exercise of gene linkage in genetics

Question

Could you help me solve this problem:

AAbbCC is crossed with aaBBcc and the offspring re-crossed with aabbcc. Considering that A and B are linked and their distance is 12 mapping units, while the gene C is not linked to them, find the percentage of the offspring that will have phenotype aabbcc.

I used this method to solve it (I know it's not correct but I don't know where I am mistaken):

P: AAbbCC x aaBBcc

F1: AaBbCc

AbBbCc x aabbcc

Since the distance between A and B is 12, their recombinant frequency is 12%

88% then is the recombinant frequency of the parental

Dividing by 2, 44% is the RF of parental AB/ab and 44% of parental aa/bb

In conclusion, since gene C is not linked, I calculated the probability to obtain cc from CC x cc, and that's 1/2

So my final result is: 0,44 x 0,5 = 0,22 = 22%

Answer 1

You did right for most of the exercise, but you got confused on one of the (apparently) easiest things.

Try to reason in terms of chromosomes, and not merely of genotypes:

F1 inherit their chromosomes from two homozygous parents, so you know that they are AbC and aBc.

You can visualize the cross with the recessive homozygote as follows:

Ab C     ab c

-- -  X  -- -

aB c     ab c

The second parent will always pass abc, so in order to have an offspring with genotype aabbcc you only need that parent 1 passes abc.

Given that the parentals are Ab and aB and that RF is 12% (0.12), regarding the first two loci, the frequencies of the gametes are

Ab = 0.44 
aB = 0.44 
AB = 0.06 
ab = 0.06

Including locus C, the frequency of the gamete abc is then simply 0.06 * 0.5 (because c is unlinked), and that corresponds to 3%.