Chapter 2, Problem 16CONQ

In cocker spaniels, solid coat color is dominant over spotted coat color. If two heterozygous dogs were crossed to each other, what would be the probability of the following combinations of offspring?

A. A litter of five pups, four with solid fur and one with spotted fur.

B. A first litter of six pups, four with solid fur and two with spotted fur, and then a second litter of five pups, all with solid fur.

C. A first litter of five pups, the firstborn with solid fur, and then among the next four, three with solid fur and one with spotted fur, and then a second litter of seven pups in which the firstborn is spotted, the second born is spotted, and the remaining five are composed of four solid and one spotted animal.

D. A litter of six pups, the firstborn with solid fur, the second born spotted, and among the remaining four pups, two with spotted fur and two with solid fur.

Summary Introduction

To review:

The probability of the following combinations of offspring on crossing two heterozygous cocker spaniel dogs:

Five pups, four with solid fur and one with spotted fur.

First litter with six pups, out of which four have solid fur and two have spotted fur, followed by a second litter with five pups, all having solid fur.

First litter with five pups, in which the firstborn has solid fur and among the rest, three have solid fur and one has spotted fur, followed by a second litter with seven pups, in which the first and second born pups are spotted and among the rest, four have solid fur and one has spotted fur.

Six pups, in which the first born has solid fur, the second born has spotted fur, and among the rest, two have solid fur and two have spotted fur.

Introduction:

The coat color in the cocker spaniel can be either solid or spotted. The coat colors are dependent on what genes are passed through the generations in the cocker spaniel. There is one gene responsible for the coat color, which has one dominant allele (B) and one recessive allele (b).

Explanation of Solution

A cross between two heterozygous cocker spaniel dogsproduced the following offspring in the F₁ generation.

Alleles involved: B (dominant), b (recessive)

Parental genotypes: Parent 1 = Bb (solid coat color), Parent 2 = Bb (solid coat color)

Gametes	B	b
B	BB (solid coat color)	Bb (solid coat color)
b	Bb (solid coat color)	Bb (spotted coat color)

Five pups, four with solid fur and one with spotted fur.

The probability can be calculated using the binomial expansion method which uses the following equation:

$\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}$

According to the question, the total number of pups is five. So, n = 5.

The number of solid pups is four. So, x = 4.

From the cross between the heterozygous parents, the probability of having an offspring with solid coat coloris $\frac{3}{4}$. So, p is $\frac{3}{4}$. Therefore, the probability of having an offspring with spotted coat coloris $\frac{1}{4}$. So, q = $\frac{1}{4}$.

Putting these values in the binomial equation,

$\begin{array}{c}\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}\\ =\frac{5!}{4!(5-4)!}\times {\frac{3}{4}}^4\times {\frac{1}{4}}^{5-4}\\ =\frac{405}{1024}\\ =0.395\text{ or 39}\text{.5\%}\end{array}$ $$

First litter with six pups, out of which four have solid fur and two have spotted fur, followed by a second litter with five pups, all having solid fur.

In this question, for the first litter, the values that are to be substituted are as follows:

n = 6, x = 4, p = $\frac{3}{4}$, q = $\frac{1}{4}$

Putting the values in the binomial equation,

$\begin{array}{c}\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}\\ =\frac{6!}{4!(6-4)!}\times {\frac{3}{4}}^4\times {\frac{1}{4}}^{6-4}\\ =\frac{1215}{4096}\\ =0.296\text{ or 29}\text{.6\%}\end{array}$

For the second litter, the values that are to be substituted are as follows:

n = 5, x = 5, p = $\frac{3}{4}$, q = $\frac{1}{4}$

Putting the values in the binomial equation,

$\begin{array}{c}\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}\\ =\frac{5!}{5!(5-5)!}\times {\frac{3}{4}}^5\times {\frac{1}{4}}^{5-5}\\ =\frac{243}{1024}\\ =0.237\text{ or 23}\text{.7\%}\end{array}$

Multiplying the probability of offspring in both the litters, the total probability comes out to be:

$\begin{array}{c}\text{P = 0}\text{.296}\times \text{0}\text{.237}\\ \text{= 0}\text{.070 or 7}\text{.01\%}\end{array}$

First litter with five pups, in which the first-born has solid fur and among the rest, three have solid fur and one has spotted fur, followed by a second litter with seven pups, in which the first and second-born pups are spotted and among the rest, four have solid fur and one has spotted fur.

In the first litter, the probability of having a pup with solid fur is $\frac{\text{3}}{\text{4}}\text{ or 0}\text{.75}$.

The values to be substituted in the equation for the rest of the pups are as follows:

n = 4, x = 3, p = $\frac{3}{4}$, q = $\frac{1}{4}$

$\begin{array}{c}\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}\\ =\frac{4!}{3!(4-3)!}\times {\frac{3}{4}}^3\times {\frac{1}{4}}^{4-3}\\ =\frac{27}{64}\\ =0.421\text{ or 42}\text{.1\%}\end{array}$

Total probability of offspring in the first litter:

$\begin{array}{c}\text{P = 0}\text{.75}\times \text{0}\text{.421}\\ \text{= 0}\text{.315 or 31}\text{.5\%}\end{array}$

In the second litter, the probability of the first two pups having spotted fur is $\frac{1}{4}\times \frac{1}{4}=\frac{1}{8}=0.125$.

The values to be substituted in the equation for the rest of the pups are as follows:

n = 5, x = 4, p = $\frac{3}{4}$, q = $\frac{1}{4}$

$\begin{array}{c}\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}\\ =\frac{5!}{4!(5-4)!}\times {\frac{3}{4}}^4\times {\frac{1}{4}}^{5-4}\\ =\frac{405}{1024}\\ =0.395\text{ or 39}\text{.5\%}\end{array}$

Total probability of offspring in the second litter:

$\begin{array}{c}\text{P = 0}\text{.125}\times \text{0}\text{.395}\\ \text{= 0}\text{.049 or 4}\text{.93\%}\end{array}$

Final probability of both the litters combined comes out to be:

$\begin{array}{c}\text{P = 0}\text{.315}\times \text{0}\text{.049}\\ \text{= 0}\text{.015 or 1}\text{.54\%}\end{array}$

Six pups, in which the first-born has solid fur, the second-born has spotted fur and among the rest, two have solid fur and two have spotted fur.

The probability of the first-born pup having solid fur = $\frac{3}{4}=0.75$

The probability of the second-born pup having spotted fur = $\frac{1}{4}=0.25$

The values to be substituted in the equation for the rest of the pups are as follows:

n = 4, x = 2, p = $\frac{3}{4}$, q = $\frac{1}{4}$

$\begin{array}{c}\text{P = }\frac{n!}{x!\left(n-x\right)!}\times p^x\times q^{n-x}\\ =\frac{4!}{2!(4-2)!}\times {\frac{3}{4}}^2\times {\frac{1}{4}}^{4-2}\\ =\frac{27}{128}\\ =0.211\text{ or 21}\text{.1\%}\end{array}$

The probability of having the given offspring becomes:

$\begin{array}{c}\text{P = 0}\text{.75}\times \text{0}\text{.25}\times \text{0}\text{.211}\\ \text{= 0}\text{.039 or 3}\text{.95\%}\end{array}$

Conclusion

Therefore, it can be concluded that the probability of the given combination of offspring is:

Five pups, four with solid fur and one with spotted fur = 39.5%

First litter with six pups, out of which four have solid fur and two have spotted fur, followed by a second litter with five pups, all having solid fur = 7.01%

First litter with five pups, in which the firstborn has solid fur and among the rest, three have solid fur and one has spotted fur, followed by a second litter with seven pups, in which the first and second born pups are spotted and among the rest, four have solid fur and one has spotted fur = 1.54%

Six pups, in which the first born has solid fur, the second born has spotted fur and among the rest, two have solid fur and two have spotted fur = 3.95%