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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 + seh. 6.8 m.u. 0.5 m.u. jv. se. h. jv +. se +. h +. Single Crossovers Region II (SCO II) jv se h + - PowerPoint PPT Presentation
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Longer regions have more crossovers and thus higher recombinant frequencies.
jvjv+sese+hh+6.8 m.u.Region: I0.5 m.u. IIProb(XO in Region I) = 0.068 Prob(XO in Region II) = 0.005 Prob(XOs in Region I and II) = (0.068) x (0.005) = 0.00034Three Point Mapping: Forward Analysis Single Crossovers Region I (SCO I)
jv se+ h+jv+ sehSingle Crossovers Region II (SCO II)
jv se h+jv+ se+hDouble Crossovers (DCO)
jv se+ hjv+ seh+No Crossovers, Parental (NCO)
jv se hjv+ se+ h+
jvjv+sese+hh+6.8 m.u.Region: I0.5 m.u. IIProb(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) =0.92734Three Point Mapping: Forward Analysis Single Crossovers Region I (SCO I)
jv se+ h+jv+ sehSingle Crossovers Region II (SCO II)
jv se h+jv+ se+hDouble Crossovers (DCO)
jv se+ hjv+ seh+No Crossovers, Parentals (NCO)
jv se hjv+ se+ h+
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
Three Point Mapping: Backward Analysis wing morphology
cucurled: wings curled upwardcu+ (+) straight wings
wing angle
txtaxi: wings 75 from axistx+ (+) wings not held out
bristles
grogroucho: clumps above eyesgro+ (+) 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 + + +cu tx grocu tx grocu tx gro + + + + + +cu tx gro(P)(F1)Linkage TestcrossGenerate progeny.Suppose n = 1000
Three Point Mapping: Backward AnalysisProgeny Phenotypes/Gametes from Heterozygote Phenotypecu tx groCounts269ClassNCO+ + +324NCOcu tx +1?+ + gro2?Phenotypecu + +Counts201Class?+ tx gro196?cu + gro3?+ tx +4?GENE ORDER: The locus in the middle is the one that differs from the parentals. A B Ca b cA b Ca B cA b Ca B cDCO class is expected to have the fewest number of progeny.59333977
Three Point Mapping: Backward Analysis
Phenotypecu gro txCounts269ClassNCO+ + +324593cu + tx1DCO+ gro +23Phenotypecu + +Counts201ClassSCO I+ gro tx196397cu gro +3SCO II+ + tx47Region
cu - grogro - txcu - txRecombination 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