Question

Task- genetics

The genes for body color (A), eye color (B) and wing length (C) are linked to the same autosomal chromosome of D. melanogaster, in the order of A - B - C. For body color, alleles encode 'A +' for wild type and 'a' for black body; for eye color, alleles encode 'B +' for wild type and 'b' for pink eyes; For wing length, alleles code 'C +' for wild type and 'c' for short wings. Wild-type alleles ('A +', 'B +', 'C +') dominate the mutant alleles (a, b, c) and the genetic distances between loci are determined at 10 cM (A-B) and 20 cM (B - C).

A pure-eyed female with black body color, short wings and wild-type eyes is crossed to a male with pink eyes and wild-type phenotype for the two other characteristics of the parent generation. The F2 generation is created by crossing F1 females to males from a strain expressing all three recessive mutant properties (analysis crossing).

a) What is the expected frequency of double crossover among the F2 offspring?

b) What is the allelion in the F1 female (indicate as a chromosome sketch)?

c) What phenotypes will be observed among the offspring in F2 and in what conditions if the interference is 1?

d) What phenotypes will be observed among the offspring in F2 and in what conditions if the interference is 0?

Answer 1

a) Please find the DCO proportions in part c and part d.

b)

c)

The steps involved in this problem:

1. Calculation of DCO from COC
2. Calculation of SCO and
3. Calculation of PCO.

1. Calculation of DCO freequency :

As, I + COC = 1, if I =1 then the COC = 1-I = 1-1 = 0.
COC = 0.0.
COC = Prop of ODCO/ Prop of EDCO.
=> 0 = Prof of ODCO/ (10/100 X 20/100) => DCO = 0 X 0.1 X 0.2 = 0

2. Calculation of SCO:.
The Distiance between two genes= Prop of SCO+ Prop of DCO.
So, Prop of SCO = Distiance between two genes - Prop of DCO.
The freequency = Prop of SCO * Total no of progeny

Between first pair of genes =
The Distance = 10, The proportion = 10/100 = 0.1
Prop of SCO = 0.1- 0 = 0.1
The Freequency of each SCO category = 0.1/2 = 0.05

Between second pair of genes =
The Distance = 20, The proportion = 20/100 = 0.20
Prop of SCO = 0.2- 0 = 0.2
The Freequency of each SCO category = 0.2/2 = 0.1

3. Calculation of PCO:
Parental combinations = Total freequency - SCO - DCO
1- 0.1 - 0.2 = 0.7
The Freequency of each Parental category = 0.7/2 = 0.35

d)

The steps involved in this problem:

1. Calculation of DCO from COC
2. Calculation of SCO and
3. Calculation of PCO.

1. Calculation of DCO freequency :

As, I + COC = 1, if I =0 then the COC = 1-I = 1-0 = 1.
COC = 1.0.
COC = Prop of ODCO/ Prop of EDCO.
=> 1.0 = Prof of ODCO/ (10/100 X 20/100) => DCO = 1 X 0.1 X 0.2 = 0.02

As each category contains two genotypes, the Freequency of each DCO category = 0.02/2= 0.01

2. Calculation of SCO:.
The Distiance between two genes= Prop of SCO+ Prop of DCO.
So, Prop of SCO = Distiance between two genes - Prop of DCO.
The freequency = Prop of SCO * Total no of progeny

Between first pair of genes =
The Distance = 10, The proportion = 10/100 = 0.1
Prop of SCO = 0.1- 0.02 = 0.08
The Freequency of each SCO category = 0.08/2 = 0.04

Between second pair of genes =
The Distance = 20, The proportion = 20/100 = 0.20
Prop of SCO = 0.2- 0.02 = 0.18
The Freequency of each SCO category = 0.18/2 = 0.09

3. Calculation of PCO:
Parental combinations = Total freequency - SCO - DCO
1-0.08-0.18-0.02 = 0.72
The Freequency of each Parental category = 0.72/2 = 0.36