Consider a diploid fish in which tail length is solely controlled by the additive action of two unlinked loci,F (f)and G (g),without any effect of the environment.A female of genotype F/F;G/G,with a 3 cm long tail,is crossed to a male of genotype f/f;g/g,with a 2 cm long tail.All the F₁ fish have 2.5 cm long tails. a)If the F₁ fish are crossed to each other,how many different tail length phenotypes will be present in F₂? A)16 B)5 C)4 D)3 E)2

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Quizplus · Accepted Answer

The F₁ fish have the same phenotype, regardless of their genotype, which is 2.5 cm long tail. When two of these fish are crossed to each other, their offspring will have the following possible genotypes: F/F;G/G, F/F;g/g, f/f;G/G, f/f;g/g (in a 1:2:2:1 ratio, respectively). Using a Punnett square, we can determine the tail lengths of these offspring: - F/F;G/G: 3 cm - F/F;g/g: 2.5 cm - f/f;G/G: 2.5 cm - f/f;g/g: 2 cm Therefore, there are four different tail length phenotypes present in F₂ offspring: 3 cm, 2.5 cm (two genotypes), and 2 cm.

Consider a Diploid Fish in Which Tail Length Is Solely