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 distance between gene a and b is estimated by the recombination frequency in a genetic cross, which is 30% or 30 map units, corresponding to option C.

Step-by-step explanation:

The distance between two linked genes can be estimated by the frequency of recombination events occurring between them. In the scenario presented, we have 700 out of 1000 individuals displaying the parental type in a cross between homozygous recessive individuals for genes a and b (aabb), and the wild type (AB).

The proportion of recombinant offspring, which shows the nonparental types, is given by subtracting the number of parental individuals from the total number of offspring, then dividing by the total. In this case, the calculation is (1000 - 700) / 1000 = 0.30, meaning 30% of the offspring are recombinants. Since the recombination frequency corresponds to the genetic distance in map units or centimorgans (cM), the answer is that the distance between a and b is 30 map units.

Therefore, the correct option is:

C. 30 map units.