Question

In Guinea pigs, short hair (S) is dominant over long hair (s), and black coat color (B) is dominant over
albino (b). Both genes assort independently. A female of a true breeding line with a black coat and long
hair is mated with a short haired albino male (also from a true breeding line). (a) What are the
phenotypes of the F1 generation? Explain. (b) If F1 is mated, which percentage of the offspring will be
homozygous for both characters? Give all genotypes and phenotypes of these homozygotes.​

Answer 1

Answer:

Explanation:

Short hair (S) is dominant over long hair (s), and black coat color (B) is dominant over albino (b).

A female of a true breeding line with a black coat and long  hair is mated with a short haired albino male (also from a true breeding line)

parents genotype      BBss  x    SSbb

gametes                       Bs     x    Sb

F1 generation geenotype    BbSs

Phenotype- They are all with black coat and short hair

Mating of F1   BbSs  x  BbSs

   Percentage of offspring homozygous for both characters:

  Bb  x   Bb           Ss   x   Ss

 BB Bb Bb bb     SS  Ss Ss ss

 1/4             1/4         1/4       1/4

1/4*1/4 = 1/16 = 0.0625 = 6% for each homozygous characters (BBSS-6% and bbss-6%)

The genotypes of the homozygotes are BBSS and bbss

The phenotype of BBSS is black coat with long hair

The phenotype of bbss is an albino with short hair