Chapter 2, Problem 28P

Summary Introduction

a.

To determine:

The probability of one offspring having normal phenotype for both the characters if female having genotype Ttnn is mated with a male having genotype TtNn.

Introduction:

Gametes carry genes from parents. Each parent passes on a haploid set of genes to the progeny during the sexual reproduction. The fusion of gametes carrying the haploid gene set from each of the parents gives rise to a diploid progeny.

Explanation of Solution

In the given case, dominant characters are normal wings (T) and normal/oval eyes (N), whereas the recessive characters have tiny wings (t) and narrow eyes (n). Cross between genotypes Ttnn (male) and TtNn (female) can be shown as given below.

	TN	Tn	tN	tn
Tn	TTNn	TTnn	TtNn	Ttnn
tn	TtNn	Ttnn	ttNn	ttnn

Based on the above table, the probability of each of the genotypes is found to be:

Normal wings =3/4

Tiny wings = 1/4

Oval eyes = 1/2

Narrow eyes = 1/2

Hence, the total probability of an offspring having normal characters can be calculated by multiplying the respective probabilities of both the normal characters.

$\begin{array}{rll}\text{Probability}\text{of}\text{normal}\text{characters}& =& \left[\begin{array}{l}\text{probability}\text{of}\text{normal}\text{wings}\\ \text{×}\text{probability}\text{of}\text{normal}\text{eyes}\end{array}\right]\\ & =& \left(\frac{3}{4}\right)\times \left(\frac{1}{2}\right)\\ & =& \left(\frac{3}{8}\right)\end{array}$

Hence, 3/8 of the total offspring is having normal genotype (TTNN, TtNn, TTNn, or TtNN).

Summary Introduction

b.

To determine:

The expected phenotype in the offspring from the above cross along with the number of offsprings in each phenotypic class if there are a total of 200 progeny obtained.

Introduction:

During sexual reproduction, the offspring somewhat resembles the parents because they carry the genes that are passed on from the parents to them through the gametes. Sexual reproduction brings genes from both the parents into the progeny.

Explanation of Solution

Based on the cross between genotypes Ttnn (male) and TtNn (female), probabilities are found to be as follows:

Normal wings = 3/4

Tiny wings = 1/4

Oval eyes = 1/2

Narrow eyes = 1/2

The expected phenotype in the offspring from the above cross can be calculated as follows:

$\begin{array}{rll}\text{normal}\text{wings}\text{and}\text{oval}\text{eyes}& =& \left(\frac{\text{3}}{\text{4}}\right)\text{×}\left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{\text{3}}{\text{8}}\right)\end{array}$

$\begin{array}{rll}\text{normal}\text{wings}\text{and}\text{narrow}\text{eyes}& =& \left(\frac{\text{3}}{\text{4}}\right)\text{×}\left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{\text{3}}{\text{8}}\right)\end{array}$

$\begin{array}{rll}\text{tiny}\text{wings}\text{and}\text{oval}\text{eyes}& =& \left(\frac{\text{1}}{\text{4}}\right)\text{×}\left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{\text{1}}{\text{8}}\right)\end{array}$

$\begin{array}{rll}\text{tiny}\text{wings}\text{and}\text{narrow}\text{eyes}& =& \left(\frac{\text{1}}{\text{4}}\right)\text{×}\left(\frac{\text{1}}{\text{2}}\right)\\ & =& \left(\frac{\text{1}}{\text{8}}\right)\end{array}$

The total number of offsprings is given as 200. Based on this, the total number of offspring with each of the character can be calculated as follows:

$\begin{array}{rll}\left(\frac{3}{8}\right)\times 200& =& 75\\ \left(\frac{1}{8}\right)\times 200& =& 25\end{array}$

This gives the total number of offsprings as follows:

Normal wings and oval eyes: 75

Normal wings and narrow eyes: 75

Tiny wings and oval eyes: 25

Tiny wings and narrow eyes: 25