In a certain diploid plant, three loci A/a , B/b , and C/c are linked as follows
ID: 32215 • Letter: I
Question
In a certain diploid plant, three loci A/a, B/b, and C/c are linked as follows:
A/a is linked to B/b by 40 m.u.
B/b is linked to C/c by 30 m.u.
A/a is linked to C/c by 70 m.u.
One plant is available to you (call it the parental plant). It has the constitution A b c /a B C.
(a) Assuming no interference between the regions, if the parental plant is crossed with the a b c/a b c plant, how many progeny will have the phenotype of A B c if there are 1000 total progeny?
(b) Assuming 20 percent interference between the regions, if the parental plant is crossed with thea b c/a b c plant, how many progeny will have the phenotype of A B c if there are 1000 total progeny?
(c) Assuming 20 percent interference between the regions, if the parental plant is crossed with thea b c/a b c plant, how many progeny will have the phenotype of A B C if there are 1000 total progeny?
Explanation / Answer
Answer
A/a linked to B/b = 40m.u (0.4%)
B/b linked to C/c = 30 m.u (0.3%)
A/a linked to C/c = 70 m.u
a) Coefficient of coincidence (C.O.C) = observed double crossovers/expected double cross overs (1000)
1= observed double crossovers/ (0.4) (0.3) (1000)
Observed double crossover = 1 x (0.4) * (0.3) * (1000)
Each group of DCO = 120/2 = 60
The % of Single crossover of 2nd pair = 120/2 = 60
Out of 1000 plants, 60 plants would have ABc genotype.
b) Interference = 20% (0.2)
C.O.C = 1 – Interference
C.O.C = 1 – 0.2
C.O.C = 0.8
C.O.C = observed double crossovers/expected double cross overs (1000)
0.8 = x/0.4 * 0.3*1000
= 96
Each group of DCO = 96/2 = 48
Out of 1000 plants, 48 plants would have ABc genotype
c) Interference = 20% (0.2)
C.O.C = 1 – Interference
C.O.C = 1 – 0.2
C.O.C = 0.8
0.8 = x/0.4 * 0.3*1000
= 96
The SCo’s of first pair = 200 – 96
= 104
Out of 100 plants, 104 pants would have the genotype ABC genotype.