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Lecture 11

BIOC15Fall2013 Lecture 11 Notes.docx

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Department
Biological Sciences
Course
BIOC15H3
Professor
Karen Williams
Semester
Fall

Description
BIOC15Fall2013 Lecture 11 Notes: Genetic mapping and mutations Problem: Ichthyosis a+ a o The Xg locus on the X-chromosome has two alleles Xand Xg (let’s call them “+” normal and “a” lacking antigen). The Xg locus is about 10 map units away from the X-linked ichthyosis (STS) locus. A man with X-linked ichthyosis (scaly skin) and no Xg antigen has a normal daughter with Xg antigen who marries a man normal skin and no Xg antigen. What is the probability that a son will have normal skin and no Xg antigen? o o so we have 2 classes of parentals and classes of recombinants o if Xg antigen is 10 m.u from STS o RF = (number of recombinants)/(total number of progeny) x 100 o But 10 m.u = 10% RF o Linked in cis conformation  all wildtype alelles are on one chromosome o STS+ a+ / sts a o 2 classes of recombinants and 10% RF  so 5% per class of recombinants o 100-10% = 90% for parentals, 45% per class of parentals Polytene chromosome in the salivary glands of Drosophila larvae o 10 o in Drosophila, interphase chromosomes replicate 10 x without going through mitosis  so each chromosome has 2 double helices o banding patterns are reproducible and provide detailed physical guide to gene mapping  about 5000 bands, size ranging from 3-150 kb o distance is measured in RF (map units) o 1 o o most of the flies are parental combinations of all three genes + + o among the recombinants the smallest number is for vg b pr and vg b pr o o o but the distance separating the outside of vg and b genes does not equal 18.7 which is the sum of the 2 intervening distances  discrepancy due to the rare occurrence of double crossovers – interference: the phenomenon of crossovers not occurring independently o to correct for this we have to add the double cross overs 2x because each individual in the double cross over groups is the result of two exchanges between vg and b o o coefficient of coincidence  ratio between the actual number of observed double crossovers in an experiment and the number of double crossovers expected on the basis of independent possibilities o o NOTE: to figure out which gene is the middle one in a trihybrid coss look at the double cross over, in this case they are vg b pr+ and vg+ b+ pr. In the parents, you always have vg an b together and vg+ b+ together which is mainting in these double crossovers. Which means that the middle gene must then be pr. Working backwards o purple eyes and verstigial wings are recessive to red eyes and long wings in Drosophila o parents are purple eyed and vestigial winged x normal red eyes and normal long winged flies o the resulting F1 is all normal eyes and wings o the F1 generation is test crossed (pr vb/ pr vb) 2 o o parentals = 965 + 1067 = 2032 o recombinants = 157 + 146 = 303 o RF = 303/ 2335 x 100% o RF = 13% o Linked in trans conformation  pr vg+ / pr+ vg o So the parents must have been + + + + o pr /pr · vg / vX pr / pr · vg/ vg + + o and F1 i
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