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BIOL 2000 (29)
Lecture

# Chapter 4a-3.docx

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School
University of Lethbridge
Department
Biology
Course
BIOL 2000
Professor
Joseph Rasmussen
Semester
Winter

Description
Chapter 4a-3 Recombinants  Aa X Bb x aabb Cis-dihybrid o AB 150 Parental o Ab 60 Recombinants o aB 50 Recombinants o ab 155 Parental  frequency of recombinant products = map distance  60+50 / 415 = 26.5%  The genes curly win and ebony are are 20 map units apart on chromosome 3. A wild type fly is crossed by a fly homozygous recessive for mutation at both genes, If the F1 is test crossed, what frewuecny of wild type winged and ebony-eyed flies will be seen in the progeny? o Map distance = frequency of recombinants  20% recombinants  80% parental o + + / + + X e cy / e cy [ plus is wildtype and recessive is small letters]  F1 ++/ e cy x ecy/ ecy  Cis-di-hybrid  ++ Parental 40%  + cy Recombinant 10%  e + recombinant 10%  e cy Parental 20%  TTG and ttg are the same distance as they code for the same gene, so they are in the same spot on the chromosome Clicker:  In tomoato R and W are 15 cm apart. Gene R controls fruit (R=red r=orange) and W controls flower color (W=yellow w=white). If it is cis-dihybrid for R and W is test crossed. In test cross progeny, what is the frequency of orange fruit and white plant? o 42.5%  15 cM = 15% recombinants  RW / rw  RW 42.5% P  Rw 7.5% R  rW 7.5% R  rw 42.5% P  If there was no test cross, but an F2 generation. What is the probability of the rrww? o Self cross RW/rw  Two rw must fuse.  .425 x .425  = .18062 Objective 3: Use recombination frequency to map three segregating genes Three point test cross  Three point test cross-mapping 3 genes relative to one another o Consider 3 Arabidopsis genes  LFY- flowers / lfy- leaves  CER3+ epidermal wax / cer3 no wax  YI green inflorescence / yi yellow inflorescence o Following test crossing to tri-hybrid individual [heterozygous for all three genes], the following phenotypes were scored in the progeny [number of classes is determined by 2 ] o If they were not linked you would get equal segregation. But we do not see that so we assume they are linked  S LFY and CER3 CER3 YI
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