Bacteriophage Genetics

T2 and its close relative T4 are viruses that infect the bacterium E. coli. The infection ends with destruction (lysis) of the bacterial cell so these viruses are examples of bacteriophages ("bacteria eaters").

Each virus particle (virion) consists of:

Life Cycle:

Occasionally, new phenotypes appear such as a change in the appearance of the plaques or even a loss in the ability to infect the host.

Examples: As with so many organisms, the occurrence of mutations provides the tools to learn about such things as

Mapping by Recombination Frequencies

As we have seen, E. coli strain B can be infected by both h+ and h strains of T2. In fact, a single bacterial cell can be infected simultaneously by both.

Let us infect a liquid culture of E. coli B with two different mutant T2 viruses

When this is done in liquid culture, and then plated on a mixed lawn of E. coli B and B/2, four different kinds of plaques appear.

Genotype Phenotype Number of Plaques
hr+ clear, small 460
h+r turbid, large 460
h+r+ turbid, small 40
hr clear, large 40
Total = 1000

The most abundant (460 each) are those representing the parental types; that is, the phenotypes are those expected from the two infecting strains. However, small numbers (40 each) of two new phenotypes appear. These can be explained by genetic recombination having occasionally occurred between the DNA of each parental type within the bacterial cell.

Just as in higher organisms [Link], one assumes that the frequency of recombinants is proportional to the distance between the gene loci. In this case, 80 out of 1000 plaques were recombinant, so the distance between the h and r loci is assigned a value of 8 map units or centimorgans (cM).

Now coinfect E. coli B with two other strains of T2:
hm+ 470
h+m 470
h+m+ 30
hm 30
Total = 1000

Again, 4 kinds of plaques are produced: parental (470 each) and recombinant (30 each).

The smaller number of recombinants indicates that these two gene loci (h and m) are closer together (6 cM) than h and r (8 cM).

But the order of the three loci could be either To find out which is the correct order, perform a third mating using
mr+ 440
m+r 440
m+r+ 60
mr 60
Total = 1000

This makes it clear that the order is m—h—r, not h—m—r.

But why only 12 cM between the outside loci (m and r) instead of the 14 cM produced by adding the map distances found in the first two matings?

A Three-Point Cross

The answer comes from performing a mating between T2 viruses differing at all three loci:

(Note: this time one parent has all mutant; the other all wild-type alleles — don't be confused!)

Group 1 hmr 435
Group 2 h+m+r+ 435
Group 3 h+mr+ 25
Group 4 hm+r 25
Group 5 hmr+ 35
Group 6 h+m+r 35
Group 7 hm+r+ 5
Group 8 h+mr 5
Total = 1000

The result: 8 different types of plaques are formed.

Analyzing these data shows how the two-point cross between m and r understated the true distance between them.

Let's first look at single pairs of recombinants as we did before (thus ignoring the third locus).
The three-point cross is also useful because it gives the gene order simply by inspection:
Link to another example of a three-point cross.
Try a three-point cross problem.
There is another mapping technique — deletion mapping — that was used with T4, another "T-even" bacteriophage. Link to a discussion.
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19 February 2011