Question

For a heterozygous individual with the genetic map and gene arrangement below:

A--------18 m.u.---------b-----------20 m.u.------------C

a---------------------------B---------------------------------c

What percentage of gametes would have the following genotypes?

1. AbC
2. ABC
3. Abc

Answer 1

Since nothing is given about the interference or Coefficient of coincidence So we will take the Coefficient of coincidence as 1.
Coefficient of coincidence = Observed frequency of double recombinant /expected frequency of double recombinant
So, The observed frequency of double recombinant =expected frequency of double recombinant.
Map distance between A and b is 18 cM
Map distance between b and C is 20 cM
Expected frequency of double recombinant is given by
Map distance between A and b /100 * map distance b and C/100
18/100 * 20/100 = 0.036
The observed frequency of double recombinant =expected frequency of double recombinant.
So observed frequency of double recombinant = 0.036
The observed frequency of double recombinant = sum of the frequency of both the double recombination.
Double recombinants are the progeny whose middle gene is exchanged.
So genotype of double recombinants is ABC and abc.

Frequency of ABC = observed frequency of double recombinant/ 2 = 0.036 /2 = 0.018.

The distance between A and b is 18 cM.
Total number of recombinant/total number of progeny = frequency of recombinant.
The frequency of recombination between A and b will be a sum of frequencies of aBC, AbC, ABC and abc
Linkage distance = Frequency of recombinants * 100
18 = Frequency of recombinants * 100
Frequency of recombinants = 18/100 = 0.18.
Frequency of aBC, AbC = 0.18 - Freqeucny of double recombiants (ABC and abc) = 0.18 - 0.036 = 0.144
Frequency of aBC, AbC = 0.144

The distance between A and b is 18 cM.
Total number of recombinant/total number of progeny = frequency of recombinant.
The frequency of recombination between A and b will be a sum of Frequencies of Abc, aBC, ABC and abc
Linkage distance = Frequency of recombinants * 100
20= Frequency of recombinants * 100
Frequency of recombinants = 20/100 = 0.20.
Frequency of aBC, AbC = 0.20 - Freqeucny of double recombiants (ABC and abc) = 0.20- 0.036 = 0.164
Frequency of Abc, aBC, = 0.164.
Frequency of Abc = 0.164 / 2 = 0.082

Frequencies of AbC and aBc = 1 - Sum of the frequency of all recombinants = 1 - (0.036 + 0.144+0.164)
= 1 - 0.344
Frequencies of AbC and aBc = 0.656
Frequency of AbC = 0.656/2 = 0.328.

% of gametes = Frequency of genotype * 100
AbC = 0.328 * 100 = 32.8%
ABC = 0.018 * 100 = 1.8%
Abc = 0.082*100 = 8.2%

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