In Labrador retrievers, black color coat (B/–) is dominant to brown color coat (

Question

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:

what is the probability that the first born will be brown?

what is the probability all of them will be brown?

what is the probability that the first born will be black?

what is the probability all of them will be black?

what is the probability that at least one of them will be brown (not all them are black)?

Explanation / Answer

Since, this is the case of complete dominance, we will abide by the Mendelian Laws of inheritance. By the given data, it is obvious that the parent is heterozygous for Black coat color as it has produced Brown puppies in the past. Therefore, the possible genotypic and phenotypic outcomes can be predicted using Punett square method. Here, Bb is genotypic representation for heterozygous black and bb for homozygous brown(recessive) . B b B BB Bb b Bb bb The probability that the puppy will be brown is 1/4 while the probability that the puppy is black is 3/4. (a) The probability that first puppy born is brown is 1/4 since, only one ovum matures out of four gametes formed. (b) The probability that all of them are brown is 1/4*1/6=1/24 (c) The probability that the first born is black is 3/4. The logic applied is same as (a) . (d) The probability that all of the them will be black is 3/4*1/6= 3/24. The probability that at least one of them will be brown is 1/6.

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