Download 1 Chromosome Mapping in Eukaryotes

William S. Klug
Michael R. Cummings
Charlotte A. Spencer
Concepts of Genetics
Chromosomes consist of linear sequences of genes.
STUDY UNIT 4
Chromosome mapping:
•
Chapter 5
Chromosome Mapping in
Eukaryotes
•
Genes located on the same chromosome are said to be linked
and demonstrate linkage in genetic crosses.
Copyright © 2006 Pearson Prentice Hall, Inc.
•Genes are not
cytologically linked
•Genes are
cytologically linked,
complete linkage.
•
•
•Gametes form at
equal frequencies
•Gametes form at
equal frequencies
Figure 5-1a
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•Genes are
cytologically and
genetically linked
(<50% recombination)
Figure 5-1b
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5.1 Genes Linked on the Same
Chromosome Segregate
Together
•
5.1.1 The Linkage Ratio
If complete linkage exists between 2 genes because of
close proximity, the F2 phenotypic ratio of a dihybrid
cross will be unique, designated the
.
•Four genetically
different gametes are
formed
Genes located on the same chromosome will show
evidence of linkage to one another. A
can be established for each chromosome (corresponds
to n number of chr)
•Frequency of crossing
over between genes
generally proportional
to distance between
them
Figure 5-1c
Eg fig 5-2 (Klug et al, 2006)
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Drosophila melanogaster:
Recessive mutants:
brown eye (bw) < bw+ wildtype red eye
heavy wing vein (hv) < hv+ wildtype thin wing vein
The two genes are cytologically linked.
Flies with:
Brown eyes, normal thin veins X wildtype red eyes, heavy veins
P1: bw
bw
hv+ X bw + hv
hv+
bw + hv
F1: bw hv+
bw + hv
Figure 5-2-01
5.2
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Crossing Over Serves as the Basis of
Determining the Distance between
Genes during Chromosome Mapping
5.2.1 Morgan and Crossing Over
• Usually crossing over occurs between genes on the
same chr
• Result: percentage of offspring arise from recombinant
gametes
•
• 1911, Thomas Morgan and Alfred Sturtevant
Figure 5-2ab
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5.2.2 Sturtevant and Mapping
•
•
Sturtevant first to realize that Morgan’s results could
be used to map the sequence of linked genes
Compiled data on recombination between the genes
representing yellow, white and miniature
(1) yellow, white
(2) white, miniature
(3) yellow, miniature
0.5%
34.5%
35.4%
(1) + (2) approx. = (3) – Recombination frequencies
between linked genes are additive
Figure 5-3
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5.2.3 Single Crossovers
Frequency of exchange gives an estimate of relative
distance between loci
• A limited number of crossovers occur in each tetrad
during meiosis
• Occur randomly along the length of the tetrad
In honour of Morgan’s work, map units referred to as
centimorgans (cM)
•
• The farther apart two linked loci are from each other,
the more likely a random crossover event will occur
between them.
Figure 5-4
Copyright © 2006 Pearson Prentice Hall, Inc.
•
• The other 2 nonsister chromatids enter the gamete unchanged
(parental/noncrossover gametes)
•If a single crossover occurs 100% of the time between 2 loci,
recombination will only be observed in 50% of the potential
gametes formed.
Figure 5-5a & b Copyright © 2006 Pearson Prentice Hall, Inc.
Figure 5-6
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Consider only single crossovers:
• Means crossing over occurs in 40% of the tetrads
When two linked genes are more than 50 map units
apart, a crossover can be expected to occur between
them in 100% of the tetrads.
General rule for single crossovers between two loci:
•
• Observe 20% recombinant gametes
Expect equal proportions of the 4 gametes
•
Therefore, the theoretical limit of recombination due to crossing
over is 50%
If there is 50% or less recombination between 2 loci, the 2 loci are
said to be genetically linked.
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How do we use this information for mapping
purposes?
TWO-POINT MAPPING
Linkage analysis of two genes
Linkage notation:
Genetic map distances measured in map units or
centimorgans (cM)
where 1% recombination = 1 mu / 1 cM
Map distances between many linked genes are combined
to construct a genetic map.
Consider individual AaBb.
Ratio of parental gametes : recombination gametes
May produce the following gametes:
depends on how frequently crossing over occurred.
For mapping?
AB
ab
Want to determine the freq of recombination gametes
as an indication of distance between A and B.
Impossible to analyse the genotype of a gamete,
therefore,
Ab
aB
Linkage best analysed in testcross:
Reason?
•
heterozygote
x
homozygote recessive
•
∴ each of the 4 phenotypes in progeny represents the
alleles received from one parent (AaBb)
• Cross-over in aabb parent not detected
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Example
• can distinguish recombinant from parental progeny
1. Perform testcross
• phenotypic freq of progeny reflects the frequency of
C1 c1 Wx wx
gametes formed by heterozygote parent,
x
c1c1 wx wx
coloured, starchy
colourless, waxy
• i.e. recombination frequency of heterozygote
Progeny:
coloured, starchy
(C1 c1 Wx wx)
coloured, waxy
(C1 c1 wx wx)
310
colourless, starchy (c1 c1 Wx wx)
311
colourless, waxy
858
(c1 c1 wx wx)
781
Total
2. Determine if loci are indeed linked
2260
3. Identify recombinants in progeny
- measure deviation from independent assortment
parental phenotypes
expect ratio of
= 2
classes
recombinant phenotypes = 2 less abundant classes
in progeny
coloured, starchy
(C1 c1 Wx wx)
if independent assortment had occurred in
coloured, waxy
(C1 c1 wx wx)
310
gametes of heterozygous parent.
colourless, starchy (c1 c1 Wx wx)
311
colourless, waxy
-
use chi – squared analysis
(c1 c1 wx wx)
Total
___
2260
- if ratio is not 1 : 1 : 1 : 1
⇒ indicates linkage
4.
Identify coupling or repulsion conformation
5.
Calculate recombination frequency (or % )
where 1% recombination = 1 cM
Coupling (cis):
Rf =
Repulsion (trans): 1 dominant and 1 recessive allele
number recombinants
= 310 + 311
total number progeny
2260
linked in parental chromosomes
= 0.275
C1
W1
c1
w1
= 27.5 %
6.
∴
Coupling phase
⇒ 27.5 cM
Compile genetic map
C1
27.5
Wx
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1.
In maize, the recessive allele of a locus causes brownmidrib (bm) and the recessive allele of another locus brevis (bv)
(dwarf). A double heterozygote was crossed with a brevis plant
with a brown midrib and 200 seeds were germinated. The following
results are obtained:
+ +
93
bm +
7
+ bv
5
bm bv
95
2. In Drosophila, ebony (e) is recessive and the dominant allele (e+) results in
tan body colour. The recessive mutation (d) causes dumpy (shortened)
wings while the dominant allele (d+) causes long, normal wings. A tan female
with normal wings is crossed with an ebony male with dumpy wings. The
resulting progeny are as follows:
44 tan, long
41 ebony, dumpy
15 tan, dumpy 17 ebony, long
a) Give the genotypes of the parents.
(a)
Show the crossing and progeny symbolically.
b) Give the gamete ratios of the parents.
(b)
What is the % recombination between the 2 loci?
c) Set up a chromosome linkage map.
5.3 Determining the Gene
Sequence during Mapping
Relies on the Analysis of
Multiple Crossovers
5.3.1 Multiple Exchanges
• Study single crossovers between 2 linked genes =
distance
Figure 5-7ab
Copyright © 2006 Pearson Prentice Hall, Inc.
In a single tetrad – possible to have 2, 3 or more crossovers
Double exchanges of genetic material are as a result of double
crossovers (DCOs)
•
P of a single exchange between A and B and B and C is directly related to the
physical distance between them
5.3.2 Three-Point Mapping in Drosophila
Double crossover – 2 separate and individual events must occur
simultaneously
Criteria
Product law
Example:
SCO between A and B 20% ~ p = 0.20
SCO between B and C 30% ~ p = 0.30
DCO between A and B and B and C is predicted to be (0.20)(0.30) = 0.06 (6%)
1. Organism being mapped must be heterozygous at
all loci
2.
test cross – progeny reflect heterozygote parents
gametes
hemizygous male must carry recessive alleles
3. Large number of progeny must be examined
If 3 linked genes relatively close to one another, expected frequency of DCO
gametes is very low
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5.3 Determining the Gene
Sequence during Mapping
Relies on the Analysis of
Multiple Crossovers
yellow body colour (y)
white eye colour (w)
echinus eye shape (ec)
5.3.3 Determining the Gene Sequence
The gene order must be determined before proceeding
with the problem
2 methods…
Figure 5-8
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Method I
Based on the fact that there are only 3 possible orders
3 loci: y, w and ec
Method II
According to fig 5-8,
NCO: y w ec
DCO: y w+ ec
&
&
y+ w+ ec+
y+ w ec+
–
– Identify the single allele that has been
switched so that it no longer is associated
with its original neighboring alleles.
– This allele is in the middle.
Figure 5-9
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THREE-POINT MAPPING
8 types of gametes possible in progeny:
Linkage analysis of 3 genes in a single testcross
Crossovers can be detected in progeny of an
individual that is heterozygous for the 3 linked
genes.
• parental class P
= most abundant
NCO
• crossover in one
interval = single
recombination
class 1
SCO I
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Note:
• crossover in other
If genes are not linked, all 8 gametes will be
•
interval = single
recombination
equally frequent
class 2
∴ would expect
1:1:1:1:1:1:1:1
phenotypic ratio (independent assortment)
• double crossover =
•
double recomb.
class
1.
DCO
2.
Perform testcross
C1c1 Shsh Wxwx
x
coloured, full, starchy
Determine if loci are indeed linked
Ratio ≠ 1:1:1:1:1:1:1:1 ⇒ indicates linkage
c1c1 shsh wxwx
colourless, shrunken, waxy
3.
Phenotypes
Gametes heterozygote
coloured, full, starchy
C1 Sh Wx
coloured, shrunken, starchy
C1 sh Wx
coloured, full, waxy
C1 Sh wx
coloured, shrunken, waxy
C1 sh wx
colourless, full, starchy
c1 Sh Wx
colourless, shrunken, starchy
c1 sh Wx
colourless, full, waxy
c1 Sh wx
colourless, shrunken, waxy
c1 sh wx
Total:
Number
4
2538
113
601
626
116
2708
2
6708
Identify coupling or repulsion phases of the 3 genes
Consider parental class (NCO) = most
coloured, shrunken, starchy
C1 sh Wx
2538
colourless, full, waxy
c1 Sh wx
2708
∴ noncrossover chrs are
and
c1 Sh wx
∴ C1 and Wx are in coupling phase and sh is in
repulsion to them.
4. Determine gene order :
Parental
compare P/NCO class with DCO
C1 sh Wx
(order not yet known)
C1
sh
Wx
C1
c1
Sh
wx
c1
sh
Wx
- this indicates
which gene is in the middle.
Sh
wx
Parental / No Crossing Over class
coloured, shrunken, starchy C1 sh Wx
2538
colourless, full, waxy
2708
c1 Sh wx
Double Crossover class (least) :
DCO
C1
c1
coloured, full, starchy
C1 Sh Wx
4
colourless, shrunken, waxy
c1 sh wx
2
sh
Sh
Wx
wx
C1
c1
Sh
sh
Wx
wx
Sh in the middle produces the correct double
crossover phenotypes.
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5. Identify single recombination classes as being SCOI
or
6. Calculate recombination frequency (or %)
( compare to NCO )
% recomb in region I =
coloured, shrunken, starchy
coloured, full, waxy
coloured, shrunken, waxy
C1 sh Wx
C1 Sh wx
2538
113
c1 sh Wx
c1 Sh wx
+ % DCO
(113 + 116) + (4+2)
6708
= 0.035 = 3.5 %
colourless, full, starchy
colourless, shrunken, starchy
colourless, full, waxy
% SCO I
=
⇒ 3.5 cM
∴ distance between C1 and Sh = 3.5 cM
116
2708
% recomb in region 2 = % SCO II
Single crossover class I:
=
+ % DCO
(601 + 626) +
(4+2)
6708
- crossover between C1 and Sh
= 0.184
= 18.4%
⇒ 18.4 cM
- crossover between Sh and Wx
∴ distance between Sh and Wx = 18.4 cM
5.4
6. Compile genetic map
C1
3.5
Sh
Interference Affects the Recovery
of Multiple Exchanges
INTERFERENCE
18.4
Wx
Until now we have assumed that a crossover at one
position is independent of another in the same
bivalent. Not necessarily true.
Interference:
one chiasma interferes with the formation of a second
chiasma in the same vicinity.
∴
From example:
C1
3.5
Sh
18.4
Wx
Interference measured as coefficient of coincidence (C)
observed dco was
C
=
expected dco
observed DCO
expected DCO
∴C
(4+2) / 6708
= 0.00089
= 0.035 x 0.184 = 0.00644
0.00089 / 0.00644
= 0.139
Interference = 1 – C
Expected dco (DCOexp)?
Observed freq crossover
in area I
= 1 – 0.139
x Observed freq crossover
in area II
= 0.861
If Interference > 0
Interference will be = 0 if C = 1 (when observed dco =
expected dco)
Interference = 1
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5.5 As the Distance between
Two Genes Increases,
Mapping Experiments
Become More Inaccurate
• When two genes are close together, the accuracy
of mapping is high. As the distance between them
increases, the accuracy of mapping decreases.
• In most cases experimentally derived distance is
an underestimate.
•
Figure 5-12 & 5-14 Copyright © 2006 Pearson Prentice Hall, Inc.
5.12 Gene Mapping Is Now
Possible Using Molecular
Analysis of DNA
DNA markers
•
Followed in pedigrees
•
Associated with specific chr
•
Human genes analyzed in relationship to
markers
Gene Mapping Using Annotated Computer
Databases
•
Genome projects – physical maps in bp
3. A mielie, which was homozygous for the recessive genes a ("green"), d
("dwarf") and rg (normal leaves) was crossed with a mielie homozygous
for the dominant genes A (red), D (tall) en Rg ("ragged" leaves). The
offspring were then test crossed and the following results obtained:
Pt
No.
Pt
No.
Rg A D
265
rg A d
90
rg a d
275
Rg a D
70
rg A D
24
Rg A d
120
Rg a d
16
rg a D
140
4. Given a chromosome map:
R
M
8
T
20
Accept the given linkage setup.
Accept CC = 0.5
Calculate the size of each phenotypic class of the offspring if a trihybrid
testcross is performed.
(a) Which classes represent the single crossover classes?
(b) Set up a chromosome map.
(c) Calculate the coefficient of coincidence.
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