Transcript CHAPTER 13 Gene Mapping in Eukaryotes

Peter J. Russell
CHAPTER 13
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
Morgan’s Linkage 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 13.1).
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. 13.1 Morgan’s experimental crosses of white-eye and miniature-wing variants of
Drosophila melanogaster
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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
Fig. 13.2 Mechanism of crossing-over
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Chapter 13 slide 6
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 7
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 13.3):
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 8
Fig. 13.3 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 9
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 13.4).
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 10
+ (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 13.1).
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Chapter 13 slide 11
Fig. 13.4 Stern’s experiment to demonstrate the relationship between genetic
recombination and chromosomal exchange in Drosophila melanogaster
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Chapter 13 slide 12
Box Fig. 13.1 Holliday model for reciprocal genetic recombination
Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.
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Chapter 13 slide 13
Crossing-Over at the Four-Chromatid Stage of
Meiosis
1. Neurospora crassa (orange bread mold) forms eight haploid spores.
Their arrangement in the ascus reflects the orientation of chromatids in
the metaphase tetrad of meiosis I (ordered tetrads).
2. To determine when crossing-over occurs, crosses were made between
haploid Neurospora strains of different mating types (A and a) (Figure
13.5).
a. Mating type A, with the genotype met his+ , can make the amino acid
histidine, but not the amino acid methionine.
b. Mating type a, with the genotype met+ his, can make the amino acid
methionine, but not the amino acid histidine.
c. The met locus and the his locus are on the same chromosome.
3. Mating produces a diploid nucleus, heterozygous for both genes. The
diploid nucleus undergoes meiosis, forming four haploid spores. Each
spore is then duplicated by mitosis, for a total of eight haploid spores.
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Chapter 13 slide 14
Fig. 13.5 Experiment showing that crossing-over occurs at the four-chromatid stage of
meiosis
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Chapter 13 slide 15
4. Crossover would produce haploid recombinant progeny with the
genotypes met +his+ (wild-type) and met his (requires both in the
medium).
a. if no crossover occurred, the ascus would contain four spores of each
parental type, and no recombinant ones.
b. If crossover occurred in interphase before DNA replication, the ascus
would contain four spores of each recombinant type, and no parental
ones.
c. If crossover occurred at the tetrad stage of prophase I, the ascus would
contain two spores of each parental type, and two of each recombinant
type.
5. The overall result of this experiment was an ascus with two spores of
each possible phenotypic class. This represents crossover between two
chromatids, no crossover between the other two, and then mitotic
duplication of each. This is evidence that:
a. Crossover occurs at the tetrad stage of prophase I.
b. Crossover is a reciprocal process, with an equal exchange of
chromosome material.
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Chapter 13 slide 16
Constructing Genetic Maps
1.Genetic recombination experiments can be
used in genetic mapping.
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Chapter 13 slide 17
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 18
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 19
Fig. 13.6 The relationship between crossing-over and map distance
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台大農藝系 遺傳學 601 20000
Chapter 13 slide 20
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.
2. Mapping of genes with other mechanisms of inheritance is also done with two-point
testcrosses:
a. For autosomal dominants, a double heterozygote (A B/A+ B+) is testcrossed with a homozygous
recessive individual (A+ B+/A+ B+). The only difference from the experiment above is that when the
mutant alleles are dominant, the recessive alleles are wild-type.
b. For X-linked recessives, a female double heterozygote (a+ b+/a b) is crossed with a male
hemizygous for the recessive alleles (a b/Y).
c. For X-linked dominants, a female double heterozygote (A B/A+ B+) is crossed with a male
hemizygous for the wild-type alleles (A+ B+/Y).
d. 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.
e. Phenotypes obtained in any of these crosses will depend on whether the alleles are arranged in
coupling (cis) or repulsion (trans).
f.
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.
g. Mapping in all types of organisms shows genes arranged with a 1-to-1 correspondence between
linkage groups and chromosomes.
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Chapter 13 slide 21
Fig. 13.7 Testcross to show that two genes are linked
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Chapter 13 slide 22
Generating a Genetic Map
1. 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 13.8).
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 23
Fig. 13.8 Demonstration that the recombination frequency between two genes located
far apart on the same chromosome cannot exceed 50 percent
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Chapter 13 slide 24
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 13.9).
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 25
Fig. 13.9 Three-point mapping, showing the testcross used and the resultant progeny
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Chapter 13 slide 26
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 13.9, 13.10
and 13.11).
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Chapter 13 slide 27
Fig. 13.10 Consequences of a double crossover in a triple heterozygote for three linked
genes
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Chapter 13 slide 28
Fig. 13.11 Rearrangement of the three genes in Figure 13.9 to p j r
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Chapter 13 slide 29
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 13.12).
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 13.13).
4. Recombination frequencies are not identical to crossover
frequencies, and typically underestimate the true map distance.
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Chapter 13 slide 30
Fig. 13.12 Rewritten form of the testcross and testcross progeny in Figure 13.9, based
on the actual gene order p j r
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Chapter 13 slide 31
Fig. 13.13 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
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Chapter 13 slide 32
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 33
Fig. 13.14 Progeny of single and double crossovers
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Chapter 13 slide 34
Calculating Accurate Map Distances
1. Recombination frequency generally underestimates the true map
distance:
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
13.15).
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Chapter 13 slide 35
Fig. 13.15 A mapping function for relating map distance and recombination frequency
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Chapter 13 slide 36
Tetrad Analysis in Certain Haploid
Eukaryotes
1. In some haploid eukaryotic organisms (fungi or single-celled algae)
products of a single meiosis, the meiotic tetrad, are contained within
one structure. Tetrad analysis provides insight into meiotic events.
2. In haploid organisms, the phenotype correlates directly with the
genotype of each member of the tetrad (no dominance or
recessiveness occurs).
3. Life cycles of organisms typically used in tetrad analysis:
a. Saccharomyces cerevisiae (baker’s yeast), has two mating types, a and α.
i. Asexual reproduction occurs mitotically (vegetative life cycle) in the
haploid yeast.
ii. Sexual reproduction, fusion of a haploid a cell with a haploid a one,
produces a diploid cell (a/α) that also reproduces mitotically, giving rise
to identical diploid cells.
iii. Diploid cells sporulate by meiosis, producing four haploid ascospores
contained in an ascus. Of the ascospores, two will be type a and two
type a. In yeast they are unordered tetrads, arranged randomly in the
ascus (Figure 13.16).
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Chapter 13 slide 37
Fig. 13.16 Life cycle of the yeast Saccharomyces cerevisiae
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Chapter 13 slide 38
b. Chiamydomonas reinhardtii is a single-celled green algae with haploid
vegetative cells and two mating types.
i. Nitrogen limitation causes the cells to become gametes, and the two
opposite mating types (mt+ and mt-) fuse to produce a zygote.
ii. Meiosis of the zygote produces an unordered tetrad of haploid cells,
two of type mt+ and two mt-.
iii. Mitosis of each haploid cell results in new haploid algae cells.
c. Neurospora crassa is similar, but its ascospores are arranged in an
ordered tetrad.
i. The ordered tetrad reflects the orientation of the fur chromatids of the
tetrad at the metaphase plate in meiosis I. Spores can be isolated in
order or randomly.
ii. The ascus contains eight spores, because each haploid cell duplicates by
mitosis before spore maturation.
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Chapter 13 slide 39
Using Random-Spore Analysis to Map Genes in
Haploid Eukaryotes
1. In these organisms, three-point crosses have been
used effectively for mapping. Haploidy simplifies
interpretation of the results.
2. Analysis of random-spore data is the same as for
diploid eukaryotes. It is used for determining
linkage and constructing genetic maps (Figure
13.17).
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Chapter 13 slide 40
Fig. 13.17 Typical genetic cross for mapping three genes in a haploid organism such as
yeast or Neurospora
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Chapter 13 slide 41
Using Tetrad Analysis to Map Two Linked
Genes
1. In a two-point cross, mating produces a diploid heterozygous for both
genes, and then meiosis makes haploid spores. In the fungi
(Saccharomyces and Neurospora), the spores of the tetrad are
micromanipulated for separate germination and analysis.
2. In the cross a+ b+/a b, in which a and b are linked, three different tetrad
types can result (Figure 13.18):
a. Parental-ditype (PD) tetrad has only the two parental types (a+ b+ and a b).
A PD tetrad results either if no crossing-over occurs between the two genes,
or if a double crossover involving the same two chromatids occurs.
b. Tetratype (T) has two parentals (a+ b+ and a b) and two recombinants (a+ b
and a b+). A T tetrad results either from a single crossover between the two
genes, or if a double crossover involving three chromatids occurs.
c. Non-parental-ditype (NPD) has only recombinants (a+ b and a b+). A NPD
tetrad results from a double crossover that involves all four chromatids.
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Chapter 13 slide 42
Fig. 13.18 Three types of tetrads
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Chapter 13 slide 43
3. Segregation of alleles is 2:2. Rarely, 3:1 or 1:3
ratios are seen, due to gene conversion (Box
13.2).
4. For genes on different chromosomes, crossover is
not involved. PD and NPD tetrads are produced
with equal frequency; and no T tetrads are
expected (Figure 13.19).
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Chapter 13 slide 44
Box Fig. 13.2 Gene conversion by mismatch repair at two sites
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Chapter 13 slide 45
Fig. 13.19 Origin of tetrad types for a cross in which the two genes are located on
different chromosomes
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台大農藝系 遺傳學 601 20000
Chapter 13 slide 46
5. When genes are linked (Figure 13.20):
a. A single crossover produces a T tetrad.
b. Double crossovers vary depending on the strands involved:
i. If the same two chromatids are involved in both crossovers, a
PD tetrad results.
ii. If three chromatids are involved, a T tetrad results.
iii. If all four chromatids are involved, an NPD tetrad is
produced.
c. Genes are considered linked if the PD frequency is far greater
than the NPD frequency.
d. Genetic distance between the genes correlates with the
recombination frequency; and they are mapped accordingly,
e. For crosses involving more than two genes, they are considered
in pairs, and mapped two at a time relative to each other.
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Chapter 13 slide 47
Fig. 13.20a-c Origin of tetrad types for a cross in which both genes are located on the
same chromosome
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 48
Fig. 13.20d, e Origin of tetrad types for a cross in which both genes are located on the
same chromosome
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 49
Calculating Gene-Centromere Distance
in Organisms with Linear Tetrads
1. Neurospora’s eight spores represent the result of meiotic division followed by mitosis, and
are considered as four pairs. Their order reflects the orientation of the chromatids at
metaphase I, allowing the distance between genes and centromere to be mapped (Figure
13.21).
2. Centromeres separate just before the second meiotic division, and so spores in the top of
the ascus have the centromere from one parent, while those below have the other
parent¡¦s centromere.
3. In this example, mating type (A and a) is one locus, and the centromere is another.
a. If no crossover occurs between them, they show first-division segregation. After meiosis I, both
copies of A are at one pole and both copies of a at the other. The final result is a 4 : 4
segregation in the ascus.
b. Single crossover shows second-division segregation. A and a are each being present in two
nuclear areas until the second division, and their pattern of gene segregation depends on the
chromatids involved. Both patterns are distinguishable from the 4:4 seen in first-division
segregation:
i. A 2 : 2 : 2 :2 ratio results from AAaaAAaa and aaAAaaAA.
ii. A 2 : 4 : 2 ratio results from AAaaaaAA and aaAAAAaa.
c. The distance from the gene of interest (here the mating type locus) to the centromere is the
percentage of second-division tetrads.
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Chapter 13 slide 50
Fig. 13.21a Determination of gene-centromere distance of the mating-type locus in
Neurospora
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 51
Fig. 13.21b Determination of gene-centromere distance of the mating-type locus in
Neurospora
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 52
Mitotic Recombination
Discovery of Mitotic Recombination
1. Crossing-over during mitosis was first observed by Stern (1936) in Drosophila
(Figure 13.22).
a. The alleles involved are sex-linked and recessive to wild-type:
i. y produces yellow body color instead of wild-type grey.
ii. sn produces short, twisty bristles (“singed”) rather than the wild-type long,
curved ones. Bristles follow body color (y+/ - are black, and y / y are yellow).
b. Female progeny from the cross y+ sn / y+ sn X y sn+ I Y generally have the wild-type
phenotype of grey bodies and normal bristles, corresponding to their genotype (y+
sn / y sn+). But exceptions were seen:
i. Some flies had patches of yellow and/or singed bristles. This could be explained
by nondisjunction or chromosomal loss.
ii. Other flies had twin spots, adjacent regions of bristles, one yellow and the other
singed, a mosaic phenotype. The spots are reciprocal products of the same
genetic event, a mitotic crossing over.
c. Mitotic crossover occurred either between the centromere and the sn locus, or
between the sn and the y locus (Figure 13.23).
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Chapter 13 slide 53
Fig. 13.22 Body surface phenotype segregation in a Drosophila strain
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 54
Fig. 13.23 Production of the twin spot and single yellow spot shown in Figure 13.22 by
mitotic crossing-over
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 55
Mechanism of Mitotic Crossing-Over
1. A rare event occurring only in diploid cells, mitotic crossover can result
when replicated chromatids come together to form a structure similar to
the four-strand stage in meiosis (Figures 13.24 and 13.25).
2. If the starting genotype is d+ e / d e+ the two possible orientations of the
resulting chromatids are:
a. One cell with d+ e+ / d+ e+ and one with d e / d e. These are the ones
that are useful for mapping, because the recessive phenotype can be
observed in progeny of the d e / d e cells.
b. Reversal of the alleles, d e+ / d+ e. Phenotypically indistinguishable
from non-recombinant cells, these are not useful for mapping, but are
nonetheless derived from a crossover event.
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Chapter 13 slide 56
Fig. 13.24 Normal mitotic segregation of genes in a theoretical diploid cell with one
homologous pair of 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 57
Fig. 13.25 Result of a mitosis of the same cell type as the cell in Figure 13.24 but in
which a rare mitotic crossing-over occurs
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 58
Retinoblastoma, a Human Tumor That Can Be
Caused by Mitotic Recombination
1. Retinoblastoma is the most common childhood eye cancer, occurring
from birth to 4 years of age. Two types are known:
a. The sporadic (nonhereditary) form occurs in an individual with no
family history of the disease, and affects only one eye (unilateral).
b. The hereditary form affects both eyes (bilateral) and usually occurs at
an earlier age than sporadic.
2. A single gene (Rb) on chromosome 13q14 is involved.
a. In hereditary retinoblastoma, tumor cells have mutations in both copies
of this gene, while other cells in the same individual are heterozygous.
The disease is caused by a second mutation that affects the normal RB
allele.
b. The second mutation is often identical to the one on the other
chromosome, strong circumstantial evidence that the wild-type copy of
the gene is somehow replaced by the inherited mutated allele. One
possible explanation is mitotic recombination (Figure 13.25).
台大農藝系 遺傳學 601 20000
Chapter 13 slide 59