1 / 7

Longer regions have more crossovers and thus higher recombinant frequencies.

Longer regions have more crossovers and thus higher recombinant frequencies. Three Point Mapping: Forward Analysis. Single Crossovers Region I (SCO I) jv se + h + jv + se h. 6.8 m.u. 0.5 m.u. jv. se. h. jv +. se +. h +. Single Crossovers Region II (SCO II) jv se h +

gavril
Télécharger la présentation

Longer regions have more crossovers and thus higher recombinant frequencies.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Longer regions have more crossovers and thus higher recombinant frequencies.

  2. Three Point Mapping: Forward Analysis Single Crossovers Region I (SCO I) jv se+ h+ jv+ se h 6.8 m.u. 0.5 m.u. jv se h jv+ se+ h+ Single Crossovers Region II (SCO II) jv se h+ jv+ se+ h Region: I II Double Crossovers (DCO) jv se+ h jv+ se h+ Prob(XO in Region I) = 0.068 Prob(XO in Region II) = 0.005 Prob(XOs in Region I and II) = No Crossovers, Parental (NCO) jv se h jv+ se+ h+ (0.068) x (0.005) = 0.00034

  3. Three Point Mapping: Forward Analysis Prob(DCO) = (0.068) x (0.005) = 0.00034 Prob(SCO I) = 0.068 - 0.00034 = 0.06766 Prob(SCO II) = 0.005 - 0.00034 = 0.00466 Prob(NCO) = 1 - Prob(SCO I) - Prob(SCO II) - Prob(DCO) = 0.92734 6.8 m.u. 0.5 m.u. jv se h jv+ se+ h+ Region: I II Double Crossovers (DCO) jv se+ h jv+ se h+ Single Crossovers Region I (SCO I) jv se+ h+ jv+ se h Single Crossovers Region II (SCO II) jv se h+ jv+ se+ h No Crossovers, Parentals (NCO) jv se h jv+ se+ h+

  4. Three Point Mapping Experiment: Backward Analysis In a three-point cross, the following are essential: Knowledge of whether genes/loci are autosomal or X-linked, dominance relationship. Heterozygous individual (heterozygous at all loci) to produce gametes. Accurate determination of genotypes of gametes that are produced by the heterozygous individual. Sufficient number of progeny from the linkage testcross so that all crossover classes are represented. Determine: gene order, map distances between genes, coefficient of coincidence and interference

  5. Three Point Mapping: Backward Analysis wing morphology cu curled: wings curled upward cu+ (+) straight wings wing angle tx taxi: wings 75° from axis tx+ (+) wings not held out bristles gro groucho: clumps above eyes gro+ (+) no extra bristles Three autosomal loci are linked, but suppose gene order is unknown. Conduct linkage crosses as shown on the right. cu tx gro + + + (P) cu tx gro + + + Linkage Testcross cu tx gro cu tx gro (F1) + + + cu tx gro Generate progeny. Suppose n = 1000

  6. Three Point Mapping: Backward Analysis Progeny Phenotypes/Gametes from Heterozygote Phenotype Counts Class Phenotype Counts Class cu tx gro 269 NCO cu + + 201 ? 593 397 + + + 324 NCO + tx gro 196 ? cu tx + 1 ? cu + gro 3 ? 3 7 + + gro 2 ? + tx + 4 ? GENE ORDER: The locus in the middle is the one that differs from the parentals. A B C A b C A b C DCO class is expected to have the fewest number of progeny. a b c a B c a B c

  7. Three Point Mapping: Backward Analysis Phenotype Counts Class Phenotype Counts Class cu gro tx 269 NCO cu + + 201 SCO I + + + 324 593 + gro tx 196 397 cu + tx 1 DCO cu gro + 3 SCO II + gro + 2 3 + + tx 4 7 Region cu - gro gro - tx cu - tx Recombination Frequency (397 + 3)/1000 = 0.40 or 40% (7 + 3)/1000 = 0.01 or 1% Map Distance 40 m.u. 1 m.u. 41 m.u. Coefficient of Coincidence (C) # observed DCO = 3 = 0.75 # expected DCO (0.4)(0.01)(1000) I > 0 means fewer obs DCO than expected. I < 0 means more obs DCO than expected. Interference (I) = 1 - C I= 1 - 0.75 = 0.25

More Related