Question

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

The correct option is C

30 map unit

The correct option is C

Explanation of the correct option:

1. 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.
2. 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 Mendel's law of independent assortment.
3. This is the reason why the frequency of offspring is different from the parental phenotype.

The distance between a and b = $no.ofrecombinants/totalno.ofoffsprings\times 100$

Step 1: Given data

Total number of offsprings=1000

Number of offspring of parental types=700

Number of recombinants=(Total number of offsprings-number of offsprings of parental type)

=(1000-700)

=300.

Step 2: Applying the formula

Formula used:

Step 3: Solving the equation to get the required value

Solving the equation to get the required value:

=(300/1000) x 100

=30 map units.

Final answer: The distance between a and b is 30 map units.