problem set 3, 2010 answers

problem set 3, 2010 answers - 1. a) Because the markers are...

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1. a) Because the markers are very closely linked we can assume that there is no recombination occurring. b) Because the unaffected parents have affected children but are unaffected themselves, the gene must be recessive and each parent must be a carrier. D= disease and * = wild-type allele of the CF gene Father's genotype is 1ac 2aa 3ab D* Mother's genotype is 1bc 2aa 3bc D* Affected children are 1ac 2aa 3bc DD The affected children can only inherit 1a from the father (the mother doesn't have this allele), and therefore get 1c from the mother. Hence, these alleles must be linked to the disease allele in the respective parents. Similarly, they can only inherit 3b from the father and 3c from the mother. Therefore, 1a and 3b must be on the same chromosome and come from the father. 1c and 3c must come from the mother. Both these chromosomes must carry the D allele of the CF gene. Since there is no recombination around these markers, this tells us the phase of linkage in the parents and children. Father 1 2 3 CF Mother 1 2 3 CF Affected Child 1 2 3 CF a a b D c a c D a a b D c a a * b a b * c a c D c) The man is a carrier of CF. He is 1ab 3bb. The a allele must come from the father and it is linked to the 3b allele and the D allele. The 1b and the other 3b comes from the
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This note was uploaded on 02/09/2011 for the course BIOS 20182 taught by Professor Lahn during the Spring '11 term at UChicago.

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problem set 3, 2010 answers - 1. a) Because the markers are...

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