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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 testcr
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 double crossovers if there is no interference?
а. 10;
b. 20;
c. can’t be determined
d. 30;
e. 60;
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Biology
4 years
2021-07-18T23:22:30+00:00
2021-07-18T23:22:30+00:00 1 Answers
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Answers ( )
Answer:
20 ( B )
Explanation:
Given data:
a and b are 10 map units apart
b and c are 20 map units apart
a and c = 30 map units apart
condition ; Triple heterozygote testcrossed
number of progeny = 1000
Determine the number of double crossover result
P( crossover between a and b ) = 10/100 = 0.1
P( crossover between b and c ) = 20/100 = 0.2
p( double crossover ) = 0.1 * 0.2 = 0.02
hence number of double crossovers = number of progeny * 0.02
= 1000 * 0.02 = 20