Question

2. You cross a homozygous wildtype female Drosophila (fruit fly) with a male that has a black body (bb). You mate two of the F1 (first generation) flies and examine the F2 (second generation) offspring. The phenotypes can be seen in the table below. a. Draw a pedigree to show the P, F1, and F2 generations. You only need to draw one fly with each phenotype for the F2 generation. seen in this cross. the expected values for a cross done with your hypothetical type of b. Come up with a hypothesis about the type of inheritance c. Use a chi-squared test to see if your hypothesis is correct by comparing inheritance and the observed values from the table below Phenotype Number of F2s 379 123 wildtype black body

Answer 1

Answer:

Wildtype (BB) x Black (bb)----P

Bb (wildtype)---------------------F1

Bb x Bb --------------------------F1 selfcross

	B	b
B	BB (wildtype)	Bb (wildtype)
b	Bb (wildtype)	bb (blakck)

This is an autosomal inheritance patterns. Wild type is dominant over blackbody.

Total progeny = 502

Expected wildtype = ¾ * 502 = 376.5

Expected black body = ¼ * 502 = 125.5

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
Wildtype	379	376.5	2.5	6.25	0.02
Black body	123	125.5	-2.5	6.25	0.05
	502	502			0.07

X^2 = 0.07

Degrees of freedom = number of categories – 1

Df = 2-1=1

The X^2 value of 0.07 is less than the critical value of 3.84. So we accept null hypothesis. The observed values are almost correct with expected values.