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The map of a chromosome interval is:A——10 m.u.——B——40 m.u.——CFrom the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny, if no interference is present?Question 11Answera.20b.40c.5d.10e.80

Question

The map of a chromosome interval is:A——10 m.u.——B——40 m.u.——CFrom the cross A b c / a B C × a b c / a b c, how many double crossovers would be expected out of 1000 progeny, if no interference is present?Question 11Answera.20b.40c.5d.10e.80

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Solution 1

The calculation of double crossovers is based on the product of the probabilities of the single crossovers.

The map shows that the distance between A and B is 10 map units (m.u.) and the distance between B and C is 40 m.u.

The probability of crossover between A and B is 10/100 = 0.1 (since 1 m.u. = 1% recombination) and the probability of crossover between B and C is 40/100 = 0.4.

The probability of double crossover (crossover between A and B and between B and C) is the product of the individual probabilities, which is 0.1 * 0.4 = 0.04.

Therefore, out of 1000 progeny, the expected number of double crossovers is 0.04 * 1000 = 40.

So, the answer is b. 40.

This problem has been solved

Solution 2

The calculation of double crossovers is based on the product of the probabilities of the single crossovers that it comprises.

In this case, the single crossovers are A-B and B-C.

The probability of an A-B crossover is 10 m.u., or 10%, and the probability of a B-C crossover is 40 m.u., or 40%.

To find the expected number of double crossovers, multiply these probabilities together and then multiply by the total number of progeny.

So, (0.10) * (0.40) * 1000 = 40.

Therefore, the answer is b. 40.

This problem has been solved

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