Lecture Note
Linkage & Genetic Mapping in Eukaryotes
-
University:
California State University, Northridge -
Course:
BIOL 360 | Genetics Academic year:
2007
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Views:
111
Pages:
38
Author:
azianvtecboid0h
Linkage & Genetic Mapping in
Eukaryotes
Ch. 6
1 LINKAGE AND CROSSING OVER
! In eukaryotic species, each linear
chromosome contains a long piece of DNA
– A typical chromosome contains many hundred
or even a few thousand different genes
! The term linkage has two related meanings
– 1. Two or more genes can be located on the
same chromosome
– 2. Genes that are close together tend to be
transmitted as a unit
2 Linkage Groups
!
Chromosomes are called linkage groups
– They contain a group of genes that are linked together
!
The number of linkage groups is the number of
types of chromosomes of the species
– For example, in humans
" 22 autosomal linkage groups
" An X chromosome linkage group
" A Y chromosome linkage group
!
Genes that are far apart on the same chromosome
can independently assort from each other
– This is due to crossing-over or recombination
3 Linkage and Recombination
Genes nearby on the same chromosome tend to stay together
during the formation of gametes; this is linkage.
The breakage of the chromosome, the separation of the
genes, and the exchange of genes between chromatids is
known as recombination. (we call it crossing over)
4 Independent assortment:
Genes on different chromosomes
A
A
a
a
B
B
b
b
Gametes
A
A
B
b
a
a
B
b
5 Linkage:
Two genes on same
chromosome segregate
together
A
A
a
a
B
B
b
b
Gametes
A
B
A
B
a
b
a
b
6 Linkage and Crossing over leads to
separation of linked genes
A
A
a
a
B
B
x
b
b
Gametes
Parental
A
Recombinant a
B
a
b
B
A
b
7 Crossing Over Can Produce Recombinant
Phenotypes
! In diploid eukaryotic species
– linkage can be altered during meiosis as a result
of crossing over (recombination)
! Crossing over
– Occurs during prophase I of meiosis
– Non-sister chromatids of homologous
chromosomes exchange DNA segments
8 In 1905 Bateson and Punnett Discovered
Two Traits That Did Not Assort Independently
!
This is a dihybrid cross that is expected to yield a 9:3:3:1
phenotypic ratio in the F2 generation
A much greater
proportion of the two
types found in the
parental generation
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display
9 Some genes on the same chromosome assort
together more often than not
!
In dihybrid crosses, departures from a 1:1:1:1 ratio of F1
gametes indicate that the two genes are on the same
chromosome
" Ex 1: For AaBb, four gametes = AB, Ab, aB, ab = 1:1:1:1 thus
no linkage.
" Ex 2: For AaBb, the observed four gametes are Ab aB only. The
ratio is 2:0:2:0, thus linkage is present.
" We use a testcross of F1 and the chi square analysis to
determine whether genes are unlinked or likely to be linked
10 Chi square test pinpoints the probability that ratios are
evidence of linkage
!
Transmission of gametes is based on chance events
– Deviations from 1:1:1:1 ratios can represent chance events OR
linkage
– Ratios alone will never allow you to determine if observed data
are significantly different from predicted values.
– The larger your sample, the closer your observed values are
expected to match the predicted values.
!
Chi square test measures “goodness of fit” between
observed and expected (predicted) results
– Accounts for sample size, or the size of the experimental
population
11 Applying the Chi Square Test
!
Framing a hypothesis
– Null hypothesis – “observed values are not different from
the expected values”
" For linkage studies – no linkage is null hypothesis
" Expect a 1:1:1:1 ratio of gametes or offspring from testcross
!
Degrees of Freedom: for this example, DOF is the
amount of classes which data can be grouped into,
minus 1.
!
We expect two classes for the gametes
1. Parental class (genotype of parent cells)
2. Non-parental (recombinant) genotype
DOF = classes (parental & recombinant) -1
For linkage analysis, the DOF will ALWAYS be 1.
12 Applying the chi square test to a linkage study
Genotype
Experiment 1
Experiment 2
AB
17
34
ab
14
28
Ab
8
16
aB
11
22
Total
50
100
Observed/Expected
31
25
19
25
Observed/Expected
62
50
38
50
Class
Parentals
Recombination
13 Applying the Chi Square Test
!
We ALWAYS ALWAYS ALWAYS group the data
into two PHENOTYPIC groups or classes for
linkage analysis:
!
Parental Class
!
!
AB and ab
Recombinant (Non-parental) class
!
Ab and aB
14 Chi Square – Experiment 1 & 2
!2 = " (observed – expected)2
number expected
(Parental)
(Recombinant)
Experiment 1
!2 = " (31 – 25)2 + (19 – 25)2
25
25
= 2.88
Experiment 2
!2 = " (62 – 50)2 + (38 – 50)2
50
50
= 5.76
Degrees of Freedom = 2 (parental & recombinant classes) – 1 = 1
15 Chi Square – Experiment 1 & 2
!
Experiment 1:
– Based off of the Chi-square value of 2.88, we determine
that it is less than the critical value of 3.84, so we CAN
NOT reject the null hypothesis. Genes are not linked.
!
Experiment 2:
– A chi-square of 5.76 is greater than the critical value, so
we CAN reject the null hypothesis. Genes are linked.
16 Size Does Matter
!
Sample size makes a difference
– Larger the better
!
Use real numbers not percentages
!
P-value
– The difference between expected and observed is
statistically significant: ! .05
– Meaning difference is less than or equal to ± 5%
– Null hypothesis is true and not due to chance
17 Example if testing for linkage
!
In corn the glossy trait (AA) gives glossy leaves and
the ramosa trait (BB) determines branching of ears, a
test cross produced the following results:
–
–
–
–
Normal leaved and normal branches
Glossy leaved and ramosa branching
Normal leaved with ramosa branching
Glossy leaved with normal branching
395
382
223
247
" For a total of 1247 offspring.
– Are the glossy and ramosa genes linked?
" Null hypothesis: Genes are NOT linked
18 Example if testing for linkage
! Expected
values if 1:1:1:1 if NOT linked
– 1247/2 =
" 623.5
! Set
up !2
– (777-623.5)2/623.5 + (470-623.5)2/623.5=
" 75.6
– Degrees of freedom =
"1
– Accept or reject?
19 df = 1
!2 = 75.5
•Reject
•Therefore: genes are linked
20 GENETIC MAPPING IN PLANTS
AND ANIMALS
!
Genetic mapping is also known as gene mapping
or chromosome mapping
!
Its purpose is to determine the linear order of
linked genes along the same chromosome
!
Figure 5.8 illustrates a simplified genetic linkage
map of Drosophila melanogaster Each gene has its
own unique locus
at a particular site
within a
chromosome
Figure 5.8 Genetic Maps
!
Genetic maps are useful in many ways: they
– 1. allow us to understand the overall complexity and
genetic organization of a particular species
– 2. improve our understanding of the evolutionary
relationships among different species
– 3. can be used to diagnose, and perhaps, someday to
treat inherited human diseases
– 4. can help in predicting the likelihood that a couple will
produce children with certain inherited diseases
– 5. provide helpful information for improving agriculturally
important strains through selective breeding programs Sturtevant, an undergrad student of
Morgan’s (21 years old), invented a
way to quantify the relationship
between the rate of recombination
and the distance between genes.
# recombinants
# total
= RF
(RF=Recombination Frequency)
RF X 100 = % recombination
1% recombination = 1 map unit (mu)
= 1 cM (centiMorgan) !
Genetic maps allow us to estimate the relative distances
between linked genes, based on the likelihood that a
crossover will occur between them
!
Experimentally, the percentage of recombinant offspring is
correlated with the distance between the two genes
– If the genes are far apart # more recombinant offspring
– If the genes are close # fewer recombinant offspring
!
Number of recombinant offspring X 100
Map distance =
Total number of offspring
!
The units of distance are called map units (mu)
– They are also referred to as centiMorgans (cM)
!
One map unit is equivalent to 1% recombination frequency Genetic Mapping and Testcrosses
!
Genetic mapping experiments are typically accomplished
by carrying out a testcross
– A mating between an individual that is heterozygous for two or more
genes and one that is homozygous recessive for the same genes
!
Genes that are located on DIFFERENT chromosomes
show a recombination frequency of 50%
!
Genes that are located FAR APART on the SAME
chromosome show a recombination frequency of 50%
!
The closer two genes are to each other on a chromosome,
the small the recombination frequency will be
– (approaches 0%). The Hypothesis
– The distance between genes on a chromosome can be
estimated from the proportion of recombinant offspring
" This provides a way to map the order of genes along a
chromosome
!
Crossing Over creates recombinant offspring
– An event where homologous chromosomes exchange parts,
creating a new combination of gene alleles.
– The exchange of genetic material between the two homologous
chromosomes is termed Recombination.
!
Example
– Before Crossover:
" Maternal Chromosome Genes: ABCD
" Paternal Chromosome Genes: abcd
– After crossing over:
"
"
Maternal Chromosome Genes: ABcd
Paternal
Chromosome
Genes:
abCD
Copyright
©The McGraw-Hill Companies,
Inc. Permission
required for reproduction or display The frequency of
recombination is
proportional to the
genetic distance
(based on the
assumption that
recombination is random) Recombination and genetic distance
random cuts between (A and B) < (B and C)
Gene A
Gene B
Gene C
% recombination is proportional
to distance between genes.
Quantification of the relationship between the rate of
recombination between two genes and the distance between
them:
# recombinant offspring / total # offspring = recombination
frequency between two genes
1% recombination = 1 map unit (m.u.) = 1 cM (centiMorgan)
29 Calculating the Distance Between y and w Genes
F1
X
Xyw+
Xy+w
F2
males:
Xy w+/Y
Xy+w/Y
Xy+w+/Y
Xy w/Y
4484
4413
76
53
9026
Xy+w
Y
- Parental
- Parental
- Recombinant
- Recombinant
total
# recombinants = recombination frequency between
total #
two genes
76 + 53
9026
X 100 = 1.43%
= 1.43 map units between the y and w genes
30 Another example: Linkage in an autosomal gene
• genotypes of F1 female revealed by test cross
• Parental class outnumbers recombinant class demonstrating linkage
23 mu
31 Summary of linkage and recombination “rules”
!
!
!
!
!
!
!
Genes close together on the same chromosome are linked and do
not segregate independently
Linked genes lead to a larger number of parental class than
expected in double heterozygotes
Mechanism of recombination is crossing over
Chiasmata are the visible signs of crossing over
Farther away genes are, the greater the opportunity for chiasmata
to form
Recombination frequencies reflect physical distance between genes
Recombination frequencies between two genes vary from 0% to
50%
32 Mapping:
Locating genes along a chromosome
!
Two-point crosses:
comparisons help
establish relative gene
positions
!
Genes are arranged in a
line along a
chromosome
Fig. 5.11
33 Mapping: Locating genes along a chromosome
Fig. 5.11
Genes are arranged in a line along a chromosome
34 Example problem
!
An individual has the genotype Aa/Bb. Gene loci (A) and
(B) are 15 cM apart. Indicate all the possible gametes this
individual can produce, and the proportions of expected
progeny genotypes if a testcross is performed on this
individual.
!
What do we know?
35 Example problem
!
An individual has the genotype Ab/aB. Gene loci (A) and (B) are 15 cM
apart. Indicate all the possible gametes this individual can produce, and
the proportions of expected progeny genotypes if a testcross is
performed on this individual.
!
What do we know?
–
–
–
–
15% recombination frequency
Means 15% recombinant progeny, 85% parental progeny
Testcross is Ab/aB X abab
Results
"
"
"
"
Ab = 42.5% (half of 85% parentals)
parentals)
aB = 42.5% (half of 85% parentals)
parentals)
AB = 7.5% (half of 15% recombinants)
ab = 7.5% (half of 15% recombinants)
36 Question
What if we artificially blocked crossing over?
!
An individual has the genotype Ab/aB.
Ab/aB. Gene loci (A) and (B) are 15 cM apart. Indicate the proportions
of expected progeny genotypes if a testcross is performed on this individual.
!
What do we know?
– 0% recombination frequency
– Testcross is Ab/aB X ab/ab
– 1)
"
"
"
"
Ab/ab = 42.5% (half of 85% parentals)
parentals)
aB/ab = 42.5% (half of 85% parentals)
parentals)
AB/ab = 7.5% (half of 15% recombinants)
Ab/ab = 7.5% (half of 15% recombinants)
– 2)
" Ab/ab = 50%
" aB/ab = 50%
– 3)
" AB/ab = 50%
" Ab/ab = 50%
37 Limitations of two point crosses
!
Difficult to determine gene order if two genes are
close together
!
Actual distances between genes do not always add
up
!
Pairwise crosses are time and labor consuming
38
Linkage & Genetic Mapping in Eukaryotes
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