## 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
2 weeks
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 testcrossednumber of progeny = 1000

Determine the number of double crossover resultP( 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