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A newly identified mutant improves marbling in Angus cattle. An Angus stud breeder in Wagga has...

A newly identified mutant improves marbling in Angus cattle. An Angus stud breeder in Wagga has been selecting animals for improved marbling, and wants to find out whether he has inadvertently selected for this newly reported mutant. Accordingly, he genotypes animals for this mutant, and finds 200 cattle homozygous for the superior marbling allele (MM), 200 to be heterozygous (Mm), and another 100 to be homozygous wild-type (mm). You are a breeding consultant, and the stud breeder bring you this data for advice. Using Chi-square analysis, determine if the cattle population is at Hardy-Weinberg equilibrium, and based on results, state whether the breeder has inadvertently selected for the newly reported marbling allele. (3 Marks)

1 marks for deriving expectations (Explain how expectations have been derived)
1 mark for chi-square test (Show all working steps)
1 mark for correct interpretation of the chi-square statistic, and correct advice to the stud breeder (Explain your interpretation; A Chi-square table is available in you subject site - in 'Resources')

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Answer #1

Answer:

200 MM = 2 * 200 = 400 “M” alleles

100 mm = 200 “N” alleles

Mm 200 = 200 “M” and 200“m” alleles

Total alleles = 1000

Frequency of “M” allele = 400+200 /1000 = 0.6

Frequency of “m” allele = 200+200/1000 = 0.4

Three expected genotypes are produced as follows.

Total population = 500

Frequency of genotype, MM = 0.6*0.6 = 0.36

MM individuals = 0.36 * 500 = 180

Frequency of genotype, mm = 0.4*0.4 = 0.16

mm individuals = 0.16 * 500 = 80

Frequency of genotype, Mm = 2* 0.6 * 0.4 = 0.48

Mm individuals = 0.48 * 500 = 240

Phenotype

Observed(O)

Expected (E)

O-E

(O-E)2

(O-E)2/E

MM

200

180

20

400.00

2.2222

Mm

200

80

120

14400.00

180.0000

mm

100

240

-140

19600.00

81.6667

500

500

263.8889

Chi-square value = 263.89

Degrees of freedom = number of categories – 1

Df = 3-1 = 2

Critical value = 5.99

The chi-square value of 263.89 is greater than the critical value of 5.99. So we can reject the null hypothesis. Hence the population is not in Hardy-Weinberg equilibrium

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