Question

In: Biology

A fly was found that had two mutations, Q&Z. Reciprocal crosses between wild type flies and...

A fly was found that had two mutations, Q&Z. Reciprocal crosses between wild type flies and Q&Z mutants gave all Q&Z mutant F1 progeny in all cases. These results suggest that the Q&Z mutations have an autosomal dominant inheritance pattern. The genes X, Q, and Z appear to be linked. To map the genes the following crosses were done. A homozygous Q, Z mutant female was crossed to a homozygous X mutant males shown in the diagram below. X, Q, Z mutant females from the F1 generation were then crossed to homozygous, wildtype males. X, Q, and Z are dominant mutations. There were 8 different types of F2 progeny produced from this cross as listed in the table below. Using this information and that given above provide the information requested below.

1st complete the cross diagram below for this cross showing the complete genotypes for all of the flies, using standard conventions.

P: X x Q, Z Genotypes:

↓

F1 : X, Q, Z Genotype

↓

X,Q,Z x +++ Genotypes:

↓

F2 Genotypes:

2nd- Complete the table and necessary information below.

Phenotype	#	Crossover Category	Loci Involved	% Recombination
X	540
Q, Z	560
Z	230
X, Q	270
Q	140
X, Z	160
X, Q, Z	55
+++	45

3rd - On the line below, indicate the relative positions of the loci and the map distances between them in map units.

--------------------------------------------------------------------------------------------------------------------------

Solutions

Expert Solution

Answer:

1).

P: X + + / X+ + (parent 1) x (parent 2) + Q Z / + Q Z

F1: X + + / + Q Z

X + + / + Q Z x + + + / + + +

2).

Phenotype	#	Crossover Category	Loci Involved	% Recombination
X	540	Parentals	NA	NA
Q, Z	560
Z	230	Single crossovers	X& Z	(500/2000)100 = 25%
X, Q	270
Q	140	Single crossovers	Q & Z	(300/2000)100 = 15%
X, Z	160
X, Q, Z	55	Double crossovers	Q&X and X & Z	(100/2000)100 = 5%
+++	45

3).

Order of genes = Q---X---Z

Gene map = Q----------30mu--------X----------20mu--------------Z

Explanation:

Hint: Always non-recombinant genotypes are large numbered than the recombinant genotypes. Hence, the parental (non-recombinant) triple heterozygous genotypes is X + + / + Q Z

1).

If single crossover occurs between x&q..

Normal combination: X + / + Q

After crossover: X Q / + +

X Q progeny= 270+55 = 325

+ + progeny = 45+230 = 275

Total this progeny = 600

Total progeny = 2000

The recombination frequency between x & q = (number of recombinants/Total progeny) 100

RF = (600/2000)100 = 30%

2).

If single crossover occurs between Q&Z..

Normal combination: + + / Q Z

After crossover: + Z / Q +

+ Z progeny= 230+160 = 390

Q + progeny = 270+140 = 410

Total this progeny = 800

The recombination frequency between q&z = (number of recombinants/Total progeny) 100

RF = (800/2000)100 = 40%

3).

If single crossover occurs between x&z..

Normal combination: X + / + Z

After crossover: X Z / + +

X Z progeny= 160+55 = 215

+ + progeny = 45+140 = 185

Total this progeny = 400

Total progeny = 2000

The recombination frequency between x & z = (number of recombinants/Total progeny) 100

RF = (400/2000)100 = 20%

Recombination frequency (%) = Distance between the genes (mu)

Order of genes = Q---X---Z

Gene map = Q----------30mu--------X----------20mu--------------Z

gladiator answered 1 year ago

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