Twogenes,
a and b, are located 10 maps units fromeach
other. a third
gene,
c, is located 15 map units from b and 5 mapunits
from a. a parental
generation
consists of aabbcc and aabbccindividuals. the f1
are then test-
crossed to
aabbcc individuals. what percentage ofthe offspring would
you
expect to be
aabbcc?

5%

Explanation:

We have the following loci map:

The parental cross was between the individuals:

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.

The homozygous recessive can only produce cba gametes.

The F1 can produce 8 types of gametes:

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:

i think the answer is g1

hope i

-lavira: )