Lecture: Linkage and Mapping

1

• Lecture Linkage and Mapping
• another modification of Mendels Genetic Law
• incomplete independent assortment of two
  genes
• due to physical linkage

2
Outlines

• basic concepts of genetic linkage and mapping
• mapping exercise I
• different situations of linkage
• mapping exercise II

3
Basic concepts

• Mendels second law independent assortment of
  two genes
• the basic unit of segregation is individual
  genes
• Segregating units at the cellular level is not
  individual genes
• but chromosomes
• Each human chromosome contains average 1000
  genes,
• which segregate together------gt against Mendels
  2nd Law
• - Genetic linkage

4
Genetic linkage
Lets have two hypothetical genes (fly
behaviors) Football-maniac (fm) ------gt
recessive Beer-drinker (bd) ------gt
recessive What if two genes are located on
different chromosomes, P FM/FM BD/BD x
fm/fm bd/bd F1 FM/fm BD/bd x
fm/fm bd/bd (testcross) F2
FM/fm BD/bd ( ) fm/fm bd/bd
( ) Fm/fm bd/bd ( )
fm/fm BD/bd ( )
5
Independent assortment Genes on different
chromosomes
BD
FM
bd
fm
Meiosis of F1
Gametes
fm
Fm
BD
BD
Fm
fm
bd
bd
6
Genetic linkage

• Mendels second law says independent sorting of
  two genes
• based on one assumption that two genes are
  different chr.
• What if two genes are located on a same
  chromosome,
• P FM, BD / FM, BD x fm, bd / fm, bd
• F1 FM, BD /fm, bd x fm, bd / fm, bd
  (testcross)
• F2 FM, BD / fm, bd ( ) parental
  type
• fm, bd / fm, bd ( )
  parental type
• Fm, bd / fm, bd ( )
  recombinant type
• fm, BD / fm, bd ( )
  recombinant type

7
Linkage Two genes on same chromosome segregate
together
FM
BD
Meiosis of F1
fm
bd
Gametes
FM
BD
FM
BD
fm
fm
bd
bd
8
Crossing-over and linkage leads to separation of
linked genes
FM
BD
x
Meiosis of F1
fm
bd
Gametes
Parental
fm
FM
BD
bd
Recombinant
fm
BD
FM
bd
9
Frequency of crossing-over ---gt
an unit of genetic mapping

• The closer two genes are located on a given
  chromosome,
• the less No. of crossing-over between two genes
  during meiosis
• Mapping distance ()
• Total number of recombinants / Total number
  of offspring
• What is the possible maximum mapping distance
  for any given pair of genes?

10
Things to be considered for genetic mapping

• Genetic linkage and crossing-over are
• about one chromosome behavior during
  meiosis!
• Haploid (one chromosome) behavior during meiosis
• The first discovery was from X-linked genes of
  flies (TH Morgan)
• Genotypes need to be inferred from
  phenotypes!
• usually require testcross
• Genetic mapping requires animal breeding
  (meiosis) and
• scores of the phenotypes of at least two
  generations

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The first discovery of genetic linkage

• red eye (w)
• white eye (w)
• normal body color (y)
• yellow body color (y)
• F2 offspring -----gt
• two dominant categories
• parental types
• (genetic linkage)
• F2 offspring ------gt
• two minor categories
• recombinant types
• (crossing-over)
• genetic analyses done only
• in male. Why?

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Mapping Exercise I
We have three very unusual fly mutant
phenotypes fm (football-maniac) bd
(beer-drinker) hs (heavy-smoker) When we cross
a fly (triply homozygote) with a normal fly, we
have all normal F1 ----------------gt recessive
Now, think about what we need to do to test
whether these three genes determining unusual
behaviors are genetically linked. F1 with
triply heterozygote (fm,bd,hs/fm,bd,hs) x
a fly with triply homozygote (fm,bd,hs/fm,bd,hs
) What if all three genes are not linked?

13
Mapping Exercise I
What if all three genes are not linked? (238
different categories are all equal)
Independent assortment!!!!
hs
fm,bd,hs 1
bd
hs fm,bd,hs 1
fm
hs fm,bd,hs
1
bd
hs fm,bd,hs 1
hs
fm,bd,hs 1
bd
hs fm,bd,hs 1
fm
hs fm,bd,hs
1
bd
hs fm,bd,hs 1

14
Mapping Exercise I
What if all three genes are linked? (located in
a single chromosome) (two parental categories
are dominant!) phenotype
in symbols observed No.
hs fm,bd,hs 2000
bd hs fm,bd,hs
150 fm hs
fm,bd,hs 50 bd
hs fm,bd,hs
300 hs
fm,bd,hs 300 bd
hs fm,bd,hs 50
fm hs fm,bd,hs
150 bd
hs fm,bd,hs 2000
Total
5000
15
Mapping Exercise I
How to do genetic mapping (convert the data to
three sets of two point data)
phenotype in symbols observed No. bd
vs hs hs fm,bd,hs
2000 2000 300
parental bd hs
fm,bd,hs 150
150 50 recombinant fm
hs fm,bd,hs 50
50 150 recombinant
bd hs
fm,bd,hs 300
300 2000 parental
hs fm,bd,hs 300
bd hs fm,bd,hs
50 Map distance Rec.
type/Total fm
()
hs fm,bd,hs
150 bd
hs fm,bd,hs 2000
Total
5000
16
Mapping Exercise I
How to do genetic mapping (convert the data to
three sets of two point data)
phenotype in symbols observed No. fm
vs bd hs fm,bd,hs
2000 2000 150
parental bd hs
fm,bd,hs 150 fm
hs fm,bd,hs 50
50 300 recombinant
bd hs
fm,bd,hs 300
hs fm,bd,hs 300
300 50 recombinant
bd hs fm,bd,hs
50 fm

hs
fm,bd,hs 150 150
2000 parental bd
hs fm,bd,hs
2000 Mapping distance Total

5000 Rec./Total ()
17
Mapping Exercise I
How to do genetic mapping (convert the data to
three sets of two point data)
phenotype in symbols observed No. fm
vs hs hs fm,bd,hs
2000 2000 50
parental bd hs
fm,bd,hs 150
150 300 recombinant fm
hs fm,bd,hs 50
bd hs
fm,bd,hs 300 hs
fm,bd,hs 300
300 150 recombinant bd
hs fm,bd,hs
50 50 2000
parental fm

hs fm,bd,hs
150 bd hs
fm,bd,hs 2000
Mapping distance Total
5000
Rec./Total ()
18
Mapping Exercise I

If three genes are all linked (on a same chro.),
use the following method 1) the pair yielding
the longest genetic distance should be put in the
outermost position bd - hs
400/5000 8 fm - bd
700/5000 14 fm - hs
900/5000 18 fm - bd - hs 2)
the correct distances are derived from two
smaller genetic distances 14
8 fm - bd - hs 3) the total
distance is derived from the combined value of
two smaller genetic distances fm - hs
22 Why is the combined distance of two
smaller ones is not equal to the longest one?
double crossing-over!
19
Mapping Exercise I
phenotype in symbols
observed No. fm vs hs
hs fm,bd,hs 2000
2000 50 parental bd
hs fm,bd,hs
150 150 300
recombinant fm hs
fm,bd,hs 50
double crossing-over (x2) bd
hs fm,bd,hs
300 hs
fm,bd,hs 300 300
150 recombinant bd
hs fm,bd,hs 50
double crossing-over (x2) fm


hs fm,bd,hs 150
bd hs
fm,bd,hs 2000 50
2000 parental Total
5000
The corrected distance between fm - vs
Rec. type/Total (150300)(300150)2x(5050)
1100/5000 22
20

• What to do with genetic data for mapping
• 1) Traits are segregating?
• dominant/recessive ----gt one locus
• 2) Traits are X-linked?
• phenotype ratios differ between different
  sexes
• 3) Traits are all linked?
• case I not linked (11111111)
• case II only two out of three are linked
• (10010010010050505050
  )
• case III all three are linked
• (100100 2020101055)