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Title: Gene Linkage and Genetic Mapping


1
4 Gene Linkage and Genetic Mapping
2
If two genes are on different chromosomes
3
If two genes are on different chromosomes
Half look like they got a set of the parents
chromosomes
And half look like they got a mix of both parents
chromosomes
4
Now suppose both gene A and B were next to each
other on the same chromosome.
What happens to the ratios in this diagram?
5
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7
Gene Mapping
  • Gene mapping determines the order of genes and
    the relative distances between them in map units
  • 1 map unit 1 cM (centimorgan)
  • Gene mapping methods use recombination
  • frequencies between alleles in order to
    determine the relative distances between them
  • Recombination frequencies between genes are
    inversely proportional to their distance apart
  • Distance measurement 1 map unit 1 percent
    recombination (true for short distances)

8
Fig. 4.6
9
Gene Mapping
  • Genes with recombination frequencies less than 50
    percent are on the same chromosome linked)
  • Linkage group all known genes on a chromosome
  • Two genes that undergo independent assortment
    have recombination frequency of 50 percent and
    are located on nonhomologous chromosomes or far
    apart on the same chromosome unlinked

10
Mapping the distance between two genes
Starting with pure breeding lines, Cross Parent
1(AA BB) with Parent 2(aa bb) So Parental
chromosomes in the F1 have to be AB and ab
Now cross (AB ab) F1 progeny with (ab ab) tester
to look for recombination on these
chromosomes. Suppose you Get AB
ab 583 ltparental ab ab 597 ltparental Ab
ab 134 ltrecombinant aB ab 134 ltrecombinant
total 1448 so. 268 recombinants /1448
progeny 0.185 recombinants/progeny 18
.5 recombinants 18.5 mu

11
Mapping (and ordering) three genes
Starting with pure breeding lines, Cross Parent
1(AA BB DD) with Parent 2(aa bb dd) So you know
the Parental chromosomes in the F1 have to be ABD
and abc
Cross (ABD abd) F1 progeny with (abd abd)
tester Suppose you Get ABD
abd 580 ltparental ABd abd 3 abD
abd 5 ltparental abd abd 592 AbD
abd 45 ltrecombinant Abd abd 89 aBD
abd 94 aBd abd 40 ltrecombinant
total 1448
Ab aB (4589)(9440) recom 268 recom/1448
total 0.185 A-B 18.5mu
Bd bD (340)(545) 93 recom/1448 total
0.064 B-D 6.4mu

Ad aD (389)(594) 191 recom/1448 total
0.132 A-D 13.2mu
so the order must be A-----D---B
-13.2--6.4- ----18.5----
12
So How come 13.2 6.4 does not equal 18.5?
Cross (ABD abd) F1 progeny with (abd abd)
tester Suppose you Get ABD
abd 580 ltparental ABd abd 3 abD
abd 5 ltparental abd abd 592 AbD
abd 45 ltrecombinant Abd abd 89 aBD
abd 94 aBd abd 40 ltrecombinant
total 1448
Ab aB (4589)(9440) recom 268 recom/1448
total 0.185 A-B 18.5mu

A-----D---B -13.2--6.4- ----18.5----
13
  • Chromosome interference crossovers in one region
    decrease the probability of a second crossover
    close by
  • Coefficient of coincidence observed number of
    double recombinants divided by the expected
    number
  • \
  • Interference 1-Coefficient of coincidence

If the two crossovers were independent, we would
expect that the probability of seeing two
recombination events occur would be 0.132
between A-D AND 0.064 between D-B 0.132 X 0.064
0.008 For every 1448 progeny, this would be
(1448x0.008)12.23 double recombinants We
actually observed only (53) 8 double
recombinants So the Coefficient of coincidence
observed / expected 8/12.23 0.65
Interference 1-Coefficient of coincidence
1- 0.65 0.35
14
Mapping Genes in Human Pedigrees
  • Methods of recombinant DNA technology are used to
    map human chromosomes and locate genes
  • Genes can then be cloned to determine structure
    and function
  • Human pedigrees and DNA mapping are used to
    identify dominant and recessive disease genes
  • Polymorphic DNA sequences are used in human
    genetic mapping.

15
Genetic Polymorphisms
  • The presence in a population of two or more
    relatively common forms of a gene or a chromosome
    is called polymorphism
  • Changes in DNA fragment length produced by
    presence or absence of the cleavage sites in DNA
    molecules are known as restriction fragment
    length polymorphism (RFLP)
  • A prevalent type of polymorphism is a single base
    pair difference, simple-nucleotide polymorphism
    (SNP)
  • A genetic polymorphism resulting from a tandemly
    repeated short DNA sequence is called a simple
    sequence repeat (SSR)

16
RFLPs
  • Restriction endonucleases are used to map genes
    as they produce a unique set of fragments for a
    gene
  • There are more than 200 restriction endonucleases
    in use, and each recognizes a specific sequence
    of DNA bases
  • EcoR1 cuts double-stranded DNA at the sequence
  • 5-GAATTC-3 wherever it occurs

17
Fig. 4.18
18
RFLPs
  • Differences in DNA sequence generate different
    recognition sequences and DNA cleavage sites for
    specific restriction enzymes
  • Two different genes will produce different
    fragment patterns when cut with the same
    restriction enzyme due to differences in DNA
    sequence

Fig. 4.19
19
SNPs
  • Single-nucleotide polymorphisms (SNPs) are
    abundant in the human genome
  • Rare mutants of virtually every nucleotide can
    probably be found, but rare variants are not
    generally useful for family studies of heritable
    variation in susceptibility to disease
  • For this reason, in order for a difference in
    nucleotide sequence to be considered as an SNP,
    the less-frequent base must have a frequency of
    greater than about 5 in the human population.
  • By this definition, the density of SNPs in the
    human genome averages about one per 1300 bp

20
SSRs
  • A third type of DNA polymorphism results from
    differences in the number of copies of a short
    DNA sequence that may be repeated many times in
    tandem at a particular site in a chromosome
  • When a DNA molecule is cleaved with a restriction
    endonuclease that cleaves at sites flanking the
    tandem repeat, the size of the DNA fragment
    produced is determined by the number of repeats
    present in the molecule
  • There is an average of one SSR per 2 kb of human
    DNA

21
Mapping Genes in Human Pedigrees
  • One source of the utility of SNPs and SSRs in
    human genetic mapping is their high density
    across the genome
  • Additional source of utility of SSRs in genetic
    mapping is the large number of alleles that can
    be present in any human population.

22
Mapping Genes in Human Pedigrees
  • Human pedigrees can be analyzed for the
    inheritance pattern of different alleles of a
    gene based on differences in SSRs and SNPs
  • Restriction enzyme cleavage of polymorphic
    alleles that are different in RFLP pattern
    produces different size fragments by gel
    electrophoresis

23
A rare recessive disease that affects people late
in life is 90 linked to the RFLP marker on the
gel below.
Whats the probability that the grandchildren
(row III) are carriers?
90 10 9010 10 90 90
24
A rare recessive disease that affects people late
in life is 90 linked to the RFLP marker on the
gel below.
Whats the probability that the grandchildren
(row III) are carriers?
10 90 10 90 90 10 10
25
Tetrad Analysis
  • In some species of fungi, each meiotic tetrad is
    contained in a sac-like structure, called an
    ascus
  • Each product of meiosis is an ascospore, and all
    of the ascospores formed from one meiotic cell
    remain together in the ascus
  • Several features of ascus-producing organisms are
    especially useful for genetic analysis
  • They are haploid, so the genotype is expressed
    directly in the phenotype
  • They produce very large numbers of progeny
  • Their life cycles tend to be short

26
Tetrad Analysis
  • In tetrads when two pairs of alleles are
    segregating, three patterns of segregation are
    possible
  • parental ditype (PD) two parental genotypes
  • nonparental ditype (NPD) only recombinant
    combinations
  • tetratype (TT) all four genotypes observed

27
Tetrad Analysis
  • When genes are unlinked, the parental ditype
    tetrads and the nonparental ditype tetrads are
    expected in equal frequencies PD NPD
  • Linkage is indicated when nonparental ditype
    tetrads appear with a much lower frequency than
    parental ditype tetrads PD NPD
  • Map distance between two genes that are
    sufficiently close that double and higher levels
    of crossing-over can be neglected, equals
  • 1/2 x (Number TT / Total number of tetrads) x 100

28
Neurospora Ordered Tetrads
  • Ordered asci also can be classified as PD, NPD,
    or TT with respect to two pairs of alleles, which
    makes it possible to assess the degree of linkage
    between the genes
  • The fact that the arrangement of meiotic products
    is ordered also makes it possible to determine
    the recombination frequency between any
    particular gene and its centromere

29
Tetrad Analysis Ordered Tetrads
  • Homologous centromeres of parental chromosomes
    separate at the first meiotic division
  • The centromeres of sister chromatids separate at
    the second meiotic division
  • When there is no crossover between the gene and
    centromere, the alleles segregate in meiosis I
  • A crossover between the gene and the centromere
    delays segregation alleles until meiosis II

30
  • The map distance between the gene and its
    centromere equals
  • 1/2 x (Number of asci with second division
    segregation/ Total number of asci) x 100
  • This formula is valid when the gene is close
    enough to the centromere and there are no
    multiple crossovers

31
Gene Conversion
  • Most asci from heterozygous Aa diploids
    demonstrate normal Mendelian segregation and
    contain ratios of
  • 2A 2a in four-spored asci, or 4A 4a in
    eight-spored asci
  • Occasionally, aberrant ratios are also found,
    such as
  • 3A 1a or 1A 3a and 5A 3a or 3A 5a. The
    aberrant asci are said to result from gene
    conversion because it appears as if one allele
    has converted the other allele into a form like
    itself
  • Gene conversion is frequently accompanied by
    recombination between genetic markers on either
    side of the conversion event, even when the
    flanking markers are tightly linked
  • Gene conversion results from a normal DNA repair
    process in the cell known as mismatch repair
  • Gene conversion suggests a molecular mechanism of
    recombination

32
  • One of two ways to resolve the resulting
    structure, known as a Holliday junction, leads to
    recombination, the other does not
  • The breakage and rejoining is an enzymatic
    function carried out by an enzyme called the
    Holliday junction-resolving enzyme

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34
5 Human Chromosomes and Chromosome Behavior
35
Karyotype stained and photographed preparation
of metaphase chromosomes arranged according to
their size and position of centromeres
36
Human Chromosomes
  • Each chromosome in karyotype is divided into two
    regions (arms) separated by the centromere
  • p short arm (petit) q long arm
  • p and q arms are divided into numbered bands and
    interband regions based on pattern of staining
  • Within each arm the regions are numbered.

37
Centromeres
  • Chromosomes are classified according to the
    relative position of their centromeres
  • In metacentric it is located in middle of
    chromosome
  • In submetacentriccloser to one end of chromosome
  • In acrocentricnear one end of chromosome
  • Chromosomes with no centromere, or with two
    centromeres, are genetically unstable

38
Abnormal Chromosome Numbers
  • Aneuploid unbalanced set of chromosomes
    relative gene dosage is upset (example trisomy
    of chromosome 21)
  • Monosomic loss of a single chromosome copy
  • Polysomic extra copies of single chromosomes
  • Most chromosome abnormalities lethal, frequently
    in spontaneous abortions.
  • Exceptions are trisomy 13, trisomy 18, and
    trisomy 21 (Down syndrome), and the Sex
    chromosomes

39
An extra X or Y chromosome usually has a
relatively mild effect. Why? 1) X chromosme
inactivation/Dosage Compensation 2) Not much
(essential) on the Y
  • EXAMPLES
  • Trisomy-X 47, XXX (female)
  • Double-Y 47, XYY (male)
  • Klinefelter Syndrome 47, XXY (male, sterile)
  • Turner Syndrome 45, X (female, sterile)

40
  • Chromosome Abnormalities
  • Deletions/Duplications
  • Inversions
  • Translocations

41
Deletions Duplications
42
  • Inversions genetic rearrangements in which the
    order of genes in a chromosome segment is
    reversed
  • Inversions do not alter the genetic content but
    change the linear sequence of genetic information
  • In an inversion heterozygote, chromosomes twist
    into a loop in the region in which the gene order
    is inverted

Chromosome Inversions
43
  • Paracentric inversion
  • Does not include centromere
  • Crossing-over produces one acentric (no
    centromere) and one dicentric (two centromeres)
    chromosome
  • Pericentric inversion
  • Includes centromere
  • Crossing-over results in duplications and
    deletions of genetic information

44
Reciprocal Translocations
  • Adjacent-2 segregation homologous centromeres
    stay together at anaphase I gametes have a
    segment duplication and deletion
  • Alternate segregation half the gametes receive
    both parts of the reciprocal translocation and
    the other half receive both normal chromosomes
    all gametes are euploid, i.e have normal genetic
    content, but half are translocation carriers

45
Polyploidy
  • Polyploid species have multiple complete sets of
    chromosomes
  • The basic chromosome set, from which all the
    other genomes are
  • formed, is called the monoploid set
  • The haploid chromosome set is the
  • set of chromosomes present in a gamete,
    irrespective of the chromosome number in the
    species.
  • Polyploids can arise from genome duplications
    occurring before or after fertilization
  • Through the formation of unreduced gametes that
    have double the normal complement of chromosomes
    or
  • Through abortive mitotic division, called
    endoreduplication.

46
Polyploids can generate new species
47
A seedless watermelon is a sterile hybrid which
is created by crossing male pollen for a
watermelon, containing 22 chromosomes per cell,
with a female watermelon flower with 44
chromosomes per cell.
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49
The karyotype of the Chinook salmon has been
characterized as 2N 68, with 16 pairs of
metacentric chromosomes and 18 pairs of
acrocentric chromosomes (Simon 1963)
50
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