A spotted rabbit, when crossed with a solid-colored rabbit produced all spotted

Question

A spotted rabbit, when crossed with a solid-colored rabbit produced all spotted offspring. When these F1 rabbits were crossed among themselves they produced 32 spotted rabbits and 10 solid rabbits.
a) Which of these characters is dependend upon a dominant gene?
b) how many of the rabbits which were spotted in the F1 would be expected to be homozygous?
c) how many of the rabbits which were spotted in the F2 would be expected to be homozygous?
d) how many of the solid-colored rabbits in the F2 would be homozyous?
e) How could you most easily determine which of the spotted rabbits in the F2 generation were homozygous and which were heterozygous?

Explanation / Answer

In order for all the F1 to have the same phenotype, the parents have to have to be homozygous. Since they are all spotted, the spotted allele must be dominant, so the spotted parent has athe genotype SS and produces S gametes. The solid parent has the genotype ss and produces s gametes, so the Punnett square is: _____S___ s___Ss___?all ofspring are heterozygous, spotted. Crossing these, the Punnett square is: _____S_____s___ S___SS____Ss___ s___Ss____ss___ Since the final numbers are 32 and 10, the ratio is about 3:1, which is expected in the F2. The spotted rabbits are SS and Ss; 1/3 of the spotted rabbits are homozygous, SS; all of the solid rabbits in the F2 would be homozygous, ss.

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