Transcript Linkage and Mapping 2

LINKAGE AND
GENETIC MAPPING
IN EUKARYOTES
genetic linkage map


Based on recombination data
What we will study now
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
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
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
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What is crossing over?

When non-sister chromatids of
homologous chromosomes exchange
DNA segments
Crossing Over May Produce
Recombinant Phenotypes

Occurs during prophase I of meiosis at the
bivalent stage

In diploid eukaryotic species, linkage can
be altered during meiosis as a result of
crossing over
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B
A
B b
A a
B
A
b
a
Diploid cell after
chromosome replication
Diploid cell after
chromosome replication
Meiosis
Meiosis
B
A
b
a
b
a
B
A
B
A
B
a
b
a
b
a
b
A
Possible haploid cells
(a) Without crossing over, linked alleles
segregate together.
Possible haploid cells
(b) Crossing over can reassort linked
alleles.
These haploid cells contain a
combination of alleles NOT
found in the original
chromosomes
These are
termed
parental or
nonrecombinant
cells
Figure 5.1
This new combination of
alleles is a result of
genetic recombination
These are termed
nonparental or recombinant
cells
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Bateson and Punnett Discovered Two
Traits That Did Not Assort Independently
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In 1905, William Bateson and Reginald Punnett
conducted a cross in sweet pea involving two
different traits


Flower color and pollen shape
This is a dihybrid cross that is expected to yield
a 9:3:3:1 phenotypic ratio in the F2 generation

However, Bateson and Punnett obtained surprising
results
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Figure 5.2
A much greater proportion
of the two types found in
the parental generation
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Morgan Provided Evidence for the
Linkage of Several X-linked Genes
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The first direct evidence of linkage came from
studies of Thomas Hunt Morgan
Morgan investigated several traits that followed
an X-linked pattern of inheritance in?
You guessed it
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5-11
His experiment involved three traits: Body color, Eye color,
Wing length
x
y+ w+ m+ Y
yy ww mm
F1 generation contains wild-type
females and yellow-bodied,
white-eyed, miniature-winged
males.
F1 generation
x
y+y w+w m+m
ywmY
P Males
P Females
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Morgan observed a much higher proportion of the
combinations of traits found in the parental generation
Morgan’s explanation:
 All three genes are located on the X chromosome
 Therefore, they tend to be transmitted together as a
unit
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Morgan Provided Evidence for the
Linkage of Several X-linked Genes

1. Why did the F2 generation have a
significant number of nonparental
combinations?

2. Why was there a quantitative difference
between the various nonparental
combinations?
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Let’s reorganize Morgan’s data by considering the pairs of
genes separately

Gray body, red eyes
1,159
Yellow body, white eyes
1,017
Gray body, white eyes
Yellow body, red eyes
Total
17
12
2,205
Red eyes, normal wings
770
White eyes, miniature wings
716
Red eyes, miniature wings
White eyes, normal wings
Total
401
318
2,205
But this nonparental
combination was rare
It was fairly common
to get this nonparental
combination
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
Morgan made three important
hypotheses to explain his results

1. The genes for body color, eye color and
wing length are all located on the Xchromosome
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2. Due to crossing over, the homologous X
chromosomes (in the female) can exchange
pieces of chromosomes

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They tend to be inherited together
This created new combination of alleles
3. The likelihood of crossing over depends
on the distance between the two genes

Crossing over is more likely to occur between two
genes that are far apart from each other
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Figure 5.4
These parental phenotypes are
the most common offspring
These recombinant offspring
are not uncommon
because the genes are far apart
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Figure 5.4
These recombinant offspring
are fairly uncommon
because the genes are very close together
These recombinant offspring
are very unlikely
1 out of 2,205
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Creighton and McClintock Experiment
They demonstrated physical evidence of cross-overs.
C
wx
Normal
chromosome 9
Parental
chromosomes
c
Abnormal
chromosome 9
Knob
Interchanged
piece from
chromosome 8
(a) Normal and abnormal chromosome 9
C = Colored
c = colorless
Wx = Starchy endosperm
wx = waxy endosperm

Wx
Crossing over
c
wx
Nonparental
chromosomes
C
Wx
(b) Crossing over between normal and
abnormal chromosome 9
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
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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
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Genetic linkage map

Based on recombination data to
determine the relative position of
genes on the chromosome
Physical Maps


Use nucleotide sequences to map genes
Will talk about this later

Experimentally, the percentage of recombinant
offspring is correlated with the distance between
the two genes

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If the genes are far apart  many recombinant
offspring
If the genes are close  very few 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
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Chromosomes are
the product of a
crossover during
meiosis in the
heterozygous parent
Recombinant
offspring are fewer
in number than
nonrecombinant
offspring
Figure 5.9
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
The data at the bottom of Figure 5.9 can be used to
estimate the distance between the two genes
Number of recombinant offspring X 100
Map distance =
Total number of offspring
=
76 + 75
542 + 537 + 76 + 75
X 100
= 12.3 map units
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Alfred Sturtevant’s Experiment

The first genetic map was constructed in 1911 by
Alfred Sturtevant
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He was an undergraduate who spent time in the
laboratory of Thomas Hunt Morgan
Sturtevant wrote:
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“In conversation with Morgan … I suddenly realized that
the variations in the length of linkage, already attributed
by Morgan to differences in the spatial orientation of the
genes, offered the possibility of determining sequences
[of different genes] in the linear dimension of the
chromosome. I went home and spent most of the night
(to the neglect of my undergraduate homework) in
producing the first chromosome map, which included the
sex-linked genes, y, w, v, m, and r, in the order and
approximately the relative spacing that they still appear
on the standard maps.”
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Figure 5.10
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The Data
Alleles
Concerned
Number Recombinant/ Percent Recombinant
Total Number
Offspring
y and w/w-e
214/21,736
1.0
y and v
1,464/4,551
32.2
y and r
115/324
35.5
y and m
260/693
37.5
w/w-e and v
471/1,584
29.7
w/w-e and r
2,062/6,116
33.7
w/w-e and m
406/898
45.2
v and r
17/573
3.0
v and m
109/405
26.9
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Interpreting the Data
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In some dihybrid crosses, the percentage of
nonparental (recombinant) offspring was rather low
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For example, there’s only 1% recombinant offspring in
the crosses involving the y and w or w-e alleles
This suggests that these two genes are very close
together
Other dihybrid crosses showed a higher
percentage of nonparental offspring


For example, crosses between the v and m alleles
produced 26.9% recombinant offspring
This suggests that these two genes are farther apart
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
Sturtevant assumed that the map distances
would be more accurate among genes that are
closely linked.
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Therefore, his map is based on the following distances
 y – w (1.0), w – v (29.7), v – r (3.0) and v – m (26.9)
Sturtevant also considered map distances amongst
gene pairs to deduce the order of genes
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Percentage of crossovers between w and r was 33.7
Percentage of crossovers between w and v was 29.7
Percentage of crossovers between v and r was 3.0
Therefore, the gene order is w – v – r
 Where v is closer to r than it is to w
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
Sturtevant collectively considered all these data
and proposed the following genetic map

Sturtevant began at the y gene and mapped the
genes from left to right
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5-56

A close look at Sturtevant’s data reveals two points
that do not agree very well with his genetic map


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The y and m dihybrid cross yielded 37.5% recombinants
 But the map distance is 57.6
The w and m dihybrid cross yielded 45.2% recombinants
 But the map distance is 56.6
So what’s up?

As the percentage of recombinant offspring
approaches a value of 50 %
 This value becomes a progressively more
inaccurate measure of map distance
 Refer to Figure 5.11
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5-57
Figure 5.11

When the distance between two genes is large
 The likelihood of multiple crossovers increases
 This causes the observed number of recombinant
offspring to underestimate the distance between the
two genes
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Figure 5-12a
Copyright © 2006 Pearson Prentice Hall, Inc.
Chi Square Analysis

This method is frequently used to determine
if the outcome of a dihybrid cross is
consistent with linkage or independent
assortment
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5-20
Trihybrid Crosses


Data from trihybrid crosses can also yield information
about map distance and gene order
The following experiment outlines a common strategy for
using trihybrid crosses to map genes
 In this example, we will consider fruit flies that differ in
body color, eye color and wing shape
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
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
b = black body color
b+ = gray body color
pr = purple eye color
pr+ = red eye color
vg = vestigial wings
vg+ = normal wings
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Trihybrid Crosses
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
Step 1: Cross two true-breeding strains that differ
with regard to three alleles.
Order of genes
not important
here.
Female is mutant
for all three traits

Male is homozygous
wildtype for all three
traits
The goal in this step is to obtain aF1 individuals that
are heterozygous for all three genes
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
Step 2: Perform a testcross by mating F1 female
heterozygotes to male flies that are homozygous
recessive for all three alleles

During gametogenesis in the heterozygous female F1
flies, crossovers may produce new combinations of
the 3 alleles
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
Step 3: Collect data for the F2 generation
Phenotype
Number of Observed Offspring
Gray body, red eyes, normal wings
+++
411
Gray body, red eyes, vestigial wings
+ + vg
61
Gray body, purple eyes, normal wings
+ pr +
2
Gray body, purple eyes, vestigial wings
+ pr vg
30
b/pr
Black body, red eyes, normal wings
b++
28
b/pr
Black body, red eyes, vestigial wings
B + vg
1
Black body, purple eyes, normal wings
B pr +
60
Black body, purple eyes, vestigial wings
412
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B pr vg
parental
pr/vg
b/pr and pr/vg
b/pr and pr/vg
pr/vg
parental
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
The three genes exist as two alleles each
Therefore, there are 23 = 8 possible combinations
of F2 offspring
If the genes assorted independently, all eight
combinations would occur in equal proportions
It is obvious that they are far from equal

In the offspring of crosses involving linked genes,
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Parental phenotypes occur most frequently
Double crossover phenotypes occur least frequently
Single crossover phenotypes occur with
“intermediate” frequency
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
The combination of traits in the double crossover tells us
which gene is in the middle

A double crossover separates the gene in the middle from the
other two genes at either end
Notice pr is no longer with the b and vg
And pr+ is not with the b+ and vg+.

In the double crossover categories, the recessive purple
eye color is separated from the other two recessive
alleles
 Thus, the gene for eye color lies between the genes
for body color and wing shape
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Which are the double cross-overs?

The ones with the least amount.
Step 4: Calculate the map distance between pairs of
genes
 Number of recombs between pr and vg: 61 + 60+ 2 +
1 = 124
 Number of recombs between b and pr: 30 + 28 + 2 +
1 = 61
 Number of recombs between b and vg, all but double
cross-overs: 61 + 60 + 30 + 28 = 178
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Map Distance



pr/vg = 124/1005 x 100 = 12.3
b/pr = 61/1005 x 100 = 6
b and vg = 179/1005 x 100 = 17.8
_____6____________12.3____________
b
pr
vg
The distance between b and vg was found to
be 17.8. The actual distance is 18.3 mu.
Interference


The slightly smaller lower value was a small
underestimate because we did not consider
the double crossovers in the calculation
between b and vg.
The lower than expected value is due to a
common genetic phenomenon, termed
positive interference. The first
crossover decreases the probability
that a second crossover will occur
nearby.
GENETIC MAPPING IN HAPLOID
EUKARYOTES

Much of our earliest understanding of genetic
recombination came from the genetic analyses of fungi

Fungi may be unicellular or multicellular organisms
They are typically haploid (1n)
They reproduce asexually and, in many cases,
sexually



The sac fungi (ascomycetes) have been particularly
useful to geneticists because of their unique style of
sexual reproduction
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Meiosis produces
four haploid cells,
termed spores
These are
enclosed in a sac
termed an ascus
Figure 5.12
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

The cells of a tetrad or octad are contained
within a sac
In other words, the products of a single
meiotic division are contained within one sac
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Types of Tetrads or Octads

The arrangement of spores within an ascus
varies from species to species

Unordered tetrads or octads


Ascus provides enough space for the spores to
randomly mix together
Ordered tetrads or octads

Ascus is very tight, thereby preventing spores from
randomly moving around
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Ascus provides
space for spores to
randomly mix
together
Tight ascus
prevents mixing
of spores
Mold
Yeast
Figure 5.13
Unicellular alga
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Ordered Tetrad Analysis

Ordered tetrads or octads have the following
key feature



The position and order of spores within the ascus is
determined by the divisions of meiosis and mitosis
In crosses of tan and black Neurospora cultures, the
spores appear tan or black in a certain order.
All black spores or all tan spores indicate no
hybridization.
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Pairs of daughter
cells are located
next to each other
Figure 5.13
All eight cells are
arranged in a linear,
ordered fashion
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020
No hybridization
Non-crossovers
Cross-overs
Non cross-overs
Non cross-overs
Cross-overs

To calculate this distance, the experimenter must count
the number of cross-over asci, as well as the total
number of asci
 In cross-over asci, only half of the spores are actually
the product of a crossover
 Therefore
Map distance = (1/2) (Number of cross-over asci) X 100
Total number of asci
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Linkage on human chromosomes


Hampered by the inability to perform
desired crosses and small number of
progeny in most human families
Geneticists mostly use pedigrees
which are often incomplete.
Nail-Patella Syndrome


One of the first documented
demonstrations of linkage in humans
Linkage between this syndrome and
ABO blood types
Pedigree of Nail-Patella Syndrome

More on physical maps later……