CHAPTER 5
LINKAGE AND
GENETIC MAPPING
IN EUKARYOTES
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INTRODUCTION

Each species of organism must contain
hundreds to thousands of genes


Yet most species have at most a few dozen
chromosomes
Therefore, each chromosome is likely to
carry many hundred or even thousands of
different genes

The transmission of such genes will violate
Mendel’s law of independent assortment
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5.1 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

Linkage influences inheritance patterns
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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
Genes that are far apart on the same chromosome
may independently assort from each other

This is due to crossing-over


A dihybrid cross studies crossing-over between two genes
A trihybrid cross studies crossing-over between three genes
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Crossing Over May Produce
Recombinant Phenotypes

In diploid eukaryotic species, linkage can be altered
during meiosis as a result of crossing over

Crossing over

Occurs during prophase I of meiosis



replicated sister chromatid homologues associate as bivalents
Non-sister chromatids of homologous chromosomes
exchange DNA segments
Figure 5.1 illustrates the consequences of crossing
over during meiosis
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The haploid cells contain
the same combination of
alleles as the original
chromosomes
The arrangement of linked
alleles has not been altered
Figure 5.1
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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

In 1905, William Bateson and Reginald Punnett
conducted a cross in sweet pea involving two
different traits


This is a dihybrid cross that is expected to yield a
9:3:3:1 phenotypic ratio in the F2 generation


Flower color and pollen shape
However, Bateson and Punnett obtained surprising
results
Refer to Figure 5.2
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Figure 5.2
A much greater proportion
of the two types found in
the parental generation
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Bateson and Punnett Discovered Two
Traits That Did Not Assort Independently

They suggested that the transmission of the two
traits from the parents was somehow coupled


The two traits are not easily assorted in an independent
manner
However, they did not realize that the coupling is
due to the linkage of the two genes on the same
chromosome
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Morgan Provided Evidence for the
Linkage of Several X-linked Genes



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
Figure 5.3 illustrates an experiment involving three
traits



Body color
Eye color
Wing length
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Figure 5.3
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P Males
P Females


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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

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  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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
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
Figure 5.9 provides an example of a testcross

This cross concerns two linked genes affecting bristle length and
body color in fruit flies


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
e = ebony body color
e+ = gray body color
One parent displays both recessive traits


s = short bristles
s+ = normal bristles
It is homozygous recessive for the two genes (ss ee)
The other parent is heterozygous for the two genes


The s and e alleles are linked on one chromosome
The s+ and e+ alleles are linked on the homologous chromosome
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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

Therefore, the s and e genes are 12.3 map units apart
from each other along the same chromosome
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Alfred Sturtevant’s Experiment

The first genetic map was constructed in 1911 by
Alfred Sturtevant


He was an undergraduate who spent time in the
laboratory of Thomas Hunt Morgan
Sturtevant wrote:

“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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
Sturtevant considered the outcome of crosses
involving six different mutant alleles

All of which are known to be recessive and X-linked
Mutant allele
(Phenotype)
Wild-type allele
(Phenotype)
y (yellow body color)
y+ (gray body color)
w (white eye color)
w+ (red eye color)
w-e (eosin eye color)
w-e+ (red eye color)
Alleles of the
same gene
Alleles of different
genes that affect
eye color
v (vermillion eye color) v+ (red eye color)
m (miniature wings)
m+ (normal wings)
r (rudimentary wings)
r+ (normal wings)

Alleles of different
genes that affect
wing length
Therefore, his genetic map contained only 5 genes

y, w, v, m and r
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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
Testing the Hypothesis

Refer to Figure 5.10
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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

In some dihybrid crosses, the percentage of
nonparental (recombinant) offspring was rather low



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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
To construct his map, Sturtevant assumed that the
map distances would be more accurate among
genes that are closely linked


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 other features of the
data to deduce the order of genes


For example,
 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
Do not have a JPG for this

Sturtevant began at the y gene and mapped the
genes from left to right
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
A close look at Sturtevant’s data reveals two points
that do not agree very well with his genetic map



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% The likelihood of multiple crossovers increases



Even numbers of crossover won’t be seen as recombination
So the observed recombinations tend to underestimate
the actual measure of map distance
Refer to Figure 5.11
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5-57
Figure 5.11

Multiple crossovers set a quantitative limit on measurable
recombination frequencies as the physical distance increases
 A testcross is expected to yield a maximum of only 50%
recombinant offspring
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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






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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
Step 1: Cross two true-breeding strains that differ
with regard to three alleles.
Do not have a JPG for this
Female is mutant
for all three traits

Male is homozygous
wildtype for all three
traits
The goal in this step is to obtain F1 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
Do not have a JPG for this

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
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
Analysis of the F2 generation flies will allow us to
map the three genes
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
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The three genes exist as two alleles each
Therefore, there are 23 = 8 possible combinations of
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,



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
Do not have a JPG for this

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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
Step 4: Calculate the map distance between pairs of
genes
 To do this, one strategy is to regroup the data
according to pairs of genes


From the parental generation, we know that the
dominant alleles are linked, as are the recessive alleles
This allows us to group pairs of genes into parental and
nonparental combinations



Parentals have a pair of dominant or a pair of recessive alleles
Nonparentals have one dominant and one recessive allele
The regrouped data will allow us to calculate the map
distance between the two genes
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Parental offspring
Gray body, red eyes
(411 + 61)
Black body, purple eyes
(412 + 60)
Total
Nonparental Offspring
Total
472
Gray body, purple eyes
(30 + 2)
32
472
Black body, red eyes
(28 + 1)
29
944

61
The map distance between body color and eye color is
Map distance =
61
944 + 61
X 100 = 6.1 map units
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Parental offspring
Gray body, normal wings
(411 + 2)
Black body, vestigial wings
(412 + 1)
Total
Nonparental Offspring
Total
413
Gray body, vestigial wings
(30 + 61)
91
413
Black body, normal wings
(28 + 60)
88
826

179
The map distance between body color and wing shape is
Map distance =
179
826 + 179
X 100 = 17.8 map units
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Parental offspring
Red eyes, normal wings
(411 + 28)
Purple eyes, vestigial wings
(412 + 30)
Total
Nonparental Offspring
Total
439
Red eyes, vestigial wings
(61 + 1)
62
442
Purple eyes, normal wings
(60 + 2)
62
881

124
The map distance between eye color and wing shape is
Map distance =
124
881 + 124
X 100 = 12.3 map units
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
Step 5: Construct the map

Based on the map unit calculation the body color and
wing shape genes are farthest apart
 The eye color gene is in the middle
Do not have a JPG for this

The data is also consistent with the map being drawn
as vg – pr – b (from left to right)

In detailed genetic maps, the locations of genes are
mapped relative to the centromere
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
To calculate map distance, we have gone
through a method that involved the separation of
data into pairs of genes (see step 4)

An alternative method does not require this
manipulation
 Rather, the trihybrid data is used directly

This method is described next
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Phenotype
Number of
Observed
Offspring
Gray body,
purple eyes,
vestigial wings
30
Black body,
red eyes,
normal wings
28
Gray body,
red eyes,
vestigial wings
61
Black body,
purple eyes,
normal wings
60
Gray body,
purple eyes,
normal wings
2
Black body,
red eyes,
vestigial wings
1
Single crossover
between b and pr
30 + 28
= 0.058
1,005
Single crossover
between pr and vg
61 + 60
Double crossover,
between b and pr,
and between
pr and vg
1+2
= 0.120
1,005
= 0.003
1,005
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
To determine the map distance between the genes, we
need to consider both single and double crossovers

To calculate the distance between b and pr
 Map distance = (0.058 + 0.003) X 100 = 6.1 mu

To calculate the distance between pr and vg
 Map distance = (0.120 + 0.003) X 100 = 12.3 mu

To calculate the distance between b and vg


The double crossover frequency needs to be multiplied by two
 Because both crossovers are occurring between b and vg
Map distance = (0.058 + 0.120 + 2[0.003]) X 100
= 18.4 mu
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
Alternatively, the distance between b and vg can be
obtained by simply adding the map distances between
b and pr, and between pr and vg
 Map distance = 6.1 + 12.3 = 18.4 mu

Note that in the first method (grouping in pairs), the
distance between b and vg was found to be 17.8 mu.
 This slightly lower value was a small underestimate
because the first method does not consider the double
crossovers in the calculation
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