Linkage Mapping

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Linkage Mapping

1. Physical basis of linkage mapping
2. Mapping by the 2-factor testcross method
3. Mapping by the 3-factor testcross method

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A. 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

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A. Physical Basis
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A. Physical Basis
µ
Genes

e
between th

Distance

Frequency
ion
Recombinat

• 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)

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B. 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
  parent
• Each phenotype in a testcross has a unique
  genotype (unlike in the F2 of dihybrid cross)
• So, to map the distance between two genescross
  an individual that is heterozygous for each gene
  with an individual that is homozygous recessive
  for each gene

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B. 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 39
• Smooth long 463
• Peachy round 451
• Peachy long 47

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B. 2-factor Testcross

• 2-F STEP 1 Arrange the phenotypic classes into
  pairs, with each different phenotype represented
  in each pair
• Smooth round
• Peachy long
• Smooth long
• Peachy round

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B. 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 39
• Peachy long 47
• Smooth long 463
• Peachy round 451

Recombinant
Nonrecombinant
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B. 2-factor Testcross

• 2-F STEP 3 Calculate the map distanceR P
  gene distance 86/1000 X 100 8.6 m.u.
• Smooth round 39
• Peachy long 47
• Smooth long 463
• Peachy round 451

Recombinant
Nonrecombinant
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B. 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 39
• Peachy long 47
• Smooth long 463
• Peachy round 451

Recombinant
Nonrecombinant
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B. 2-factor Testcross

• Cis linkage When two dominant alleles are linked
  together in the original heterozygote

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B. 2-factor Testcross

• Trans linkage When a dominant allele is linked
  to a recessive allele in the original
  heterozygote

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B. 2-factor Testcross

• The double crossover problem
• Double crossovers occur whenever two crossover
  events occur between two genes
• If this occurs, then the recombinant progeny will
  not be counted, because each allele goes back
  to its original linkage
• For this reason, the map distance given by a
  2-factor testcross often is too low

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C. 3-factor Testcross

• By performing a testcross with 3 genes, we can
  estimate how many double crossovers are occurring
• Example Maize
• Green (Y) vs. yellow (y) plant color
• Full (S) vs. shrunken (s) seed shape
• Colored (C) vs. colorless (c) seed color
• Cross Yy Ss Cc X yy ss cc

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C. 3-factor Testcross

• 3-F CROSS STEP 1Arrange the phenotypic classes
  into pairs, with each different phenotype
  represented
• Green Full Colored
• Yellow Shrunk Colorless
• Green Full Colorless
• Yellow Shrunk Colored
• Green Shrunk Colored
• Yellow Full Colorless
• Green Shrunk Colorless
• Yellow Full Colored

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C. 3-factor Testcross

• 3-F CROSS STEP 2Identify the nonrecombinant
  (largest) and double crossover (smallest) classes
• Green Full Colored 100
• Yellow Shrunk Colorless 95
• Green Full Colorless 25
• Yellow Shrunk Colored 20
• Green Shrunk Colored 380
• Yellow Full Colorless 375
• Green Shrunk Colorless 2
• Yellow Full Colored 3

NR
DC
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C. 3-factor Testcross

• 3-F CROSS STEP 3Compare the NR DC classes to
  determine which gene is in the middle (Its
  Y-C-S)
• Green Full Colored 100
• Yellow Shrunk Colorless 95
• Green Full Colorless 25
• Yellow Shrunk Colored 20
• Green Shrunk Colored 380
• Yellow Full Colorless 375
• Green Shrunk Colorless 2
• Yellow Full Colored 3

NR
DC
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C. 3-factor Testcross

• 3-F CROSS STEP 4Determine the identity of the
  two single crossover classes (compare with NR
  class)
• Green Full Colored 100
• Yellow Shrunk Colorless 95
• Green Full Colorless 25
• Yellow Shrunk Colored 20
• Green Shrunk Colored 380
• Yellow Full Colorless 375
• Green Shrunk Colorless 2
• Yellow Full Colored 3

S-C Sing.
Y-C Sing.
NR
DC
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C. 3-factor Testcross

• 3-F CROSS STEP 5Calculate the distances between
  each pair of genes
• Y-C distance (252023)/1000 X 100 5 m.u.
• S-C distance (9510023)/1000 X 100 20 m.u.

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C. 3-factor Testcross

• 3-F CROSS STEP 6Calculate
• The expected double crossover frequencyExpected
  d.c. freq. (0.05)(0.2) 0.01
• The obtained double crossover frequencyObtained
  d.c. freq. (23)/1000 0.005
• The coefficient of coincidenceCoincidence
  (Obtained d.c.)/(Expected d.c.) 0.005 / 0.01
  0.5
• Interference 1 Coincidence 1 0.5 0.5

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C. 3-factor Testcross

• Interference
• The occurrence of one crossover event may
  interfere with a second crossover event
• If the obtained d.c. expected d.c.
  thenCoincidence 1Interference 0
• If the obtained d.c. lt expected d.c.
  thenCoincidence lt 1Interference is a positive
  number
• If the obtained d.c. gt expected d.c.
  thenCoincidence gt 1Interference is a negative
  number