In Labrador retrievers black color coat B is dominant to bro

In Labrador retrievers, black color coat (B/-) is dominant to brown color coat (b/b). A breeder crosses two black individuals who have previously produced some brown puppies. If the cross produces six puppies:

a) What is the probability that the first born will be brown?

b) What is the probability that at least one of them will be brown?

Solution

Assume that the gene coding for black colour coat is B and the gene coding for brown color coat is b; B is dominant over b. Black individuals are heterozygous because they previously produced some brown puppies.

Cross between two black individuals (Bb) will have the progeny with the following genotypes,

Bb* Bb -à B_ (3/4, black), bb (1/4, brown)

(a). The probability of first born will be brown is, ¼

(b). The probability of at least one of them will be brown (of the total six puppies) is = 1 - probability of having no brown puppy = 1-(3/4)^6 = 1- 729/4096 = 3367/4096