(3) With the loss of habitat increasing, a population of African wild dogs (Lyca

Question

(3) With the loss of habitat increasing, a population of African wild dogs (Lycaon pictus) in South Africa has become a unit of concern for wildlife management officials. To assess diversity in this population (Kruger National Park), the following genetic data were collected. Genotype D1D1 D1D2 D1D3 D2D2 D2D3 D3D3 Kruger NP 8 4 12 5 6 b) What is the rate of inbreeding in this population? [2 marks] c) What are the predicted genotype frequencies in the next generation assuming the same deviation from Hardy-Weinberg, as calculated in part b)? [2 marks]

Explanation / 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
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)

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
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)

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

note - redo the calculaions using the given formulae to cross check the answers

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

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