in this population, the following genetic data were collected S. Africa Genotype D1D1 D1D2 D1D3 11 17 14 D2D2 D2D3 D3D3 a) Does this population deviate significantly from Hardy-Weinberg? (2 marks] X-end df=1-3.841; df=25.991; df=37.815, df=4-9.488 b) What is the rate of inbreeding in this population? [2 marks]

Answer 1

Answer:

• This is a one locus three allele case.
• There are 3 alleles in the population -> D1 D2 D3
• let the corresponding allelic frequencies be -> p q r
• The number of indviduals in each genotype are given in the question,
• the total number of individual are, 8 + 4 + 12 + 5 + 9 + 6 = 44

$\rightarrow$Therefore the genotypic frequencies are as follows,

• f(D1D1) = 8/44 = 0.1818, (Homozygote)
• f(D1D2) = 4/44 = 0.0909, (Heterozygote)
• f(D1D3) = 12/44 = 0.2727, (Heterozygote)
• f(D2D2) = 5/44 = 0.1136, (Homozygote)
• f(D2D3) = 9/44 = 0.2045, (Heterozygote)
• f(D3D3) = 6/44 = 0.1363, (Homozygote)

$\rightarrow$the allelic frequencies will be as follows,

• p = f(D1D1) + f(D1D2)/2 + f(D1D3)/2 = 0.1818 + 0.0454 + 0.1363 = 0.3635, p = 0.3635
• q = f(D2D2) + f(D1D2)/2 + f(D2D3)/2 = 0.1136 + 0.0454 + 0.1022 = 0.2612, q = 0.2621
• r = f(D3D3) + f(D2D3)/2 + f(D1D3)/2 = 0.1363 + 0.1022 + 0.1363 = 0.3748, r = 0.3748

b):

• Rate of inbreeding in the population -
• the formula for inbreeding coefficient F is, F = 1 - H0/He
• where, H0 = observed heterozygote frequency
• He = expected heterozygote frequency under random mating
• here H0 = f(D1D2) + f(D2D3) + f(D1D3) = 0.0909 + 0.2045 + 0.2727 = 0.5681, H0 = 0.5681
• assumning random mating the expected heterozygote frequencies are 2pq, 2qr and 2pr
• (Reason, allelic frequencies of D1, D2 and D3 are p, q and r respectively. Hence frequency of D1D2 = p*q,
• The factor of '2' comes because D1D2 = D2D1)
• therefore, He = 2pq + 2qr + 2pr = (2*0.3635*0.2621) + (2*0.2621*0.3748) + (2*0.3635*0.3748)
• = 0.1905 + 0.1964 + 0.2725
• = 0.6594, He = 0.6594
• now the rate of inbreeding, F = 1 - H0/He = 1 - (0.5681/0.6594) = 1 - 0.8615 = 0.1385, F = 0.1385

$\rightarrow$Hence the rate of inbreeding is 0.1385

C):

• To calculate this part we assume that F is the fraction in the population which does inbreeding, then (1-F) is the fraction which does random mating,
• Hence (1-F) will produce offsprings in Hardy Weinberg proportions and F will produce offsprings which have both alleles identical by descent.
• (Look at the following table to understand it better)

Genotype	Probability of Random mating	Genotypic probability (by random mating)	prabibility of inbreeding	Genotypic probability (by inbreeding)
D1D1	(1-F)	p2	F	p
D1D2	(1-F)	2pq	F	0
D1D3	(1-F)	2pr	F	0
D2D2	(1-F)	q2	F	q
D2D3	(1-F)	2qr	F	0
D3D3	(1-F)	r2	F	r

• Now the predicted genotypic frequencies in the next generation will be, (denoted by f ' () )
• f ' (D1D1) = (1-F)*p² + F*p = (0.8615 * 0.3635 * 0.3635) + (0.1385 * 0.3635) = 0.1138 + 0.0503 = 0.1641
• f ' (D1D2) = (1-F)*2pq + F*0 = (0.8615 * 2 * 0.3635 * 0.2621) + 0 = 0.1641
• f ' (D1D3) = (1-F)*2pr + F*0 = (0.8615 * 2 * 0.3635 * 0.3748) + 0 = 0.2347
• f ' (D2D2) = (1-F)*q² + F*q = (0.8615 * 0.2621 * 0.2621) + (0.1385 * 0.2621) = 0.0592 + 0.0363 = 0.0955
• f ' (D2D3) = (1-F)*2qr + F*0 = (0.8615 * 2 * 0.2621 * 0.3748) + 0 = 0.1693
• f ' (D3D3) = (1-F)*r² + F*r = (0.8615 * 0.3748 * 0.3748) + (0.1385 * 0.3748) = 0.1210 + 0.0519 = 0.1729

$\rightarrow$note - redo the calculaions using the given formulae to cross check the answers.

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