Download CHAPTER 16 Advanced Gene Mapping in Eukaryotes

Peter J. Russell
A molecular Approach 2nd Edition
CHAPTER 16
Advanced 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
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, MATa and MATα
(Figure 16.1).
i. Asexual reproduction occurs mitotically (vegetative life cycle) in the haploid
yeast.
ii. Sexual reproduction, fusion of a haploid a cell with a haploid α 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 2
Fig. 16.1 Life cycle of the yeast Saccharomyces cerevisiae
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台大農藝系 遺傳學 601 20000
Chapter 13 slide 3
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 four 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 4
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 5
Calculating Gene-Centromere Distance
in Organisms with Ordered 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
16.2).
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 6
Fig. 16.2a 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 7
Fig. 16.2b 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 8
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 16.3):
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 9
Fig. 16.3 Three types of tetrads
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台大農藝系 遺傳學 601 20000
Chapter 13 slide 10
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 16.4).
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Chapter 13 slide 11
Box Fig. 13.2 Gene conversion by mismatch repair at two sites
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
Fig. 16.4 Origin of tetrad types for a cross in which the two genes are located on
different 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 13
5. When genes are linked (Figure 16.9):
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 14
Fig. 16.5 a-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 15
Fig. 16.5 d, 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 16
Mitotic Recombination
Discovery of Mitotic Recombination
1. Mitotic crossing-over (mitotic recombination) produces a progeny cell with
different gene combinations than the diploid parental cell has.
a. Mitotic crossing-over occurs at a stage similar to the 4-strand stage of meiosis.
b. This stage is rare, so mitotic crossover is much less frequent than meiotic crossover.
2. Crossing-over during mitosis was first observed by Stern (1936) in Drosophila
(Figure 16.6).
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+ / 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 16.7).
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Chapter 13 slide 17
Fig. 16.6 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 18
Fig. 16.7 Production of the twin spot and single yellow spot shown in Figure 16.6 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 19
3. Mitotic crossing-over makes all genes distal to the
crossover point homozygous if the chromatids
align properly at the metaphase plate.
a. Proper alignment occurs one half of the time.
b. Homozygosity applies only to genes on the same
chromosome arm, with no effect on genes of the other
arm.
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Chapter 13 slide 20
Mitotic Recombination in the Fungus
Aspergillus nidulans
1. Meiotic recombination analysis is not used with Aspergillus because the fungus selfs. It is
used for studies of mitotic recombination because:
a. Its asexual spores are uninucleate.
b. Phenotype of each asexual spore is controlled by its nucleus.
c. Two haploid strains can be fused by mixing.
2. Heterozygous strains are constructed for study by fusing two haploid strains of different
genotype, resulting in a heterokaryon (mycelium with two nuclear types that coexist and
divide mitotically).
3. Rarely, haploid nuclei of the heterokaryon fuse to form a diploid nucleus (diploidization)
that will form diploid spores. Diploid spores are larger than haploid spores, and may be
isolated for study.
4. An example is given in Figure 16.8:
a. Allele designations are as follows:
i. w (with either y or y+) produces white colonies, w+y produces yellow, and w+y+ produces
green.
ii. ad cells require adenine in the growth medium, ad+ do not.
iii. pro cells require proline in the growth medium, pro+ do not.
iv. paba cells require para-aminobenzoic acid (PABA) in the growth medium, paba+ do not.
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v. bi cells require biotin in the growth medium,
bi+ do not.遺傳學 601 20000
Chapter 13 slide 21
Fig. 16.8 Photo of a green diploid Aspergillus colony, genotype w ad+ pro paba+ y+
bi/w+ ad pro+ paba y bi+ with a yellow and white sector.
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Chapter 13 slide 22
b. In this example, strain 1 is crossed with strain 2:
i. Strain 1 is wad+propaba+y+bi
ii. Strain 2 is w+adpro+pabaybi+
iii. The diploid strain produced will have wild-type alleles for all these
traits, and so will not require growth supplements.
c. Diploid Aspergillus spores will produce green colonies with haploid or
diploid sectors of white and yellow (green are also produced but are not
detectable). Spore size indicates whether the sector is haploid or
diploid.
d. Haploidization (mitosis without chromosome replication) produces
haploid segregants (haploid progeny nuclei). Haploid white sectors
provide an example:
i. Half the haploid white sectors have the genotype wad+propaba+y+bi
and the other half w+adpro+pabaybi+. Except for the w allele, these are
reciprocals.
ii. This indicates that w is on one chromosome, and the five others are
on a different chromosome, although gene order cannot be determined
from haploid segregants.
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Chapter 13 slide 23
e. Diploid (mitotic) segregants are rare, and so are usually single crossovers. All genes
distal to the crossover point on the chromosome arm are homozygous, allowing the
recessive phenotypes to be seen. Diploid yellow sectors are an example (Figure
16.9).
i. Mitotic crossover between pro and paba will produce a segregant
homozygous for the y allele (y+/y+ twin spot is green and so not detected). The
yellow segregant will require PABA acid to grow (Figure 16.10).
ii. Mitotic crossover between paba and y will also produce a yellow sector, but
these will not require PABA for growth.
iii. Gene order, therefore, may be determined by mitotic crossover experiments
with markers distal to the chromosome. Counting diploid segregants allows
computation of map distance between genes (similar to meiotic recombination
studies).
5. Genetic recombination without regular alternation of meiosis and fertilization
are parasexual systems. In fungi such as Aspergillus, the parasexual cycle is:
a. Formation of heterokaryon.
b. Fusion in the heterokaryon of haploid nuclei of different genotypes to form
heterozygous diploid nucleus.
c. Mitotic crossing-over within the diploid nucleus.
d. Haploidization of the diploid nucleus.
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Chapter 13 slide 24
Fig. 16.9 Possible mitotic crossing-over event between the pro and paba loci that can
give rise to a diploid yellow sector in the green diploid Aspergillus strain of Figure 16.8
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Chapter 13 slide 25
Fig. 16.10 Production of diploid yellow sector in a green diploid Aspergillus strain by
mitotic crossing-over between the paba and y loci.
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Chapter 13 slide 26
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.
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Chapter 13 slide 27
Mapping Human Genes
Mapping Human Genes by Recombination
Analysis
1. Experimental crosses are not done in humans and so genetic mapping relies on pedigree
analysis, but only a few multigenerational pedigrees are appropriate for analyzing
autosomal linkage. Recombination analysis in humans is easier for X-linked genes.
2. In a theoretical example (Figure 16.11):
a. A man with rare X-linked recessive alleles a and b marries a woman who expresses neither of
these traits and is likely wild type (a+b+/a+b+).
b. A daughter of this couple would be doubly heterozygous (a+b+/ab). She would produce
recombinant gametes at a frequency related to the distance between the genes.
c. If she pairs with a normal (a+b+/Y) man:
i. All daughters will be wild type, due to inheriting a+b+ from their father.
ii. Sons will express four phenotypic classes (parental a+b+ or ab, or recombinant a+b or
ab+).
d. If a large number of pedigrees are available, an estimate of genetic map distance may be
obtained for these genes.
3. This approach has shown that the distance between the green weakness gene (g)
responsible for a form of color blindness, and the hemophilia A gene (h) is eight map
units.
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Chapter 13 slide 28
Fig. 16.11 Calculation of recombination frequency for the X-linked human genes by
analyzing the male progenies of a woman doubly heterozygous for the two genes.
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Chapter 13 slide 29
lod Score Method for Analyzing Linkage of
Human Genes
1. The lack of suitable human pedigrees showing segregation of defined
linked traits makes mapping autosomes especially difficult, and so
usually the lod (logarithm of odds) score method is used for statistical
analysis of pedigree data.
a. A lod score compares the expected distributions of traits if they are
linked, and if they are not linked.
b. The lod score is the log10 of the ratio of the two probabilities. The
higher the lod score, the closer the two genes.
2. The map distance for linked markers is computed from the recombination frequency given by the highest lod score, by solving lod scores
for a range of proposed map units. For the human genome, 1mu
is approximately 1 megabase (Mb).
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Chapter 13 slide 30
High-Density Genetic Maps of the Human
Genome
1. Human genetic mapping was revolutionized by:
a. Discovery of many polymorphic DNA markers.
b. Development of molecular tools to type the markers (determining
molecular phenotypes).
2. Hundreds of markers may be typed in a given cross, and computer
algorithms are used to determine linkage relationships.
3. A high-density human genetic map was completed in 1994.
a. A consortium of laboratories worked on the same set of DNA samples
(mapping panel), so their data could be combined.
b. Localized 5,264 STR (short tandem repeat) markers to 2,335
chromosomal loci, with an average density of one marker per 599kb
(Figure 16.12).
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Chapter 13 slide 31
Fig. 16.12 A high density genetic map with 5,264 microsattelites localized to 2,335
chromosomal loci
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Chapter 13 slide 32
Physical Mapping of Human Genes
FISH (Fluorescent in situ Hybridization) Maps
1. Individual metaphase chromosomes are probed in
situ with specific fluorescently labeled DNA
sequences, identifying homologous sequences in
the chromosome (Figure 16.13).
2. Different probes labeled with different fluorescent
dyes may be used in the same experiment.
Fluorescence microscopy provides data for
computer imaging analysis to determine the
binding site(s) for each probe.
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Chapter 13 slide 33
Fig. 16.13 An example of the results of fluorescent in situ hybridization (FISH) in which
fluorescently tagged DNA probes were hybridized to human metaphase chromosome
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Chapter 13 slide 34
Radiation Hybrid Maps
1. A radiation hybrid (RH) is a rodent cell line carrying a
small genomic DNA molecule from another organism
(e.g., a human). In this technique (Figure 16.14):
a. Exposure to X-rays breaks the DNA in human cells. The
fragments become smaller with more X-ray exposure, and
fragment length determines the map resolution.
b. Irradiation kills the human cells, which are then fused
with rodent cells, rescuing chromosomal fragments that
are typically a few Mb in length.
c. Human DNA in the RH is analyzed for gene and/or DNA
markers. The closer two markers are to each other on the
chromosome, the more likely they are to be found
together in an RH.
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Chapter 13 slide 35
Fig. 16.14 Making the radiation hybrid.
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Chapter 13 slide 36