Question

9. You are studying the inheritance of three genes in the scorpion fly Panorpa communis. You have identified three different autosomal recessive traits, lavender eyes, green body and pointed wings. These phenotypes are caused by alleles in thee different genes, x, y, and z respectively. Wild type flies have red eyes, brown body, and straight wings (x+, y+ and z+). You cross two different pure breeding and obtain a wild-type F1. You then cross the F1 to a true-breeding line that has all three recessive traits and obtain 100 progeny whose numbers and phenotype are listed below:

Phenotype	Number
Wild type	3
Lavender eyes, green body, pointed wings	7
Lavender eyes, pointed wings	34
Green body	36
Green body, pointed wings	8
Lavender eyes	12

1. Using the genetic notation given in the question, write the proper genotypes of the two parental pure breeding lines? (1 marks)
2. How are the three genes located relative to each other (2 marks)
3. Using the map distances you calculated above in part [B], if F1 flies were intercrossed, what frequency of the resulting F2 would have lavender eyes, green body and pointed wings? (2 marks)

Answer 1

A).

Phenotype	Number	Genotype
Wild type	3	x+ y+ z+
Lavender eyes, green body, pointed wings	7	x y z
Lavender eyes, pointed wings	34	x y+ z
Green body	36	x+ y z+
Green body, pointed wings	8	x+ y z
Lavender eyes	12	x y+ z+

Always non-recombinant genotypes are large numbered than the recombinant genotypes.

Hence, the parental (non-recombinant) triple heterozygous genotypes is x y+ z / x+ y z+

x y+ z / x y+ z (paent 1) x (parent 2) x+ y z+/x+ y z+ --Paents

x y+ z / x+ y z+ (wild type)-----------------------------------F1

x y+ z / x+ y z+ (F1) x (tester paent) x y z / x y z ---Testcross

B).

Order of genes = y—x----z

Gene map = y----------10mu--------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 y+ z / x+ y z+

x y+ z / x y+ z (paent 1) x (parent 2) x+ y z+/x+ y z+ --Paents

x y+ z / x+ y z+ (wild type)-----------------------------------F1

x y+ z / x+ y z+ (F1) x (tester paent) x y z / x y z ---Testcross

1).

If single crossover occurs between x&y..

Normal combination: xy+/x+y

After crossover: xy/x+y+

xy progeny= 7

x+y+ progeny =3

Total this progeny = 10

Total progeny = 100

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

RF = (10/100)100 = 10%

2).

If single crossover occurs between y&z..

Normal combination: y+z/yz+

After crossover: y+z+/yz

y+z+ progeny= 3+12 = 15

yz progeny = 7+8 = 15

Total this progeny = 30

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

RF = (30/100)100 = 30%

3).                                        

If single crossover occurs between x&z.

Normal combination: xz/x+z+

After crossover: xz+/x+z

xz+ progeny= 12

x+z progeny = 8

Total this progeny = 20

Total progeny = 100

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

RF = (20/100)100 = 20%

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

Order of genes = y—x----z

Gene map = y----------10mu--------x-----------20mu--------------z

C).

If F1 progeny are selfcross, then the probability of lavender eyes, green body and pointed wings = y x z/ y x z = 4% * 4% = 0.0016 = 0.16%

Explanation:

Gene map = y----------10mu--------x-----------20mu--------------z

Expected double crossover frequency = Frequency between y&x * frequency between x & z

= 10% * 20 = 0.02 = 2%

Triple zygote (y+ x z/ y x+ z+) produces xyz gametes when single corss over occurs between y & x. Then the recombination frequency between y&x = 10 – 2 = 8%

When single corss over occurs between y & x, it produces two types of gametes, y x z & y+ x+ z+ in 8%. Each gamete is in 4%.

If F1 progeny are selfcross, then the probability of lavender eyes, green body and pointed wings = y x z/ y x z = 4% * 4% = 0.0016 = 0.16%