Question

Genes x, y and z are linked on a particular chromosome in this order. The distance between x and y is 7.5 map units, while y and z are 23.1 map units apart. What is the probability of a double crossover in this chromosomal region? (Assume no interference)

a) 7.5%
b) 23.1%
c) 30.6%
d) 177.6%

Answer 1

The probability of a double crossover in the specified chromosomal region is 0.825% (none of the answers options provided). This value is calculated by multiplying the individual crossover probabilities for each of the two intervals.

Step-by-step explanation:

The probability of a double crossover in this chromosomal region, assuming no interference, is calculated by multiplying the probability of each crossover event. The distance between genes x and y is 7.5 map units, and the distance between y and z is 23.1 map units. Since 1 map unit (cM) corresponds to a 1% recombination frequency, we can use these distances to estimate the crossover probabilities.

Evaluating a double crossover: the chance of a crossover between x and y is 7.5%, and between y and z is 23.1%. Multiplying these gives (7.5/100) * (23.1/100) = .00825, or 0.825%, probability of a double crossover is c) 0.825% after converting to percentage. The options might have been based on incorrect calculations or assumptions, thus it's important to address this with your teacher.