Question

E10 from Brooker: In fruit flies, curved wings are recessive to straight wings, and ebony body is recessive to gray body. A cross was made berween true-breeding flies with curved wings and pray bodies and flies with straight wings and ebony bodies. The F1 offspring were then mated to flies with curved wings and ebony bodies to produce an F2 generation. A. Diagram the genotypes of this cross starting with the parental generation and ending with the Frencration B. What are the predicted phenotypic ratios of the F2 generation? C. Let's suppose the following data were obtained for the F2 generation: 114 curved wings. ebony body 105 curved wings. gray body Il 1 straight wings. gray body 114 straight wings. ebony body Conduct a chi square analysis to determine if the experimental data are consistent with the expected outcome based on Mendel's laws. El from Brooker: Mexican hairless dogs have little hair and few teeth. When a Mexican hairless is mated to another breed of dog, about half of the puppies are hairless. When two Mexican hairless dogs are mated to each other, about 1/3 of the surviving puppies have hair and about 2/3 of the surviving puppies are hairless. However, about two out of eight puppies from this type of cross are bom grossly deformed and do not survive. Explain this pattern of inheritance.

Answer 1

E10. for homozygous- Curved Wings (WW); Straight Wings (ww); Gray bodies (BB); Ebony Bodies (bb)

First Generation- (curved wings and gray bodies) X (straight wings and ebony bodies)

	wb	wb	wb	wb
WB	WwBb	WwBb	WwBb	WwBb
WB	WwBb	WwBb	WwBb	WwBb
WB	WwBb	WwBb	WwBb	WwBb
WB	WwBb	WwBb	WwBb	WwBb

So, 1st generation, total 16 containing Curved Wings and Gray Bodies (WwBb).

Second Generation (F2): Curved Wings and Gray Bodies (WwBb) X Curved Wings and Ebony Bodies (WWbb)

	Wb	Wb	Wb	Wb
WB	WWBb	WWBb	WWBb	WWBb
Wb	WWbb	WWbb	WWbb	WWbb
wB	WwBb	WwBb	WwBb	WwBb
wb	Wwbb	Wwbb	Wwbb	Wwbb

So, 2nd generation, total 16 containing- 4 Curved Wings Gray Bodies (WWBb); 4 Curved Wings Ebony Bodies (WWbb); 4 Curved Wings Gray Bodies (WwBb); 4 Curved Wings E bony bodies (Wwbb).

So, total phenotypes: Curved Wings Gray Bodies- 8 and Curved Wings E bony bodies- 8.

Total number of flies after 16 X1 will be 256 flies. But here in the observed values total number of flies are- 114 + 105 + 111 + 114= 444.

So, the values are different, but for chi square test: degree of freedom =4-1 = 3 and chi square value at 5% for degree of freedom 3 -= 7.82.

The table for chi square test-

Type	Observed Frequency (O)	Expected Frequency (E)	(O-E)2	(O-E)2/E
Curved and ebony	114	128	196	1.72
Curved and gray	105	128	529	5.03
Straight and gray	111	0	111 * 111	1
Straight and ebony	114	0	114 * 114	1

so, chi square value = 8.75, which is greater than the table value 7.82.