Question

E. coli strain B is doubly infected with two rII mutants of phage T4. 0.1 ml of a 10^6 dilution of the progeny is plated on E. coli B and 0.1 ml of a 10⁵ dilution of the progeny is plated on E. coli K. 400 plaques appeared on strain B, 10 on strain K. Calculate the map distance between these two mutations.

A. 0.05 mu

B. 0.1 mu

C. 0.2 mu

D. 0.5 mu

E. 1.0 mu

Answer 1

The map distance between the two rII mutations is calculated as 2.5 mu, using the number of recombinant plaques formed on E. coli strain K and the total number of progeny from plaques on E. coli strain B. The provided options do not include the correct value, as the closest value of 0.1 mu is inaccurate.

Step-by-step explanation:

The map distance between two rII mutations in the T4 phage can be calculated by analyzing the number of recombinants and the total number of progeny. We take plaques formed on E. coli strain B and strain K and use them to calculate the recombination frequency. The rII mutants can only form plaques on E. coli strain B because strain K restricts growth of rII mutant phages.

We find that 400 plaques formed on strain B indicate the total number of progeny (as both mutants can grow on B). The number of plaques on strain K (10 plaques) represents the number of recombinants since K will only allow growth of wild-type phage, which must come from recombinant events between the two rII mutants. Using these figures, we can calculate the recombination frequency, which is the number of recombinants divided by the total number of progeny. Therefore, recombination frequency = 10/400 = 0.025 or 2.5%. The recombination frequency is equivalent to the map distance in map units (mu).

The correct map distance between these two rII mutations is therefore 2.5 mu, which makes B (0.05 mu), C (0.2 mu), D (0.5 mu), and E (1.0 mu) incorrect. The closest option provided, B (0.1 mu) is still incorrect as it is four times smaller than our calculated value.