Question:

The linkage map of X-chromosome of fruitfly has $66$ units, with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be:-

Updated On: Jun 26, 2024
  • 0.66
  • > 50%
  • $\leq\, 50 \%$
  • 1
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The Correct Option is C

Solution and Explanation

The linkage map is a chromosome map which is determined by the recombination relations. The map distances are expressed by recombination frequencies and are given by recombination frequencies and are sometimes represented in map unit. X-chromosome has 66 crossover units with yellow and bobbed genes at two extreme ends of the map. Recombination frequency never exceed 50% between any two loci, but these 66 units will be actually obtained by making use of a mapping function.
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Concepts Used:

Non-Mendelian Genetics

The term - non-mendelian inheritance refers to any pattern of heredity in which features do not separate according to Mendel's laws. These principles describe how features linked with single genes on chromosomes in the nucleus are passed down through generations.

Types of Non-Mendelian Inheritance

Codominance Inheritance

It is a form of incomplete dominance in which both alleles for the same feature are expressed in the heterozygote at the same time. For example, the MN blood types of humans.

Incomplete Dominance

In a heterozygote, the dominant allele does not always completely cover the phenotypic expression of the recessive gene, resulting in an intermediate phenotype which is referred to as "incomplete dominance”.