Crossing Over of John Edward’s Chromosomes
Linkage & Mapping
Linkage



Eukaryotic chromosomes are long ds DNA
molecules
Typical chromosome contains thousands of
genes (loci)
Linkage


loci located on the same chromosome
linked loci tend to be transmitted as a unit
Linkage

Because they are a group of genes linked together,
chromosomes are functionally linkage groups

# of linkage groups = the # of types of chromosomes

Crossing over causes loci that are far apart on the
same chromosome to sometimes independently
assort

known as incomplete linkage
Crossing Over Produce Recombinant
Phenotypes


Crossing over (meiotic recombination)
Occurs during prophase I of meiosis at the
bivalent stage


zygotene - pachytene - diplotene
Non-sister chromatids of homologous
chromosomes exchange DNA segments
Linkage Prevents
Independent Assortment
Figure 5.1
Crossing Over May Produce
Recombinant Phenotypes
These are
termed
parental
gametes
Figure 5.1
gametes with a combination
of alleles NOT found in the
original chromosomes as a
result of meiotic
recombination
These are called nonparental
or recombinant gametes
Example of Linkage

Bateson and Punnett conducted a cross in sweet
pea involving two traits


Flower color and pollen shape
Dihybrid cross expected to give 9:3:3:1 phenotypic
ratio of F2 phenotypes


Observed linkage
Called it coupling
Example of Linkage
x
Purple flowers,
long pollen (PPLL)
Red flowers,
round pollen (ppll)
F1 offspring
Purple flowers,
long pollen (PpLl)
Self-fertilization
F2 offspring phenotypes
Observed Observed Expected Expected
Ratio
Ratio
number
number
(9:3:3:1)
P
NP
NP
P
Purple flowers, long pollen
Purple flowers, round pollen
Red flowers, long pollen
Red flowers, round pollen
296
19
27
85
15.6
1.0
1.4
4.5
240
80
80
27
9
3
3
1
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



Body color
Eye color
Wing length
Linkage to a
Particular
Chromosome
x
/
y+ w+ m+ Y
y w m/y w m
F1 generation
Sex linkage of all traits
places them all on X
chromosome
x
y+w+ m+/ y w m
F2 generation
P
P
F1 generation contains wild-type
females and yellow-bodied,
white-eyed, miniature-winged
males.
ywm/Y
Females
Tan body, red eyes, normal wings
Tan body, red eyes, miniature wings
Tan body, white eyes, normal wings
Tan body, white eyes, miniature wings
Yellow body, red eyes, normal wings
Yellow body, red eyes, miniature wings
Yellow body, white eyes, normal wings
Yellow body, white eyes, miniature wings
439
208
1
5
7
0
178
365
Males
Total
319
193
0
11
5
0
139
335
758
401
1
16
12
0
317
700
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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5-13
Morgan Provided Evidence for the
Linkage of Several X-linked Genes

However, Morgan still had to interpret two
key observations

Why did the F2 generation have a significant
number of nonparental combinations?

Why was there a quantitative difference between
the various nonparental combinations?
Reorganize Morgan’s data considering 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
Figure 5.4
These parental phenotypes are
the most common offspring
These recombinant offspring
are not uncommon
because the genes are far apart
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
it is product of single cross over probabilities
Incomplete Linkage

Linked loci that are sometimes separated by
recombination are inherited in a pattern between
linked and independently assorting



not always together
not present in normal dihybrid ratios
The percentage of offspring with the loci linked vs
those with the loci separated is a measure of the
physical distance separating the loci on the
chromosome
Alfred Sturtevant’s Analysis



An undergraduate in the laboratory of T. H. Morgan
Constructed first genetic map in 1911
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.”
Linkage and Genetic Maps

Estimating the relative distances between linked
genes, based on the amount of recombination
occuring between them allows us to generate
genetic maps





If the genes are far apart  many recombinant offspring
If the genes are close  very few recombinant offspring
Number of recombinant offspring X 100
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
Map distance =
Linkage Analysis and Mapping

Genetic mapping experiments are typically
accomplished by carrying out a testcross

Example of a two-point mapping cross

Cross of two linked genes affecting bristle length
and body color in fruit flies




s = short bristles
+ = normal bristles


e = ebony body color
+ = gray body color
One parent double recessive (homozygous recessive at
both loci) – s/s ; e/e
Other parent is heterozygous at both loci (+/s ; +/e)
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
5-47
Linkage Analysis and Mapping


The phenotype data are used to estimate the distance
between the two loci
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
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+ = grey body color
pr = purple eye color
pr+ = red eye color
vg = vestigial wings
vg+ = normal wings
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5-59

Step 1: Cross two true-breeding strains that differ at
three loci.
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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5-60

Step 2: Perform a testcross by mating F1 female
heterozygotes to homozygous recessive, male flies

During gametogenesis in the heterozygous female F1 flies,
crossovers may produce new combinations of the 3 alleles
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5-61

Step 3: Collect data for the F2 generation
Number of Observed Offspring
(males and females)
Phenotype
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
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5-62

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
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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5-63

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

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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5-64

Step 4: Calculate the map distance between pairs of
genes
 To do this, one strategy is to group the data
according to pairs of phenotypes resulting from
non-crossovers, single crossovers, & double
crossovers
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5-65
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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5-72

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

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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5-69

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
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5-73
Interference

The product rule allows us to predict the likelihood of a
double crossover from the individual probabilities of each
single crossover
P (double crossover) = P (single crossover X P (single crossover
between b and pr)
between pr and vg)
= 0.061 X 0.123 = 0.0075

Based on a total of 1,005 offspring the expected number of
double crossover offspring is
= 1,005 X 0.0075 = 7.5
Interference

Therefore, we would expect seven or eight offspring to be
produced as a result of a double crossover

However, the observed number was only three!



Two with gray bodies, purple eyes, and normal sings
One with black body, red eyes, and vestigial wings
This 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
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5-75


Interference (I) is expressed as
 I = 1 – C
 where C is the coefficient of coincidence

Observed number of double crossovers
C=
Expected number of double crossovers

C=
3
7.5
= 0.40
I = 1 – C = 1 – 0.4
= 0.6 or 60%
 This means that 60% of the expected number of
crossovers did not occur

Since I is positive, this interference is positive interference

Rarely, the outcome of a testcross yields a negative value
for interference
 This suggests that a first crossover enhances the rate of
a second crossover

The molecular mechanisms that cause interference are not
completely understood
 However, most organisms regulate the number of
crossovers so that very few occur per chromosome
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5-77
Somatic Cell Hybridization


Fusion of human cell and rodent cell
Eventually, most human sequences lost


Lines carrying sub-sets of human chromosomes are
compared to determine which express the given gene
product



entire chromosomes, portions (arms) of chromosomes
have to have a detectible DNA sequence or protein
Process of elimination determines on which chromosome
(region of chromosome) a gene is located
Hybrid Panel
Other Mapping Techniques

Cytogenetic Mapping



Have a mutant phenotype
Perform karyotype
Observe altered chromosomes




aneuploidies
translocations
deletions
normal
Aberrant chromosome probably
location of gene
translocation
t(14;17)
Patau Syndrome - Trisomy 13 Karyotype: 47, 13+
Partial Chromosomal Deletions

cri-du-chat - 46, 5pFragile X mental retardation – 46, Xp-

Monosomy




loss of an entire chromosome
lethal in all cases for autosomes
XO – Turner’s syndrome is only viable
monosomy
Mapping Strategies
Deletion Mapping


Cross recessive heterozygote to deletion
heterozygote lines
appearance of recessive phenotype
localizes recessive gene within deletion
interval
Complementation

Determines if two phenotypes are
caused by same mutant gene
Complementation Analysis
Complementation Analysis
Complementation Analysis of Eye Color Mutations

Complementation groups

white, cherry, coral, apricot, buff





W1, wch , wc , wa , wb Alleles of white gene
garnet
ruby
vermillion
carnation
Complementation Analysis
Mapping with Unknown Phase of Linkage
When you have heterozygotes, but do not know what the
phenotypes of the parents were.
12 possible diploid arrangements for 3 linked loci
bm
+
pr
+
+
+
bm
pr
bm
+
bm
brown midrib, virescent seedling, purple aleurone
pr
Use Results of Cross to Determine Phase of Linkage
NCOs = parental
chromosomes = phase
DCOs = fewest
Do not directly
detect middle locus
NCO Gives 3 Possibilities for Phase
12 possible diploid arrangements for 3 linked loci
bm
+
pr
+
+
+
bm
pr
bm
+
bm
pr
DCO Gives Loci Order
DCO Defines 1 of these 3 Possibilities for Phase
Because a DCO was required to group all mutant loci on same homologue
12 possible diploid arrangements for 3 linked loci
bm
+
pr
+
+
+
bm
pr
bm
+
bm
pr