Question

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

In a cross between individuals homozygous for (a.b) and wild type (+ +), it was seen that 700 out of 1000 offspring were of parental type. According to the problem the number of recombinant offsprings are (1000-700)=300, which helps us find the recombinant frequency by the formula (No of recombinant progeny/ Total number of offsprings)= 300/1000 = 0.3. Thus the distance between the genes is 0.3 X 100 = 30 map units.

So, the correct option is '30 map units'.