In horses, chestnut ( ‘HBHB’ ) and white ( ‘HWHW’ ) coat colours represent homoz

Question

In horses, chestnut (‘HBHB’) and white (‘HWHW’) coat colours represent homozygous genotypes constituted by autosomal alleles that are codominant. Heterozygous horses have a blend of both colours, which appears golden tan, and is called palomino (‘HBHW’). A horse stud farm in Wagga has 95 chestnut, 50 palomino and 5 white horses. Using Chi-square analysis, determine if the horse population is at Hardy-Weinberg equilibrium.

1.5 marks for deriving expectated number of each genotype (Explain how expectations have been derived)

1 mark for performing chi-square analysis (Show all working steps)

0.5 mark for correct interpretation of the chi-square statistic derived in context of the stipulated hypothesis (Explain your interpretation; A Chi-square table is available in you subject site - in 'Resources')

Explanation / Answer

Expected number of genotype

Let the frequency of HB allele = p

frequency of HW allele = q

Now from given population, p2 = 95/150 = 0.6333

q2 = 5/150 = 0.033

2pq = 50/150 = 0.333

For codominant alleles, allelic frequency of p = frequency of HBHB + 1/2 of frequency of HBHW

allelic frequency of p = 0.6333 + 1/2 (0.333) = 0.799 = 0.8

allelic frequency of q = 1-p = 0.2

Chi square analysis

Interpretation:

Chi square is 0.2604

degrees of freedom = 2

Table value is 5.991 at 95% level of confidence and 4.605 at 99% level of confidence

Our value is 0.2604 which is less than both table values

So, null hypothesis is accepted and the given population is in Hardy Weinberg equilibrium.

Genotype HBHB HBHW HWHW Total Observed horses (O) 95 50 5 150 Expected horses (E) p2 x 150 = (0.8)2 x 150 = 0.64 x 150 = 96 2pq x 150 = 2 x 0.8 x 0.2 x 150 = 48 q2 x 150 = (0.2)2 x 150 = 0.04 x 150 = 6 150 (O-E)2/E (-1)2/96 = 0.0104 (2)2/48 = 0.0833 (-1)2/6 = 0.1667 Chi square   0.2604

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