Biology 352 Activity Section 3 Spring 201 To create a linkage map, we will begin

Question

Biology 352 Activity Section 3 Spring 201 To create a linkage map, we will begin with the end products of the experiment: the phenotypes of the progeny from the testcross. All examples used in the book and lecture will concern genes with complete dominance. We will consider the example from your textbook, which concerns a breeding experiment with Drosophila (Figure 7.13). The three genes are: 7) st' st and st st e e and ee ss+ ss and ss ss normal eyes -normal body color = normal bristles st st scarlet eyes ee ebony body color ss ss spineless Wild type Scarlet, ebony ste ss st e ss e ss Testcross In this example and others, the experimental data you have access to are the phenotynes, not the genotypes. If you look at Figure 7.13, you'll see the following shorthand notation used for the phenotypes, just under each fly picture: st normal eyes e normal body color ss+ = normal bristles st scarlet eyes e ebony body color ss spineless Note that in principal, each dominant phenotype in the left column could be homozygous or heterozygous. But you will never have a homozygous dominant genotype in the progeny of a 3-point testcross. Why? a. 36

Explanation / Answer

The parent is heterozygous so it has both wild type and mutant type. Then it undergoes test cross in this also the cross occurs between wildtype and recessive genotype

                                      st+ e+ ss+       and      st    e    ss

                                      st    e    ss                     st    e    ss

So, the offsprings are,

      st+ e+ ss+     / st e ss     and st e ss / st e ss       No recombinant similar to parent

      st e+ ss+      /   st e ss    and st+ e ss / st e ss      Single cross over between st and e

      st+ e+ ss     / st e ss     and st e ss+ / st e ss       Single cross over between e and ss

      st+ e ss+     / st e ss     and st e+ ss / st e ss       Double recombinant

In all the offspring we have the recessive genotype with it st e ss . That's why it not a homozygous dominant genotype.

                                     

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