Sum7LS4Week2b(1slide)

# Sum7LS4Week2b(1slide) - Announcements Midterm I-Time Franz...

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• Midterm I (07/08/08) -Time: Franz 1178; 8- 9:50 am (Lecture following) -Lectures 1-4 (100 points) • Use Black or Blue Pen - no regrade if written in pencil • Bring your own calculator, - No Graphing Calculators - No cell phones or translators • Bring your Bruin card or some form of photo ID • Place backbacks and bags in the front of the class. Announcements

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Linkage and Mapping Chapter 7 read pp. 130 - 142 Problems 3,5,7,11,13,15,17,21,23,25,29
• Sturtevant, one of Morgan’s undergrad students (21 years old), invented a way to quantify the relationship between the rate of recombination and the distance between genes. # recombinants/# total = Recombination frequency (Rf) between two genes Rf X 100 = % recombination. Recombination and Genetic Distance 1% recombination = 1 Map unit (mu) = 1 cM (centiMorgan)

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Gene A Gene B Gene C • The frequency of recombination is proportional to the genetic distance between two genes. Random cuts between A and B Random cuts between A and B < < B and C . B and C . Recombination and Genetic Distance 1% recombination = 1 Map unit (mu) = 1 cM (centiMorgan)
Linkage Mapping • Recombination is the basis of genetic mapping • Each gene has a place, or locus (latin for place), on its chromosome. One can use two point crosses (for two genes) and three point crosses (for three genes) to… 1) calculate the distances between genes 2) order genes relative to one another on the chromosome.

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X F1 Two Point Cross: Our Previous Example pr + vg + / pr vg pr vg/pr vg F2 purple eye, normal wing red eye, vestigial wing 1339 1195 151 154 2839 recombinant parental pr vg/pr vg purple eye, vestigial wing red eye, normal wing Total pr+ vg+/ pr vg pr vg+/ pr vg pr+ vg/ pr vg
• In Morgan’s cross, the testcross produced the following F2 progeny: pr+ vg+ 1339 Parental pr vg 1195 Parental pr+ vg 154 Recombinant pr vg+ 151 Recombinant 2839 Rf = # of recombinant progeny Total # of progeny = 2839 • Therefore, the purple and vestigial genes are map units (mu) or centiMorgans (cM) apart. Two Point Cross: Our Previous Example ???

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If two pea genes AB are 28 mu apart, what percent of the progeny from a test cross of the heterozygote AB/ab to ab/ab will have the phenotype Ab? A B a b A b a B Parental Recombinant 28% 72% A B 36% a b 36% A b 14% a B 14% Parental Recombinant • Map distances can be used to predict the types of progeny that result from the cross of individuals carrying different alleles Using Map Distances • A single crossover generates two reciprocal recombinant products, which explains why the reciprocal recombinant classes are generally equal in frequency.
Meiotic recombination: Recombinants are those products of meiosis with allele combinations different from those of haploid cells that formed the meiotic diploid. Recombinant Alleles A B a b n n AaBb Input Dihybrid Meiosis 2n Output A B a b A b a B n n n n Parental (input) type Parental (input) type Recombinant Recombinant (gametes) (gametes) -We detect recombination by comparing the output genotypes of meiosis with the input genotypes.

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