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.

B. 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 . Cearly 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

E. Determine the degrees of freedom.

F. Are the experimental data consistent with the expected outcome based on Mendel’s laws?

Answer 1

•Wings :- straight wings (C); curved wings (c)

•Body :- gray body (G); ebony body (g)

•true-breeding flies:-

1.straight wings and gray bodies :- CCGG

2.curved wings and ebony bodies:- ccgg

3.All progeny of F1 is :- CcGg : straight wings and gray bodies

•Test cross of F1 with recessive parent should produced the offspring in ratio 1:1:1:1.

Answer A:- Complete a Punnett square representing the test-cross .

A test cross having the parental genotype :- CcGg   X ccgg.

•Parent - CcGg X ccgg

•Gametes -                    CG, Cg, cG, cg     x      cg

•Let us keep in Punnett square .

	CG	Cg	cG	cg
cg	CcGg straight wings and gray bodies	Ccgg straight wings and ebony bodies	ccGg curved wings and gray bodies	ccgg curved wings and ebony bodies

Answer B :- according to Mendel’s law of independent assortment , the above test cross , resulted four different type of progeny in 1:1:1:1 ratio.

straight wings & gray bodies : straight wings & ebony bodies: curved wings &gray bodies : curved wings & ebony bodies.

CcGg :                     Ccgg                     :                     ccGg                :                       ccgg

Answer C :- the hypothesis are-

C1 :- null hypothesis :- The given cross follow the law of independent assortment and the test cross resulted progeny are   in the ratio 1:1:1:1 ratio . [ 1/4: 1/4: 1/4: 1/4 ]

C2:- alternative hypothesis :- the test cross result is not followed the Mendel’s law of independent assortment. So the test cross progeny are not in 1:1:1:1 ratio.

answer D :- chi squre analysis - see the photo :-

Answer E:- the degree of freedom is 3 for this problem .[see photo.]

Answer F :- for the 3 degree of freedom and 0.05 probability of critical exceeding valeu [from standard table], :- 7.815.... when we compare our calculatede valuse [0.48], we found that our calculated value is less than the table value so we can say that our null hypothesis is accepted. this cross is following mendel's law.