Chapter 5. Chapter 5: Mapping Bacterial Genes Using Interrupted Mating and Recombination During Bacterial Conjugation

Unpacking the Problem

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You must read each slide, and complete any questions on the slide, in sequence.

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

A cross is made between two E. coli strains: Hfr arg⁺ bio⁺ leu⁺ × F^- arg^- bio^- leu^-. Interrupted mating studies show that arg⁺ enters the recipient last, and so any arg⁺ recombinants are selected on a medium containing biotin and leucine only. These recombinants are tested for the presence of bio⁺ and leu⁺. The following numbers of individuals are found for each genotype:
arg⁺ bio⁺ leu⁺ 320
arg⁺ bio⁺ leu^- 8
arg⁺ bio^- leu⁺ 0
arg⁺ bio^- leu^- 48

Unpack the Problem: Break this problem into several parts and arrive at a solution using this guided, step-by-step approach.

• Part A (step 1): Understand what an Hfr strain is and how it functions during conjugation.
• Part B (steps 2-7): What is the gene order?
• Part C (step 8): What are the map distances in recombination percentages?

General Hints

The map of the three genes is:

A variety of different double crossover exconjugants can form by different combinations of crossing-over in different regions. The probability of crossing over increases as the distance between selectable markers increases. The distance between selectable markers is determined by calculating the percent recombinants relative to total exconjugants scored. The total number of exconjugants is 320+8+0+48=376. We can determine the recombination units between selected markers using the genotype data provided. Assume a crossover to the left (distal to arg⁺) of the arg⁺ marker is constant, and the second crossover can occur in any other region between arg⁺ and the origin of transfer (arrow of Hfr donor DNA). A crossover between arg⁺ and bio⁺ will generate arg⁺bio^-leu^- exconjugants. The number of these from the data is: 48/376 = 12.8% or 12.8 recombination units. If the second crossover occurs between bio⁺ and leu⁺, then arg⁺bio⁺leu^- exconjugants are formed (8/376= 2.1% or 2.1 r.u.). If the second crossover occurs between leu⁺ and the origin of transfer on the Hfr (the arrow in the diagram above), then arg⁺bio⁺leu⁺ exconjugants are formed (320/376 = 85.1% or 85.1 recombination units). This demonstrates that most of the crossovers actually took place proximal to the first selected marker (leu⁺).