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

d. 80

Explanation:

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

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)

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.

21 stacks

step-by-step explanation:

divide 630/30

21 stacks

enjoy!

after making a punnet square using bv bv bv bv you will find that the ratio will be 9: 7 or a 128.6% probability.

hope this