# Genetics Based Question - HW equation and Chi square?

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 brings you this data for advice.

1. derive expectations (Explain how expectations have been derived)

2. chi-square test (Show all working steps)

### 1 Answer

- jacob sLv 71 month agoFavourite answer
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 2.222

Mm 200 80 120 14400 180.000

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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