Concept 11 Genes get shuffled when chromosomes exchange pieces.
Cross pure-bred pea plants to identify dominant flower color.
HI! Morgan and his lab determined that Drosophila melanogaster, the fruit fly, has four linkage groups. By 1915, they had mapped almost 100 genes to these four linkage groups – see partial list on the right. The famous white eye mutation is mapped to group I, the X chromosome. Ebony body color is a recessive trait and ebony is not linked to the X chromosome (group I). What do you do in order to see if ebony body color is linked to any of the other groups? Ebony is not linked, it's a trait all on its own. Ebony is an inherited recessive trait, it will be linked to one of the linkage groups. Ebony is linked to all the groups except for group I. One gene can't be linked to multiple linkage groups. Mate an ebony fly with flies that have mutations already mapped to each of the linkage groups II, III, IV. That is correct. In order to see if one gene is linked to another, you have to introduce both genes into one animal. You have to cross the ebony fly with a fly carrying a mutation from each of the other linkage groups. You choose the apterous wing mutation from Group II; the sepia eye color mutation from Group III and the eyeless mutation from Group IV to cross with your ebony flies. [II. Apterous III. Sepia IV. Eyeless] These mutations are all recessive; apterous flies have almost no wings, sepia flies have brown eyes, and eyeless flies have no eyes. You mate ebony flies with each of these flies. What is the phenotype of the F1 progeny of each of these crosses? (Remember all the traits are recessive.) All the F1 progeny will be wild type. That is correct. If ebony is linked, the F1 progeny will be ebony. No, the F1 progeny is heterozygous for both recessive traits. All the F1 progeny will be ebony. No, the F1 progeny is heterozygous for both recessive traits. If ebony is linked, the F1 progeny will be ebony and display the phenotype of the linked mutation. No, the F1 progeny is heterozygous for both recessive traits. The F1 progeny will all have wild type phenotypes because they are heterozygous for each recessive trait. You cross the F1 progeny of each set to each other. Let's start with the apterous-ebony cross. In the F2 progeny, you see a variety of phenotypes – there are wild type; apterous; ebony; and apterous, ebony flies. The recessive traits are back. If apterous wing and ebony body color are unlinked, which of the following is most likely? Out of 100 flies you get: All wild type. No, offspring can inherit two recessive alleles; some of the flies will show the recessive phenotypes. 49 wild type; 51 ebony/apterous. No, these are only the dominant and the doubly recessive phenotypes. 24 wild type; 28 apterous 26 ebony; 22 ebony/apterous. No, these are the right phenotypes, but you won't see equal numbers of these flies. 49 wild type; 26 apterous; 25 ebony. No, you're missing one of the phenotype classes. 56 wild type; 19 ebony; 19 apterous; 6 ebony, apterous. That is correct. The 56:19:19:6 outcome in the F2 is most likely since it approximates the 9:3:3:1 Mendelian ratio of recessive dihybrid crosses. Ebony is not linked to group II with apterous wing. Now let's look at the progeny from the cross using sepia from group III. Your F1 progeny are all wild type because both ebony and sepia are recessive traits. You cross the F1 progeny to each other and in the F2 progeny you see a variety of phenotypes. What types of flies can you expect to see if sepia is not linked to ebony? ebony; sepia; wildtype; ebony, sepia. That is correct. dark brown bodies with dark pink eyes. No, unlinked or linked, genes don't blend. wild type; ebony, sepia. No, if ebony and sepia are unlinked, you should see some ebony, sepia flies. sepia; ebony, sepia. No, whether or not ebony and sepia are linked, you should expect some wild type flies. ebony; sepia. No, whether or not ebony and sepia are linked, you should expect some wild type flies. If sepia is not linked to ebony, then in the F2 progeny, you would expect to see 4 phenotypes in a 9:3:3:1 ratio. To your surprise, out of 100 flies, you get 48 wild type, 26 ebony, 24 sepia and only 2 ebony, sepia. What does this mean? Neither ebony nor sepia are on group III. No, you know sepia is linked on III. Sepia is on group III but ebony isn't. No, if ebony isn't on Group III, you would get the 9:3:3:1 ratio. Ebony is on group III but sepia isn't. No, you know sepia is on Group III. Ebony and sepia are linked, and both are on Group III. That is correct. Ebony and sepia are linked on group III. The ratio of phenotypes does not follow the 9:3:3:1 rule of independently assorting genes. Now, when you check the F2 progeny from the ebony and eyeless test cross, do you expect to see the 9:3:3:1 ratio? Yes. That is correct. No. No, by process of elimination, ebony is not linked to eyeless. You should see the 9:3:3:1 ratio again. You mapped ebony to linkage group III. In a test cross with eyeless from group IV, ebony and eyeless will segregate independently and the F2 flies will show the 9:3:3:1 ratio.