Chapter 10, Problem 5GP

Summary Introduction

To describe:

The proportion of offspring being heterozygous for all the genes.

Concept introduction:

Mating between two ferns, one having genotype $AABbCcddEe$ and another fern having genotype $aaBbCCDdee$ takes place. All the genes assort independently. The product rule states that the probability (P) of occurrence of two or more independently occurring events occurring together can be calculated by multiplying their individual probabilities.

Explanation of Solution

Tabular representation:

Table.1: The progeny obtained by crossing AA and aa.

Table.2: The progeny obtained by crossing Bb and Bb.

Table.3: The progeny obtained by crossing Cc and CC.

Table.4: The progeny obtained by crossing dd and Dd.

Table.5: The progeny obtained by crossing Ee and ee.

Explanation:

The genotypes for the first traits are AA and aa. The gametes formed by these will be as follows:

$\begin{array}{cccc}\text{Genotype}& AA& \times & aa\\ & & \downarrow & \\ \text{Gametes}& A,A& & a,a\end{array}$

The offspring obtained from crossing can be observed in the Punnett square in Table.1: “The progeny obtained by crossing AA and aa”. All the offspring would be heterozygous. The probability of having heterozygous offspring from first trait is $\frac{4}{4}$.

The genotypes for the second traits are Bb and Bb. The gametes formed by these will be as follows:

$\begin{array}{cccc}\text{Genotype}& Bb& \times & Bb\\ & & \downarrow & \\ \text{Gametes}& B,b& & B,b\end{array}$

The probability of offspring produced from crossing can be observed in the Punnett square in Table.2: “The progeny obtained by crossing Bb and Bb”. There will be $\frac{1}{4}$ offspring which are homozygous dominant and $\frac{1}{4}$ which are homozygous recessive. The probability of having heterozygous offspring from second trait is $\frac{2}{4}$.

The genotypes for the third traits are Cc and CC. The gametes formed by these will be as follows:

$\begin{array}{cccc}\text{Genotype}& Cc& \times & CC\\ & & \downarrow & \\ \text{Gametes}& C,c& & C,C\end{array}$

The offspring obtained from crossing can be observed in the Punnett square in Table.3: “The progeny obtained by crossing Cc and CC”. There will be $\frac{2}{4}$ offspring which are homozygous dominant the probability of having heterozygous offspring from third trait is $\frac{2}{4}$.

The genotypes for the fourth traits are dd and Dd. The gametes formed by these will be as follows.

$\begin{array}{cccc}\text{Genotype}& dd& \times & Dd\\ & & \downarrow & \\ \text{Gametes}& d,d& & D,d\end{array}$

The offspring obtained from crossing can be observed in the Punnett square in Table.4: “The progeny obtained by crossing dd and Dd”. There will be $\frac{2}{4}$ offspring which are homozygous recessive and the probability of having heterozygous offspring from fourth trait is $\frac{2}{4}$.

The genotypes for the fifth traits are Ee and ee. The gametes formed by these will be as follows.

$\begin{array}{cccc}\text{Genotype}& Ee& \times & ee\\ & & \downarrow & \\ \text{Gametes}& E,e& & e,e\end{array}$

The offspring obtained from crossing can be observed in the Punnett square in Table.5: “The progeny obtained by crossing Ee and ee”. There will be $\frac{2}{4}$ offspring which are homozygous recessive and the probability of having heterozygous offspring from fourth trait is $\frac{2}{4}$.

The probability of offspring heterozygous for all the traits can be calculated using the product rule that is by multiplying individual probabilities of heterozygotes for each trait. The probability of offspring heterozygous for all genes can be calculated as follows:

$\begin{array}{c}\left(\text{P}\right)\text{of offspring heterozygous for all genes}=\left(\text{P}\right){\text{ of offspring heterzygous for 1}}^{\text{st}}\text{ gene}\\ \times \left(\text{P}\right){\text{ of offspring heterzygous for 2}}^{\text{nd}}\text{ gene}\\ \times \left(\text{P}\right){\text{ of offspring heterzygous for 3}}^{\text{rd}}\text{ gene}\\ \times \left(\text{P}\right){\text{ of offspring heterzygous for 4}}^{\text{th}}\text{ gene}\\ \times \left(\text{P}\right){\text{of offspring heterzygous for 5}}^{\text{th}}\text{ gene}\\ =\frac{\text{4}}{\text{4}}\times \frac{\text{2}}{\text{4}}\times \frac{\text{2}}{\text{4}}\times \frac{\text{2}}{\text{4}}\times \frac{\text{2}}{\text{4}}\\ =\frac{\text{16}}{\text{256}}\\ =\frac{\text{1}}{\text{16}}\end{array}$

Conclusion

The probability of offspring heterozygous for all the genes is $\frac{1}{16}$.