linkage mapping a. physical basis of linkage mapping b. mapping by the 2-factor testcross method c....
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Linkage MappingLinkage Mapping
A. Physical basis of linkage mappingB. Mapping by the 2-factor testcross methodC. Mapping by the 3-factor testcross method
A. Physical BasisA. Physical Basis
If two genes are located on the same chromosome, their alleles can recombine only when there is crossing over during meiosis
The probability that crossover will occur is proportional to the distance between the genes
Typically, there are fewer recombinant (crossover) gametes than nonrecombinant gametes
A. Physical BasisA. Physical Basis
A. Physical BasisA. Physical Basis
One “map unit” (or “morgan”) of distance is the distance that produces a recombination frequency of 1%; Therefore:
Map distance (in map units) = recombination frequency X 100
= (# Recombinant gametes) X 100(# Recombinant gametes) + (# nonrecombinant gametes)
Genes ebetween th Distance Frequency ion Recombinat
B. 2-factor TestcrossB. 2-factor Testcross
A testcross lets us “count” the number of recombinant and nonrecombinant gametes
The phenotype of the testcross progeny is determined by the gametes from the heterozygous parentEach phenotype in a testcross has a unique genotype (unlike in the F2 of dihybrid cross)
So, to map the distance between two genes:cross an individual that is heterozygous for each gene with an individual that is homozygous recessive for each gene
B. 2-factor TestcrossB. 2-factor Testcross
Example: tomato plants, fruit shape & texture genes:A heterozygous round, heterozygous smooth plant (Rr Pp) was crossed with a long, peachy (rr pp) plant. The results are given in the table below
Smooth round 39Smooth long 463Peachy round 451Peachy long 47
B. 2-factor TestcrossB. 2-factor Testcross
2-F STEP 1: Arrange the phenotypic classes into pairs, with each different phenotype represented in each pair
Smooth roundPeachy longSmooth longPeachy round
B. 2-factor TestcrossB. 2-factor Testcross
2-F STEP 2: Look at the numbers to determine which class is recombinant (lesser numbers) and which is nonrecombinant (greater numbers)
Smooth round 39Peachy long 47Smooth long 463Peachy round 451
Recombinant Nonrecombinant
B. 2-factor TestcrossB. 2-factor Testcross
2-F STEP 3: Calculate the map distance:R – P gene distance = 86/1000 X 100 = 8.6 m.u.
Smooth round 39Peachy long 47Smooth long 463Peachy round 451
Recombinant Nonrecombinant
B. 2-factor TestcrossB. 2-factor Testcross
2-F STEP 4: Determine the linkage (cis or trans) of the alleles in the nonrecombinant heterozygote parent.In this particular cross, the linkage is trans
Smooth round 39Peachy long 47Smooth long 463Peachy round 451
Recombinant Nonrecombinant
B. 2-factor TestcrossB. 2-factor Testcross
Cis linkage: When two dominant alleles are linked together in the original heterozygote:
B. 2-factor TestcrossB. 2-factor Testcross
Trans linkage: When a dominant allele is linked to a recessive allele in the original heterozygote:
B. 2-factor TestcrossB. 2-factor Testcross
The double crossover problem:Double crossovers occur whenever two crossover events occur between two genesIf this occurs, then the recombinant progeny will not be counted, because each allele “goes back” to its original linkageFor this reason, the map distance given by a 2-factor testcross often is too low
C. 3-factor TestcrossC. 3-factor Testcross
By performing a testcross with 3 genes, we can estimate how many double crossovers are occurring
Example: MaizeGreen (Y) vs. yellow (y) plant colorFull (S) vs. shrunken (s) seed shapeColored (C) vs. colorless (c) seed colorCross Yy Ss Cc X yy ss cc
C. 3-factor TestcrossC. 3-factor Testcross 3-F CROSS STEP 1:
Arrange the phenotypic classes into pairs, with each different phenotype represented
Green Full ColoredYellow Shrunk ColorlessGreen Full ColorlessYellow Shrunk ColoredGreen Shrunk ColoredYellow Full ColorlessGreen Shrunk ColorlessYellow Full Colored
C. 3-factor TestcrossC. 3-factor Testcross 3-F CROSS STEP 2:
Identify the nonrecombinant (largest) and double crossover (smallest) classes
Green Full Colored 100Yellow Shrunk Colorless 95Green Full Colorless 25Yellow Shrunk Colored 20Green Shrunk Colored 380Yellow Full Colorless 375Green Shrunk Colorless 2Yellow Full Colored 3
NRDC
C. 3-factor TestcrossC. 3-factor Testcross 3-F CROSS STEP 3:
Compare the NR & DC classes to determine which gene is in the middle (It’s Y-C-S)
Green Full Colored 100Yellow Shrunk Colorless 95Green Full Colorless 25Yellow Shrunk Colored 20Green Shrunk Colored 380Yellow Full Colorless 375Green Shrunk Colorless 2Yellow Full Colored 3
NRDC
C. 3-factor TestcrossC. 3-factor Testcross 3-F CROSS STEP 4:
Determine the identity of the two single crossover classes (compare with NR class)
Green Full Colored 100Yellow Shrunk Colorless 95Green Full Colorless 25Yellow Shrunk Colored 20Green Shrunk Colored 380Yellow Full Colorless 375Green Shrunk Colorless 2Yellow Full Colored 3
S-C Sing. Y-C Sing.
NRDC
C. 3-factor TestcrossC. 3-factor Testcross 3-F CROSS STEP 5:
Calculate the distances between each pair of genes:Y-C distance = (25+20+2+3)/1000 X 100 = 5 m.u.
S-C distance = (95+100+2+3)/1000 X 100 = 20 m.u.
C. 3-factor TestcrossC. 3-factor Testcross 3-F CROSS STEP 6:
Calculate:The expected double crossover frequencyExpected d.c. freq. = (0.05)(0.2) = 0.01The obtained double crossover frequencyObtained d.c. freq. = (2+3)/1000 = 0.005The coefficient of coincidenceCoincidence = (Obtained d.c.)/(Expected d.c.)
= 0.005 / 0.01 = 0.5Interference = 1 – Coincidence = 1 – 0.5 = 0.5
C. 3-factor TestcrossC. 3-factor Testcross Interference
The occurrence of one crossover event may interfere with a second crossover eventIf the obtained d.c. = expected d.c. then:Coincidence = 1Interference = 0If the obtained d.c. < expected d.c. then:Coincidence < 1Interference is a positive numberIf the obtained d.c. > expected d.c. then:Coincidence > 1Interference is a negative number