17. Two mutations were isolated in bacteriophage, one causing clear plaque (c) and the other causing minute plaque (m). The genes responsible for these two mutations are 9 CM apart. The plaques with genotype c⁺m^– and c^–m⁺ were mixed to infect bacterial cells. The progeny plaques were collected, cultured and plated on bacteria. The expected number of the different types of plaques are shown below:
A. c⁺m⁺455, c⁺m^– 45, c^–m⁺ 45, c^–m^– 455
B. c⁺m⁺ 455, c⁺m^– 455, c^–m⁺ 45, c^–m^– 45
C. c⁺m⁺45, c⁺m^– 455, c^–m⁺ 455, c^–m^– 45
D. c⁺m⁺65, c⁺m^– 680, c^–m⁺ 685, c^–m^– 70
Which one of the following options represents the combination of all correct statements?
(1) A only        (2) B only
(3) C only        (4) Cand D

The correct answer is option (3) C only.

Explanation:

• Given:

  • Two mutations in bacteriophage: clear plaque (c) and minute plaque (m).

  • Genes responsible are 9 cM apart, which means a 9% recombination frequency.

  • Initial genotypes are c+m- and c-m+.

  • Total progeny plaques are 1000.

• Genetic recombination between these linked genes will produce parental and recombinant types as follows:

  • Recombinant frequency (RF) = 9% = 0.09.

  • Recombinants occur in equal proportions: 0.09/2 = 0.045 or 4.5% each.

  • Parental types occur in equal proportions = 1 – 0.09 = 0.91.

  • Each parental type proportion = 0.91/2 = 0.455 or 45.5%.

• Expected numbers for each phenotype in 1000 plaques:

  • c+m+ (recombinant) = 0.045 × 1000 = 45.

  • c+m- (parental) = 0.455 × 1000 = 455.

  • c-m+ (parental) = 0.455 × 1000 = 455.

  • c-m- (recombinant) = 0.045 × 1000 = 45.

• Comparing with options:

  • C matches these expected numbers exactly.

  • A, B, and D do not match these frequencies for the 1000 total plaques.

Therefore, option C correctly represents the expected distribution based on recombination frequency between linked genes 9 cM apart.

Introduction:
Genetic recombination plays a key role in bacterial virus (bacteriophage) genetics when mutations affect traits like plaque size or clarity. In bacteriophage populations with linked mutations such as clear plaque (c) and minute plaque (m), the distance between genes in centiMorgans (cM) helps predict the recombination frequency and the resulting proportions of progeny plaque types. Here, we explore how to solve a classic recombination problem involving these mutations separated by 9 cM.

Solution Details:

• Two mutations, clear plaque (c) and minute plaque (m), are located 9 cM apart.

• When bacteriophages with genotypes c+m- and c-m+ infect bacteria together, both parental and recombinant plaques are produced.

• Recombination frequency corresponds directly to gene distance: 9 cM = 9% recombinants.

• Each recombinant class (c+m+ and c-m-) will appear with a frequency of 4.5%, while parental genotypes (c+m- and c-m+) will each appear at 45.5%.

• For 1000 total plaques, the expected numbers are 45 c+m+, 455 c+m-, 455 c-m+, and 45 c-m- plaques.

• This calculation matches option C, confirming its correctness.

This understanding is critical in genetic mapping and phage biology studies where phenotypic mutations are linked to gene loci. Calculating recombinant frequencies based on plaque phenotypes aids in gene mapping and functional genomics of bacteriophages.

References:
This solution is based on classic bacterial genetics principles of recombination frequency, gene linkage, and plaque phenotype analysis in bacteriophages.​

Let me know if you want the complete step-by-step math or additional explanations on gene linkage and recombination mechanisms.