32. In Drosophila, Bar eye (B) is a dominant mutation whileminiature wing (m) and yellow body colour (y) arerecessive mutations. eterozygous females for thesemutations were crossed to normal eyed miniaturewinged and yellow body coloured males. Assume thefollowing progeny was obtained:

Based on the results obtained, the order of genes will be:
A. B m y         B. m B y
The genetic distance between B and y will be:
C. 40 cM         D. 17.1 cM
The correct combination of answer is
(1) A and C        (2) B and C
(3) A and D        (4) B and D

Three‑Point Test Cross in Drosophila: Gene Order and Map Distance Analysis

Introduction

Three‑point test crosses in Drosophila are powerful tools to determine gene order and distances between linked genes such as Bar eye (B), miniature wing (m) and yellow body (y). In this article, the given progeny data are analysed step by step to identify parental, single‑crossover and double‑crossover classes, deduce the order of genes, calculate recombination frequencies and then evaluate every option in the MCQ.

Given Cross and Data

Mutations: Bar eye (B, dominant), miniature wing (m, recessive), yellow body (y, recessive) on the X‑chromosome of Drosophila.

Cross: Heterozygous female for these three genes × male with normal eye, miniature wings and yellow body (i.e., B⁺ m y).

Progeny Phenotypes and Numbers

• B⁺ m⁺ y⁺ = 30
• B m y⁺ = 25
• B m⁺ y⁺ = 165
• B⁺ m y⁺ = 120
• B m y = 20
• B⁺ m y = 185
• B m⁺ y = 110
• B⁺ m⁺ y = 45

Total progeny = 700.

Identifying Parental and Double‑Crossover Classes

• Parental (most frequent) classes: B⁺ m y (185) and B m⁺ y⁺ (165)
• Double‑crossover (least frequent) classes: B m y (20) and B⁺ m⁺ y⁺ (30)

From the parental classes, the chromosomal arrangements are:

• One chromosome: B⁺ m y
• Other chromosome: B m⁺ y⁺

From the DCO classes:

• One DCO: B m y
• Other DCO: B⁺ m⁺ y⁺

Between parental and DCO, the gene that changes its coupling relative to the other two is the middle one.

Comparison

• Parental: B⁺ m y → DCO: B m y (only B changes)
• Parental: B m⁺ y⁺ → DCO: B⁺ m⁺ y⁺ (only B changes)

Therefore, B is the middle gene. Gene order must be m – B – y (or equivalently y – B – m).

Evaluating Options

• Option A: B m y → B at one end (incorrect)
• Option B: m B y → B in the middle (correct)

Hence, the correct gene order is m – B – y, corresponding to Option B.

Calculation of Map Distances

Step 1: Classify Crossovers (Order: m – B – y)

• Parental: B m⁺ y⁺ (165), B⁺ m y (185)
• DCO: B m y (20), B⁺ m⁺ y⁺ (30)
• SCO (m–B): B m y⁺ (25), B⁺ m⁺ y (30)
• SCO (B–y): B m⁺ y (110), B⁺ m y⁺ (120)

Step 2: Recombination Frequencies

For region m–B:

Recombinants = (SCO m–B) + (DCO) = 55 + 50 = 105

Distance (m–B) = (105 / 700) × 100 = 15 cM

For region B–y:

Recombinants = (SCO B–y) + (DCO) = 230 + 50 = 280

Distance (B–y) = (280 / 700) × 100 = 40 cM

Total distance between m and y = 15 + 40 = 55 cM.

Question: “The genetic distance between B and y will be:”

• Option C: 40 cM (correct)
• Option D: 17.1 cM (incorrect)

Evaluating Combination Options

The final question: “The correct combination of answer is”

1. A and C
2. B and C
3. A and D
4. B and D

From analysis:

• Correct gene order: m B y → Option B
• Correct distance between B and y: 40 cM → Option C

Therefore, the correct combination is (2) B and C.

Option Analysis

• Option (1) A and C: Wrong, gene order incorrect.
• Option (2) B and C: Correct, both gene order and distance right.
• Option (3) A and D: Both incorrect.
• Option (4) B and D: Correct order but wrong distance.

Quick Revision Points for Exams

• In a three‑point test cross, the two most frequent phenotypic classes are parental and the two least frequent are double crossovers.
• The gene whose allele arrangement flips between parental and double‑crossover classes is the middle gene, revealing gene order.
• Recombination frequency for each interval = (SCO + DCO) / Total progeny × 100.
• Distance between terminal genes equals the sum of the two adjacent interval distances.