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Principles of Inheritance and Variation - Linkage and Recombination

Grade 12CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Linkage is the physical association of genes on a chromosome. It was discovered by T.H. Morgan through his experiments on DrosophilaDrosophila melanogastermelanogaster.

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Genes located on the same chromosome are called linked genes. They tend to be inherited together, which leads to a higher frequency of parental phenotypes compared to recombinant phenotypes in the offspring.

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Recombination is the process of forming new combinations of alleles in the offspring, different from those found in the parents, due to crossing over during ProphaseProphase II of meiosis.

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The strength of linkage is inversely proportional to the distance between the genes. Tightly linked genes show very low recombination (e.g., 1.3%1.3\% for yellow body and white eye in DrosophilaDrosophila), while loosely linked genes show higher recombination (e.g., 37.2%37.2\% for white eye and miniature wing).

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Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position on the chromosome (Genetic Mapping).

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The maximum possible recombination frequency between any two genes is 50%50\%, which indicates that the genes are either very far apart on the same chromosome or located on different chromosomes (Independent Assortment).

📐Formulae

Recombination Frequency (RF)=Number of Recombinant ProgenyTotal Number of Progeny×100\text{Recombination Frequency (RF)} = \frac{\text{Number of Recombinant Progeny}}{\text{Total Number of Progeny}} \times 100

1 Map Unit (m.u.)=1 centiMorgan (cM)=1% Recombination Frequency1 \text{ Map Unit (m.u.)} = 1 \text{ centiMorgan (cM)} = 1\% \text{ Recombination Frequency}

Frequency of Parental types+Frequency of Recombinant types=100%\text{Frequency of Parental types} + \text{Frequency of Recombinant types} = 100\%

💡Examples

Problem 1:

In a test cross between a dihybrid (AaBbAaBb) and a double recessive parent (aabbaabb), the following offspring were observed: 4545 AaBbAaBb, 4545 aabbaabb, 55 AabbAabb, and 55 aaBbaaBb. Calculate the recombination frequency and the distance between genes AA and BB.

Solution:

First, identify the recombinants: AabbAabb (55) and aaBbaaBb (55). Total recombinants = 5+5=105 + 5 = 10. Total offspring = 45+45+5+5=10045 + 45 + 5 + 5 = 100. RF=10100×100=10%\text{RF} = \frac{10}{100} \times 100 = 10\% Since 1% RF=1 cM1\% \text{ RF} = 1 \text{ cM}, the distance is 10 cM10 \text{ cM}.

Explanation:

The recombination frequency represents the percentage of offspring that have non-parental gene combinations. This percentage is directly translated into map units to determine genetic distance.

Problem 2:

Three genes XX, YY, and ZZ are located on the same chromosome. The recombination frequency between XX and YY is 5%5\%, between YY and ZZ is 10%10\%, and between XX and ZZ is 15%15\%. Determine the linear order of the genes.

Solution:

Distance X−Y=5 unitsX-Y = 5 \text{ units}, Distance Y−Z=10 unitsY-Z = 10 \text{ units}, Distance X−Z=15 unitsX-Z = 15 \text{ units}. Since X−ZX-Z is the largest distance (1515), XX and ZZ must be the flanking genes. Checking the middle gene: X−Y(5)+Y−Z(10)=15 unitsX-Y (5) + Y-Z (10) = 15 \text{ units}, which matches the X−ZX-Z distance.

Explanation:

By placing the genes with the highest recombination frequency at the ends, we can verify if the sum of the internal distances matches the total distance, confirming the sequence as X−Y−ZX-Y-Z.