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BIO 184
Fall 2006
LECTURE 12
Lecture 12:
Linkage and Genetic Mapping in Eukaryotes
Map of human chromosome 21 indicating the relative locations of known disease-causing genes.
These genes were originally mapped using genetic mapping (linkage mapping) as described in this
set of lecture notes.
http://www.ornl.gov/sci/techresources/Human_Genome/posters/chromosome/chromo21.shtml
The genomes of eukaryotic organisms contain hundreds to thousands of genes (an
estimated 30,000-50,000 in humans). Yet there are only a handful of
chromosomes. Thus, each chromosome in a eukaryotic genome must contain a large
number of genes.
The transmission of genes located on the same chromosome may violate Mendel’s
Law of Independent Assortment, particularly if they are located very close
together along the same arm of a chromosome.
This set of lecture notes will explain why, and provide the theoretical basis for
mapping genes along a chromosome by following the degree to which they violate
Mendel’s Law of Independent Assortment during genetic crosses.
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LECTURE 12
I. 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
o 1. Two or more genes can be located on the same chromosome
o 2. Genes that are close together tend to be transmitted as a unit
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
o 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 due to crossing-over during meiosis.
 Occurs during prophase I of meiosis
 Homologous chromosomes exchange DNA segments
See Figure 5.1, Brooker
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
See Figure 5.2, Brooker
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
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Fall 2006
LECTURE 12
However, they did not realize that the coupling is due to the linkage of the
two genes on the same chromosome
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, Brooker, illustrates an experiment involving three traits
o Body color
o Eye color
o Wing length
However, Morgan still had to interpret two key observations:
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?
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
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But this nonparental
combination was rare
It was fairly common
to get this nonparental
combination
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BIO 184
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LECTURE 12
To explain these data, Morgan considered the previous studies of the cytologist
F.A Janssens


Janssens had observed chiasmata microscopically and proposed that crossing
over involves a physical exchange between homologous chromosomes
Morgan shrewdly realized that crossing over between homologous X
chromosomes was consistent with his data and that:
o Crossing over did not occur between the X and Y chromosome
o The three genes were not found on the Y chromosome
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 X-chromosome. Thus, they tend to be inherited together.
2. Due to crossing over, the homologous X chromosomes (in the
female) can exchange pieces of chromosomes
 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
See Figure 5.4, Brooker
II. 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, Brooker, illustrates a simplified genetic linkage map of Drosophila
melanogaster
Genetic maps are useful in many ways:
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BIO 184
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LECTURE 12
1. They allow us to understand the overall complexity and genetic
organization of a particular species
2. They improve our understanding of the evolutionary relationships among
different species
3. They can be used to diagnose, and perhaps, someday to treat inherited
human diseases
4. They can help in predicting the likelihood that a couple will produce
children with certain inherited diseases
5. They provide helpful information for improving agriculturally important
strains through selective breeding programs
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
Map distance =
Number of recombinant offspring X 100
Total number of offspring
The units of distance are called map units (mu) or centiMorgans (cM).
One map unit is equivalent to 1% recombination frequency.
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
A. Dihybrid Test Cross Mapping

Figure 5.9, Brooker, provides an example of a testcross involving two genes
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LECTURE 12
o This cross concerns two linked genes affecting bristle length and body
color in fruit flies:






s = short bristles
s+ = normal bristles


e = ebony body color
e+ = gray body color
One parent displays both recessive traits – i.e. 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 same homologous chromosome
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
Total number of offspring
76 + 75
X 100
542 + 537 + 76 + 75
=
= 12.3 map units (cM)
Therefore, the s and e genes are 12.3 cM apart from each other along the same
chromosome.
B. Trihybrid Test Cross Mapping
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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

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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LECTURE 12
Step 1: Cross two true-breeding strains that differ with regard to three alleles.
Male is homozygous
wildtype for all three
traits
Female is mutant
for all three traits
 The goal in this step is to obtain F1 individuals that are
heterozygous for all three genes
Step 2: Perform a testcross by mating F1 female heterozygotes to male flies that
are homozygous recessive for all three alleles
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LECTURE 12
(Produced by cross above)
During gametogenesis in the heterozygous female F1 flies, crossovers may produce
new combinations of the 3 alleles.
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
61
Gray body, purple eyes, normal wings
2
Gray body, purple eyes, vestigial wings
30
Black body, red eyes, normal wings
28
Black body, red eyes, vestigial wings
1
Black body, purple eyes, normal wings
60
Black body, purple eyes, vestigial wings
412
Analysis of the F2 generation flies will allow us to map the three genes
 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
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LECTURE 12

It is obvious that they are far from equal

In the offspring of crosses involving linked genes:
o Parental phenotypes occur most frequently
o Double crossover phenotypes occur least frequently
o Single crossover phenotypes occur with “intermediate” frequency

The combination of traits in the double crossover tells us which gene is in
the middle
o A double crossover separates the gene in the middle from the other
two genes at either end

In the double crossover categories, the recessive purple eye color is
separated from the other two recessive alleles
o Thus, the gene for eye color lies between the genes for body color
and wing shape
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
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BIO 184
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Fall 2006
LECTURE 12
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
Parental offspring
Total Nonparental Offspring Total
Gray body, red eyes
(411 + 61)
472
Gray body, purple eyes
(30 + 2)
32
Black body, purple eyes
(412 + 60)
472
Black body, red eyes
(28 + 1)
944
61
29
The map distance between body color and eye color is therefore:
61
X 100 = 6.1 cM
944 + 61
Map distance =
Parental offspring
Total Nonparental Offspring Total
Gray body, normal wings
(411 + 2)
413
Gray body, vestigial
wings
(30 + 61)
91
Black body, vestigial
wings
(412 + 1)
Black body, normal
wings
(28 + 60)
88
826
413
179
And the map distance between body color and wing shape is:
Map distance =
179
X 100 = 17.8 cM
826 + 179
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Parental offspring
Fall 2006
LECTURE 12
Total Nonparental Offspring Total
Red eyes, normal wings
(411 + 28)
439
Red eyes, vestigial wings
(61 + 1)
62
Purple eyes, vestigial
wings
(412 + 30)
Purple eyes, normal
wings
(60 + 2)
442
881
62
124
And, finally, the map distance between eye color and wing shape is
Map distance =
124
881 + 124
X 100 = 12.3 cM
Step 5: Construct the map
Based on the map unit calculation, the body color and wing shape genes are
farthest apart and the eye color gene is in the middle:
?
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