:1 asked in Science & MathematicsBiology · 1 month ago

# 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)

Relevance
• 1 month ago

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