Black coat colour in Labrador Retrievers is produced by the dominant B allele. Brown labs are homozygous recessive (bb). Duke, a black male, whose mother was brown, is mated with Shadow, a black
female, who in a previous mating with another father produced brown offspring. What is the probability
that when Duke and Shadow are mated…
a. Their first pup is brown?
b. They have 3 male pups and at least one is black?
c. They have 3 black pups (regardless of sex)?
Please help me, I have answers but I am completely unsure if they are right. My answers are a)1/4 chance, b) 7/8 chance, and c)3/4. Please show how you get your answers so I know what I am doing wrong. Please and thank you
To determine the probabilities, we need to analyze the genetic principles involved in coat color inheritance in Labrador Retrievers.
Let's break down each question and calculate the probabilities step-by-step:
a. Their first pup is brown?
Duke is a homozygous dominant black male (BB) since black coat color is produced by the dominant B allele. His mother was brown, meaning she must have been homozygous recessive (bb).
For Duke to be black, he must have received a dominant B allele from his mother, and for a pup to be brown, it must inherit two recessive b alleles (one from each parent). Therefore, Duke must be heterozygous (Bb).
Shadow, the black female, must also be heterozygous (Bb) since her previous mating produced brown offspring when coupled with another father.
When Duke and Shadow mate, there is a 25% chance that both will pass on their recessive b alleles to their first pup. Therefore, the probability that their first pup is brown is 1/4 or 25%.
So, your answer for part a) is correct: 1/4 chance or 25%.
b. They have 3 male pups, and at least one is black?
To calculate this probability, we need to consider different combinations of genotypes in male pups that result in at least one black pup.
There are three possible ways to have at least one black male pup:
1) Black, Black, Black
2) Black, Black, Brown
3) Black, Brown, Black
Let's calculate the probability for each case:
1) The probability that the first male pup is black is 100% or 1.
2) The probability that the first male pup is black is 100%, and the probability that the second male pup is also black is 100%. The probability that the third male pup is brown is 1/4 or 25%. So, the probability that at least one of the three male pups is black is 100% * 100% * 75% = 75%.
3) The probability that the first male pup is black is 100%, and the probability that the second male pup is brown is 1/4 or 25%. The probability that the third male pup is black is 100%. So, the probability that at least one of the three male pups is black is 100% * 75% * 100% =75%.
Now, we add up the probabilities of each case: 1 + 75% + 75% = 1 + 0.75 + 0.75 = 2.5.
The combined probability is 2.5. However, probability ranges from 0 to 1, so the final probability is capped at 1. Therefore, the probability that they have 3 male pups, and at least one is black, is 1.
Your answer for part b) is incorrect. The correct probability is 1, not 7/8.
c. They have 3 black pups (regardless of sex)?
To calculate the probability of having 3 black pups, we need to calculate the probability of having black pup for each birth, assuming each birth is an independent event.
The probability of having a black pup from Duke and Shadow is 75% (Bb x Bb), as there is a 75% chance of passing the dominant B allele.
To calculate the probability of having 3 black pups, we multiply the probabilities of each birth:
75% * 75% * 75% = (3/4) * (3/4) * (3/4) = 27/64 ≈ 0.422.
Therefore, the probability that they have 3 black pups (regardless of sex) is approximately 0.422 or 42.2%.
Your answer for part c) is incorrect. The correct probability is approximately 0.422 or 42.2%, not 3/4.
I hope this explanation clarifies the correct probabilities for each scenario. Let me know if you have any further questions!