Answer:
5%
Step-by-step explanation:
We have the following loci map:
C/c -------------A/a--------------------------B/b
5 m.u. 10 m.u.
The parental cross was between the individuals:
- CCbbAA, which can be written as CAb/CAb.
- ccBBaa, which can be written as caB/caB.
Each parental individual can produce only 1 type of gamete, so the F1 will be homogeneous with the genotype: CAb/caB.
The F1 are test crossed to cba/cba individuals.
CAb/caB X cab/cab
The homozygous recessive can only produce cba gametes.
The F1 can produce 8 types of gametes:
- The parentals: CAb and caB
- The crossovers between the genes C/c and A/a: CaB and cAb
- The crossovers between the genes A/a and B/b : CAB and cab
- The double crossovers: Cab and cAB
The question is asking about the percentage of offspring that will have the genotype CAB/cab
The cab chromosome comes from the homozygous recessive individual with a probability of 1.
The CAB chromosome comes from the F1 individual, and was a result of crossing over between the genes A/a and B/b.
The formula to relate genetic distance with recombination frequency is:
Genetic Distance (m.u.)= Recombination Frequency X 100.
In our problem, Genetic distance between A/a and B/b loci is 10 map units.
Therefore:
10 m.u. = (Recombination Frequency between A/a and B/b) x 100.
0.1 = Recombination Frequency between A/a and B/b
When crossing over happens between the A/a and B/b genes, both CAB and cab are generated, so each of them will appear in a frequency of half the total recombination frequency between those genes, to add a total of 0.1.
The gamete CAB will appear with a frequency of 0.05.
The gamete cab will appear with a frequency of 1.
The percentage of the offspring that will be CAB/cab is:
0.05 x 1 x 100% = 5%