Question

In the Jubjub bird, the gene G controls feather color. The dominant allele G gives rise to black pigmentation, and the recessive allele g gives rise to a pale gray color when homozyous. The gene D regulates size. The dominant allele D gives rise to giant jubjubs with wingspans over 5 meters, and dwarfism, with wingspans averaging 3 meters, is recessive. The ability to talk is also recessive. The dominant allele of the T locus results in mute birds. A single female jubjub bird can lay a clutch of over 900 eggs. A fully heterozygous female was mated with a dwarf, gray male who could talk. The offspring were counted. Please indicate whether any or all of these genes assort independently of one another. Map any linked loci.

Color	Adult Size	Speech	Number of Offspring
Black	Giant	Mute	5
Black	Giant	talking	17
Black	dwarf	Mute	0
Black	dwarf	talking	361
gray	Giant	Mute	379
gray	Giant	talking	0
gray	dwarf	Mute	14
gray	dwarf	talking	6

Answer 1

Answer:

All genes are present in the same chromosome and they are not at all assorted independently.

Gene map:

G----------1.22 mu--------T-----------3.44 mu--------------D

Explanation:

Based on the information, the cross is test cross. If the genes are not linked and they are assorted independently, the trihybrid testcross ratio is “1:1:1:1:1:1:1:1”

But the number of offspring produced is not in the ratio mentioned above. Therefore the genes are linked.

Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.

Hence, the parental (non-recombinant) genotypes is Gdt/gDT

Color	Adult Size	Speech	Number of Offspring	Genotype
Black	Giant	Mute	5	GDT
Black	Giant	talking	17	GDt
Black	Dwarf	Mute	0	GdT
Black	Dwarf	Talking	361	Gdt
gray	Giant	Mute	379	gDT
gray	Giant	Talking	0	gDt
gray	Dwarf	Mute	14	gdT
gray	Dwarf	Talking	6	gdt

1).

If single crossover occurs between G&d..

Normal combination: Gd/gD

After crossover: GD/gd

GD progeny= 5+17=22

gd progeny = 6+14=20

Total this progeny = 42

The recombination frequency between G&d = (number of recombinants/Total progeny) 100

RF = (42/900)100 = 4.67%

2).

If single crossover occurs between d & t..

Normal combination: dt/DT

After crossover: dT/Dt

dT progeny= 14+0=14

Dt progeny = 17+0=17

Total this progeny = 31

The recombination frequency between d&t = (number of recombinants/Total progeny) 100

RF = (31/900)100 = 3.44%

3).

If single crossover occurs between G&t

Normal combination: Gt/gT

After crossover: GT/gt

GT progeny= 5+0=5

gt progeny = 6+0=6

Total this progeny = 11

The recombination frequency between G&t = (number of recombinants/Total progeny) 100

RF = (11/900)100 = 1.22%

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

G----------1.22 mu--------T-----------3.44 mu--------------D

All genes are present in the same chromosome and they are not at all assorted independently.