Correct option is D
The problem involves two mutations in a bacteriophage: one causing a clear plaque (c) and the other causing a minute plaque (m). These mutations are located in genes that are 9 centimorgans (cM) apart, indicating a recombination frequency of 9%. When bacteriophages with genotypes c⁺ m⁻ and c⁻ m⁺ are mixed to infect bacterial cells, the progeny plaques are collected and plated. The expected number of different plaque types is provided, and we need to identify the combination of options that represents all correct statements.
Key Concepts:
- A recombination frequency of 9% means that 9% of the progeny will be recombinants, while 91% will be parental types.
- The parental genotypes are c⁺ m⁻ and c⁻ m⁺.
- The recombinant genotypes will be c⁺ m⁺ and c⁻ m⁻.
- The expected progeny should reflect both parental and recombinant types, with the proportion of recombinants being approximately 9% each, and parental types making up the remaining 82% (split between the two parental combinations).
Analyzing the Options:
Conclusion:
- A. c⁺ m⁺ 455, c⁺ m⁻ 45, c⁻ m⁺ 45, c⁻ m⁻ 455
- This suggests equal numbers of recombinants (c⁺ m⁺ and c⁻ m⁻) and parental types (c⁺ m⁻ and c⁻ m⁺), which would imply a 50% recombination frequency (far higher than the 9% expected). This is incorrect.
- B. c⁺ m⁺ 455, c⁺ m⁻ 455, c⁻ m⁺ 45, c⁻ m⁻ 45
- This also suggests a high recombination frequency (approximately 10% for c⁻ m⁺ and c⁻ m⁻, but the numbers are inconsistent with the 9% recombination rate). The parental types (c⁺ m⁻ and c⁺ m⁺) are overrepresented. This is incorrect.
- C. c⁺ m⁺ 45, c⁺ m⁻ 455, c⁻ m⁺ 455, c⁻ m⁻ 45
- Here, c⁺ m⁻ (455) and c⁻ m⁺ (455) are the parental types, making up the majority (91% of 1000 = 910, or 455 each), which aligns with the 91% parental expectation. The recombinants c⁺ m⁺ (45) and c⁻ m⁻ (45) make up 9% (90 of 1000, or 45 each), consistent with the 9 cM distance. This is correct.
- D. c⁺ m⁺ 65, c⁺ m⁻ 680, c⁻ m⁺ 685, c⁻ m⁻ 70
- Here, c⁺ m⁻ (680) and c⁻ m⁺ (685) are the parental types, totaling approximately 1365 out of 1500 (91%), which fits the 91% parental expectation. The recombinants c⁺ m⁺ (65) and c⁻ m⁻ (70) total about 135 out of 1500 (9%), aligning with the 9 cM distance. This is also correct.
Both options C and D provide combinations of plaque types that are consistent with a 9% recombination frequency between the c and m genes. The slight variation in numbers (e.g., 45 vs. 65 for recombinants) could be due to rounding or sample size differences, but both reflect the expected parental and recombinant ratios. Therefore, the combination of all correct statements is represented by option 4 (C and D).
