When a true-breeding black fish and a true-breeding yellow fish mate, all of the

Question

When a true-breeding black fish and a true-breeding yellow fish mate, all of their (F, ) offspring are yellow. If two of the F_1 offspring mate with each other, what phenotypic and genotypic ratio would you expect among their F_2 offspring? Give each allele an upper- or lower case letter & say what it stands for: black = & yellow = What are the genotypes of the two true-breeding parents? x What are the genotypes of the two _1, parents? x Punnett square for the two F_1 parents (show fractions!): Genotypic ratio for this square (must include genotypes with fractions): Phenotypic ratio for this square (must write either black or yellow): If a heterozygous fish mates with a true-breeding black fish and the female lays 182 eggs, how many of the eggs are expected to develop into black fish? What are the genotypes of the two parents? x Punnett square (show fractions):

Explanation / Answer

Answer:

(a) All the offspring are yellow. So according to law of dominance, Yellow (Y) is dominant. So the genotype of F1 is Yy.

The genotypes of the true breeding parents are: YY (Yellow) and yy (Black)

Cross: Yy * Yy

F2 genotypic ratio: YY (1) : Yy (2) : yy (1)

F2 phenotypic ratio: Yellow : Black = 3 : 1

Black : yy

Yellow: Yy or YY

(b) Yy (Heterozygous) * yy (True breeding Black)

F1:

Genotypes: Yy (1) : yy (1)

Phenotypes: Yellow : Black = 1 : 1

Total eggs = 182

Number of eggs which are expected to develop into black fish = 182 / 2 = 91

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