Suppose that a, b and c are recessive mutations that are linked to each other in
ID: 187237 • Letter: S
Question
Suppose that a, b and c are recessive mutations that are linked to each other in mice, but the gene map order is unknown. A female (named Minnie) who is heterozygous at all three loci was mated to a homozygous recessive male (named Mickey). Since Minnie and Mickey are “super-mice”, they were able to have 1000 progeny. The number of progeny with each phenotype was as follows: a b c 23 a b + 46 a + + 427 + + + 26 + b c 424 + + c 54
What are the distances between genes a, b and c?
a to b = 5% and b to c = 10% and a to c = 15%
a to c =10% and c to b = 5% and a to b = 15%
a to c = 5% and c to b = 10% and a to b = 15%
a to b = 5% and b to c = 10% and a to c = 15%
a to c =10% and c to b = 5% and a to b = 15%
a to c = 5% and c to b = 10% and a to b = 15%
Explanation / Answer
Answer - C a to c = 5% and c to b = 10% and a to b = 15%
The order of gene - a c b
Parent genotype - a + + = 427 and + b c = 424
Cross between a and c = + + + = 26 and a b c = 23
Cross between b and c = a b + = 46 and a b c = 54
a to c = 26 + 23 / 1000 = 5%
b to c = 46 + 54 / 1000 = 100 / 1000 = 10 %
a to b = 5 + 10 = 15%
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