7 Gene MappingGene mapping determines the order of genes and the relative distances between them in map units1 map unit = 1 cM (centimorgan)Gene mapping methods use recombinationfrequencies between alleles in order to determine the relative distances between themRecombination frequencies between genes are inversely proportional to their distance apartDistance measurement: 1 map unit = 1 percent recombination (true for short distances)
9 Gene MappingGenes with recombination frequencies less than 50 percent are on the same chromosome = linked)Linkage group = all known genes on a chromosomeTwo 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 abNow cross (AB ab) F1 progeny with (ab ab) tester to look for recombination on these chromosomes.Suppose you Get……AB ab 583 <parentalab ab 597 <parentalAb ab 134 <recombinantaB ab 134 <recombinant total= 1448so…. 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 abcCross (ABD abd) F1 progeny with (abd abd) testerSuppose you Get……ABD abd 580 <parentalABd abd 3abD abd 5 <parentalabd abd 592AbD abd 45 <recombinantAbd abd 89aBD abd 94aBd abd 40 <recombinant total= 1448Ab + aB = (45+89)+(94+40) recom268 recom/1448 total =0.185A-B =18.5muBd + bD = (3+40)+(5+45)93 recom/1448 total= 0.064B-D =6.4muAd + aD = (3+89)+(5+94)191 recom/1448 total= 0.132A-D =13.2muso the order must be A-----D---B
12 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 =19.6muCross (ABD abd) F1 progeny with (abd abd) testerSuppose you Get……ABD abd 580 <parentalABd abd 3abD abd 5 <parentalabd abd 592AbD abd 45 <recombinantAbd abd 89aBD abd 94aBd abd 40 <recombinant total= 1448Ab + aB = (45+89)+(94+40) recom268 recom/1448 total =0.185A-B =18.5muA-----D---B
13 Interference = 1-Coefficient of coincidence Chromosome interference: crossovers in one region decrease the probability of a second crossover close byCoefficient of coincidence = observed number of double recombinants divided by the expected number\Interference = 1-Coefficient of coincidenceIf the two crossovers were independent,we would expect that the probability of seeing two recombination events occur would bebetween A-D AND between D-B0.132 X = 0.008For every 1448 progeny, this would be (1448x0.008)=12.23 double recombinantsWe actually observed only (5+3)= 8 double recombinantsSo the Coefficient of coincidence = observed / expected = 8/12.23 =0.65Interference = 1-Coefficient of coincidence== 0.35
14 Mapping Genes in Human Pedigrees Methods of recombinant DNA technology are used to map human chromosomes and locate genesGenes can then be cloned to determine structure and functionHuman pedigrees and DNA mapping are used to identify dominant and recessive disease genesPolymorphic 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 polymorphismChanges 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 RFLPsRestriction endonucleases are used to map genes as they produce a unique set of fragments for a geneThere are more than 200 restriction endonucleases in use, and each recognizes a specific sequence of DNA basesEcoR1 cuts double-stranded DNA at the sequence5’-GAATTC-3’ wherever it occurs
18 RFLPsDifferences in DNA sequence generate different recognition sequences and DNA cleavage sites for specific restriction enzymesTwo different genes will produce different fragment patterns when cut with the same restriction enzyme due to differences in DNA sequenceFig. 4.19
19 SNPsSingle-nucleotide polymorphisms (SNPs) are abundant in the human genomeRare 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 diseaseFor 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 SSRsA 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 chromosomeWhen 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 moleculeThere 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 genomeAdditional 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 SNPsRestriction 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.What’s the probability that II1 is a carrier of the disease?ddDDDd100%What’s the probability that the grandchildren (row III) are carriers?90% 10% 90%10% 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.What’s the probability that II1 is a carrier of the disease?ddDDDd100%What’s the probability that the grandchildren (row III) are carriers?10% 90% 10% 90% 90% 10% 10%
25 Tetrad AnalysisIn some species of fungi, each meiotic tetrad is contained in a sac-like structure, called an ascusEach product of meiosis is an ascospore, and all of the ascospores formed from one meiotic cell remain together in the ascusSeveral features of ascus-producing organisms are especially useful for genetic analysis:They are haploid, so the genotype is expressed directly in the phenotypeThey produce very large numbers of progenyTheir life cycles tend to be short
26 Tetrad AnalysisIn tetrads when two pairs of alleles are segregating, three patterns of segregation are possibleparental ditype (PD) = two parental genotypesnonparental ditype (NPD) = only recombinant combinationstetratype (TT) = all four genotypes observed
27 Tetrad AnalysisWhen genes are unlinked, the parental ditype tetrads and the nonparental ditype tetrads are expected in equal frequencies: PD = NPDLinkage is indicated when nonparental ditype tetrads appear with a much lower frequency than parental ditype tetrads: PD » NPDMap 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
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 genesThe 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 divisionThe centromeres of sister chromatids separate at the second meiotic divisionWhen there is no crossover between the gene and centromere, the alleles segregate in meiosis IA 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 100This formula is valid when the gene is close enough to the centromere and there are no multiple crossovers
31 Gene ConversionMost asci from heterozygous Aa diploids demonstrate normal Mendelian segregation and contain ratios of2A : 2a in four-spored asci, or 4A : 4a in eight-spored asciOccasionally, aberrant ratios are also found, such as3A : 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 itselfGene conversion is frequently accompanied by recombination between genetic markers on either side of the conversion event, even when the flanking markers are tightly linkedGene conversion results from a normal DNA repair process in the cell known as mismatch repairGene 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 notThe breakage and rejoining is an enzymatic function carried out by an enzyme called the Holliday junction-resolving enzyme
34 Human Chromosomes and Chromosome Behavior 5Human Chromosomes and Chromosome Behavior
35 Karyotype = stained and photographed preparation of metaphase chromosomes arranged according to their size and position of centromeres
36 Human ChromosomesEach chromosome in karyotype is divided into two regions (arms) separated by the centromerep = short arm (petit); q = long armp and q arms are divided into numbered bands and interband regions based on pattern of stainingWithin each arm the regions are numbered.
37 Centromeres• Chromosomes are classified according to the relative position of their centromeresIn metacentric it is located in middle of chromosomeIn submetacentric—closer to one end of chromosomeIn acrocentric—near one end of chromosomeChromosomes 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 copyPolysomic = extra copies of single chromosomesMost 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 Compensation2) Not much (essential) on the YEXAMPLES:Trisomy-X = 47, XXX (female)Double-Y = 47, XYY (male)Klinefelter Syndrome = 47, XXY (male, sterile)Turner Syndrome = 45, X (female, sterile)
42 Chromosome Inversions Inversions = genetic rearrangements in which the order of genes in a chromosome segment is reversedInversions do not alter the genetic content but change the linear sequence of genetic informationIn an inversion heterozygote, chromosomes twist into a loop in the region in which the gene order is invertedChromosome InversionsA B CC B ABAC
43 Paracentric inversion Does not include centromereCrossing-over produces one acentric (no centromere) and one dicentric (two centromeres) chromosomePericentric inversionIncludes centromereCrossing-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 deletionAlternate 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 PolyploidyPolyploid species have multiple complete sets of chromosomesThe basic chromosome set, from which all the other genomes areformed, is called the monoploid setThe haploid chromosome set is theset of chromosomes present in a gamete, irrespective of the chromosome number in the species.Polyploids can arise from genome duplications occurring before or after fertilizationThrough the formation of unreduced gametes that have double the normal complement of chromosomes orThrough abortive mitotic division, called endoreduplication.
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.