Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 55P

a) All of the 4 genes are linked. This can be determined by separately examining each pair:

Between A and B :

• AB - 140+305 = 445

• ab - 145 + 310 = 455

• aB – 42+6 = 48

• Ab – 43+9 = 52

Here the parental and recombinant classes are clearly seen. Hence, the two genes are linked. The distance between the genes can be determined by the recombination frequency:

Between A and D :

• AD - 0

• ad - 0

• Ad – 43+140+9+305 = 497

• aD – 42+145+6+310 = 503

Here only the parental class is seen. No recombination is seen and from this result we can say that they are 0 m.u. apart.

Between B and C :

• BC – 42+140 = 182

• bc - 43+145 = 188

• Bc – 6+305 = 311

• bC – 9+310 = 319

Here the parental and recombinant classes are clearly seen. Hence, the two genes are linked. The distance between the genes can be determined by the recombination frequency:

Between C and D :

• CD– 42+310 = 350

• cd - 43+305 = 348

• Cd – 140+9 = 149

• cD – 145+6 = 151

Here the parental and recombinant classes are clearly seen. Hence, the two genes are linked. The distance between the genes can be determined by the recombination frequency:

b) If two pure lines have been crossed to produce the heterozygous individual their genotypes would be:

BAdc x baDC

c) Since A and D show no recombination we can remove the D gene and rewrite the progeny:

Now the data looks like a trihybrid cross with 3 genes. The class with the highest number of progeny is the parentals while the class with the lowest number is the double crossovers. By comparing these two classes we can deduce the gene order as BAC.

Now putting the progeny in proper order with proper category:

To construct the map we need map distances. The formula used to find out map distance between two genes is:

Between A and B :

Between A and C :

The linkage map can be drawn in this way:

d) Interference can be calculated with the following formula: