- When F1 female Drosophila of he genotype a+a b+ b c+ c is test crossed, the following progenies were obtained:
| Progeny classes | No. of progenies |
| a+ b+ c+ | 22 |
| a+ b+ c | 28 |
| a b c+ | 26 |
| a b c | 24 |
| a+ b c+ | 230 |
| a+ b c | 220 |
| a b+ c+ | 225 |
| a b c | 225 |
| Total | 1000 |
The progeny has been shown as classes derived from the female gamete. Statements A to F as given below are conclusions derived from the above result.
A) Genes a and b are linked in cis.
B) Genes a and b are linked in trans.
C) Genes a and b are linked in cis while b and c are linked in trans.
D) The genotype of the parents are a+ a+ b+ b+ and aabb
E) The genotype of the parents are a+ a+ b b and aab+b+
F) Genes a and b are 10cM apart.
Which of the above statements are correct?
(1) C alone (2) A, E and F
(3) B, E and F (4) A, D and F
Drosophila Three-Point Test Cross Question Analysis
Question:
An F₁ female Drosophila with genotype a⁺a b⁺b c⁺c is test-crossed to an abc/abc recessive male. The 1000 offspring show the following maternal gamete classes:
| Phenotype | Count |
|---|---|
| a⁺ b⁺ c⁺ | 22 |
| a⁺ b⁺ c | 28 |
| a b c⁺ | 26 |
| a b c | 24 |
| a⁺ b c⁺ | 230 |
| a⁺ b c | 220 |
| a b⁺ c⁺ | 225 |
| a b⁺ c | 225 |
Step 1: Identify Parental and Recombinant Classes
The largest two classes represent parental (non-recombinant) gametes, and the smallest two represent double crossovers.
- Largest (parental): a⁺ b c⁺ (230) and a b⁺ c (225)
- Smallest (double crossovers): a⁺ b⁺ c⁺ (22) and a b c (24)
Hence parental female gametes are:
Chromosome 1: a⁺ b c⁺
Chromosome 2: a b⁺ c
Step 2: Determine Gene Order
Compare each double crossover with parents: only the middle gene changes.
Comparing parental a⁺ b c⁺ with DCO a⁺ b⁺ c⁺, only b changes (b → b⁺). Thus gene b is in the middle.
Gene order: a – b – c
Step 3: Linkage Phase Between Gene Pairs
Parental Chromosomes:
- Chromosome I: a⁺ b c⁺
- Chromosome II: a b⁺ c
a–b Pair: a⁺ with b on one, a with b⁺ on the other → trans (repulsion).
b–c Pair: b with c⁺, b⁺ with c → trans (repulsion).
Both gene pairs a–b and b–c are linked in trans configuration.
Step 4: Calculate Map Distance Between a and b
Recombinants between a and b occur in any gamete not matching the parental combinations (a⁺ b or a b⁺).
Recombinant classes for a–b:
- a⁺ b⁺ c⁺ (22)
- a⁺ b⁺ c (28)
- a b c⁺ (26)
- a b c (24)
Total recombinants = 22 + 28 + 26 + 24 = 100
Map distance a–b = (100 / 1000) × 100 = 10 cM
Step 5: Evaluate Statements (A–F)
| Statement | Evaluation |
|---|---|
| A) Genes a and b are linked in cis | False – actual configuration is trans |
| B) Genes a and b are linked in trans | True |
| C) a–b in cis while b–c in trans | False – both pairs are in trans |
| D) Parental genotypes are a⁺a⁺ b⁺b⁺ and a a b b | False – does not fit F₁ configuration |
| E) Parental genotypes are a⁺a⁺ b b and a a b⁺b⁺ | True – produces F₁ with a⁺ b / a b⁺ (trans) |
| F) Genes a and b are 10 cM apart | True |
Final Answer
Correct statements: B, E, and F.
Final Answer: Option (3) – B, E, and F
1 Comment
Juber Khan
February 23, 2026Order is a+b c+ or a b c
Distance byw ab 100/1000=.1✖️100= 10 cm
And a and b in trans condition
Option B E F RIGHT