Questions In flies, three 16 through 18 are based on the chromosome map below. a

Question

Questions In flies, three 16 through 18 are based on the chromosome map below. autosomal genes have the following map distances: 10 mu 15 mu a. ales are crossed to homozygous abe males. Assuming that 1 flies are counted, answer the questions that follow. a. 765 b. 150 c. 135 d. 100 e. 85 Question 16. In total, how many progeny would you expect with the phenotypes tac' and 'b' Question 17 In total, how many progeny would you expect with the phenotypes 'a' and 'be Question 18. In total, how many progeny would you expect with the phenotypes ab' and 'c

Explanation / Answer

16 - ac and b are the original parental genotype that was crossed with. Parental types are always the maximum number of progeny in a recombination. so they will have the most number of progenies.

So the answer will be 765

17 - progeny a and bc will be formed when there is recombination between b and c. the distance between b and c is 10 map unit. So the maximum number of progenies that have bc and a genotype will be 100. but 100 progenies include the progeny which are the double crossover and have abc genotype.

So the answer is 85

18- progeny ab and c will be formed when there is recombination between a and b. the distance between a and b is 15 map unit. So the maximum number of progenies that have ab and c genotype will be 150. but 150 progenies include the progeny which are the double crossover and have abc genotype.

So the answer is 135

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