Chapter 16, Problem 42P

The researchers who investigated bioluminescence and quorum sensing found that E. coli transformed with a plasmid containing a 9 kb fragment of V.fischeri DNA could glow when the cell population was dense. They mutagenized these E. coli cells and isolated many mutations that mapped to the 9 kb fragment and prevented the cells from glowing. They then performed complementation testing on these mutants by transforming E. coli cells simultaneously with two plasmids, each containing the 9 kb fragment with one of the mutations. To ensure the E.coli cells were transformed with both plasmids, one of the two plasmids had a gene conferring resistance to ampicillin, while the other plasmid had a gene conferring resistance to tetracycline, and cells were selected on petri plates that had both antibiotics.

a. Construct a 9 × 9 complementation table for the nine mutations list that follows, using + to indicate cells that would glow and - to indicate cells that would remain dark. (You only need to fill in half the table.) 1. Mutation 1: Encodes a LuxA protein that cannot bind a substrate for the luciferase enzyme 2. Mutation 2: Encodes a LuxA protein that cannot associate with the LuxB protein 3. Mutation 3: Encodes a LuxB protein that cannot associate with the LuxA protein 4. Mutation 4: A null mutation in the luxI gene 5. Mutation 5: Encodes a LuxR protein that cannot bind DNA 6. Mutation 6: Encodes a LuxR protein that cannot bind to the autoinducer 7. Mutation 7: A mutation in the luxR promoter that prevents transcription 8. Mutation 8: A mutation in the luxICDABE promoter that prevents transcription 9. Mutation 9: A mutation in the luxICDABE promoter region that blocks binding of the LuxR protein

b. How many complementation groups exist among these nine mutations?

c. Is your answer to part (b) also the number of different genes? Explain

Summary Introduction

a.

To construct:

A 9 $\times$ 9 non complementation table that involves the given list of mutants.

Introduction:

The researchers investigating bioluminescence and quorum sensing found out that the E.coli that is transformed with a plasmid containing a 9 kb fragment of Vibrio fischeri glows when the cell population becomes dense. Many of the E.coli cells were mutagenized and isolated that mapped to the 9 kb fragment and prevented the cells from glowing.

Explanation of Solution

The following table represents the 9 $\times$ 9 complementation table for the nine mutations given in the list.

	Mut1	Mut2	Mut3	Mut4	Mut5	Mut6	Mut7	Mut8	Mut9
Mut1	-	+	+	+	+	+	+	-	-
Mut2		-	+	+	+	+	+	-	-
Mut3			-	+	+	+	+	-	-
Mut4				-	+	+	+	-	-
Mut5					-	+	-	+	+
Mut6						-	-	+	+
Mut7							-	+	+
Mut8								-	-
Mut9									-

The genes that play an important role in the bioluminescence of Vibrio fischeri are luxR, luxICDABE, and luxI. luxR produces LuxR protein, which is a transcriptional activator required for the transcription of luxICDABE mRNA. luxICDABE involves a polycistronic region producing the proteins LuxI, LuxC, LuxD, LuxA, LuxB, and LuxE. The LuxI is a type of synthase enzyme that synthesizes an autoinducer. The autoinducer binds to LuxR protein and enables it to bind DNA. With reference to figure 1, the negative sign indicates that the combination of mutations does not complement each other; and due to this reason, the cells remain dark. The presence of positive sign indicates that the mutation complement each other, and the cells tend to glow.

Summary Introduction

b.

To determine:

The number of complementation groups that exist among the nine mutations.

Introduction:

The complementation test was performed by transforming E.coli cells with two plasmids, each containing the 9kb fragment with one of the mutations. One plasmid had a gene conferring resistance to ampicillin, while others had a gene conferring resistance to tetracycline. The cells on two plates were selected on pertri plates containing both the antibiotics.

Explanation of Solution

A positive sign indicates that the given mutations complement each other, and the negative sign indicates that the mutation does not complement each other. In totality, 25 complement pairs are existing among the nine mutations.

Summary Introduction

c.

To determine:

Whether the number of complementation groups existing among the nine mutations also describes the number of different genes.

Introduction:

The complementation test aids in determining the association between the specific phenotype and the difference of two variable genes. This test is significant for recessive traits that are masked by the dominant allele.

Explanation of Solution

The complementation groups also describe the number of different genes. It is because the event of complementation tends to occur only when the mutations are present in different genes. As there are 25 complementation groups, it means there are 25 different genes as well.