Two color traits of chickens are black ie black colored feat

Two color traits of chickens are black, i.e. black colored feathers, and “splash” in which the bird is mostly white with just a few black feathers. These two traits are due to alleles of the same gene; B1B1 chickens have splash phenotype and B2B2 chickens have black phenotype. When you cross true-breeding black and true-breeding splash all of the offspring are “blue” indicating that there is an incomplete dominance relationship between black and splash. In other words B1B2 has a blue phenotype.

Suppose that F1 generation chickens are mated with each other to produce 800 F2 generation offspring. Among these, the following is observed in the F2 generation: 225 black, 395 blue,180 splash

A. How many chickens of each color are expected in the F2 generation if the trait is governed by one gene with two alleles having incomplete dominance as described above?

B. Use the Chi-squared test to see if the observed data fit these expectations, and give all of the following:

State the null hypothesis and alternative hypothesis:

Calculate the Chi-squared value:

Estimate the p-value:

Based on these results do you reject or fail to reject the null hypothesis?

Finally, the interpretation: Do you accept that the observed data fit expectations?

Solution

A) Gneotype frequency using the given data (total population size= 800)
B1B1 (p²)= 225/800=0.281
B2B2 (q²)= 180/800=0.225
B1B2 (2pq)= 395/800=0.493
Allele frequency=
Frequency of B1= p = p² + 1/2 (2pq) = 0.281 + 0.493/2= 0.5275
q= 1-p= 1- 0.5275= 0.4725
According to Hardy-Weinberg law, expected genotype frequency:
B1B1 (p2)= (0.5275)²= 0.278
B1B2 (2pq)= 2x 0.5275 x 0.4725=0.498
B2B2 (q2)= (0.4725)²=0.223
Expected genotype numbers
B1B1= 0.278 x 800=222.4
B1B2= 0.498 x 800=398.4
B2B2= 0.223 x 800=178.4