In rabbits,there is a gene that controls ear length,with a dominant allele "T" for long ears and a recessive allele "t" for short ears.At another gene locus,there are alleles "B" for black coat and "W" for white coat.Neither the B or W allele is dominant,and BW produces a gray coat.These two allele pairs assort independently.If a gray rabbit that is heterozygous at the gene locus controlling ear length is mated with a white rabbit that is also heterozygous at the gene locus controlling ear length,what is the probability that their first offspring will be gray with long ears?
A)1/16
B)3/16
C)1/8
D)3/8
E)1/2

Question

Quizplus · Accepted Answer

The gray rabbit has the genotype B/W T/t, and the white rabbit has the genotype W/W T/t. When these genotypes are crossed, the probability of getting a gray (BW) offspring is 1/2 (since the gray rabbit can pass either B or W, but the white rabbit can only pass W, so the only way to get gray is if the gray rabbit passes B). The probability of getting an offspring with long ears (T/_ where _ can be either T or t) is 3/4, since there are three combinations that result in long ears (TT, Tt, tT) out of four possible combinations (TT, Tt, tT, tt). Multiplying these probabilities gives 1/2 * 3/4 = 3/8.

In Rabbits,there Is a Gene That Controls Ear Length,with a Dominant