- E coli cells were simultaneously infected with two strains of phage λ, One strain of λ had a mutant host range, is temperature sensitive and known to produce clear plaques (genotype h st c); another strain of λCarried the wild type alleles (genotype h+st+ c+), progeny phage were collected from the lysed cells and were plated on bacteria. The following numbers of different progeny were obtained:
| Progeny phage genotype | Number of plaques |
| h+st+ c+ | 350 |
| h+ c st | 86 |
| h+ c+st | 4 |
| h c st | 300 |
| h+ c st+ | 90 |
| h c st+ | 6 |
| h c+st+ | 114 |
| h c+st | 50 |
What will be the order of the three genes and the map distance between them?
(1) h 36cM c 15cM st
(2) c21cMh15cMst
(3) h21cM st15cM c
(4) h36cM c∞cM st
Lambda Phage Three-Gene Recombination Analysis: h, st, c
Two λ strains infect [translate:ई. कोलाई] simultaneously:
- Mutant: h st c (mutant host range, temperature sensitive, clear plaques)
- Wild type: h⁺ st⁺ c⁺
Progeny phage genotypes and plaque numbers (total 1000):
| Genotype | Count | Frequency (%) |
|---|---|---|
| h⁺ st⁺ c⁺ | 350 | 35.0 |
| h⁺ c st | 86 | 8.6 |
| h⁺ c⁺ st | 4 | 0.4 |
| h c st | 300 | 30.0 |
| h⁺ c st⁺ | 90 | 9.0 |
| h c st⁺ | 6 | 0.6 |
| h c⁺ st⁺ | 114 | 11.4 |
| h c⁺ st | 50 | 5.0 |
Step 1: Identify Parental and Double Crossover Classes
- Parental (largest classes): h⁺ st⁺ c⁺ (350), h c st (300)
- Double crossovers (smallest classes): h⁺ c⁺ st (4), h c st⁺ (6)
Step 2: Determine Gene Order
Comparing parental de>h⁺ st⁺ c⁺ with double crossover de>h⁺ c⁺ st shows de>st changes while de>h and de>c remain the same.
Similarly, comparing parental de>h c st with double crossover de>h c st⁺, only de>st changes.
Therefore, gene order is de>h – st – c.
Step 3: Classify Single Crossovers (SCO) for Each Interval
- Interval 1 (de>h ↔ st): de>h⁺ c st (86), de>h c⁺ st⁺ (114)
- Interval 2 (de>st ↔ c): de>h⁺ c st⁺ (90), de>h c⁺ st (50)
- Double crossovers: de>h⁺ c⁺ st (4), de>h c st⁺ (6)
Step 4: Calculate Recombination Frequencies and Map Distances
Total progeny = 1000
- Recombination between h and st: SCO (86 + 114) + DCO (4 + 6) = 200 + 10 = 210 → 21% → 21 cM
- Recombination between st and c: SCO (90 + 50) + DCO (10) = 140 + 10 = 150 → 15% → 15 cM
Map order: h — 21 cM — st — 15 cM — c
Step 5: Evaluate Map Options
- Option (1): h 36 cM c 15 cM st — Incorrect distances
- Option (2): c 21 cM h 15 cM st — Incorrect gene order (st not in middle)
- Option (3): h 21 cM st 15 cM c — Correct gene order and distances
- Option (4): h 36 cM c … st — Incorrect placement and unclear intervals
Final Answer
The correct gene order is h – st – c with map distances 21 cM between h and st, and 15 cM between st and c.
Correct option: (3) h 21 cM st 15 cM c
1 Comment
Juber Khan
February 23, 2026Option 3 is right
Order is h+st+c or h st c