Transcript CHAPTER 15 Gene Mapping in Eukaryotes

Peter J. Russell
A molecular Approach 2nd Edition
CHAPTER 15
Gene Mapping In Eukaryotes
edited by Yue-Wen Wang Ph. D.
Dept. of Agronomy,台大農藝系
NTU
遺傳學 601 20000
Chapter 13 slide 1
Discovery of Genetic Linkage
1. Genes on non-homologous chromosomes assort independently,
but genes on the same chromosome (syntenic genes) may
instead be inherited together (linked), and belong to a linkage
group.
2. Classical genetics analyzes the frequency of allele recombination
in progeny of genetic crosses.
a. New associations of parental alleles are recombinants, produced
by genetic recombination.
b. Testcrosses determine which genes are linked, and a linkage
map (genetic map) is constructed for each chromosome.
c. Genetic maps are useful in recombinant DNA research and
experiments dealing with genes and their flanking sequences.
3. Current high-resolution maps include both gene markers from
testcrosses, and DNA markers composed of genomic regions
that differ detectably between individuals.
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Chapter 13 slide 2
Early Study of Genetic Linkage: Morgan’s
Experiments with Drosophila
1. Both the white eye gene (w) and a gene for miniature wings (m) are on
the Drosophila X chromosome. Morgan (1911) crossed a female white
miniature (w m/w m) with a wild-type male (w + m+/ Y)
(Figure 15.1).
a. In the F1, all males were white-eyed with miniature wings (w m/Y), and all
females were wild-type for both eye color and wing size (w+m+/w m).
b. F1 interbreeding is the equivalent of a testcross for these X-linked genes,
since the male is hemizygous recessive, passing on recessive alleles to
daughters and no X-linked alleles at all to sons.
i. In the F2, the most frequent phenotypes for both sexes were the
phenotypes of the parents in the original cross (white eyes with
miniature wings, and red eyes with normal wings).
ii. Non-parental phenotypes (white eyes with normal wings or red eyes
with miniature wings) occurred in about 37% of the F2 flies. Well below
the 50% predicted for independent assortment, this indicates that nonparental flies result from recombination of linked genes.
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Chapter 13 slide 3
Fig. 15.1 Morgan’s experimental crosses of white-eye and miniature-wing variants of
Drosophila melanogaster
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 4
c. Morgan proposed that:
i. During meiosis alleles of some genes assort together because they
are near each other on the same chromosome.
ii. Recombination occurs when genes are exchanged between the X
chromosomes of the F1 females.
d. A series of experiments supported Morgan’s hypothesis. In each case,
parental phenotypes were the most frequent, while recombinant phenotypes
occurred less frequently.
e. Some relevant terminology:
i. A chiasma (plural chiasmata) is the site on the homologous
chromosomes where crossover occurs.
ii. Crossing-over is the reciprocal exchange of homologous chromatid
segments, involving the breaking and rejoining of DNA.
iii. Crossing-over is also the event leading to genetic recombination
between linked genes in both prokaryotes and eukaryotes.
f. Crossing-over occurs at the four-chromatid stage of prophase I in meiosis.
Each crossover event involves two of the four chromatids. All chromatids
may be involved in crossing-over, as chiasmata form along the aligned
chromosomes (Figure 13.2).
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Chapter 13 slide 5
Gene Recombination and the Role of
chromosomal Exchange
1. Morgan’s results were only circumstantial
evidence. Proof that physical exchange between
chromosomes results in genetic recombination
came in the 1930s.
Animation: Relationship Between Genetic
Recombination and Chromosomal Exchange
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Chapter 13 slide 6
Corn Experiments
1. Creighton and McClintock (1931) worked with corn (Zea mays) plants in which
the two chromosomes under study differed cytologically.
2. The study used a corn strain heterozygous for two genes on chromosome 9
(Figure 15.2):
a. One gene determines seed color (C for colored seeds, c for colorless).
b. The other gene is involved in starch synthesis. The wild-type allele (Wx) produces
amylose, and the combination of amylose and amylopectin forms normal starch in a
corn seed. The waxy mutant (wx) lacks amylose, and has waxy starch containing
only amylopectin.
3. In this corn strain, the appearance of each chromosome 9 homolog correlated
with its genotype:
a. One chromosome 9 had the genotype c Wx, and a normal appearance.
b. Its homolog had the genotype C wx, and cytological markers at each end of the
chromosome. The end near the C locus had a darkly staining knob, and the other
end, nearer the wx locus, had a translocated piece of chromosome 8.
4. When testcrossed, recombinant phenotypes were evident, and could be
correlated with cytological features:
a. Whenever the genes had recombined, the cytological features had also recombined.
b. In the parental (non-recombinant.) type progeny, no exchange of cytological markers
was evident.
5. This was direct evidence of physical exchange between homologs resulting in
genetic recombination.
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Chapter 13 slide 7
Fig. 15.2 Evidence of the association of gene recombination with chromosomal
exchange in corn
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 8
Drosophila Experiments
1. Identical observations were made by Stern a short time later in Drosophila
melanogaster, using a similar approach involving crosses between strains with
defined genetic and cytological markers on their X chromosomes (Figure 15.3).
2. The two linked gene loci were:
a. The car (carnation) gene is recessive. Homozygotes have carnation colored eyes,
rather than wild-type red. The car locus is near the “left” end of the X chromosome.
b. The B (bar-eye) gene is incompletely dominant. Homozygotes (B/B) have a barshaped eye rather than wild-type non-bar (round). Heterozygotes (B/+) have a widebar (kidney shaped) eye. The B locus is farther from the “left” end of the X
chromosome (thus nearer the centromere) than the car locus.
3. In Stern’s crosses:
a. Male parents carried recessive alleles for both eye-color (car) and eye-shape (+) on a
single X chromosome. Phenotype is carnation, non-bar eyes.
b. Female parent carried two abnormal and cytologically distinct X chromosomes, with
a genotype of + + / B car, and a phenotype of wide-bar red eyes.
i. One X chromosome had a translocated fragment of Y chromosome. It carried
the wild-type alleles (+ +, red and non-bar) for both traits.
ii. The second X chromosome had lost a region by translocation to chromosome 4.
This chromosome was visibly shorter than a normal X chromosome (Figure
13.4). Its alleles were the two mutants, car and B.
c. Gamete formation would produce two types in males, X with both recessive alleles,
and Y with neither of the alleles. Females produce
four types, two non-recombinant
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and two recombinant. A Punnett square shows the segregation of alleles: Chapter 13 slide 9
+ (non-bar)
car
Normal X
Y
+ (non-bar) + (red)
X with Y translocation
No recombination
+ + / + car
non-bar red
++/Y
non-bar red
B car
Short X
No recombination
B car / + car
bar
carnation
B car / Y
bar
carnation
+ (non-bar) car
Normal X
Recombination
+ car / + car
non-bar
carnation
+ car / Y
non-bar carnation
B+(red)
Short X with
Y translocation
Recombination
B+/+car
bar
red
B+/Y
bar red
d. Cytological examination of progeny showed:
i. Both males and females with nonbar carnation eyes had a normal X
chromosome, along with a second normal X in females, or a Y in males.
ii. Female flies with wide-bar red eyes and males with bar red eyes had a
short X chromosome with the Y translocation, along with a normal X or Y.
4. This confirmed that physical crossing-over between chromosomes results in
genetic recombination (Box 15.1).
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Chapter 13 slide 10
Fig. 15.3 Stern’s experiment to demonstrate the relationship between genetic
recombination and chromosomal exchange in Drosophila melanogaster
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 11
Box Fig. 15.1 Holliday model for reciprocal genetic recombination
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 12
Constructing Genetic Maps
1. Genetic recombination experiments can be used
in genetic (or linkage) mapping.
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Chapter 13 slide 13
Detecting Linkage through Testcrosses
1.
Linked genes are used for mapping. They are found by looking for deviation from the frequencies expected from
independent assortment.
2.
A testcross (one parent is homozygous recessive) works well for analyzing linkage:
3.
a.
If the alleles are not linked, and the second parent is heterozygous, all four possible combinations of traits will be present in equal
numbers in the progeny.
b.
A significant deviation in this ratio (more parental and fewer recombinant types) indicates linkage.
Chi-square analysis is used to analyze testcross data and determine whether a deviation is “significant.” A null
hypothesis (“the genes are not linked”) is used because it is not possible to predict phenotype frequencies produced
by linked genes.
a.
If two genes are not linked, a testcross should yield a 1:1 ratio of parentals : recombinants.
b.
The formula is: χ2 = Σ d2 / e
i. Σ is “sum of.”
ii. d = deviation value = (o - e).
iii. o = the observed number.
iv. e = the expected number.
c.
χ2
The value and the degrees of freedom (df) for the data set are used with a table of chi-square probabilities to determine the
probability (P) that the deviation of observed from expected values is due to chance.
i. If P ＞ 0.05 (probability of more than 5 in 100 that deviation was by chance alone) the deviation is not considered
statistically significant.
ii. If P ＜ 0.05, the deviation is considered statistically significant, and not due to chance. The null hypothesis is likely to be
invalid.
iii. If P ≦ 0.01, deviation is highly statistically significant, and the data are not consistent with the hypothesis, which must
be rejected. If the hypothesis “the genes are not linked” is rejected, the only remaining option is that the genes are linked.
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Chapter 13 slide 14
The Concept of a Genetic Map
1.
2.
In an individual heterozygous at two loci, there are two arrangements of alleles:
a.
The cis (coupling) arrangement has both wild-type alleles on one homologous chromosome, and both mutants on the other (e.g.,
w+m+ and w m).
b.
The trans (repulsion) arrangement has one mutant and one wild-type on each homolog (e.g., w+m and w m+).
c.
A crossover between homologs in the cis arrangement results in a homologous pair with the trans arrangement. A crossover
between homologs in the trans arrangement results in cis homologs.
Drosophila crosses showed that crossover frequency for linked genes (measured by recombinants) is characteristic
for each gene pair. The frequency stays the same, whether the genes are in coupling or in repulsion.
a.
Morgan and Sturtevant (1913) used recombination frequencies to make a genetic map.
i. A 1% crossover rate is a genetic distance of 1 map unit (mu). A map unit is also called a centimorgan (cM).
ii. Geneticists use recombination frequency as a way to estimate crossover frequency. It is not an exact measure, however.
iii. The farther apart the two genes are on the chromosome, the more likely it is that crossover will occur between them, and
therefore the greater their crossover frequency (Figure 13.6).
b.
The first genetic map was based on crosses in Drosophila involving the three sex-linked genes:
i. w gives white eyes.
ii. m gives miniature wings.
iii. y gives yellow body.
c.
The crosses gave the following recombination frequencies:
i. w x m was 32.6.
ii. w x y was 1.3.
iii. m x y was 33.9.
d.
Map is therefore: m————————————————w—y.
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Chapter 13 slide 15
Gene Mapping Using Two-Point Testcrosses
1. With autosomal recessive alleles, when a double heterozygote is testcrossed,
four phenotypic classes are expected. If the genes are linked, the two parental
phenotypes will be about equally frequent and more abundant than the two
recombinant phenotypes(Figure 15.4).
2. Mapping of genes with other mechanisms of inheritance is also done with twopoint testcrosses:
a. For X-linked recessives, a female double heterozygote (a+b+/ab) is crossed with a
male hemizygous for the recessive alleles (ab/Y).
b. For either X-linked case, it is possible to cross the females with males of any type.
As long as only male progeny are analyzed, the father’s X will be irrelevant.
c. Phenotypes obtained in any of these crosses will depend on whether the alleles are
arranged in coupling (cis) or repulsion (trans).
d. Recombination frequency is used directly as an estimate of map units.
i. The measure is more accurate when the alleles are close together.
ii. Scoring large numbers of progeny increases the accuracy.
e. Mapping in all types of organisms shows genes arranged with a 1;1 correspondence
between linkage groups and chromosomes.
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Chapter 13 slide 16
Fig. 15.4 Testcross to show that two genes are linked
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 17
Generating a Genetic Map
1. A Genetic map is generated from estimating the crossover rate in a particular segment of
the chromosome. It may not exactly match the physical map because crossover is not
equally probable at all sites on the chromosome.
2. Recombination frequency is also used to predict progeny in genetic crosses. For
example, a 20% crossover rate between two pairs of alleles in a heterozygote (a+ b+ /a b)
will give 10% gametes of each recombinant type (a+ b and a b+ ).
3. A recombination frequency of 50% means that genes are unlinked. There are two ways
in which genes may be unlinked:
a. They may be on separate chromosomes.
b. They may be far apart on the same chromosome.
4. If the genes are on the same chromosome, multiple crossovers can occur. The further
apart two loci are, the more likely they are to have crossover events take place between
them. The chromatid pairing is not always the same in crossover, so that 2, 3, or 4
chromatids may participate in multiple crossover (Figure 15.5).
5. To determine whether the genes are on the same chromosome, or different ones, other
genes in the linkage group may be mapped in relation to a and b, and used to deduce
their locations.
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Chapter 13 slide 18
Fig. 15.5 Demonstration that the recombination frequency between two genes located
far apart on the same chromosome cannot exceed 50 percent
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 19
Gene Mapping Using Three-Point Testcrosses
1. Typically, geneticists design experiments to
gather data on several traits in 1 testcross. An
example of a three-point testcross would be
p+ r+ j+/p r j X p r j/p r j (Figure 15.6).
2. In the progeny, each gene has two possible
phenotypes. For three genes there are (2)3 = 8
expected phenotypic classes in the progeny.
Animation: Three-Point Mapping
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Chapter 13 slide 20
Fig. 15.6 Three-point mapping, showing the testcross used and the resultant progeny
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 21
Establishing the Order of Genes
1. The order of genes on the chromosome can be
deduced from results of the cross. Of the eight
expected progeny phenotypes:
a. Two classes are parental (p+ r+ j+/p r j and p r j/p r j)
and will be the most abundant.
b. Of the six remaining phenotypic classes, two will be
present at the lowest frequency, resulting from
apparent double crossover (p+ r+ j/p r j and p r j+/p r j).
This establishes the gene order as p j r (Figures 15.7).
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Chapter 13 slide 22
Fig. 15.7 Consequences of a double crossover in a triple heterozygote for three linked
genes
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 23
Calculating the Recombination Frequencies for
Genes
1. Cross data is organized to reflect the gene order, and in this example
the region between genes p and j is called region I, and that between j
and r is region II (Figure 15.8).
2. Recombination frequencies are now calculated for two genes at a time.
It includes single crossovers in the region under study, and double
crossovers, since they occur in both regions.
3. Recombination frequencies are used to position genes on the genetic
map (each 1% recombination frequency = 1 map unit) for the
chromosomal region (Figure 15.9).
4. Recombination frequencies are not identical to crossover frequencies,
and typically underestimate the true map distance.
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Chapter 13 slide 24
Fig. 15.8 Rewritten form of the testcross and testcross progeny in Figure 13.9, based
on the actual gene order p j r
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 25
Fig. 15.9 Genetic map of the p-j-r region of the chromosome computed from the
recombination data in Figure 13.12
iActivity: Crossovers and Tomato Chromosomes
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 26
Interference and Coincidence
1. Characteristically, double crossovers do not occur as often as expected
from the observed rate of single crossovers. Crossover appears to
reduce formation of other chiasmata nearby, producing interference.
Interference = 1 is total interference, with no other crossovers occurring
in the region.
2. The coefficient of coincidence expresses the extent of interference.
a. Interference = 1 - coefficient of coincidence. The values are inversely
related.
b. A value of 1 means the number of double crossovers that occurs is what
would be predicted on the basis of two independent events, and there is
no interference.
c. A value of 0 means that none of the expected crossovers occurred, and
interference is total.
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Chapter 13 slide 27
Fig. 13.14 Progeny of single and double crossovers
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 28
Calculating Accurate Map Distances
1. Recombination frequency generally underestimates the true map
distance (Figure 15.10):
a. Double crossovers between two loci will restore the parental genotype,
as will any even number of crossovers. These will not be counted as
recombinants, even though crossovers have taken place.
b. A single crossover will produce recombinant chromosomes, as will any
odd number of crossovers. Progeny analysis assumes that every
recombinant was produced by a single crossover.
c. Map distances for genes that are less than 7 mu apart are very accurate.
As distance increases, accuracy declines because more crosses go
uncounted.
2. Mapping functions are mathematical formulas used to define the
relationship between map distance and recombination frequency. They
are based on assumptions about the frequency of crossovers compared
with distance between genes (Figure 15.11).
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Chapter 13 slide 29
Fig. 15.11 A mapping function for relating map distance and recombination frequency
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
台大農藝系 遺傳學 601 20000
Chapter 13 slide 30