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4Gene Linkage and Genetic

Mapping

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If two genes are on different chromosomes…

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And half look like they got a mix of both parents chromosomes…

Half look like they got a set of the parents chromosomes…

If two genes are on different chromosomes…

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Now suppose both gene A and B were next to each other on the same chromosome.

What happens to the ratios in this diagram?

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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)

8Fig. 4.6

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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

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Now cross (AB ab) F1 progeny with (ab ab) tester to look for recombination on these chromosomes.Suppose you Get……

AB ab 583 <parental

ab ab 597 <parental

Ab ab 134 <recombinant

aB ab 134 <recombinant total= 1448

so…. 268 recombinants /1448 progeny = 0.185 recombinants/progeny=

18.5% recombinants=18.5 mu

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

Mapping the distance between two genes

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Cross (ABD abd) F1 progeny with (abd abd) tester

Suppose you Get……

ABD abd 580 <parentalABd abd 3

abD abd 5 <parental abd abd 592

AbD abd 45 <recombinant Abd abd 89

aBD abd 94aBd abd 40 <recombinant total= 1448

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

Ab + aB = (45+89)+(94+40) recom268 recom/1448 total =0.185A-B =18.5mu

Bd + bD = (3+40)+(5+45) 93 recom/1448 total= 0.064B-D =6.4mu

Ad + aD = (3+89)+(5+94) 191 recom/1448 total= 0.132 A-D =13.2mu

so the order must be A-----D---B -13.2--6.4- ----18.5----

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Cross (ABD abd) F1 progeny with (abd abd) tester

Suppose you Get……

ABD abd 580 <parentalABd abd 3

abD abd 5 <parental abd abd 592

AbD abd 45 <recombinant Abd abd 89

aBD abd 94aBd abd 40 <recombinant total= 1448

Ab + aB = (45+89)+(94+40) recom268 recom/1448 total =0.185A-B =18.5mu

A-----D---B -13.2--6.4- ----18.5----

So How come 13.2 + 6.4 does not equal 18.5?

We missed the double recombinants on the first pass…

longer the distance, more potential to underestimate recomb freq.

(45+89)+(94+40)+2(3+5) recom284recom/1448 total = 0.196A-B =19.6mu

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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

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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 recombinantsWe actually observed only (5+3)= 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

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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.

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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)

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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

17Fig. 4.18

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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

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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

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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

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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.

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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

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A rare recessive disease that affects people late in lifeis 90% linked to the RFLP marker on the gel below.

What’s the probability that the grandchildren (row III) are carriers?

90% 10% 90%10% 10% 90% 90%

What’s the probability that II1 is a carrier of the disease?

dd DD

Dd

100%

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A rare recessive disease that affects people late in lifeis 90% linked to the RFLP marker on the gel below.

What’s the probability that the grandchildren (row III) are carriers?

10% 90% 10% 90% 90% 10% 10%

What’s the probability that II1 is a carrier of the disease?

dd DD

Dd

100%

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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

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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

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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, equals1/2 x (Number TT / Total number of tetrads) x 100

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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

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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

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• 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

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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

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•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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5Human Chromosomes and

Chromosome Behavior

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Karyotype = stained and photographed preparation of metaphase chromosomes arranged according to their size and position of centromeres

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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.

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Centromeres• Chromosomes are classified

according to the relative position of their centromeres

• In metacentric it is located in middle of chromosome

• In submetacentric—closer to one end of chromosome

• In acrocentric—near one end of chromosome

• Chromosomes with no centromere, or with two centromeres, are genetically unstable

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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

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EXAMPLES:• Trisomy-X = 47, XXX

(female)• Double-Y = 47, XYY

(male)• Klinefelter Syndrome =

47, XXY (male, sterile)• Turner Syndrome = 45, X

(female, sterile)

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

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Chromosome Abnormalities

•Deletions/Duplications

•Inversions

•Translocations

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Deletions

Duplications

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A B C

C B A

BB

AAC

C

Chromosome Inversions

• 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

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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

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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

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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.

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Polyploids can generate new species

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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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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)

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