Question

You are working with an ornamental fish that shows two color phenotypes, red or blue. The color is controlled by a single gene. These fish are hermaphrodites-meaning they can either (1) self-fertilize or (2) mate with another fish. You have three fish: fish 1, fish 2, and fish 3. While working further with this fish, you discover one other mutation (in a different gene) that leads to the recessive phenotype of long dorsal fins. Further research establishes that the genes are linked and are 5 cm apart. We now have two phenotypes: 1. 2. Color: red or blue. Alleles denoted by R and r. Dorsal Fins: long (recessive) or short (dominant). Alleles denoted by D and d. In experiment A, you cross true-breeding blue fish with short fins to true-breeding red fish with long fins and get all red fish with short fins. You then cross the F1 fish to true-breeding blue fish with long fins Experiment A: blue fish with short fins x red fish with long fins (F1) red fish with short fins x blue fish with longs fins F2: ? You obtain 1200 F2 offspring. Identify the genotype for each F2 phenotype given. Finally, give the number of offspring expected with each phenotype. Possible Phenotypes Genotypes Total F2 showing this phenotype Red fish with short fins Red fish with long fins Blue fish with long fins Rr Dd Redd rrod rrDd Blue fish with short fins

Answer 1

Given that an experiment A, the crooss take place in between troue breeding blue fish with short fins whose genotype is rr DD and troue breeding red fish with long bins whosh genotype er is RR dd Mr DD X RR dd Rr Dd (El) Red fish with short fins. Robd (ED) Fish cross with troue breeding blue fish with long fins whose genotype is world - Rr Dd x mold glas RO Rd no rd rod Rod Rodd ardd modd

Therefore the genotype ratio will be 1:1.1!). If the number of fo off sbroins are 1200 Hence Possible phenotyke Total F2 showing Genotype this phenotype Red with short fins Rood 300 Red fish with long fins Rodd 300 blue fish with long fins mold 300 blue fish with short tim and 300