Chapter 5, Problem 4P

The Punnett square in Fig. 5.4 on p. 131 shows how Mendel’s dihybrid cross results would have been altered had the two genes (A and B) been linked and had the P generation cross been AB/AB × ab/ab

a.	What would be the frequency of each F2 phenotypic class if the parentals are present at a frequency of 80%?
b.	Answer part (a) assuming that the original, P generation cross was Ab/Ab × aB/aB.

Summary Introduction

a.

To determine:

The frequency of each F₂ phenotypic class if the parents are present at a frequency of 80% in the given question.

Introduction:

Recombination frequency is used to measure the genetic linkage and for the creation of a genetic linkage map. The frequency at which a single chromosome crossover takes place between two genes during meiosis is termed as recombination frequency.

Explanation of Solution

Given:

A_B_ (Parental) $=\frac{9}{16}$

aabb (Parental) $=\frac{1}{16}$

A_bb (Recombinant) $=\frac{3}{16}$

aaB_ (Recombinant) $=\frac{3}{16}$

From the given data, since the parental frequency is 0.80 (80 cM), so the total offsprings (x) can be calculated from the given parental frequency:

$\left(\frac{10}{16}\right)=\left(\frac{80}{100}\right)\times x$

$\begin{array}{rll}x& =& \left(\frac{1000}{1280}\right)\\ & =& \left(\frac{100}{128}\right)\\ & =& 78\end{array}$

Thus, the number of total offsprings is about 78.

Number of recombination progeny is as follows:

$\begin{array}{l}\left(\frac{20}{100}\right)\times 78\\ =15.6\\ \sim 16\end{array}$

The frequency of each phenotypic class is as follows:

$\begin{array}{rll}A\_B\_& =& \left(\frac{9}{16}\right)\times 78\\ & =& >43.8\end{array}$

$\begin{array}{rll}A\_bb& =& \left(\frac{3}{16}\right)\times 78\\ & =& <14.6\end{array}$

$\begin{array}{rll}aaB\_& =& \left(\frac{3}{16}\right)\times 78\\ & =& <14.6\end{array}$

$\begin{array}{rll}aabb& =& \left(\frac{1}{16}\right)\times 78\\ & =& >4.8\end{array}$

Summary Introduction

b.

To determine:

The frequency of each F₂ phenotypic class if the parentas are present at a frequency of 80% in the given question.

Introduction:

The first set of parents crossed is termed as the parental generation (P). All the offsprings from the parents constitute the F₁ (first filial) generation. All the offspring obtained when the F₁ individuals are interbred constitute the F₂ (second filial) generation.

Explanation of Solution

The cross is as follows:

$\left(AB/AB\times ab/ab\right)$

Male/Female gametes	AB	Ab	aB	ab
AB	AABB	AABb	AaBB	AaBb
Ab	AABb	AAbb	AaBb	Aabb
aB	AaBB	AaBb	aaBB	aaBb
ab	AaBb	Aabb	aaBb	aabb

A_bb (Parental) $=\frac{3}{16}$

aaB_ (Parental) $=\frac{3}{16}$

A_B_ (recombinant) $=\frac{9}{16}$

aabb (recombinant) $=\frac{1}{16}$

Here, the offspring’s showing A and B gene linkage are more in number. They are about 80% of the F₂ phenotype.

Frequency of each phenotypic class is as follows:

$\begin{array}{rll}A\_B\_& =& \left(\frac{9}{16}\right)\times 78\\ & =& >43.8\end{array}$

$\begin{array}{rll}A\_bb& =& \left(\frac{3}{16}\right)\times 78\\ & =& <14.6\end{array}$

$\begin{array}{rll}aaB\_& =& \left(\frac{3}{16}\right)\times 78\\ & =& <14.6\end{array}$

$\begin{array}{rll}aabb& =& \left(\frac{1}{16}\right)\times 78\\ & =& >4.8\end{array}$

This cross produces the same offsprings as the previous cross indicating that A and B genes are linked, a and b genes are linked.