In bears, (F) is the dominant allele for brown fur and (f) is the recessive allele for white fur. If a heterozygous male bear is crossed with homozygous recessive female. what would be the probability of the following combinations of offspring? a. A litter of six cubs all with white fur ? b. A litter of five cubs, two with brown fur and three with white fur? c. A litter of seven cubs, and the first six cubs, one with white fur and five with brown fur, and the last born cub with white fur.
d. A first litter of eight cubs, five with brown fur and three and three with white fur, and then a second litter of six cubs, four with brown fur and two with white fur.
e. A first litter of five cubs, four with brown fur and one with white fur, and then a second litter of seven cubs in which the first born is homozygous, the second born is brown, and the remaining five cubs are four brown and one white.
To determine the probability of different combinations of offspring, we need to use Punnett squares. A Punnett square is a tool used to visualize the possible combinations of alleles from two parents.
Let's denote the heterozygous male bear as Ff and the homozygous recessive female as ff.
a. A litter of six cubs all with white fur:
The male bear's genotype is Ff, so half of his gametes will carry the f allele. The female bear's genotype is ff, so all her gametes will carry the f allele. The Punnett square for this cross will look like:
| f | f |
----|----|----|
F | Ff | Ff |
----|----|----|
f | ff | ff |
All the possible combinations in the offspring have ff genotype, which corresponds to white fur. Therefore, the probability of having a litter of six cubs all with white fur is 1.
b. A litter of five cubs, two with brown fur and three with white fur:
Using the same Punnett square as in the previous case, we can count the number of combinations that result in two cubs with brown fur (Ff) and three cubs with white fur (ff). Since there are two potential Ff combinations, the probability is 2/4 or 1/2.
c. A litter of seven cubs, with the first six cubs having one with white fur and five with brown fur, and the last-born cub with white fur:
Again, using the same Punnett square, we can calculate the probability of having one ff combination (white fur) and five Ff combinations (brown fur) in the first six cubs. The probability of having ff combination in the first-born cub is 1/4. For the other five cubs, the probability is (1/2)^5 since each cub's genotype is independent of the others. Finally, for the last-born cub, the probability of having ff combination is 1/4. Therefore, the overall probability is (1/4) * (1/2)^5 * (1/4) = 1/512.
d. A first litter of eight cubs, five with brown fur and three with white fur, and then a second litter of six cubs, four with brown fur and two with white fur:
For each litter, we can use the same Punnett square to calculate the probability. For the first litter, the probability of having five Ff combinations (brown fur) is (1/2)^5, and the probability of having three ff combinations (white fur) is (1/4)^3. Therefore, the overall probability is (1/2)^5 * (1/4)^3.
Similarly, for the second litter, the probability of having four Ff combinations (brown fur) is (1/2)^4, and the probability of having two ff combinations (white fur) is (1/4)^2. So, the overall probability is (1/2)^4 * (1/4)^2.
To find the combined probability, we multiply both probabilities together: (1/2)^5 * (1/4)^3 * (1/2)^4 * (1/4)^2.
e. A first litter of five cubs, four with brown fur and one with white fur, and then a second litter of seven cubs, with the first-born being homozygous, the second-born brown, and the remaining five cubs being four brown and one white:
For the first litter, the probability of having four Ff combinations (brown fur) is (1/2)^4, and the probability of having one ff combination (white fur) is (1/4)^1.
For the second litter, the probability of having the first-born cub as homozygous FF (brown fur) is 1/4, the probability of having the second-born cub as Ff (brown fur) is 1/2, and the probability of having the remaining five cubs as four Ff (brown fur) and one ff (white fur) combinations is (1/2)^4 * (1/4)^1.
To find the combined probability, we multiply both probabilities together: (1/2)^4 * (1/4)^1 * 1/4 * 1/2 * (1/2)^4 * (1/4)^1.
Remember that these probabilities are based on Mendelian genetics, assuming independent assortment and no genetic linkage. In real-life situations, variations may occur due to factors such as genetic recombination.