Maximum Testcross Phenotypic Recombination Frequency 0.5

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Maximum Testcross Phenotypic Recombination
Frequency 0.5

• Recomb fq 0.5 indicates genes unlinked
• Genes may be unlinked in two ways
• Genes may be
• Genes may be

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Unlinked Genes

• Genes may be on different chromosomes
• If an individual is AaBb, and genes are on
  different chromosomes, expect an equal number of
  AB, ab, Ab, and aB allelic combinations in
  gametes if homologues segregate and align
  independently of other chromosomes at metaphase
  plate
• Genes may be far apart on same chromosome
• If an individual is AaBb, and genes are on the
  same chromosome, expect an equal number of AB,
  ab, Ab, and aB allelic combinations in gametes if
  diff sections of the same chromosome are as
  likely to migrate to opposite poles as different
  chromosomes (50 recombination fq)

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Maximum Testcross Phenotypic Recombination
Frequency 0.5

• Once recomb fq reaches 0.5, like genes on
  different chromosomes, unlinked genes on the same
  chromosome are equally likely to end up in the
  same or different gametes
• At recombination frequency 0.5
• Expect equal number of recombinant and parental
  chromosomes
• If AB/ab, expect 50 gametes to have parental (AB
  or ab), and 50 gametes to have recombinant (aB
  or Ab) types
• A recomb fq gt0.5 would only occur if repulsion
  effect

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Two strand single CO

• High frequency of CO produces a recombination fq
  of 0.5 between two genes an equal of progeny
  with recombinant and nonrecombinant chromosomes
• Two strand single CO between any pair of
  nonsister chromatids results in 2 parental, 2
  recombinant (50 recomb)
• Because sister chromatids are just copies, it
  doesnt matter which copy exchanges with the
  nonsister chromatid

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Double CO also results in
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Three-Factor Linkage Analysis

• To determine the distances between three linked
  genes
• Could just add up the distances from two-factor
  linkage
• Requires two testcrosses
• It is possible to map three or more genes from
  the progeny of one testcross
• Steps to three-factor linkage analysis
• 1) Determine the order of the genes on the
  chromosome
• 2) Determine the genetic map distances among the
  three linked genes from the progeny of the same
  cross

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Three-Factor Linkage Analysis

• Determine the order of the three genes
• Three possible orders
• A B C
• B A C
• A C B
• Determine order from the proportions of
  individuals
• in each class of testcross progeny

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Three-Factor Linkage Analysis

• AaBbCc x aabbcc produces 8 genotypes (
  phenotypes) of progeny
• If independent assortment, expect 11111111
  ratio
• Linkage would show as a deviation from this ratio

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Three-Factor Linkage Analysis
How do the 8 gamete types from the trihybrid
arise? 2 arise from NCO 4 arise from SCO 2
arise from DCO
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Three-Factor Linkage Analysis

• To determine the order of three linked genes in
  testcross
• Most frequent 2 classes in offspring are NCOs
• Least frequent 2 classes in offspring are DCOs

Fig. 15.14
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Three-Factor Linkage Analysis

• Which gene order explains an observed double
  crossover phenotypic class? (for example below,
  all genes in cis)
• Three orders NCO type Observed DCO type
• 1) ABC AbC
• ____ -gt ____
• abc aBc
• 2) BAC BaC
• ____ -gt ____
• bac bAc
• 3) ACB AcB
• ____ -gt ____
• acb aCb

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Three-Factor Linkage Analysis
Testcross V v Gl gl Va va x v v gl gl va va
V_ normal, vv virescent seedlings (turning
green) Gl_ normal, gl gl glossy leaves
Va_ normal, va va variable
sterility Progeny Normal 235 Glossy,
variable sterile 62 Variably sterile 40 Varia
bly sterile, virescent 4 Glossy 7 Glossy,
virescent 48 Virescent 60 Virescent,
glossy, variably sterile 270 Total 726
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Three-Factor Linkage Analysis

• Determine the genetic map for these three genes
• Determine gene order from the possibilities
• Gl V Va Gl v Va (virescent)
• _______ -gt ______
• gl v va gl V va (glossy, variably
  sterile)
• V Gl Va V gl Va (glossy)
• _______ -gt ______
• v gl va v Gl va (virescent, variably
  sterile)
• Gl Va V Gl va V (variably sterile)
• _______ -gt ______
• gl va v gl Va v (glossy, virescent)

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Three-Factor Linkage Analysis
Once gene order is known (V Gl Va), can identify
crossover types Normal 235 No
crossover Glossy, var. ster. 62 Single crossover
(V Gl) Variably sterile 40 Single crossover
(Gl Va) Var. ster., virescent 4 Double
crossover Glossy 7 Double crossover Glossy,
virescent 48 Single crossover (Gl
Va) Virescent 60 Single crossover (V
Gl) Vir., glossy, var ster 270 No crossover
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Three-Factor Linkage Analysis
2) Determine genetic map distance Map distance
between v and gl Single crossover (glossy,
var. ster. and virescent) 62 60 Double
crossover 4 7 (62 60 4 7)/726 0.183
or 18.3 cM Map distance between gl and va
Single crossover (variable sterility and glossy,
vir) 40 48 Double crossover 4 7 (40
48 4 7)/726 0.136 or 13.6 cM V 18.3
Gl 13.6 Va
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Three-Factor Linkage Analysis
Previous example was from a cis arrangement of
dominant alleles (coupling conformation) How do
you determine which genes are in cis and which
are in trans? NCO progeny tell you which alleles
are in trans Then what? Det. gene order by
comparing DCO and NCO classes (the middle
allele changes position) Det. total crossovers
between two genes and make map
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Three-Factor Linkage Analysis
Determine which genes in cis and trans cl cl
colorless, wx wx waxy, sh sh shrunken Progeny
of a testcross Colored, starchy,
full 4 Colored, starchy, shrunken 2538 (no
crossover) Colored, waxy, full 113 Colored,
waxy, shrunken 601 Colorless, starchy,
full 626 Colorless, starchy, shrunken 116 Color
less, waxy, full 2708 (no crossover) Colorless,
waxy, shrunken 2 Colored, starchy, shrunken
(Cl _ Wx _ sh sh) Colorless, waxy, full (cl cl wx
wx Sh _)
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Map Distance Issues
Factors that alter crossover frequencies Chromoso
mal location of a gene Age Sex Genotype Enviro
nmental influences Ex Drosophila no crossovers
in male flies, crossovers in females depend on
age, temperature, diet
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Map Distance Issues

• Two-factor mapping experiments underestimate map
  distance when recombination frequencies exceed
  0.07 (true map distance exceeds 7 cM) because of
  undetected double crossovers (DCOs)
• Double crossovers
• A second CO between two genes reverses the
  effects of the first CO, restoring the original
  combination of alleles

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Mapping Issues

• As the actual map distance exceeds 7 cM, the
  observed recombination frequency p becomes
  substantially less than the actual recombination
  frequency d (due to undetected DCOs) and map
  distances become substantially underestimated
• Frequency (probability) of DCO is the square of
  the genetic map distance
• Ex A and B are separated by 20 cM
• Frequency of DCO (0.2)2 0.04

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Mapping Functions

• Equations used to correct approximate crossover
  fqs
• Haldane Function (1919 JBS Haldane) for DCO
  effect
• d -0.5 ln(1-2p)
• d estimated map distance (crossover fq)
• p phenotypic recombination fq
• Relationship doesnt account for interference
• V 18.3 Gl 13.6 Va
• d -0.5 ln(1-(20.183))
• d -0.5 ln(1-(20.136))

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Interference
Interference Presence of one crossover prevents
another in its immediate vicinity Example (Cl
Sh Wx) Given frequency of crossover between cl
and sh 0.035 and the frequency between sh
and wx 0.184 Using product rule
Expected frequency of double crossover (one
between cl and sh, and one between sh and wx)
0.035 x 0.184 0.00644 This assumes no
interference Total number of progeny 6708
Expected number of double crossover progeny
0.00644 x 6708 43.2 Observed number of
double crossover progeny 6
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Interference
Coefficient of Coincidence Quantification of
interference C observed number of double
crossovers expected number of double
crossovers From previous example C 6/43.2
0.139
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Interference
Interference I 1 - C From previous
example I 1 - 0.139 0.861 Positive
interference I gt 0 Complete interference I 1
When genes are very tightly linked Negative
interference I lt 0