Problem

Solutions For An Introduction to Genetic Analysis Chapter 5 Problem 45P

Step-by-Step Solution

Solution 1

An (High frequency recombinant) genotype of male strain is crossed with a genotype of female strain. These two strains are subjected to the interrupted mating and the cells are then plated on three types of agar.

All the three plates have streptomycin (+), so the bacteria with the gene can only grow. The Agar type-1 does not have Nutrient “C,” hence it is selected for the donor gene. The Agar type-2 does not have Nutrient “A,” hence it is selected for the donor gene. The Agar type-3 does not have Nutrient “B,” hence it is selected for the donor gene.

S.no	Agar type	Selected genes
1	1
2	2
3	3

The gene order can be discovered based on the array appearance of colonies on the agar-type. In the given tabulated data, colonies first appear on Type agar, at sampling time minutes with colonies. Since, Agar type selects for gene so should be first in order.

Colonies subsequently appear on Type agar, at sampling time minutes with colonies. Since, Agar type selects for gene so should be located next to the gene-. Finally, colonies first appear on Type agar, at sampling time minutes with colonies. Since, Agar type selects for gene so should be the last in the sequence.

Thus, the order of genes is

Lack of nutrient in the medium cannot allow the growth of strains, so only strains will grow. Form the colonies; the number of colonies survived in different types of agar without nutrient D is as follows:

• Agar type -1: 90

• Agar type -2: 52

• Agar type -3: 9

The Agar type -3 has fewer colonies, since it selects for gene the gene is close to the gene and then to gene a.

Thus, the order of genes is

Step #2 of 2

At sampling time minutes, one can the expect colonies to first appear on agar medium containing streptomycin and nutrient . This is because due to lack of neither nutrient nor in the agar would not allow the growth strains.