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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Cherry 2 weeks 2021-07-18T23:22:30+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-18T23:24:28+00:00

    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

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