Question

In fruit flies, curved wings (c) are recessive to straight wings (C), and ebony body (g) is recessive to gray body (G). A cross was made between true-breeding flies with straight wings and gray bodies and flies with curved wings and ebony bodies. The F₁offspring were then mated to flies with curved wings and ebony bodies to produce an F₂ generation (test-cross). The following offspring was observed:

114 curved wings, ebony body

105 curved wings, gray body

111 straight wings, gray body

114 straight wings, ebony body

A.    Complete a Punnett square representing the test-cross specified above. Make sure you specify the genotype and phenotype in each square of your Punnett square. Complete this by hand, take a picture and upload your work in the folder "SHOW YOUR WORK" no later than 10 minutes after you complete your exam.

B. In the space provided below and assuming independent assortment, determine the expected phenotypic ratio of the test-cross described above. 3 marks

C. Propose a null and an alternative hypothesis regarding the assortment of the genes that determine shape of the wings and color of the body . In the space below, clearly identify the null hypothesis as C.1 and the alternative hypothesis as C.2.

D. Conduct a chi-square analysis to determine if the experimental data are consistent with the expected outcome based on Mendel’s laws and your Punnett square constructed in A. Clearly specify the following:

- Equation

- Chi-square calculation

. Complete this by hand, take a picture and upload your work in the folder "SHOW YOUR WORK" no later than 10 minutes after you complete your exam. In the space below, clearly specify the Chi-square value calculated.

E. Determine the degrees of freedom. This is a simple calculation. Thus, in the space provided below, specify the equation and the result. .

F. Are the experimental data consistent with the expected outcome based on Mendel’s laws? In the space below answer yes/no and provide a one or two-sentence explanation of your results.

Answer 1

Ans:

A. The complete Punnett square representing the test cross is given below-

B. The expected phenotypic ratio is given below-

C. The null and alternative hypothesis are-

C.1 Null hypothesis: There is no significant difference between observed and expected value, so, gene C and G is assort independently.

C.2 Alternative hypothesis:There is significant difference between observed and expected value, so, gene C and G is linked.

D. Chi-square analysis is given below-

The calculated chi-square value is 0.486.

E. Degree of freedom = n - 1 (where, n= number of phenotypic classes)

So, DF= 4 - 1 = 3.

F. Yes, the experimental data are consistent with the expected outcome based on Mendel’s laws.

The calculated chi-square value 0.486 is much less than the table value 7.8 at 3 DF (5% probability level). The deviation between observed and expected value is not significant and therefore, the null hypothesis is accepted. So, the gene C and G is assort independently.