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Download Lab. Handouts

Download Lab. Handouts. Download Lab. #3 and Print out 3 files: http://www.mun.ca/biology/dinnes/B2250/B2250.html Quiz #2 Marks posted on Webpage. Mendelian Genetics. . . . Topics: -Transmission of DNA during cell division Mitosis and Meiosis - Segregation

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Download Lab. Handouts

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  1. Download Lab. Handouts Download Lab. #3 and Print out 3 files: http://www.mun.ca/biology/dinnes/B2250/B2250.html Quiz #2 Marks posted on Webpage

  2. Mendelian Genetics    Topics: -Transmission of DNA during cell division Mitosis and Meiosis - Segregation - Sex linkage (problem: how to get a white-eyed female) - Inheritance and probability - Mendelian genetics in humans - Independent Assortment - Linkage - Gene mapping - 3 point test cross  - Tetrad Analysis (mapping in fungi) - Extensions to Mendelian Genetics - Gene mutation - Chromosome mutation - Quantitative and population genetics  

  3. LinkageChapter 6 - recombination - linkage maps Ch. 6 p. 148 – 165 Prob: 1-5, 7, 8, 10, 11, 14

  4. Linkage of Genes - Many more genes than chromosomes - Some genes must be linked on the same chromosome; therefore not independent

  5. Independent Assortment Linkage Fig 6-6 Fig 6-11 Interchromosomal Intrachromosomal

  6. Two ways to produce dihybrid P X X A B a b A b a B A B a b A b a B cis A B AaBb A b trans a b (dihybrid ) a B Gametes: AB P Ab ab P aB Ab R AB aB R ab

  7. Example Test Cross AaBb X aabb ab Exp. Obs. AB AaBb 25 10 R Ab Aabb 25 40 P aB aaBb 25 40 P ab aabb 25 10 R 100 100 How to distinguish: Parental high freq. Recombinant low freq.

  8. Example (cont.) Gametes: AB R Ab P aB P ab R Therefore dihybrid: A b (trans) a B

  9. Linkage Maps Genes close together on same chromosome: - smaller chance of crossovers between them - fewer recombinants Therefore: percentage recombination can be used to generate a linkage map

  10. Linkage maps A B large # of recomb. a b C D small number of recombinants c d

  11. Linkage mapsexample 65 Testcross progeny: P AaBb 2146 R Aabb 43 R aaBb 22 P aabb 2302 Total 4513 1.4 map units = 1.4 % RF 4513 A 1.4 mu B

  12. Additivity of map distances separate maps A B A C 7 2 combine maps C A B 2 7 or Locus A C B (pl. loci) 2 5

  13. Linkage Deviations from independent assortment Dihybrid gametes 2 parent (noncrossover) common 2 recombinant (crossover) rare % recombinants a function of distance between genes % RF = map distance

  14. Gametes Number of Genes Number of Different Gametes monohybrid 1 (Aa) 2 dihybrid 2 (AaBb) 4 trihybrid 3 (AaBbCc) 8

  15. Trihybrid Three Point Test Cross AaBbCc X aabbcc ABC ABc abc AbC Abc aBC aBc abC abc 8 gamete types

  16. Three Point Test Cross Trihybrid Gametes C ABC B c ABc A C AbC b c Abc a

  17. Three Point Test Cross Trihybrid AaBbCc 3 genes: Possibilities: 1. All unlinked 2. Two linked; one unlinked 3. Three linked - order ? A---B---C B---C---A B---A---C

  18. Three Point Test Cross Three recessive mutants of Drosophila: +, v vermilion eyes +, cv crossveinless +, ct cut wing P +/+ cv/cv ct/ct X v/v +/+ +/+

  19. Three Point Test Cross P +/+ cv/cv ct/ct x v/v +/+ +/+ Gametes + cv ct v + + F1 trihybrid v/+ cv/+ ct/+

  20. Three Point Test Cross F1 v/+ cv/+ ct/+ x v/v cv/cv ct/ct v cv ct 8 gamete types one gamete type

  21. 8 gamete types Parental (most frequent) F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Recombinant

  22. 8 gamete types Parental F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Recombinant 268 Recombinant Parental 268 1448 = 18.5 %

  23. 8 gamete types Parental F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Parental Recombinant 191 Recombinant 191 1448 = 13.2 %

  24. 8 gamete types Parental F1 v/+ cv/+ ct/+ v + + 580 + cv ct 592 v cv + 45 + + ct 40 v cv ct 89 + + + 94 v + ct 3 + cv + 5 1448 Recombinant 93 Parental Recombinant 93 1448 = 6.4 %

  25. Calculate Recombination Fraction 1. v - cv R v cv 45 + 89 R + + 40 + 94 268 / 1448 = 18.5 % 2. v - ct R + + 94 + 5 R v ct 89 + 3 191/1448 = 13.2 % 3. ct - cv R ct + 40 + 3 R + cv 45 + 5 93/1448 = 6.4 %

  26. Three point test cross Observations: all 3 RF < 50 % 3 genes on same chromosome v-----cv largest distance ct in middle map v-------ct-------cv = cv-------ct-------v 13.2 + 6.4 = 19.6 > 18.5 !! Why ?

  27. Three Point Test Cross P +/+ ct/ct cv/cv x v/v +/+ +/+ gametes + ct cv v + + F1 trihybrid v + + + ct cv

  28. Three Point Test Cross X X X X Double crossover class rarest: v---cv P v + + v + P + ct cv + cv R v ct + v + R + + cv + cv

  29. Three Point test cross 1. Double crossovers not counted in v--cv RF 2. Double crossovers generate P types (with respect to v--cv) 3. Double crossovers not detected as recombinants Consequence: underestimate of v----cv map distance Greater distance of genes  greater error

  30. Double recombinant class: (3 + 5) x 2 = 16 268 + 16 = 284 284/1448 = 19.6 NOTE: double crossovers detected because of middle gene (ct)

  31. Mapping Function Genes close together on chromosome -RF good estimate of map distance Genes far apart on chromosome - RF underestimates true map distance due to undetected multiple crossovers

  32. Mapping Function m = avg. # crossovers per meiosis (linear with true map distance) if m = 1 (1 cross over for every meiosis) then 50 % recombinants produced Therefore: map units (mu) = m x 50

  33. Mapping Function Mapping function: - relates RF to true map distance (better estimate for genes separated by large distances) m = -ln (1 - 2RF) mu = m x 50 Mapping function

  34. Mapping Function m = -ln(1 - 2RF)

  35. Mapping Functionexample 1. RF = 18.5 % m = 0.46 mu = 23.1 2. RF = 6.4 % m = 0.137 mu = 6.8 Summary: - short distances: use RF - long distances: use mapping function

  36. Linkage Other Points: 1. No crossing over in male Drosophila male: AaBb A B  gametes AB, ab a b use female dihybrid: AaBb x aabb O O

  37. Linkage 2. Linkage of genes on the X chromosome: AaBb x --Y O O Male progeny: AB Y Ab Y male progeny direct aB Y measure of female meiotic ab Y products

  38. Fungal Genetics Fungi: important organisms in the ecosystem - decomposers - pathogens important for humans - food - pathogens (Biology 4040 – Mycology)

  39. Fun Facts About Fungi http://www.herbarium.usu.edu/fungi/funfacts/factindx.htm

  40. Fungi

  41. Neurospora crassa(bread mold) Morphological mutants Biochemical mutants (one gene, one enzyme)

  42. Linkage Map Neurospora crassa Linkage group I

  43. Fungus Life Cycle vegetative stage haploid +, - mating types brief diploid stage  meiosis n n + spores + meiosis - 2n n - n

  44. Gamete Pool Gametes: Products of many meioses all pooled together A B a b AB AB ab ab AB ab P A B ab ab AB ab Ab AB Gamete P a b AB aB ab ab AB AB pool R a B ab AB AB ab R A b

  45. Tetrad Analysis Some Fungi and algae: 4 products of a single meiosis can be recovered Advantages: 1. haploid organism - no dominance 2. examine a single meiosis - test cross not needed 3. small, easy to culture 4. Tetrad Analysis - map gene to centromere

  46. Ascus with ascospores

  47. Tetrad Analysis Types of Tetrads: 1. Unordered - 4 products mixed together 2. Ordered (linear) - 4 products lined up, each haploid nucleus can be traced back through meiosis 3. Octads - mitotic division after meiosis 8 products (2 x 4)

  48. Linear Tetrad Analysis + = a+ a a a a + a + Life Cycle: + Meiosis + Diploid Haploid + Mating: a x +  a /+ n n 2n 4 haploid products

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