Q.45 Heterozygous female fruit flies with gray body and purple eyes were mated with homozygous males with black body and red eyes. The number of offspring obtained and their phenotypes are shown below:
| Number of offspring | Phenotype |
|---|---|
| 300 | Gray body–purple eyes |
| 347 | Black body–red eyes |
| 61 | Gray body–red eyes |
| 55 | Black body–purple eyes |
Calculate the recombination frequency.
Recombination frequency in this Drosophila cross is 15% (15 map units) between the genes for body colour and eye colour.
Introduction
In classical genetics, calculating recombination frequency from test‑cross data is a key step for constructing linkage maps and understanding how closely two genes are located on a chromosome. This article uses the well‑known Drosophila problem involving gray body–purple eyes and black body–red eyes to show every step of the calculation for CSIR‑NET and other life‑science exams.
Step 1: Understand the cross
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Heterozygous female: gray body and purple eyes (both mutant traits are recessive to gray body and red eyes, but her phenotype tells us she carries these alleles in cis).
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Male tester: homozygous black body and red eyes, so all his gametes contribute the recessive body allele and dominant eye allele, letting the female’s gametes be “read” directly in the offspring phenotypes.
From the question, the offspring numbers are:
| Offspring phenotype | Number | Interpretation |
|---|---|---|
| Gray body – purple eyes | 300 | one parental |
| Black body – red eyes | 347 | other parental |
| Gray body – red eyes | 61 | recombinant |
| Black body – purple eyes | 55 | recombinant |
Parental classes are the two most frequent phenotypes, while the less frequent two are the recombinants.
Step 2: Apply the recombination frequency formula
The general formula for recombination frequency (RF) between two linked genes is:
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RF = (number of recombinant progeny ÷ total progeny) × 100%.
Here:
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Recombinants = 61 (gray–red) + 55 (black–purple) = 116.
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Total offspring = 300 + 347 + 61 + 55 = 763.
So,
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RF = (116 ÷ 763) × 100 ≈ 15.2%, usually rounded to 15% or 15 map units (cM).
This value being far below 50% confirms that the body‑colour and eye‑colour genes are linked on the same chromosome rather than assorting independently.
Why each option (phenotype class) matters
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Gray body–purple eyes (300): Represents one parental combination carried intact from the heterozygous female without crossing over between the two loci.
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Black body–red eyes (347): Represents the second parental combination and, together with the first, dominates the progeny count because most meioses occur without recombination between these linked genes.
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Gray body–red eyes (61): Shows gray body allele linked with red eye allele, a new combination created only when a crossover happens between the body‑colour and eye‑colour loci in the female.
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Black body–purple eyes (55): The reciprocal recombinant class, where black body is now associated with purple eyes due to the same crossover event type.
These two recombinant classes together quantify how often crossing over separates the original parental combinations, which is exactly what recombination frequency measures.
Key exam takeaways
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Always identify the two largest classes as parental and the two smallest classes as recombinants in a simple two‑point test cross.
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Sum only the recombinant progeny for the numerator; use the total progeny for the denominator to obtain RF in percent and in map units (1% ≈ 1 cM).
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An RF of about 15% means the body‑colour and eye‑colour genes in this Drosophila cross are moderately linked, with an estimated genetic distance of 15 map units.
