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.

he movement of air through earth's or any planet's atmosphere is called wind, and the main cause of earth's winds is uneven heating by the sun. this uneven heating causes changes of atmospheric pressure, and winds blow from regions with high pressure to those with low pressure.

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