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
From your data, both dogs are Bb. Using a Punnett square, there is a 25% chance of having a brown (bb) offspring.
2. What does gender have to do with it?
3. There is .75 or 3/4 probability of any one pup being Black (Bb or BB). Probability of all events occurring is found by multiplying the probability of the individual events.
.75 * .75 * .75 = ?
for number two you gave to calculate the probability of only 3 male offsprings being born and the probability of at least one being black.
To calculate the probabilities, we need to consider the genetic makeup of Duke and Shadow and use Punnett squares to analyze the possible outcomes.
First, let's determine the genotype of Duke and Shadow. Duke is a black male, which means he must have at least one dominant B allele. Since his mother was brown (bb), Duke must be heterozygous (Bb).
Shadow is a black female who, in a previous mating, produced brown offspring. This means that Shadow must be a carrier of the recessive brown allele (Bb), just like Duke.
Now, we can proceed to answer the questions:
a. Their first pup is brown?
To determine the probability of having a brown pup, we need to consider the genotype of both parents. Since both Duke and Shadow are heterozygous (Bb), they can each contribute a B or b allele to their offspring.
When mate-mixing Bb x Bb, the possible genotypes of the offspring are:
BB, Bb, Bb, bb
We can see that only one combination produces a brown pup (bb). Thus, there is a 1/4 or 25% chance that their first pup will be brown.
b. They have 3 male pups, and at least one is black?
To determine the probability of having at least one black male pup, we need to consider the possible combinations that result in a black male and add their probabilities.
When mate-mixing Bb x Bb, the possible genotypes are the same as mentioned earlier:
BB, Bb, Bb, bb
To calculate the probability, we need to find the odds of not having a male black pup (bb) for each individual birth and subtract that from 1 (since we want the probability of having at least one black male pup).
The probability of not having a black male pup (bb) at a single birth is 1/4 or 25%. So, the probability of having at least one black male pup in three births is:
1 - 0.25^3 = 1 - 1/64 = 63/64 or approximately 0.9844
Hence, the probability of having 3 male pups, with at least one being black, is 63/64 or approximately 0.9844.
c. They have 3 black pups (regardless of sex)?
To determine the probability of having 3 black pups, we need to consider the possible combinations that result in black pups.
When mate-mixing Bb x Bb, the possible genotypes are the same as mentioned earlier:
BB, Bb, Bb, bb
Out of these four possible genotypes, three of them result in black pups (BB, Bb, and Bb). Thus, the probability of having 3 black pups is 3/4 or 75%.
Therefore, the correct answers are:
a. The probability of their first pup being brown is 1/4 or 25%.
b. The probability of having 3 male pups, with at least one being black, is 63/64 or approximately 0.9844.
c. The probability of having 3 black pups, regardless of sex, is 3/4 or 75%.