Please help my Genetics homework!! Need this urgently! Thank you! a) Imagine you

ID: 61300 • Letter: P

Question

Please help my Genetics homework!! Need this urgently! Thank you!

a) Imagine you are a cattle farmer and you are performing a dihybrid cross to determine if the alleles of Gene A and Gene B assort independently. You mate a bull with the AABB genotype with a cow that has the aabb genotype (A dominant over a; B dominant over b), cross the F1’s with themselves, and produce the following ratio of phenotypes in 160 offspring from the F2 generation:

Character A Trait/Character B Trait

89 Dominant/Dominant

28 Dominant/Recessive

32 Recessive/Dominant

11 Recessive/Recessive

Remembering that Mendel’s expected phenotype ratio is based on the assumption that the two genes’ alleles do assort independently, perform a Chi square analysis to determine if these two genes assort independently. Use 7.815 as your critical value of Chi squared.

Explanation / Answer

The frequency of individuals with the dominant genotype = 89/ 160 = 0.556 = p2; p = 0.74

The proportion of individuals with the recessive genotype = 11/160 = 0.068 = q2; q = 0.26

The frequency of individuals with the dominant genotype = 89/ 160 = 0.556 = p2; p = 0.74

The proportion of individuals with the recessive genotype = 11/160 = 0.068 = q2; q = 0.26

The proportion of heterozygous individuals, 2pq = 60/ 160 = 0.375

CHI - SQUARE (X2):

X2 = (O - E)2 / E

Where O = Observed frequency

E = Expected frequency

Genotype

Expected Numbers

Observed Numbers

AA (p2)

0.556 X 160 = 88.96

Aa (2pq)

0.068 X 160 = 10.88

aa (q2)

0.375 X 160 = 60

X2 = (89- 88.96)2 / 88.96 + (11- 10.88)2 /10.88 + (60- 60)2 /60

= 0.0017 + 0.036 = 0.0377.

The calculated Chi square value is very much less than the critical value of Chi squared (Accept null hypothesis). Thus, these genes assort independently.