Question

"Ebony body, held-out wings and sepia eyes are ALL RECESSIVE AND ALL ON CHROMOSOME 3. The COEFFICIENT of COINCIDENCE for chromosome 3 is 0.5. A homozygous ebony male is mated to a homozygous held out wing and sepia eyed female. The F1 are wild-type. An F1 female is test crossed to a fly that is homozygous for all three recessive alleles and they produce 1000 progeny. If ebony and held-out are separated by 12 map units, and held-out and sepia are 14 map units apart, and if the order of these loci is ebony, held-out, and sepia: How many ebony, sepia (but not held-out) progeny are expected from this test cross?" 55.8 8.4 65.8 131.6 374.2 QUESTION 19 How many progeny expressing ebony only are expected? 55.8 8.4 4.2 65.8 131.6 374.3

Answer 1

Answer:

18). 65.8

19). 374.2

Explanation:

e + + / e + + X + h s / + h s ---Parents

e + + / + h s (wild type)--------F1

e + + / + h s (F1) x e h s / e h s (tester)-----Testcross

COC = 0.5

COC = Observed double crossover progeny / Expected double crossover progeny

Expected double crossover progeny = 12% * 14% = 0.0168

0.5 = Observed double crossover progeny / 0.0168

Observed double crossover progeny = 0.0168 * 0.5 = 0.0084 or 0.84%

Class of gametes	Phenotype	Frequency of reciprocal pair	Numbers	Frequency of each class	No. of each progeny = Frequency of each class * 900
Parental	e + +	1- all recombinant progeny	100%-(13.16% + 11.16%+0.84%) =74.84%	37.42%	374.2 (ebony body)
	+ h s			37.42%	374.2 (held out wing, sepia eye)
SCO 1	e h s	RF in region 1 = SCO1 – DCO	SCO1 = 12%- 0.84 = 11.16%	5.58%	55.8 (ebony body, held out wing, sepia eye)
	+ + +			5.58%	55.8 (wild type)
SCO 2	e + s	RF in region 2 = SCO2 – DCO	SCO2 =14% - 0.84 = 13.16%	6.58%	65.8 (ebony body, sepia eye)
	+ h +			6.58%	65.8 (held out wing)
DCO	e h +	(RF in region 1) * (RF in region 2)	0.84%	0.42%	4.2 (ebon bodyy, held out wing)
	+ + s			0.42%	4.2 (sepia eyes)

Number of offspring expected to have the following phenotypes:

Ebony body, sepia eye (but not held out wing) = 65.8

Ebnoy body only = 374.2