Bft1033 6 linkage genes_print

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Bft1033 6 linkage genes_print

  1. 1. (GENE MAPPING)
  2. 2.  Most chromosomes consist of very large numbers of genes  Genes that are part of the same chromosome are said linked  These genes demonstrate linkage in genetic crosses  During meiosis, they are not free to undergo independent assortment  They are transmitted as a unit  Crossover results in reshuffling or recombination of alleles between homolog. INTRODUCTION
  3. 3.  No crossover: two genetically different gametes are formed  Each gamete receive the alleles present on one   homolog or the other Illustrate complete linkage Produce parental or noncrossover gametes  Crossover: produce four types of gametes  Two parental gametes  Two recombinant or crossover gametes Introduction
  4. 4. Introduction
  5. 5.  Complete linkage in Drosophila melanogaster:  Mutants: brown (bw) eye and heavy (hv) wing vein  Normal alleles: bw+ (red eye) and hv+ (thin wing vein)  Cross between brown eye and thin vein with red eyes and heavy vein  P:  F1: bwhv bw hv   bwhv bw hv Red-heavy Brown-thin bwhv bw hv red, thin The Linkage Ratio
  6. 6. The Linkage Ratio
  7. 7.  When the F1 generation is interbred, the F2 generation will be produced in a 1:2:1 phenotypic and genotypic ratio.  When the F1 is tescrossed, it will produce a 1:1 ratio of brown thin and red heavy The Linkage Ratio
  8. 8. The Linkage Ratio
  9. 9. Morgan crossed yellow bodied white eyed female and wild type male P: yellow, white ♀ X wild-type ♂ F1: ♀: wild type ♂: expressed both mutant traits F2:  98.7% parental types (gray bodied, red eyed)  1.3% either: yellow bodied with red eyed , or gray bodied with white eyed Crossover and Gene Distance
  10. 10. Morgan made crosses involving other X-linkage genes P: White-eye, miniature wing ♀ F1: F2: X wild type ♂ even more puzzling phenotypes differed 62.8%: parental types 37.2%: either: white eyed or miniature wing Crossover and Gene Distance
  11. 11. Morgan crossed yellow bodied white eyed female and wild type male: yw/yw X y+w+Y Crossover and Gene Distance Morgan made crosses involving other Xlinkage genes: Whiteeye, miniature wing ♀ X wild type ♂
  12. 12.  Morgan postulated that exchange occurred between the mutant genes on the two X chromosomes of the F1 females  Lead to 1.3 and 37.2 recombinant gametes  The closer two gene are, the less likely genetic exchange will occur between them  Morgan proposed the term crossing over to describe the physical exchange leading to recombination. Crossover and Gene Distance
  13. 13. Crossover and Gene Distance
  14. 14. Crossover and Gene Distance
  15. 15.  Two arrangements of alleles exist for an individuals heterozygous at two loci: ‘cis’ or coupling ‘trans’ or repulsion w + m+ w+ m w w m m+  Cross-over of cis results in trans and vice versa  Frequency of recombinants (%) is a characteristic of each gene pair, regardless of cis or trans arrangements Concept of a genetic map
  16. 16.  In cross A  Parental types (yellow-white, wild type): 98.7%  Recombinant types (white, yellow): 1.3%  Distance between genes: 1.3 mu  In cross B  Parental types (white-miniature, wild type): 62.8%  Recombinant types (white, miniature): 37.2%  Distance between genes: 37.2 mu Concept of a genetic map
  17. 17.  Cross-over is more likely to occur between distant genes than close genes
  18. 18.  Sturtevant (1913) recognized that recombination frequencies could be used to create a map  1% cross-over rate = 1 map unit (mu) or centiMorgan (cM)  Map units (mu) and centiMorgans (cM) are relative measures. # of recombinant progeny Recombination frequency  X 100% total # of progeny Calculating Recombination frequency
  19. 19.  The test cross Ab/aB x ab/ab is performed. The following numbers of progeny of each genotype are obtained: 87 AaBb, 409 Aabb, 390 aaBb, 114 aabb.  What is the approximate distance (in map units) between the two genes in question? Recombination frequency  # of recombinant progeny X 100% total # of progeny  RF = (87 + 114)/(87 + 409 + 390 + 114) x 100%  = 201/1000 x 100% = 20.1% So the distance between the two genes is 20.1 cM Example
  20. 20.  First genetic map was for Drosophila: 3 sex-linked genes w = white-eyes m =miniature wings y = yellow body  Recombination frequencies: wxy wxm mxy = 0.5% = 34.5% = 35.4% 0.5 34.5 35.5
  21. 21.  Single crossover Single Crossover
  22. 22.  Double crossover Double Crossover
  23. 23.  The genotype of the organism producing the crossover gametes must be heterozygous at all loci  The cross must be constructed so that genotypes of all gametes can be accurately determined by observing the phenotypes of the resulting offspring  A sufficient of number of offspring must be produced in the mapping experiment to recover a representative sample of all crossover. Three-Point Mapping
  24. 24.  Males hemizygous for all three wild type alleles are crossed to female with three mutant traits (yellow body, white eyes, and echinus eye shape)  F1 consists of females heterozygous at all loci and males hemizygous for all three mutant alleles Three-Point Mapping
  25. 25.  When the F1 is intercrossed to produce F2, it produces 8 different classes:  Two classes of parental types (the biggest    proportion) Two classes from single crossover in region I Two classes from single crossover in region II Two classes from double crossover (the smallest proportion).
  26. 26. Phenotypes white, echinus yellow yellow, white echinus yellow, echinus white Determining Gene Sequence
  27. 27.  In maize, the recessive mutant genes:  bm (brown midrib), v (virescent seedling), and pr (purple aleurone) are linked on chromosome 5  A female plant is heterozygous for all three traits is crossed with a male homozygous for all three mutant alleles  F1 data:     [+ v bm] 230 [pr + +] 237 [+ + bm] 82 [pr v +] 79     [+ v +] 200 [pr + bm] 195 [pr v bm] 44 [+ + +] 42  What is the correct sequence of genes?  What is the distance between each pairs of gene? A Mapping Problem in Maize
  28. 28.  The Five Steps to Solve the Problem 1. Determine the parental and dco types 2. Examine the gene in the middle 3. Re-order the genes (if necessary) 4. Examine sco in region I and II 5. Calculate the distance A Mapping Problem in Maize
  29. 29.  The Five Steps to Solve the Problem 1. Determine the parental and dco types 2. Examine the gene in the middle 3. Re-order the genes (if necessary) 4. Examine sco in region I and II 5. Calculate the distance A Mapping Problem in Maize
  30. 30.  The parental types are the biggest number, and dco types are the smallest         [+ v bm] 230 [pr + +] 237 [+ + bm] 82 [pr v +] 79 [+ v +] 200 [pr + bm] 195 [pr v bm] 44 [+ + +] 42 Determine the parental and dco types
  31. 31.  The parental types are the biggest number, and dco types are the smallest         [+ v bm] 230  parental type [pr + +] 237  parental type [+ + bm] 82 [pr v +] 79 [+ v +] 200 [pr + bm] 195 [pr v bm] 44  dco type [+ + +] 42  dco type Determine the parental and dco types
  32. 32.  The Five Steps to Solve the Problem 1. Determine the parental and dco types 2. Examine the gene in the middle 3. Re-order the genes (if necessary) 4. Examine sco in region I and II 5. Calculate the distance A Mapping Problem in Maize
  33. 33. + v bm + + bm pr + + pr v + + bm v + + v pr + + v + bm pr v bm pr + bm + pr + + + +     [+ v bm] [pr + +] [pr v bm] [+ + +] 230  parental type 237  parental type 44  dco type 42  dco type Examine the gene in the middle
  34. 34.  The Five Steps to Solve the Problem 1. Determine the parental and dco types 2. Examine the gene in the middle 3. Re-order the genes (if necessary) 4. Examine sco in region I and II 5. Calculate the distance A Mapping Problem in Maize
  35. 35. Temporary Order         [+ v bm] [pr + +] [+ + bm] [pr v +] [+ v +] [pr + bm] [pr v bm] [+ + +] Correct Order 230 237 82 79 200 195 44 42         v + + v v + v + + bm pr + + bm pr + ++ pr bm pr bm ++ Re-order the genes (if necessary) 230 237 82 79 200 195 44 42
  36. 36.  The Five Steps to Solve the Problem 1. Determine the parental and dco types 2. Examine the gene in the middle 3. Re-order the genes (if necessary) 4. Examine sco in region I and II 5. Calculate the distance A Mapping Problem in Maize
  37. 37. v + pr + + pr + pr + bm + bm v v + I v + bm + pr + II + v v + pr + + bm + bm + + + pr bm v + bm 230  parental type + pr + 237  parental type v pr + 79  scoI type + + bm 82  scoI type v ++ 200  scoII type + pr bm 195  scoII type Examine sco in region I and II
  38. 38.  The Five Steps to Solve the Problem 1. Determine the parental and dco types 2. Examine the gene in the middle 3. Re-order the genes (if necessary) 4. Examine sco in region I and II 5. Calculate the distance A Mapping Problem in Maize
  39. 39.  The formula to calculate the distance between two genes:  In region I = scoI dco x 100 Total  In regio II = scoII dco x 100 Total Calculate the distance
  40. 40.  v + bm 230  parental type  + pr + 237  parental type  + + bm 82  scoI type  v pr + 79  scoI type  v ++ 200  scoII type  + pr bm 195  scoII type  v pr bm 44  dco type  + ++ 42  dco type   Total Calculate the1109 distance
  41. 41.  Distance between v-pr = 82  79  44  42 x 100  22.27 cM 1109  v + bm 230  parental type  + pr + 237  parental type  + + bm 82  scoI type  v pr + 79  scoI type  v ++ 200  scoII type  + pr bm 195  scoII type  Distance between pr-bm =  v pr bm 44  dco type  + ++ 42  dco type 200  195  44  42 x 100  43.37 cM 1109  Distance between v-bm = 22.27  43.37  65.64 cM  And the map is 22.27 cM v+/v Calculate the distance 43.37 cM pr+/pr bm+/bm
  42. 42.  Interference: a crossover at one spot on a chromosome decreases the likelihood of a crossover in a nearby spot I=1–c  where c: coefficient of coincidence obs dco c exp dco  obs dco: observed data  exp dco: expected dco = sco I x sco II Interference and Coincidence
  43. 43.  From the data: 22.27 cM v+/v 43.37 cM pr+/pr bm+/bm  obs dco = 86/1109 = 0.0775  exp dco = 0.2227 x 0.4337 = 0.0966  v + bm  c = 0.0775/0.0966 = 0.80 230  parental type  + pr + 237  parental type  + + bm 82  scoI type  v pr + 79  scoI type  v ++ 200  scoII type  + pr bm 195  scoII type  I = 1 – 0.80 = 0.20  v pr bm 44  dco type  + ++ 42  dco type Interference and Coincidence
  44. 44. Any questions? Thank you

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