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Gene Linkage and
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Genetic
Mapping
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CHAPTER ORGANIZATION
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4.1 SALE
LinkedOR
alleles
tend to stay
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together in meiosis.
112
4.4NOT
Polymorphic
DNA OR
sequences
are
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used in human genetic mapping. 128
The degree of linkage is measured by the
Single-nucleotide polymorphisms (SNPs)
frequency of recombination.
113
are abundant in the human genome.
The frequency of recombination is the same
SNPs in restriction sites yield restriction
for coupling and repulsion heterozygotes.
114
fragment length polymorphisms (RFLPs).
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The frequency
Simple-sequence
repeats
(SSRs) often
from one gene pair to the next.
114 FOR SALE
differ
in copy
number.
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Recombination does not occur in
Gene dosage can differ owing to copy115
number variation (CNV).
Drosophila males.
Copy-number variation has helped human
4.2 Recombination results from
populations adapt to a high-starch diet.
crossing-over between
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linked
alleles.
129
130
131
133
134
all
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116 LLC 4.5 Tetrads contain
four
products
of
meiosis.
135
Physical
often—but
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NOTdistance
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always—correlated with map distance.
One crossover can undo the
effects of another.
4.3 Double crossovers are revealed
in three-point
crosses.
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Interference
the chance of
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multiple crossing-over.
Varieties of maize.
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120
122
124
127
Unordered tetrads have no relation to the
geometry of meiosis.
136
Tetratype tetrads demonstrate that
crossing-over takes place at the fourstrand stage of meiosis and is reciprocal.
137
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analysis
affords a convenient
test for
linkage.
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The geometry of meiosis is revealed in
ordered tetrads.
139
Gene conversion suggests a molecular
mechanism of recombination.
142
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TXX.XX A Head Goes Here
4TH PAGES
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111
4.6 Recombination is initiated by a
double-stranded break in DNA.
Recombination
tends
to take place at
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preferred positions in the genome.
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the human connection
Starch Contrast
G
143
146
134
Chapter Summary
Learning Outcomes
& Ideas
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Solutions: Step by Step
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Concepts in Action: Problems for Solution
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GENETICS on the web
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tend to remain together in inheritance, a phenomenetic mapping means determining the relaenon known as linkage. Nevertheless, the linkage
tive positions of genes along a chromosome.
is incomplete. Some gametes are produced that have
It is one
of the main
experimental
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different ©combinations
of the white
and miniature
genetics. This may seem odd in organisms in which the
alleles than
those
in the
parental
chromosomes.
The
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DNA sequence of the genome has been determined. If
new combinations are produced because homologous
every gene in an organism is already sequenced, then
chromosomes can exchange segments when they are
what is the point of genetic mapping? The answer
paired. This process (crossing-over) results in recomis that a gene’s sequence does not always reveal its
bination of alleles between the homologous chro­function, nor
does a genomic
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mosomes.
The probabi­
lity of recombination
between
which genes interact in a complex biological process.
any two
genesOR
serves
as a measure of genetic distance
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When a new mutant gene is discovered, the first
between the genes and allows the construction of a
step in genetic ­analysis is usually genetic mapping to
genetic map, which is a diagram of a chromosome
determine its position in the genome. It is at this point
showing the relative positions of the genes. The linthat the genomic sequence, if known, becomes useful,
ear order of genes along a genetic map is consistent
because in some cases
the position
of the mutant
gene LLC
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with the conclusion that ©
each
gene &
occupies
a wellcoincides with a gene whose sequence suggests a role
defined position, or locus,
in the
chromosome,
with
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in the biological process being investigated. For examthe alleles of a gene in a heterozygote occupying
ple, in the case of flower color, a new mutation may
corresponding locations in the pair of homologous
map to a region containing a gene whose sequence
chromosomes.
suggests that it encodes an enzyme in anthocyanin
In discussing linked genes, it is necessary to distinsynthesis.
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is not always
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parenrevealed by its DNA sequence, and so in some cases,
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further genetic or molecular analysis is necessary to
(“/”). The alleles in one chromosome are depicted
sort out which one of the genes in a sequenced region
to the left of the slash, and those in the homologous
corresponds to a mutant gene mapped to that region.
chromosome are depicted to the right of the slash.
In human genetics, genetic mapping is important
For example, in the cross AA BB 3 aa bb, the genobecause it enables genes associated with hereditary
type&ofBartlett
the doubly
heterozygous
progeny is denoted
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diseases, such as those that predispose to breast canA B/aSALE
b because
the
A and B alleles were inherited
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cer, to be localized and correlated with the genomic
in one parental chromosome and the alleles a and b
sequence in the region.
were inherited in the other parental chromosome. In
this genotype the A and B alleles are said to be in the
coupling or cis ­configuration; likewise, the a and
4.1 Linked alleles tend to stay
b alleles are in coupling.©Among
four possible
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together in meiosis.
types of gametes, the A NOT
B andFOR
a b types
areOR
called
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parental combinations because the alleles are in the
In meiosis, homologous chromosomes form pairs in
same configuration as in the parental chromosomes,
prophase I by undergoing synapsis, and the individand the A b and a B types are called recombinants
ual members of each pair separate from one another
(­FIGURE 4.1, part A).
at anaphase I. Genes that are close enough together
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Another
possible&
configuration
of the A, a and
B, b
in©
theJones
same chromosome
therefore
be expected
allele pairs
is AFOR
b/a B.SALE
In this case
A and B alleles
toNOT
be transmitted
together.
Thomas Hunt Morgan
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NOT
OR the
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are said to be in the repulsion or trans configuration.
examined this issue using two genes present in the
Now the parental and recombinant gametic types are
X chromosome of Drosophila. One was a mutation
reversed (Figure 4.1, part B). The A b and a B types are
for white eyes, the other a mutation for miniature
the parental combinations, and the A B and a b types
wings. Morgan found that the white and miniature
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are the
recombinants.
alleles present
in each XLLC
chromosome of a female do
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CHAPTER 4 Gene Linkage and Genetic Mapping
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(A)–Parental alleles in coupling
(A)–or cis configuration
The degree of linkage is
measured by the frequency
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b
of recombination.
(B)–Parental alleles in repulsion
(B)–or trans configuration
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a
b
A
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a
Meiosis
Meiosis
Parental combinations
A
B
a
b
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Parental combinations
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A
b
a
B
Recombinants
Recombinants
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b
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a
B
a
B
b
In his early experiments with Drosophila,
Morgan found mutations in each of several
X-linked genes that provided ideal materials for studying linkage. One of these genes,
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with alleles w1 and
w, determines
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red eye color versus
whiteFOR
eyes; SALE
another such
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gene, with the alleles m1 and m, determines
whether the size of the wings is normal
or miniature. The initial cross is shown as
Cross 1 in FIGURE 4.2. It was a cross between
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females
with&white
eyes and
normal wings
and
males
withSALE
red eyesOR
and miniature
wings:
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wwmm+ +
ww+ +mm
×
×
//
??
YY
wwmm+ +
FIGURE 4.1 For any pair of alleles, the gametes produced through
meiosis have the alleles either in a parental configuration or in a recombinant
configuration. Which types are parental and which recombinant depends
In thisLearning,
way of writing
the genotypes, the
© Jones &
Bartlett
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© Jones
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on whether
the configuration
the alleles in the parent is (A) coupling
or
horizontal
line
replaces
the slash. Alleles
repulsion.
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(A)
Parents:
Parents:
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Miniature-wing
White-eyed and
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White-eyed
females
w+
w+
(B)
Cross 1
males
+m
Y
miniature-wing
females
wm
wm
F©
1:
Cross 2
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Wildtype
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males
++
Y
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+
wm
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Mutant alleles
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w m OR DISTRIBUTION
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w m Mutant alleles
in homologous
chromosomes
in the same
chromosomes
F2:
F1:
Y
+m
F2:
Progeny 1
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White
eyes,
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normal wings
(maternal gamete: w +)
++
Progeny 2
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Parental types:NOT FOR
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types:
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66.5% have
parental allele
combinations
(nonrecombinant).
202
Red eyes,
miniature wings
(maternal gamete:
+ m) & Bartlett
© Jones
37.7% have
nonparental allele
combinations
(recombinant).
Learning, LLC
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OR DISTRIBUTION
types:
Parental types:
Red eyes,
normal wings
(maternal gamete: + +)
114
White eyes,
102
Y
33.5% have
nonparental allele
combinations
(recombinant).
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miniature
wings
644
(maternal gamete: w m)
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62.3% have
parental allele
combinations
(nonrecombinant).
247
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395
382
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FIGURE 4.2 An experiment demonstrating that the frequency of recombination between two mutant alleles is independent of whether
they are present in the same chromosome or in homologous chromosomes. (A) Cross 1 produces F1 females with the genotype w 1/1 m,
and the w 2 m recombination frequency is 33.5 percent. (B) Cross 2 produces F1 females with the genotype w m/1 1, and the w 2 m
&
Bartlettfrequency
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recombination
is 37.7LLC
percent. These values are within the
of variation
expectedLearning,
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4.1 Linked Alleles Tend to Stay Together in Meiosis
4TH PAGES
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113
written above the line are present in one chromoKEY CONCEPT
some, and those written below the line are present in
Genes with recombination frequencies smaller than
homologous
chromosome.
Bartlett
Learning,
© Jones & the
Bartlett
Learning,
LLC In the females, both ©X Jones50&percent
are present
in the LLC
same c­ hromosome
1
chromosomes
carry
w
and
m
.
In
males,
the
X
chroNOT
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(linked).
Two
genes
that
undergo
independent
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mosome carries the alleles w1 and m. (The Y written
assortment, indicated by a recombination f­requency
below the line denotes the Y chromosome in the male.)
equal to 50 percent, either are in nonhomologous
Figure 4.2 illustrates a simplified symbolism, commonly
chromosomes or are located far apart in a single
used in Drosophila genetics, in which a wildtype allele
chromosome.
is denoted by a 1 sign
in the appropriate
position.
The LLC
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© Jones & Bartlett Learning, LL
1 symbolism is unambiguous
because
linked
genes
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NOT FOR
SALEtheOR
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in a chromosome are always written in the same order.
The frequency of recombination is
Using the 1 notation,
the same for coupling and
repulsion heterozygotes.
w+
w m+
means
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+
w m+
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and
1 m
Y
means
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Morgan also
studied&progeny
from
the coupling
configurationNOT
of theFOR
w1 and
m1 alleles,
which results from
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the mating designated as Cross 2 in Figure 4.2. In this
case, the original parents had the genotypes
w+ m
Y
ww m
m
××
1
11
1
Y
wm
m //
Y ??
© Jones & Bartlett
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The resulting
F1 female progeny
genotype
1/1 m (or, equivalently, w m1/w1 m). NOT
In
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The resulting F1 female progeny from Cross 2 have the
this genotype, the w1 and m1 alleles are in repulsion.
genotype
w m/1 1 (equivalently, w m/w1 m1). In this
When these females were mated with w m/Y males,
case the wildtype alleles are in the same chromosome.
the offspring denoted as Progeny 1 in Figure 4.2 were
When these F1 female progeny were crossed with
obtained. In each class of progeny, the gamete from
w m/Y males, they yielded©the
types of
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the female parent is shown in the column at the left,
lated
as
Progeny
2
in
Figure
4.2.
and the gamete from
theFOR
maleSALE
parent OR
carries
either
NOT FOR SALE OR DISTRIBUT
NOT
DISTRIBUTION
Because the alleles in Cross 2 are in the coupling
w m or the Y chromosome. The cross is equivalent
configuration, the parental-type gametes carry either
to a testcross, and so the phenotype of each class of
w
m or 1 1, and the recombinant gametes carry
­progeny reveals the alleles present in the gamete from
either
w 1 or 1 m. The types of gametes are the same
the mother.
as
those
1, but
the parental
and
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©observed
Jonesin& Cross
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results of Cross 1 show a great departure from
recombinant
types
are
opposite.
Yet
the
frequency
of
theNOT
1 : 1 FOR
: 1 : 1 SALE
ratio of OR
the four
male phenotypes that
DISTRIBUTION
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recombination
is
approximately
the
same:
37.7
percent
is expected with independent assortment. If genes
versus 33.5 percent. The difference is within the range
in the same chromosome tended to remain together
expected to result from random variation from experiin ­
inheritance but were not completely linked, this
ment to experiment. The consistent finding of equal
­pattern of deviation might be observed. In this case,
recombination
frequencies
in experiments
in which the
the combinations
of phenotypic
of Jones
© Jones & Bartlett
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mutant
alleles
are
in
the
trans
or
the
cis
configuration
the original
cross (parental phenotypes) were present
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leads to the following conclusion:
in 428/644 (66.5 percent) of the F2 males, and nonparental combinations (recombinant phenotypes) of
KEY CONCEPT
the traits were present in 216/644 (33.5 percent). The
33.5 percent recombinant X chromosomes is called
Recombination between linked genes takes place
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Jones
Bartlett
the frequency of ©
recombination,
and it Learning,
should be LLCwith the same frequency ©
whether
the&
alleles
of the Learning, LL
contrasted with theNOT
50 percent
expected
genes are in the repulsion
(trans)FOR
configuration
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FORrecombination
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with independent assortment.
in the coupling (cis) configuration; it is the same no
The recombinant X chromosomes w1 m1 and w m
matter how the alleles are arranged.
result from crossing-over in meiosis in F1 females. In
this example, the frequency of recombination between
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© Jones &ofBartlett
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the
w and
m genes was
33.5 percent.
With other
The frequency
recombination
pairs
of linked
genes, the
of recombination
NOT
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differs from one gene pair
to
ranges from near 0 to 50 percent. Even genes in the
the
next.
same chromosome can undergo independent assortment (frequency of recombination equal to 50 percent)
The principle that the frequency of recombination
if they are sufficiently far apart. This implies the followdepends on the particular pair of genes may be
© Jones & Bartlett
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ing principle:
illustrated
­
usingLearning,
the recessive
allele y of another
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CHAPTER 4 Gene Linkage and Genetic Mapping
4TH PAGES
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The results of these and other experiments give
X-linked gene in Drosophila, which results in yellow
support to two general principles of recombination:
body color instead of the usual gray color determined
The yellow
body (y) and white eye © Jones & Bartlett Learning, LLC
the y1 allele.
© Jones by
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n The recombination frequency is a characteristic of
(w)
genes
are
linked.
The
frequency
of recombina- NOT FOR SALE OR DISTRIBUTION
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a particular pair of genes.
tion between the genes is as shown in the data in
n Recombination frequencies are the same in cis
FIGURE 4.3. The layout of the crosses is like that
(coupling) and trans (repulsion) heterozygotes.
in Figure 4.2. In Cross 1, the female has y and w in
the trans configuration (1 w/y 1); in Cross 2, the
Recombination does©not
occur
alleles are in the cis©configuration
(y w/1 1).
The
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y and w genes exhibit
a
much
lower
frequency
of
in Drosophila males.NOT FOR SALE OR DISTRIBUT
NOT FOR SALE OR DISTRIBUTION
recombination than that observed with w and m in
Early experiments in Drosophila genetics also indicated
Figure 4.2. To put it another way, the genes y and w
that the organism is unusual in that recombination
are more closely linked than are w and m. In Cross
does not take place in males. The absence of recom1, the recombinant progeny are 1 1 and y w, and
bination in Drosophila males means that all alleles
they ©
account
for&130/9027
5 Learning,
1.4 percent ofLLC
the total.
Jones
Bartlett
© Jones & Bartlett Learning, LLC
located in a particular chromosome show complete
In Cross
the recombinant
are 1 w and
NOT2,FOR
SALE OR progeny
DISTRIBUTION
NOT
FOR
OR
linkage in the
male.
For SALE
example,
theDISTRIBUTION
genes cn (ciny 1, and they account for 94/7838 5 1.2 percent of
nabar eyes) and bw (brown eyes) are both in chromothe total. Once again, the parental and recombinant
some 2, but they are so far apart that in females, they
gametes are reversed in Crosses 1 and 2, because the
show 50 percent recombination. Because the genes
configuration of alleles in the female parent is trans
exhibit 50 percent recombination, the cross
in Bartlett
Cross 1 but
cis in Cross
2, yet the frequency of © Jones & Bartlett Learning, LLC
© Jones &
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cn bw
bw
cn
cn bw
bw
between the genes is within experi- NOT FOR SALEcn
NOT FORrecombination
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OR DISTRIBUTION
××
//
cn
cn bw
bw??
11
11
mental error.
Cross 1
Parents:
Parents:
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Yellow-body
White-eyed and
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White-eyed
females
+w
+w
Cross 2
males
y+
Y
yellow-body
females
yw
yw
F1©
:
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y w The trans
F2:
The cis
(coupling)
heterozygote
Y
y+
(repulsion)
heterozygote
F1:
© Jones & Bartlett Learning, LLC
Wildtype body,
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white eyes
(maternal gamete: + w)
males
++
Yellow body,
44
Y
Progeny 2
98.6% have
parental allele
combinations
(nonrecombinant).
1.2% have
nonparental allele
combinations
(recombinant).
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Recombinant
types:
Parental types:
86
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Parental types: NOT FOR
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types:DISTRIBUTION
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4605
Yellow body,
red eyes
(maternal gamete:
y +) & Bartlett
© Jones
Wildtype body,
red eyes
(maternal gamete: + +)
++
Y
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y w OR DISTRIBUTION
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y w F2:
Progeny 1
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Wildtype
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1.4% have
nonparental allele
combinations
(recombinant).
© Jones
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white&
eyes
9027
(maternal gamete: y w)
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98.8% have
parental allele
combinations
(nonrecombinant).
39
© Jones & Bartlett Learning, LL
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3946
3798
© Jones & Bartlett
Learning, LLC
7838
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FIGURE 4.3 An experiment demonstrating that the frequency of recombination between two genes depends on the genes. The
frequency of recombination between w and y is much less than that between w and m in Figure 4.2. The y2w experiment also confirms
the equal frequency of recombination in trans and cis heterozygous genotypes. (A) The trans heterozygous females, 1 w/y 1, yield
&
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percent recombination.
(B) The
cis heterozygous females, y w/11,
yield 1.2 percent
recombination.
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115
of 3.1 map units therefore equals 3.1 centimorgans and
yields progeny of genotype cn bw/cn bw and 1 1/cn bw
indicates 3.1 percent recombination between the genes.
(the nonrecombinant types) as well as cn 1/cn bw and
An example
is shown
in part ALLC
of FIGURE 4.5, which
1
bw/cn
bw
(the
recombinant
types)
in
the
proportions
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deals
with
the
Drosophila
mutants
w for white eyes and
1
:
1
:
1
:
1.
The
outcome
of
the
reciprocal
cross
is
difNOT FOR SALE OR DISTRIBUTION
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dm (diminutive) for small body. The female parent in the
ferent. Because no crossing-over occurs in males, the
testcross is the trans heterozygote, but as we have seen,
reciprocal cross
this configuration is equivalent in frequency of recomcn
cn bw
bw
cn
cn bw
bw
×
×
bination to the cis heterozygote. Among 1000 progeny
cn
cn bw
bw //
11
11 ??
there are 31 recombinants.©Using
this &
estimate,
we Learning,
can
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express
the
genetic
distance
between
w
and
dm
in
four
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yields progeny only of the nonrecombinant genotypes cn
completely equivalent ways:
bw/cn bw and 11/cn bw in equal proportions. The absence
of recombination in Drosophila males is a ­convenience
n As the frequency of recombination—in this
often exploited in experimental design; as shown in the
case 0.031
case of cn and bw, all the alleles ­present in any chromon As the©percent
recombination,
3.1 percent LLC
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As
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being
recombined
with alleles
p
­ resent in the homologous
map units
chromosome. The absence of c­ rossing-over in Drosophila
n As the distance in centimorgans, or
males is atypical; in most other animals and plants,
3.1 ­centimorgans (3.1 cM)
recombination takes place in both sexes, though not necessarily with the same frequency.
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A genetic
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4.2 Recombination results from
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line, and each gene is assigned a position on the line
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example, there are only two genes, w and dm, and they
are separated by a distance of 3.1 centimorgans (3.1 cM),
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represented in the NOT
form ofFOR
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of the Drosophila X chromosome extends considerably
the linear order of the genes along the chromosome
farther in both directions than indicated in this figure.
spaced so that the distances between adjacent genes is
Physically, one map unit corresponds to a length of
proportional to the frequency of recombination between
the chromosome in which, on the average, one crossthem. A genetic map is also called a linkage map or a
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undergoing meiosis.
chromosome
The concept
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This principle
illustrated
in FIGURE
4.6. If one meiotic
was
firstFOR
developed
Morgan’s
student Alfred H.
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cell in 50 has a crossover, the frequency of crossingSturtevant in 1913. The early geneticists understood
over equals 1/50, or 2 percent. Yet the frequency of
that recombination between genes takes place by an
recombination between the genes is 1 percent. The corexchange of segments between homologous chromorespondence of 1 percent recombination with 2 percent
somes in the process now called crossing-over. Each
crossing-over
is a Learning,
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until you consider that
crossover isLearning,
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a crossover
in two recombinant chromatids and
shapedOR
configuration,
between homologous chromoNOT FOR SALE
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two nonrecombinant chromatids (­Figure 4.6). A crosssomes; chiasmata are observed in prophase I of meiosis.
over frequency of 2 percent means that of the 200 chroEach chiasma results from the breaking and rejoining of
mosomes that result from meiosis in 50 cells, exactly
chromatids during meiosis, with the result that there is
2 chromosomes (those involved in the crossover) are
an exchange of corresponding segments between them.
recombinant for genetic markers
spanning
the particuThe theory of crossing-over
is that
each chiasma
results LLC
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lar chromosome segment. NOT
To putFOR
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in another
in a new associationNOT
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arkers. This
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way, 2 percent crossing-over corresponds to 1 percent
illustrated in FIGURE 4.4. When there is no crossingrecombination because only half of the chromatids in
over (part A), the alleles present in each homologous
each cell with a crossover are actually recombinant.
chromosome remain in the same combination. When a
In situations in which there are genetic markers
crossover does take place (part B), the outermost alleles
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recombination
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distanceOR
in aDISTRIBUTION
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marker genes takes place only when a crossover occurs
map unit; one map unit is equal to 1 percent recombibetween the genes. FIGURE 4.7 illustrates a case in
nation. For example, two genes that recombine with a
which a crossover takes place between the gene A and
frequency of 3.1 percent are said to be located 3.1 map
the centromere, rather than between the genes A and
units apart. One map unit is also called a centimorgan,
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4TH PAGES
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(A) No crossing-over
(B) Crossing-over
a
b
+
b
+
+
+
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Chiasma
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+
+
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FIGURE 4.4 Diagram illustrating crossing-over between two genes. (A) When there is no crossover between two genes, the alleles are
not recombined. (B) When there is a crossover between them, the result is two recombinant and two nonrecombinant products, because
the exchange takes place between only two of the four chromatids.
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do not result in recombinant chromosomes), there is
of segments between the innermost chromatids. Howan important distinction between the distance between
ever, because it is located outside the region between A
two genes as measured by the recombination frequency
and B, all of the resulting gametes must carry either the
and as measured in map units:
A B or the a b allele combination. These are nonrecombinant
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half of
the average
number of crossovers that take
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In some cases, the region between genetic markers
of crossing-over.
is large enough that two (or even more) crossovers can
n The recombination frequency between two genes
be formed in a single meiotic cell. One possible conindicates how much recombination is ­actually
4.8.
figuration for two crossovers
is shown
in FIGURE
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observed in a particular experiment; it is a ­measure
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The difference between map distance and recombinamarker genes, but the double crossover remains undetion frequency arises because double crossovers that do
tected because the markers themselves are not recomnot yield recombinant gametes, like the one depicted
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bined.
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or a b, both of which are nonrecombinant.
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Because double crossovers in a region between two
than one crossover can occur in the region in any one
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(A) Cross
(B) Genetic map
Parent:
w
dm
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White-eyed
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males
females
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w+
F1:
The trans
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+ dm
Y
3.1 cM
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wYOR
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Progeny
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White eyes,
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Parental types = 969/1000 = 96.9%
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Red eyes,
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Red eyes,
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19
normal size
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White eyes,
12
diminutive
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Recombinant
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Genetic distance = frequency of recombination, 0.031
Genetic distance = percent recombination, 3.1%
Genetic distance = map distance in map units, 3.1 map units
Genetic distance = map distance in centimorgans, 3.1 centimorgans (3.1 cM)
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FIGURE 4.5 An experiment illustrating how the frequency of recombination is used to construct a genetic map. (A) There is 3.1
percent recombination between the genes w and dm. (B) A genetic map with w and dm positioned 3.1 map units (3.1 centimorgans, cM)
apart, corresponding to 3.1 percent recombination. The map distance equals frequency of recombination only when the frequency of
recombination
sufficiently
small.
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are the same (because there are no multiple crossovers
that can undo each other). This is the basis for defining
a map unit as being equal to 1 percent recombination:
frequency between genes y and rb is 7.5 percent, and
that between rb and cv is 6.2 percent. The genetic map
might be any one of three possibilities, depending on
which
is in the
middle (y, LLC
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Over an interval so short that multiple c­ rossovers
should be greater than that between rb and y, and this
are precluded (typically yielding 10 percent
contradicts the observed data. In other words, map C
recombination or less), the map distance equals the
can be excluded because it implies that the frequency
recombination frequency because all crossovers
of recombination between©
y and
cv must
be negative.
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Furthermore, when adjacent chromosome regions
y and cv. Using the principle of additivity of map
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This important
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tance in BNOT
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In fact,
observed recomNOT FOR
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well as the logic used in genetic mapping, is illustrated
bination frequency between y and cv is 13.3 percent.
by the example in FIGURE 4.9. The genes are located
Map A is therefore correct. However, there are actually
in the X chromosome of Drosophila—y for yellow
two genetic maps corresponding to map A. They differ
body, rb for ruby eye color, and cv for shortened wing
only in whether y is placed at the left or at the right. One
crossvein. The
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(A)
A
B
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A
B
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B
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1 recombinant
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A
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1 meiosis with a
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b
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(C) Frequency of recombination:
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a
2
1+1
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= 1 percent = 1 map unit = 1 cM
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FIGURE
4.6 Diagram
of chromosomal
in which 1&
hasBartlett
a crossover between
two genes.
(A) The 49 cells without
a
crossover
result
in
98
A
B
and
98
a
b
chromosomes;
these
are
all
nonrecombinant.
(B)
The
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with
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yields
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FIGURE 4.7 Crossing-over outside the region between two genes is not detectable through recombination. Although a segment of
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FIGURE 4.8 If two crossovers take place between marker genes A and B, and both involve the same pair of chromatids, then neither
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(A)
y
rb
7.5 cM
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6.2 cM
rb
y
(B)
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If this
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rb
cv
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y
rb
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7.5 cM
cv
then the distance y to cv should
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If this is the correct genetic map,
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y cv
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6.2 cM
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FIGURE 4.9 In Drosophila, the genes y (yellow body) and rb (ruby eyes) have a recombination frequency of 7.5 percent, and rb and cv
(shortened wing crossvein) have a recombination frequency of 6.2 percent. There are three possible genetic maps, depending on whether
rb is in the middle (part A), cv is in the middle (part B), or y is in the middle (part C). Map (C) can be excluded because it implies that rb
and y should be closer than rb and cv, whereas the observed recombination frequency between rb and y is actually greater than that
between rb and cv. Maps
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are Bartlett
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Courtesy of M. G.
Neuffer, College of
Agriculture, Food, and
Natural Resources,
University of Missouri.
Courtesy of M. G.
Neuffer, College of
Agriculture, Food, and
Natural Resources,
University of Missouri.
R1-mb in a cross between a heterozygous genotype and
the other is cv2rb2y. The two ways of depicting the
a homozygous wildtype.
genetic map are completely equivalent because there
is no way of knowing from the recombination data
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whether
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cv Bartlett
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­chromosome; these genes constitute a linkage group.
The number of linkage groups is the same as the haploid
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number ofLearning,
chromosomes
of the species. For example,
Physical
distance
is often—but
not
cultivated
(Zea mays) has ten pairs of chromosomes
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always—correlated with map distance.
and ten linkage groups. A partial genetic map of chroGenerally speaking, the greater the physical separamosome 10 is shown in FIGURE 4.10, along with the
tion between genes along a chromosome, the greater
dramatic phenotypes caused by some of the mutations.
the map distance between them. Physical distance and
The ears of corn shown in parts C and F demonstrate
genetic map distance are©usually
correlated,
because
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the result of Mendelian
segregation.
The earLearning,
in part C LLC
a greater distance between
genetic
markers
a
shows a 3 : 1 segregation
yellow
: orange
kernels
proNOT
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greater chance for a crossover to take place; crossingduced by the recessive orange pericarp-2 (orp-2) allele in
over is a physical exchange between the chromatids of
a cross between two heterozygous genotypes.
paired homologous chromosomes.
On the other hand, the general correlation between
© Jones & Bartlett Learning, LLC
Jonesand
& genetic
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physical ©
distance
map
distance isLLC
by no
means absolute.
We have
already
noted that the freNOT FOR SALE OR DISTRIBUTION
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quency of recombination between genes may differ in
males and females. An unequal frequency of recombination means that the sexes can have different map
distances in their genetic maps, although the physical
The ear in part F shows a 1 : 1 segregation of
© Jones & Bartlett
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chromosomes
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the two sexesLLC
are the same and the
marbled:white
kernels produced
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120
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CHAPTER 4 Gene Linkage and Genetic Mapping
4TH PAGES
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Short arm (10)
Centromere
(C)
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61
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(B)
php1
glu1
50.8
50.7
Long arm (10)
Centromere
zn1
du1
li1
og1
51
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les16
NOT
y9
47.0
46.9
(D)
mgs1
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74
(A)
37
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wsm3
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rp1
13
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r1
lc1
mst1
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101.1
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(F)
Telomere
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122.1
125
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l13
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sr2
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© Jones
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Telomere
FIGURE 4.10 Genetic map of chromosome 10 of corn, Zea mays. The map distance to each gene is given in standard map units
(centimorgans) relative to a position 0 for the telomere of the short arm (lower left). (A) Mutations in the gene oil yellow-1 (oy1) result in a
yellow-green plant. The plant in the foreground is heterozygous for the dominant allele Oy1; behind is a normal plant. (B) Mutations in the
gene lesion-16
(les16)
in many small
to medium-sized,
on the leaf
blade and sheath.
The photograph
© Jones
&result
Bartlett
Learning,
LLCirregularly spaced, discolored©spots
Jones
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LLC
shows the phenotype of a heterozygote for Les16, a dominant allele. (C) The orp2 allele is a recessive expressed as orange pericarp, a maternal
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tissue that surrounds the kernels. The ear shows the segregation of orp2 in a cross between two heterozygous genotypes, yielding a 3 : 1 ratio
of yellow : orange seeds. (D) The gene zn1 is zebra necrotic-1, in which dying tissue appears in longitudinal leaf bands. The leaf on the left is
homozygous zn1, that on the right is wildtype. (E) Mutations in the gene teopod-2 (tp2) result in many small, partially podded ears and a simple
tassel. An ear from a plant heterozygous for the dominant allele Tp2 is shown. (F) The mutation R1-mb is an allele of the r1 gene, resulting in
red or purple color in the aleurone layer of the seed. Note the marbled color in kernels of an ear segregating for R1-mb. [Adapted from an
illustration by E. H. Coe. Photos courtesy of M. G. Neuffer, College of Agriculture, Food, and Natural Resources, University of Missouri.]
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4.2 Recombination Results from Crossing-Over Between Linked Alleles
4TH PAGES
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121
genes must have the same linear order.
Distance in map units on genetic map
For example, because there is no recom54.5
3.0
49.5
male Drosophila,
© Jones & Bartlett Learning, LLC
© Jones & bination
BartlettinLearning,
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any pair of genes located in
PhysicalNOT
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the same chromosome, when measured in
Euchromatin
Heterochromatin
Euchromatin
the male, is 0. (On the other hand, genes
on different chromosomes do undergo
independent assortment in males.)
Genetic map
The general correlation
between
phys- Learning, LLC
© Jones
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© Jones & Bartlett Learning, LL
pr
cn
sp
net
ical distance and genetic
distance
can
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NOTmap
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even break down in a single chromosome.
54.5
57.5
107.0
0.0
For example, crossing-over is much less
Map
position
frequent in heterochromatin, which consists primarily of gene-poor regions near
Very little recombination
the© centromeres,
than in Learning,
euchromatin.LLC
Jones & Bartlett
© Jones & takes
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place Learning,
in heterochromatin;
Consequently,
given OR
length
of heteroa small distance
in the genetic
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chromatin will appear much shorter in
map corresponds to a large
distance on the chromosome.
the genetic map than an equal length of
euchromatin. In heterochromatic regions,
therefore, the genetic map gives a disFIGURE 4.11 Chromosome 2 in Drosophila as it appears in metaphase of
torted picture
of the physical
mitosis (physical
map, &
top)Bartlett
and in the genetic
map (bottom).
© Jones & Bartlett
Learning,
LLC map. An
© Jones
Learning,
LLCThe genes pr and cn
are actually in euchromatin but are located near the junction with heterochromatin.
example
of
such
distortion
is
illustrated
in
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The total map length is 54.5 1 49.5 1 3.0 5 107.0 map units. The heterochromatin
­ IGURE 4.11, which compares the physical
F
accounts for 3.0/107.0 5 2.8 percent of the total map length but constitutes
map and the genetic map of chromosome
approximately 25 percent of the physical length of the metaphase chromosome.
2 in ­Drosophila. The physical map depicts
the appearance of the chromosome in
metaphase of mitosis.
Two genes
the tips
and two LLC
crossover can be canceled ©
byJones
another &
crossover
further
© Jones
& near
Bartlett
Learning,
Bartlett
Learning, LL
near the euchromatin–­
h
eterochromatin
junction
are
along
the
way.
If
two
exchanges
between
the
same
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indicated in the genetic map. The map distances across
two chromatids take place between the genes A and
the euchromatic arms are 54.5 and 49.5 map units,
B, then their net effect will be that all chromosomes
respectively, for a total euchromatic map distance of
are nonrecombinant, either A B or a b. Two of the
104.0 map units. However, the heterochromatin, which
products of this meiosis have an interchange of their
constitutes
25 percent LLC
of the entire
middle segments,
but&the
chromosomes
are not LLC
recom© Jonesapproximately
& Bartlett Learning,
© Jones
Bartlett
Learning,
chromosome,
has
a
genetic
length
in
map
units
of
binant
for
the
genetic
markers
and
so
are
genetically
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only 3.0 percent. The distorted length of the heteroindistinguishable from noncrossover chromosomes.
­
chromatin in the genetic map results from the reduced
The possibility of such canceling events means that the
frequency of crossing-over in the heterochromatin. In
observed recombination value is an underestimate of the
spite of the distortion of the genetic map across the
true exchange frequency and the map distance between
heterochromatin,
in
the
regions
of
euchromatin
there
the genes.
In higher
organisms,LLC
double crossing-over is
© Jones & Bartlett Learning, LLC
© Jones
& Bartlett
Learning,
is
a
good
correlation
between
the
physical
d
­
istance
effectively
precluded
in
chromosome
segments that are
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between genes and their ­distance, in map units, in the
sufficiently short, usually about 10 map units or less.
genetic map.
Therefore, multiple crossovers that cancel each other’s
effects can be avoided by using recombination data for
closely linked genes to build up genetic ­linkage maps.
One crossover can undo the effects
The minimum recombination
frequency
between
© Jones
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of another. © Jones & Bartlett Learning, LLC
two
genes
is
0.
The
recombination
frequency
NOT FOR SALEalso
ORhas
DISTRIBUT
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a maximum:
When two genes are located far apart along a chromosome, more than one crossover can be formed between
them in a single meiosis, and this complicates the
KEY CONCEPT
interpretation of recombination data. The probabilNo matter how far apart two genes may be, the
Bartlett increases
Learning,
© Jones & Bartlett Learning, LLC
ity©ofJones
multiple&crossovers
withLLC
the distance
maximum frequency of recombination between any
between
the genes.
Multiple
crossing-over complicates
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NOTis 50
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two genes
percent.
genetic mapping because map distance is based on the
number of physical exchanges that are formed, and
Fifty percent recombination is the same value that
some of the multiple exchanges between two genes do
would be observed if the genes were on nonhomolonot result in recombination of the genes and hence are
© Jones & Bartlett
Learning,
© Jones
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gous&chromosomes
and assorted
independently. The
not detected.
As we sawLLC
in Figure 4.8, the effect of one
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122
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CHAPTER 4 Gene Linkage and Genetic Mapping
4TH PAGES
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(A) Two-strand double crossover
a
a
b
+
+
+
+
+
double crossover
(B) Three-strand
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a
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a
b
+
+
+
b
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+
a
b
a
b
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Three-strand double crossover
+
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+
+
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+
b
a
Chromosomes Recovered
Parental Recombinant
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a
b
4
0
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b
a
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+
+
+
b
2
2
+
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a
b
a
+
+
+
a
+
2
2
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b
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(D) Four-strand double crossover
a
b
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a
a
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+
+
+
+
+
+
b
+
b
0
© Jones & Bartlett Learning, LL
4
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8
8
© Jones & Bartlett Learning, LLC
© Jones & Bartlett Learning, LLC
NOT
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NOT
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FIGURE
4.12
Diagram
showing
thatDISTRIBUTION
the result of two crossovers in the interval between
twoFOR
genes isSALE
indistinguishable
from
independent assortment of the genes, provided that the chromatids participate at random in the crossovers. (A) A two-strand double
crossover. (B) and (C) The two types of three-strand double crossovers. (D) A four-strand double crossover.
which &
three
chromatids
participate.
maximum
of recombination
is observed © Jones
© Jones &
Bartlettfrequency
Learning,
LLC
Bartlett
Learning,
LLCThe final possiis that
the OR
second
exchange connects the chrotheOR
genes
are so far apart in the chromosome NOTbility
NOT FORwhen
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matids that did not participate in the first exchange
that at least one crossover is almost always formed
(four-strand double crossover, part D), in which case
between them. In part B of Figure 4.6, it can be seen
all four products are recombinant.
that a single exchange in every meiosis would result
In most organisms, when double crossovers are
in half of the products having parental combinations
take part& in
the twoLearning, LL
and the other half ­having
recombinant
­combinations
© Jones
& Bartlett
Learning, formed,
LLC the chromatids that
© Jones
Bartlett
exchange events are selected
random.
In OR
this DISTRIBUT
of the genes. The NOT
occurrence
two OR
exchanges
NOTatFOR
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FOR of
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case, the expected proportions of the three types of
between two genes has the same effect, as shown
double exchanges are 1/4 four-strand doubles, 1/2
in FIGURE 4.12. Part A shows a two-strand double
three-strand doubles, and 1/4 two-strand doubles.
crossover, in which the same chromatids participate
This means that on the average, (1/4)(0) 1 (1/2)(2)
in both exchanges; no recombination of the marker
© isJones
& Bartlett
Learning,
LLC have
©2Jones
& Bartlett
Learning,
LLC
1 (1/4)(4) 5
recombinant
chromatids
will be found
genes
detectable.
When the
two exchanges
among the NOT
4 chromatids
produced
from
meioses with
one NOT
chromatid
common
(three-strand double
FOR in
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two exchanges between a pair of genes. This is the
crossover, parts B and C), the result is indistinguishsame proportion obtained with a single exchange
able from that of a single exchange; two products
between the genes. Moreover, a maximum of 50 perwith parental combinations and two with recomcent recombination is obtained for any number of
binant combinations are produced. Note that there
&
LLCdoubles, depending on © Jones
& Bartlett Learning, LLC
exchanges.
areBartlett
two typesLearning,
of three-strand
© Jones
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123
a
c
c
a
© Jones & Bartlett Learning, LLC
+
+
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+
+
simple method for determining the order of the three
genes, as we will see in the next section.
b
b
© Jones & Bartlett Learning, LLC
NOT FOR
OR crossovers
DISTRIBUTION
4.3 SALE
Double
are revealed
+
+
in three-point crosses.
The data in TABLE 4.1 result from a testcross in
corn with three genes in a single chromosome. The
a
c& Bartlett
b
© Jones
Learning, LLC
© Jonesto& interpreting
Bartlett Learning,
LL
analysis illustrates the approach
­
a
three-point cross. The recessive
allelesSALE
of the genes
in
NOT
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OR
DISTRIBUT
NOTa FOR
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+
b
this cross are lz (for lazy or prostrate growth habit), gl
(for glossy leaf), and su (for sugary endosperm), and
+
+
c
the multiply heterozygous parent in the cross had the
genotype
+
+
+
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FIGURE 4.13 Diagram showing that two crossovers that occur
© Jones & Bartlett Learning, LLC
Lz Gl SuOR DISTRIBUTION
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lz gl su
between the same chromatids and span the middle pair of alleles
in a triple heterozygote will result in a reciprocal exchange of the
middle pair of alleles between the two chromatids.
where each symbol with an initial capital letter represents the dominant allele. (The use of this type of symbolism
is customary
in corn genetics.)
© Jones & Bartlett Learning, LLC
© Jones
& Bartlett
Learning,
LLC The two classes
of
progeny
that
inherit
noncrossover
Double
crossing-over
is
detectable
in
recombinaNOT FOR SALE OR DISTRIBUTION
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gametes are therefore the wildtype plants and those
tion experiments that employ three-point crosses,
with the lazy-glossy-sugary phenotype. The number of
which include three pairs of alleles. If a third pair of
progeny in these classes is far larger than the number
alleles, c1 and c, is located between the outermost
in any of the crossover classes. Because the frequency
genetic markers (FIGURE 4.13), double exchanges in
of recombination is never ©
greater
than
percent, Learning,
the
© Jones
& Bartlett
Learning,
Jones
&50
Bartlett
LL
the region can be detected
when
the crossovers
flank LLC
very
fact
that
these
progeny
are
the
most
numerous
the c gene. The twoNOT
crossovers,
which
in
this
example
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indicates that the gametes that gave rise to them have
take place between the same pair of chromatids, would
the parental allele configurations, in this case Lz Gl Su
result in a reciprocal exchange of the c1 and c alleles
and lz gl su. Using this principle, we could have inferred
between the chromatids. A three-point cross is an
the genotype of the heterozygous parent even if the
efficient way to obtain recombination data; it is also a
© Jones & Bartlett Learning, LLC
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TABLE
© Jones & Bartlett Learning, LLC
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4.1
Interpreting
a Three-Point
Cross
© Jones & Bartlett
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Phenotype
of DISTRIBUTION
Genotype of gamete
NOT FOR SALE
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from hybrid parent
Number of
progeny
Wildtype
Lz Gl Su
286
Lazy
lz Gl Su
33
testcross progeny
Glossy
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The two most frequent
classes identify the nonrecombinant gametes.
Lz gl Su&
Jones
59
©
Bartlett Learning,
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Lz Gl su SALE OR DISTRIBUTION
4
NOT FOR
The two rarest classes iden-
Sugary
Lazy, glossy
lz gl Su
2
Lazy, sugary
lz Gl su
44
Glossy, sugary
Lz gl su
40
© Jones
&
Lazy,
glossy, sugary
tify the double-recombinant
gametes.
BartlettlzLearning,
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gl su
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740
These reciprocal classes result from
single recombination between one pair of
adjacent genes.
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124
© Jones & Bartlett Learning, LL
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© Jones & Bartlett Learning, LLC
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These reciprocal classes result from single
recombination between the other pair of
adjacent genes.
© Jones & Bartlett Learning, LLC
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CHAPTER 4 Gene Linkage and Genetic Mapping
4TH PAGES
© Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION.
genotype had not been stated. This is a point important
enough to state more generally:
the parental gamete lz gl su except for the allele Su.
The middle gene can be identified because the “odd
man out”
in the comparisons—in
this case, the alleles
© Jones
& Bartlett
Learning, LLC
© Jones & Bartlett Learning, LLC
of
Su—is
always
the
gene
in
the
middle.
CONCEPT
NOT FOR SALE OR DISTRIBUTIONThe reason is
NOT FORKEY
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that only the middle pair of alleles is interchanged by
In any genetic cross, no matter how complex, the
double crossing-over.
two most frequent types of gametes with respect to
Taking the correct gene order into account, the
any pair of genes are nonrecombinant; these progenotype
of the heterozygous parent in the cross yieldvide the linkage phase (cis versus trans) of the alleles
ing
the
progeny
in Table 4.1 should
be written
as
© Jones
& Bartlett
Learning, LLC
© Jones
& Bartlett
Learning,
of the genes in the multiply
heterozygous
parent.
LL
NOT
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Lz Su Gl
NOT FOR SALE OR DISTRIBUTION
lz su gl
In mapping experiments, the gene sequence is
­ sually not known. In this example, the order in which
u
The consequences of single crossing-over in this genothe three genes are shown is entirely arbitrary. H
­ owever,
type are shown
in FIGURE
4.15. A single
crossover
in
there©is Jones
an easy&way
to
determine
the
­
c
orrect
order
Bartlett Learning, LLC
© Jones
& Bartlett
Learning,
LLC
the
lz2su
region
(part
A)
yields
the
reciprocal
recomfrom NOT
three-point
data.
The
gene
order
can
be
deduced
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binants Lz su gl and lz Su Gl, and a single crossover in
by identifying the genotypes of the d
­ ouble-crossover
the su2gl region (part B) yields the reciprocal recomgametes produced by the heterozygous p
­
­arent and
binants Lz Su gl and lz su Gl. The consequences of
comparing these with the non­
recombinant gametes.
double crossing-over are illustrated in FIGURE 4.16.
Because the probability of two simultaneous exchanges is
There are
four different
types of LLC
double crossovers: a
considerably
than that
of either single exchange, © Jones
© Jones &
Bartlett smaller
Learning,
LLC
& Bartlett
Learning,
two-strand
double
(part
A),
two
types
of three-strand
the
double-crossover
gametes
will
be
the
least
frequent
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doubles (parts B and C), and a four-strand double (part
types. It is clear in Table 4.1 that the classes composed
D). These types were illustrated earlier in Figure 4.12,
of four plants with the sugary phenotype and two plants
where the main point was that with two genetic markwith the lazy-glossy phenotype (products of the Lz Gl su
ers flanking the crossovers, the occurrence of double
and lz gl Su gametes, respectively) are the least frequent
and therefore constitute
the double-crossover
© Jones
& Bartlett Learning, LLC
© Jones & Bartlett Learning, LL
progeny. Now we apply
another
principle:
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(A)
KEY CONCEPT
Lz
The effect of double crossing-over is to interchange the members of the middle pair of
© Jones & Bartlett Learning, LLC
alleles between the chromosomes.
NOT FOR SALE OR DISTRIBUTION
Lz
Gl
Gl
lz
gl
lz
gl
Su
Su
Lz
Gl
Su
Lz
gl
Su
su
su
lz
Gl
© Jones & Bartlett Learning,
LLC
su
gl
NOT FOR SALE OR lzDISTRIBUTION
su
This principle is illustrated in FIGURE 4.14. With
(B)
three genes there are three possible orders,
Gl
Lz
Su
depending on which gene is in the middle. If gl
Gl
Su
Lz
were
in the middle
(part A),LLC
the double-recom© Jones &
Bartlett
Learning,
© Jones
& Bartlett
Learning, LLCLz su
Su
Gl
Gl
Lz
gametes
would be Lz gl Su and lz Gl su,
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NOTlz FOR SALE
OR
DISTRIBUTION
gl
su
gl
lz
which is inconsistent with the data. Likewise,
Su
lz
if lz were in the middle (part C), the doublegl
su
su
gl
lz
recombinant gametes would be Gl lz Su and gl
Lz su, which is also inconsistent with the data.
(C)
The correct order ©
of Jones
the genes,
lz2su2gl,Learning,
& Bartlett
LLC
© Jones & Bartlett Learning, LL
Su
Gl
is given in part B, because
in this
case, OR
the DISTRIBUTION
Lz
Lz
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NOT FOR
SALE
Su
Gl
double-recombinant gametes are Lz su Gl and
Lz
Gl
lz
Su
Su
Gl
lz Su gl, which Table 4.1 indicates is actually
su
gl
the case. Although one can always infer which
lz
su
gl
Lz
gene is in the middle by going through all three
gl
su
lz
lz
su
gl
possibilities,
there
is a shortcut.
Each double-­
© Jones
& Bartlett
Learning,
LLC
© Jones & Bartlett Learning,
LLC
recombinant
gamete
will always
match one of
NOT FOR
SALE
OR DISTRIBUTION
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FIGURE 4.14 The order of genes in a three-point testcross may be
the parental gametes in two of the alleles. In
deduced from the principle that double recombination interchanges the
Table 4.1, for example, the double-recombinant
middle pair of alleles. For the genes Lu, Gl, and Su, there are three possible
gamete Lz Gl su matches the parental gamete
orders (parts A, B, and C), each of which predicts a different pair of gametes as
Lz Gl Su except for the allele su. Similarly, the
the result of double recombination. Only the order in part B is consistent with
double-recombinant
gamete
lz gl Su matches
© Jones &
Bartlett Learning,
LLC
© Jones
&suBartlett
LLC
the finding
that Lz Gl
and lz gl SuLearning,
are the double-recombinant
gametes.
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4.3 Double Crossovers Are Revealed in Three-Point Crosses
4TH PAGES
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125
(A) Single crossover in lz–su region
Lz
Su
Gl
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Gl
Su
Lz
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lz
su
gl
lz
su
gl
su–gl region
(B) Single crossover
©inJones
& Bartlett
Lz
Su
crossovers cannot be detected genetically. The
difference in the present case is that, here,
the genetic
marker su is
located in the middle
©
& Bartlett
Learning,
LLC
gl
su Jones
Lz
between
the
two
crossovers,
NOT FOR SALE OR DISTRIBUTIONso some of the
Gl
Su
lz
double crossovers can be detected genetically.
On the right in F
­ igure 4.16, the asterisks mark
su
gl
lz
the sites of crossing-over between nonsister
chromatids. In terms of recombination, the
result is that
Learning, LLC
© Jones & Bartlett Learning,
Lz
Gl
Su
Lz
Gl
Su
Gl FOR SALE OR DISTRIBUTION
NOT
LL
NOT
FOR
SALE
OR
DISTRIBUT
n
A two-strand double crossover (part A)
yields the reciprocal double-recombinant
products Lz su Gl and lz Su gl.
lz
gl
Gl
su
lz
n One three-strand double crossover (part B)
lz
su
gl
su
gl
lz
the double-recombinant ­product
© Jones & Bartlett Learning, LLC
©yields
Jones
& Bartlett Learning, LLC
Lz su Gl and two single-­recombinant
NOT
FOR
SALE
OR
DISTRIBUTION
NOT
FOR
ORlz Su
DISTRIBUTION
­products, LzSALE
Su gl and
Gl.
FIGURE 4.15 Result of single crossovers in a triple heterozygote, using
the Lz2Su2Gl region as an example. (A) A crossover between Lz and Su
n The other three-strand double crossover
results in two gametes that show recombination between Lz and Su and
(part C) yields the double-recombinant
two gametes that are nonrecombinant. (B) A crossover between Su and Gl
product lz Su gl and two single-­recombinant
results in two gametes that show recombination between Su and Gl and
products,
Lz su gl and
lz su Gl.
two gametesLearning,
that are nonrecombinant.
© Jones & Bartlett
LLC
© Jones & Bartlett
Learning,
LLC
n
T
he
four-strand
double
crossover (part C)
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NOT FOR SALE OR DISTRIBUTION
(A) Two-strand double crossover
yields reciprocal single recombinants
Lz
in the lz2su region, namely Lz su gl and
Gl
Su
Lz
Su
Gl
lz Su Gl, and reciprocal single recombi*
* su
Su
Lz
Gl
Gl
Lz
nants in the su2gl region, namely Lz Su gl
lz
*
gl
* LLC
and lz su Gl. © Jones & Bartlett Learning,
& Bartlett Learning,
su © Jones
gl
lz
Su
Lz
Su
su
lz
su
Lz
Gl
gl
Su
LL
gl FOR SALE OR DISTRIBUTION Note that the products
NOTofFOR
SALE OR
DISTRIBUT
NOT
recombination
in the
su
gl
lz
three-strand double crossovers (parts B and C)
are the reciprocals of each other. Because these
(B) Three-strand double crossover
two types of double crossovers are equally fre*
gl
Lz
Su
Gl
Lz
Su
quent, the reciprocal products of recombination
*
© Jones
&
Bartlett
Learning,
LLC
© expected
Jones to
& appear
Bartlett
Learning,
LLC
*
Su
Gl
are
in equal
numbers.
su
Lz
Gl
Lz
NOT lzFOR suSALEgl OR DISTRIBUTION *
NOT
FOR
OR the
DISTRIBUTION
We can
nowSALE
summarize
data in Table 4.1
Gl
Su
lz
in a more informative way by writing the genes
lz
su
gl
in the correct order and grouping reciprocal
su
gl
lz
gametic genotypes together. This grouping is
shown in TABLE 4.2. Note that each class of
(C) Three-strand double crossover
© Jones & Bartlett Learning,
LLC
©
Jones
&
Bartlett
Learning,consists
LLC of two reciprocal
Lz
Gl
Su
single
recombinants
Su
Gl
Lz
products
and
that
these
are found in approxiNOT FOR SALE OR DISTRIBUTION
NOT
FOR
SALE
OR
DISTRIBUTION
*
gl
Lz
su
Su
Gl
Lz
mately equal frequencies (40 versus 33 and
*
lz
su
59 versus 44). This observation illustrates an
gl
gl
lz
Su *
important principle:
su
lz
gl
*
su
lz
(D)
Gl
© Jones & Bartlett Learning, LLC
KEY CONCEPT © Jones & Bartlett Learning, LL
Four-strand double
crossover
NOT FOR
SALE
NOT
FOR SALE OR DISTRIBUTION The two reciprocal products
resulting
fromOR
any DISTRIBUT
Lz
Su
Gl
Lz
Su
su
*
Su
Lz
Su
Gl
*
Lz
lz
su
gl
lz
© Jones
&
Bartlett
Learning, LLC
lz
su
gl
lz
NOT FOR SALE OR DISTRIBUTION
su
*
gl
gl
Gl
*
Gl
FIGURE 4.16 Result of double crossovers in a triple heterozygote, using
the Lz2Su2Gl region as an example. Note that chromosomes showing
double recombination derive from the two-strand double crossover (A) or
from either type of three-strand double crossover (B and C). The four-strand
Bartlett
Learning,
LLC
© Jones
double crossover
(D) results only
in single-recombinant chromosomes.
© Jones &
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126
crossover, or any combination of crossovers,
are expected to appear in approximately
equal frequencies among the progeny.
© Jones & Bartlett Learning, LLC
NOT
FOR SALE
DISTRIBUTION
In calculating
theOR
frequency
of recombination from the data, remember that the
­double-recombinant chromosomes result from
two exchanges, one in each of the chromosome
regions defined by the three genes. Therefore,
Bartlett
Learning,
LLC
chromosomes
that are recombinant
between lz
&
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CHAPTER 4 Gene Linkage and Genetic Mapping
4TH PAGES
© Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION.
and su are represented by the following c­ hromosome
types:
© Jones & Bartlett Learning,
Lz su glLLC 40
NOT FOR SALE OR DISTRIBUTION
lz Su Gl
33
Lz su Gl
4
lz Su gl
2
79
f­requency between adjacent genes. You can keep from
falling into this trap by remembering that the doublerecombinant
chromosomes
have LLC
single recombination
© Jones
& Bartlett
Learning,
in
both
regions.
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Interference decreases the chance
of multiple crossing-over.
The total implies that©79/740,
percent,Learning,
of the
Jonesor&10.7
Bartlett
LLC
© Jones &
Bartlett
The detection of double crossing-over
makes
it pos-Learning, LL
chromosomes recovered
in
the
progeny
are
recombisible to determine whether exchanges
in two
different
NOT FOR
SALE
OR DISTRIBUT
NOT FOR SALE OR DISTRIBUTION
nant between the lz and su genes, so the map distance
regions of a pair of chromosomes are formed indepenbetween these genes is 10.7 map units, or 10.7 centidently of each other. Using the information from the
morgans. Similarly, the chromosomes that are recomexample with corn, we know from the recombination
binant between su and gl are represented by
frequencies that the probability of recombination is
© Jones &
Bartlett
Learning,
LLC
© Jones
Bartlett
Learning,
LLC
If
0.107 between
lz and su &
and
0.147 between
su and gl.
Lz Su gl
59
recombination
is
independent
in
the
two
regions
(which
NOT FORlz
SALE
OR
DISTRIBUTION
NOT
FOR
SALE
OR
DISTRIBUTION
su Gl
44
means that the formation of one crossover does not
Lz su Gl
4
alter the probability of the second crossover), the problz Su gl
2
ability of a single recombination in both regions is the
109
product of these separate probabilities, or 0.107 3 0.147
In Bartlett
this case the
recombination
© Jones &
Learning,
LLC frequency between su © Jones
& Bartlett
Learning,
LLCthat in a sample
5 0.0157
(1.57 percent).
This implies
and
gl
is
109/740,
or
14.7
percent,
so
the
map
distance
740 gametes,
number of double recomNOT FOR SALE OR DISTRIBUTION
NOTofFOR
SALE the
ORexpected
DISTRIBUTION
between these genes is 14.7 map units, or 14.7 centibinants would be 740 3 0.0157, or 11.6, whereas the
morgans. The genetic map of the chromosome segment
number actually observed was only 6 (Table 4.2). Such
in which the three genes are located is therefore
deficiencies in the observed number of double recombinants are common; they reflect a phenomenon called
lz
su© Jones & Bartlett Learning,
gl
LLC
Jones
& Bartlett
chromosome
i­ nterference, in©which
a crossover
in oneLearning, LL
region of a chromosome reduces
probability
of a DISTRIBUT
NOT the
FOR
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FOR
SALE
10.7 map units NOT 14.7
map
units OR DISTRIBUTION
second crossover in a nearby region. Over short genetic
distances, chromosome interference is nearly complete.
The most common error in learning how to interpret
The coefficient of coincidence is the observed
three-point crosses is to forget to include the dounumber of double-recombinant chromosomes divided
ble recombinants when calculating the recombination
© Jones & Bartlett Learning, LLC
© Jones
&Its
Bartlett
Learning,
LLC
by the expected
number.
value provides
a quantitative measure
of the
degree
of interference,
which is
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NOT
FOR
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OR DISTRIBUTION
defined as
TABLE
© Jones & Comparing
Bartlett Learning,
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Reciprocal
Products in a
Three-Point
Cross
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Genotype of gamete
from hybrid parent
Lz Su Gl
lz su gl
Lz su gl
lz Su Gl
i 5 Interference
5 1 2 (Coefficient of coincidence)
4.2
Number of
progeny
Intervals showing
recombination
286
272
© Jones & Bartlett Learning, LLC
From the data in the corn example, the coefficient of
NOTcoincidence
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is calculated
as follows:
n Observed number of double recombinants 5 6
n Expected number of double recombinants 5 0.107
3 0.147 3 740 5 11.6
© Jones & Bartlett Learning, LLC
© Jones & Bartlett Learning, LL
n Coefficient of coincidence 5 6/11.6 5 0.52
40
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NOT
FOR
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OR
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lz2su
33
The 0.52 means that the observed number of double
recombinants was only about half of the number
expected if crossing-over in the two regions were
independent. The value of the interference depends on
Lz su Gl
4
lz2su 1 su2gl
lz Su©
gl Jones & Bartlett
2
the distance©between
genetic markers
and onLLC
the
Learning, LLC
Jonesthe
& Bartlett
Learning,
­species. In some
the interference
increases as
740OR DISTRIBUTION
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NOT species,
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the ­distance between the two outside markers becomes smaller, until
Total number of recombinants in
Total number of recombinants in
a point is reached at which double
lz2su region:
su2gl region:
­crossing-over is ­eliminated; that is,
40 1 33 1 4 1 2 5 79
59 1 44 1 4 1 2 5 109
double recombinants
are found,
& Bartlett
Learning, LLC
© Jones & BartlettnoLearning,
LLC
Lz Su gl
lz su Gl
59
su2gl
44
© Jones
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4.3 Double Crossovers Are Revealed in Three-Point Crosses
4TH PAGES
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127
Recombination frequency (r )
0.5
© Jones & Bartlett
Learning, LLC
No interference
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(i = 0)
© Jones & Bartlett Learning, LLC
0.4
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0.3
Interference decreasing
with distance (i = 1 – 2r)
0.2
© Jones & Bartlett
CompleteLearning,
interference LLC
(i = 1)
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OR DISTRIBUTION
0.1
0
0
20
40
60
© Jones & Bartlett Learning, LLC
© Jones & Bartlett Learning, LL
NOT FOR SALE OR DISTRIBUT
80
100
120
Distance in map units (cM)
140
160
180
200
© Jones & Bartlett Learning, LLC
FIGURE
mapping function
is the relation between genetic map distance across
intervalSALE
and the observed
frequency of
NOT 4.17
FORA SALE
OR DISTRIBUTION
NOTanFOR
OR DISTRIBUTION
recombination across the interval. Map distance is defined as one-half the average number of crossovers converted into a percentage.
The three mapping functions correspond to different assumptions about interference, i.
mapping
function,Learning,
as shown by
the three examples in
and the coefficient
of coincidence
equals 0 (or, to say
© Jones & Bartlett
Learning,
LLC
© Jones
& Bartlett
LLC
Figure 4.17.
the same thing, the interference equals 1). In Drosophila
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this distance is about 10 map units.
The effect of interference on the relationship
4.4 Polymorphic DNA sequences
between genetic map distance and the frequency of
are used in human genetic
recombination is illustrated in FIGURE 4.17. Each
mapping. © Jones & Bartlett Learning, LL
curve is an example©ofJones
a mapping
function, which
is the LLC
& Bartlett
Learning,
mathematical relation between the genetic distance
Until quite recently, mapping
human OR
beings
NOTgenes
FORinSALE
DISTRIBUT
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across an interval in map units (centimorgans) and the
was very tedious and slow. Numerous practical obstacles
observed frequency of recombination across the intercomplicated genetic mapping in human pedigrees:
val. In other words, a mapping function tells one how
1. Most genes that cause genetic diseases are rare, so
to convert a map distance between genetic markers into
they are observed in only a small number of families.
a ­r©
ecombination
frequency
between
the
markers.
As
we
Jones & Bartlett Learning, LLC
© Jones & Bartlett Learning, LLC
have seen, when the map distance between the mark2. Many mutant genes of interest in human ­genetics
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ers is small, the recombination frequency equals the
are recessive, so they are not detected in heterozymap distance. This principle is reflected in the curves
gous genotypes.
in Figure 4.17 in the region in which the map distance
3. The number of offspring per human family is
is smaller than about 10 cM. At less than this distance,
relatively small, so segregation cannot usually be
all of the curves are nearly straight lines, which means
in single
sibships. LLC
© Jones & Bartlett
Learning, LLC
© Jones detected
& Bartlett
Learning,
that map distance and recombination frequency are
4.
The
human
geneticist
cannot perform testcrosses
NOT FOR SALE
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equal; 1OR
mapDISTRIBUTION
unit equals 1 percent recombination, and
or backcrosses, because human matings are not
10 map units equals 10 percent recombination. For
dictated by an experimenter.
distances greater than 10 map units, the recombinaHuman genetics has been revolutionized by the
tion frequency becomes smaller than the map distance
use of techniques for manipulating DNA. These techaccording to the pattern of interference along the chro© Jones & Bartlett Learning, LLC
© Jones
Bartlett
Learning, LL
niques have enabled investigators
to &
carry
out genetic
mosome. Each pattern of interference yields a different
NOT FOR SALE OR DISTRIBUTION
Q
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A MOMENT TO THINK
Problem: In his pioneering 1913 studies in Drosophila that resulted in the first genetic linkage map, A. H. Sturtevant included three
© Jones
&now
Bartlett
& Bartlett
Learning,
genetic
markers
known to Learning,
cover almost theLLC
entirety of the euchromatin of the©X Jones
chromosome.
The marker w
(white eyes) is LLC
near
the
tip, mFOR
(miniature
body)OR
near the
middle, and r (rudimentary wings) near the centromere.
In two-point
Sturtevant obtained
NOT
SALE
DISTRIBUTION
NOT FOR
SALEcrosses,
OR DISTRIBUTION
recombination frequencies of 0.32 for the interval w2m, 0.25 for the interval m2r, and 0.45 for the interval w2r. He noted that 0.45 is
smaller than the sum of 0.32 1 0.25 5 0.57, which is the value expected if the frequencies of recombination over such large distances
were additive. He commented that the discrepancy “is probably due to the occurrence of two breaks in the same chromosome, or
double crossing-over.” From Sturtevant’s data, calculate the coincidence and the interference across the region. (The answer can be
found at the end of the chapter.)
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128
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CHAPTER 4 Gene Linkage and Genetic Mapping
4TH PAGES
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