view.php?id-4896977 Jump to... Question 12 of 13 Map o A geneticist is using a t

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view.php?id-4896977 Jump to... Question 12 of 13 Map o A geneticist is using a three-point testcross to map three linked Drosophila recessive mutations called a b, and c, where a is associated with associated with curved bristies. She first crosses homozygous anomalous, buckled fles to homozygous curved flies. Next, she testcrosses the Fi progeny to anomalous, buckled, curved flies. She obtains 1000 progeny distributed as the following results. From this data, calculate the map distance between b and c. s gait; b is associated with buckled wings; and c is Testcross Progeny Phenotype curved anomalous, buckled wild type anomalous, buckled, curved buckled, curved Number 277 283 Number m.u. 128 132 83 87 anomalous, curved Prevous Give Up & View Sokution Check Answer NextE 1 Hint 0

Explanation / Answer

Answer: 27 m.u.

Explanation: Testcross Progeny Phenotype Genotype Number

Curved A B c 277

Anomalous, buckled a b C 283  

Wild type A B C 5

buckled, curved, anomalous a b c 5

Buckled, curved A b c 127

Anomalous a B C 132  

curved a B c 83

Buckled A b C 87

the parental cross is,aa bb CC * AA BB cc gives the genotype of F1 offspring are Aa Bb cC.

double cross over products are a b c and A B C

single cross over products between gene b and c are A b c and a B C

so the distance between b and c = 127 + 132 + 5+ 5 / 1000 * 100 = 26.9cM = 27 cM

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