Chapter 25, Problem 1NST

A homozygous plant with 20-cm-diameter flowers is crossed with a homozygous plant of the same species that has 40-cm-diameter flowers. The F₁ plants all have flowers 30 cm in diameter. In the ⊢₂ generation of 512 plants, 2 plants have flowers 20 cm in diameter, 2 plants have flowers 40 cm in diameter, and the remaining 508 plants have flowers of a range of sizes in between.

1. (a) Assuming all alleles involved act additively, how many genes control flower size in this plant?
2. (b) What frequency distribution of flower diameter would you expect to see in the progeny of a backcross between F₁ plant and the large-flowered parent?

Summary Introduction

To determine: The number of genes which control flower size in the given plant assuming all alleles involved act additively.

Introduction: The traits which show variation and often fall into a continuous range of phenotypes that are difficult to categorize into distinct categories are known to exhibit continuous variation. For example, height in humans. The genetic phenomenon that measures and explain the continuous variation in quantitative terms across a range of phenotypes is known as quantitative inheritance. The varying phenotypes are the result of the input of genes at more than one loci, and generally many loci and therefore referred to as polygenic.

Explanation of Solution

The concept of additive alleles is the basis of continuous variation. The alleles that contribute to the most observable traits like height, weight, eye color, and so on are known as additive alleles. Polygenic traits are additive. In the given question, a homozygous plant with 20 cm diameter flower is crossed with a homozygous plant with 40 cm diameter flower of the same species. The F₁ generation has all flowers with 30 cm in diameter. In the F₂ generation, out of 512 plants, two plants have flowers with 20 cm diameter, two plants have flowers with 40 cm diameter, and the remaining 508 plants have flowers with a range of diameters in between.

Following is the formula to calculate the number of polygenes:

${\text{The proportion of F}}_{\text{2}}\text{progeny expressing either of the parental phenotypes}=\frac{1}{4^{\text{n}}}$

Here, $\text{n}=\text{number of genes that control the phenotype}$

The ratio of F₂ plants with 20 cm diameter flowers

$\begin{array}{l}=\frac{2}{512}\\ =\frac{1}{256}\\ =\frac{1}{4^4}\end{array}$

The ratio of F₂ plants with 40 cm diameter flowers

$\begin{array}{l}=\frac{2}{512}\\ =\frac{1}{256}\\ =\frac{1}{4^4}\end{array}$

The F₂ ratio suggests that four genes, each with two alleles control the diameter of the flower in a given plant.

Thus, there are four pairs of genes which control flower size in the given plant.

Summary Introduction

To determine: The expected frequency distribution of flower diameter from the backcross between an F₁ plant and the large-flowered parent.

Introduction: In backcross, a hybrid is crossed with one of its parents. This cross is done to get the progeny having genetic identity closer to the parent. The application of the back cross is in horticulture, animal breeding, and so on.

Explanation of Solution

Let us assume that the four genes which control flower size in the given plant are A, B, C, and D.

The genotype of an F₁ plant: AaBbCcDd

The genotype of the large-flowered parent: AABBCCDD

The frequency distribution in the backcross would be: 1/16, 4/16, 6/16, 4/16, and 1/16.

Thus, the frequency distribution of flower diameter from the backcross between an F₁ plant and the large-flowered parent is 1/16, 4/16, 6/16, 4/16, and 1/16.