17. 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 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.
a. 1/64
b. 3/16
c. 1/64
d. 3/16 x 1/64 = 3/1024
e. 3/16 x 1/64 = 3/1024
To determine the probability of different combinations of offspring, we can use Punnett squares.
First, let's define the genotypes of the parent bears:
Male bear (heterozygous): Ff
Female bear (homozygous recessive): ff
a. A litter of six cubs all with white fur:
In this case, all the cubs need to inherit the recessive allele (f) from both parents. The probability of each cub having the genotype ff is 1/2 * 1/2 = 1/4.
The overall probability for all six cubs having white fur is (1/4)^6 = 1/4096.
b. A litter of five cubs, two with brown fur and three with white fur:
To have two cubs with brown fur and three cubs with white fur, we need to consider the different possible combinations of the genotypes.
The probability of each cub having brown fur (Ff) is 1/2, and the probability of each cub having white fur (ff) is 1/2.
The probability for this specific combination is calculated as follows:
(1/2)^2 * (1/2)^3 = 1/32.
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:
To have one cub with white fur and five cubs with brown fur, we need to consider the different possible combinations of the genotypes.
The probability of each cub having brown fur (Ff) is 1/2, and the probability of each cub having white fur (ff) is 1/2.
The probability for this combination is calculated as follows:
(1/2)^5 * (1/2)^1 * (1/2)^1 = 1/64.
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:
The probabilities for each litter are calculated separately and multiplied together to find the overall probability.
For the first litter:
To have five cubs with brown fur and three with white fur, we need to consider the different possible combinations of the genotypes.
The probability of each cub having brown fur (Ff) is 1/2, and the probability of each cub having white fur (ff) is 1/2.
The probability for the first litter is calculated as follows:
(1/2)^5 * (1/2)^3 = 1/128.
For the second litter:
To have four cubs with brown fur and two cubs with white fur, we use the same probabilities as before.
The probability for the second litter is calculated as follows:
(1/2)^4 * (1/2)^2 = 1/64.
The overall probability is obtained by multiplying the probabilities of both litters together:
(1/128) * (1/64) = 1/8192.
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 have four cubs with brown fur and one cub with white fur in the first litter, we need to consider the different possible combinations of the genotypes.
The probability of each cub having brown fur (Ff) is 1/2, and the probability of each cub having white fur (ff) is 1/2.
The probability for the first litter is calculated as follows:
(1/2)^4 * (1/2)^1 = 1/32.
For the second litter:
To have the first-born as homozygous (brown fur), the second born as brown fur, and the remaining five cubs as four brown fur and one white fur, we consider the different possible combinations of the genotypes.
The probability for the specific combination is calculated as follows:
(1/2)^2 * (1/2)^5 * (1/2)^4 = 1/512.
The overall probability is obtained by multiplying the probabilities of both litters together:
(1/32) * (1/512) = 1/16384.
So, to summarize:
a. The probability of a litter of six cubs all with white fur is 1/4096.
b. The probability of a litter of five cubs, two with brown fur and three with white fur is 1/32.
c. The probability of a litter of seven cubs, with the first six cubs, one with white fur and five with brown fur, and the last-born cub with white fur is 1/64.
d. The probability of 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 is 1/8192.
e. The probability of 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 is 1/16384.
To determine the probability of certain combinations of offspring, we need to analyze the principles of Mendelian genetics.
In bears, the (F) allele is dominant for brown fur, and the (f) allele is recessive for white fur. A heterozygous male bear has one dominant (F) allele and one recessive (f) allele, while a homozygous recessive female bear has two recessive (f) alleles.
a. To determine the probability of a litter of six cubs all with white fur, we need to calculate the probability of each individual cub having white fur. The genotype of the heterozygous male is (Ff), and the genotype of the homozygous recessive female is (ff).
When these two parents are crossed, each cub has a 50% chance of inheriting the recessive (f) allele from the male and a 100% chance of inheriting the recessive (f) allele from the female.
The probability of a cub having white fur is therefore 0.5 * 1.0 = 0.5.
Since there are six cubs, and we assume that their fur color inheritance is independent, we multiply the probabilities together: 0.5^6 = 0.015625.
So, the probability of a litter of six cubs all having white fur is approximately 0.015625, or 1.56%.
b. Similarly, to calculate the probability of a litter of five cubs, two with brown fur and three with white fur, we use the same method as above.
The probability of a cub having brown fur is 0.5 * 0.5 = 0.25 (representing the chance that it inherits the dominant (F) allele from the male and either the dominant (F) or recessive (f) allele from the female).
The probability of a cub having white fur is 0.5 * 1.0 = 0.5 (representing the chance that it inherits the recessive (f) allele from the male and the recessive (f) allele from the female).
Since there are two cubs with brown fur and three cubs with white fur, we multiply the probabilities together: (0.25^2) * (0.5^3) = 0.00390625.
So, the probability of a litter of five cubs, two with brown fur and three with white fur, is approximately 0.00390625, or 0.39%.
c. For a litter of seven cubs, where the first six cubs have one with white fur and five with brown fur, and the last-born cub has white fur, we can calculate the probability in the same way.
The probability of a cub having brown fur is 0.25, as mentioned earlier.
The probability of a cub having white fur is still 0.5.
To calculate the overall probability, we multiply the probabilities for the first six cubs (1 white, 5 brown) and the probability for the last cub (white): (0.5^6) * 0.5 = 0.015625.
So, the probability of this specific combination is also approximately 0.015625 or 1.56%.
d. When considering the probability of two litters, we need to calculate the probability for each litter separately and then multiply them together.
For the first litter of eight cubs, five with brown fur and three with white fur:
- Probability of a cub having brown fur: 0.25
- Probability of a cub having white fur: 0.5
To calculate the overall probability, we multiply the probabilities for each cub: (0.25^5) * (0.5^3) = 0.0009765625.
For the second litter of six cubs, four with brown fur and two with white fur:
- Probability of a cub having brown fur: 0.25
- Probability of a cub having white fur: 0.5
To calculate the overall probability, we multiply the probabilities for each cub: (0.25^4) * (0.5^2) = 0.00390625.
Finally, we multiply the probabilities of both litters together: 0.0009765625 * 0.00390625 = 0.0000038147.
So, the probability of this specific combination is approximately 0.0000038147 or 0.00038147%.
e. Finally, for a first litter of five cubs (four with brown fur and one with white fur) and a second litter of seven cubs (the first-born is homozygous, the second born is brown, and the remaining five cubs are four brown and one white):
First litter:
- Probability of a cub having brown fur: 0.25
- Probability of a cub having white fur: 0.5
To calculate the overall probability, we multiply the probabilities for each cub: (0.25^4) * (0.5^1) = 0.00390625.
Second litter:
- Probability of first-born cub being homozygous (FF): 0.25 * 0.25 = 0.0625
- Probability of second-born cub having brown fur: 0.25
- Probability of remaining five cubs having four brown and one white fur: (0.25^4) * (0.5^1) = 0.00390625
To calculate the overall probability, we multiply the probabilities for each cub: 0.0625 * 0.25 * 0.00390625 = 0.0000607678.
So, the probability of this specific combination is approximately 0.0000607678 or 0.00607678%.