Question

Genes a and b are 10 map units apart, b and c are 20 map units apart, and a and c are 30 map units apart. If a triple heterozygote is testcrossed, among 1,000 progeny, how many are expected to result from single crossovers between a and b if there is no interference? a. 20; b. 40; c. 60; d. 80; e. none of the above

Answer 1

Answer:

d. 80

Explanation:

If the triple heterozygote is in linkage phase ABC/abc, the gametes it will produce are:

• ABC: Parental
• abc: Parental
• Abc: SCO (a-b)
• aBC: SCO (a-b)
• ABc: SCO (b-c)
• abC: SCO (b-c)
• AbC: DCO
• aBc: DCO

Distance (ab) = 10 mu;

Distance (bc) = 20 mu.

There is no interference, so we can calculate recombination frequency (RF) between the genes a and b as the probability that a crossover happens between a and b, and no crossover happens between b and c:

RF ab = p(COab)  ×  p(no CObc)

$RF[ab]=\frac{Distance[ab]}{100} * (1-\frac{Distance[bc]}{100})\\RF[ab]=0.1*(1-0.2)\\RF=0.1*0.8\\RF=0.08$

In a progeny of the test cross of 1000 individuals, there will be 1000×0.08= 80 individuals resulting from a single crossover between a and b.