You have infected E. coli cells with two strains of T4 bacteriophage. One strain

Question

You have infected E. coli cells with two strains of T4 bacteriophage. One strain is minute (m), rapid lysis (r), and turbid (tu); the other is wild-type for all 3 markers. The lytic products of this infection are plated and classified. The resulting 10,351 plaques were distributed among eight genotypes as follows:

m r tu      3600

+ + +       3800

m r +          852

m + +        519

m + tu     75

+ + tu      965

+ r +         70

+ r tu 470

a) What is the order of these genes in the T4 genome?

b) What are the genetic distances between them?

c. How much interference is there?

Explanation / Answer

a)

To determine the order of the genes we need to find parental genotypes and double recombinants.

The genotypes found most frequently are the parental genotypes.From the figure it is clear that the + + + and m r tu genotypes were the parental genotypes.

The double-crossover gametes are always in the lowest frequency. From the table the + r + and m + tu genotypes are in the lowest frequency. An important event of double cross over is that a double-crossover event moves the middle allele from one sister chromatid to the other.

So the gene order is + + + or m r tu

b)

Calculating linkage distance between the genes.

First find the recombinants which show recombination between m and r or r and tu

Recombinants as a result of m and r recombination are : m + + and + r tu

Recombinants as a result of r and tu recombination are : m r + and + + tu

Calculate percentage of recombination (recombination frequency) between m and r (f1)

          = total number of m / r recombinants + double recombinants / total progeny

          = 519 + 470 + 75 + 70 / 10351

         = 0.1 or 10 %

Calculate percentage of recombination (recombination frequency) between r and tu (f2)

        = total number of r / tu recombinants + double recombinants / total progeny

        = 852 + 965 + 75 + 70 / 10351

        = 0.18 or 18 %

Calculate percentage of recombination (recombination frequency) between m and r

        = total number of single cross recombinants + double recombinants / total progeny

        = 0.28 or 28 %

So, we have a linkage map in which the distance between m and r = 10 cM and between r and tu = 18 cM, and between m and tu = 28 cM.

c)

Three-point crosses also allows one to measure interference (I). The concept is that given specific recombination rates in two adjacent chromosomal intervals, the rate of double-crossovers in this region should be equal to the product of the single crossovers.

In the m r tu,, the recombination frequency is 0.1 between genes m and r, and the recombination frequency between r and tu is 0.18. Therefore, we would expect double recombinants = (0.1 x 0.18) = 0.018 or 1.8 %.

Expected double recombinants in progeny = 0.018 x 10351 = 186

To measure interference, we first calculate the coefficient of coincidence (c.o.c.) which is the ratio of observed to expected double recombinants.

c.o.c = 145 / 186 = 0.77

Interference = 1 - c.o.c

                      = 1 – 0.77

                      = 0.23 or 23 %

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