Question

In a cross between individuals homozygous for (a,b) and wild type(+ +). In this cross 700 out of 1000 individuals were of parental tye. Then the distance between a and b is

• A. 70 map unit
• B. 35 map unit
• C. 30 map unit
• D. 15 map unit

Answer 1

Answer: (C)

30 map unit

Genetic linkage is a phenomenon in which the genes that are close enough on a chromosome will tend to stick together and the alleles of those genes will tend to be inherited as a pair.

When the genes are linked, genetic crosses in the genes will lead to the ratios of gametes (egg and sperm) and offspring types that are not predicted from the Mendel's law of independent assortment. This is the reason why the frequency of offsprings are different with the parental phenotype.

The distance between a and b = $\frac{\text{no. of recombinants}}{\text{total number of offsprings}}\times 100$

= $\frac{300}{1000}\times 100$

= 30 map units.