Question

A geneticist crossed pure breeding black mice with pure breeding brown mice. All the 992 mice in F1 generation had black coats. When these mice were crossed, the yielded 961 black coated mice and 317 brown coated mice. How could you account for the ratio of black coated to brown coated mice in the F2 generation?

Answer 1

Answer/Explanation:

The pure breeding (homozygous) black mice are crossed with pure breeding (homozygous) brown mice. The F1 offspring are all black. A punnett square is shown, with B representing the black allele, and b representing the brown allele. The cross is therefore BB x bb

The offspring are 100% heterozygous black (Bb)

Crossing two F1 heterozygous black mice (Bb x Bb) to give the F2 generation. The genotypes are 1 BB: 2 Bb: 1 bb. Therefore, the genotypes are 3 black : 1 brown.

This accounts for the real ratios shown in the F2 generation (992:317 is roughly 3:1

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