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Discovery of Genetic Linkage Genes on non-homologous chromosomes assort independently, but genes on the same chromosome (syntenic genes) may instead be.

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Presentation on theme: "Discovery of Genetic Linkage Genes on non-homologous chromosomes assort independently, but genes on the same chromosome (syntenic genes) may instead be."— Presentation transcript:

1 Discovery of Genetic Linkage 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 Discovery of Genetic Linkage Classical genetics analyzes the frequency of allele recombination in progeny of genetic crosses –New associations of parental alleles are recombinants, produced by genetic recombination. –Tests crosses determine which genes are linked, and a linkage map (genetic map) is constructed for each chromosome.

3 MORGAN’s EXPERIMENTS Both the white eye gene (w) and a gene for miniature wings (m) are on the X chromosome. Morgan (1911) crossed a female white miniature (w m/w m) with a wild-type male (w+ m+/ Y). –In the F1, all males were white-eyed with miniature wings (w m/Y), and all females were wild-type for eye color and wing size (w+ m+/w m).

4 MORGAN’s EXPERIMENTS –F1 interbreeding is the equivalent of a test cross 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. –What is the expected ratio of phenotypes in F2, if white and miniature are on different chromosomes? In 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). 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 non-parental flies result from recombination of linked genes.

5 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Morgan’s experimental crosses of white-eye and miniature- wing variants of Drosophila melanogaster

6 MORGAN’S PROPOSAL During meiosis alleles of some genes assort together because they are near each other on the same chromosome. Recombination occurs when genes are exchanged between X chromosomes of the F1 females Parental phenotypes occur most frequently, while recombinants less. Terminology –Chiasma: site of crossover –Crossing over: reciprocal exchange of homologous chromatid segments –Crossing-over occurs at prophase I in meiosis; each event involves two of the four chromatids. Any chromatids may be involved in crossing over.

7 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Mechanism of crossing-over

8 Detecting Linkage through Testcrosses Linked genes are used for mapping. They are found by looking for deviation from the frequencies expected from independent assortment. A testcross (one parent is homozygous recessive) works well for analyzing linkage –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. –A significant deviation in this ratio (more parental and fewer recombinant types) indicates linkage.

9 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Testcross to show that two genes are linked

10 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Testcross to show that two genes are linked

11 Chi-square for analysis of linkage A null hypothesis (‘the genes independently assort’) is used because it is not possible to predict the phenotype frequencies produced by linked genes. –If two genes are not linked, a testcross should yield a 1:1 ratio of parentals: recombinants. –Formula is X2 = sum (Obs-Exp)^2/Exp –If P>0.05, deviation between Obs and Exp is not significant –If P<=0.05, deviation is statistically significant; such that genes may be linked.

12 Concept of Genetic Map In an individual heterozygous at two loci, there are two arrangements of alleles: –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). –Trans (repulsion) arrangement: has one mutant and one wild-type on each chromosome (e.g., w+ m and w m+) –A crossover between homologs in cis arrangement results in a homologous pair with the trans arrangement. A crossover between homologs in the trans arrangement results in cis homologs.

13 Drosophila Crosses They showed that cross over frequency for linked genes (measured by recombinants) is characteristics for each gene pair. The frequency stays the same, whether the genes are in coupling or in repulsion. –Morgan and Sturtevant (1913) used recombination frequencies to make a genetic map. A 1% crossover rate is a genetic distance of 1 map unit (mu). A map unit is also called a centimorgan (cM). Geneticists use recombination frequency as a way to estimate crossover frequency. 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.

14 First Genetic Map Three X-linked genes – White (w): white eyes – Miniature (m): miniature wings – Yellow (y): yellow body Crosses gave the following recombination frequencies: –White x miniature was 32.6 –White x yellow was 1.3 –Miniature x yellow was 33.9 MAP: m w---y

15 Gene Mapping Using Two-Point Testcrosses 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.

16 For autosomal dominants, a double heterozygotes (A B/A+B+) is testcrossed with a homozygous wildtype (recessive) individual (A+B+/A+B+) For X-linked recessives, a female double heterozygote (a+ b+/a b) is crossed with a hemizygous recessive male (a b/Y). For X-linked dominants, a female double heterozygote (A B/A+ B+) is crossed with a male hemizygous for the wild-type (A+ B+). Phenotypes obtained in these crosses will depend on whether the alleles are in cis or trans position.

17 GENETIC MAP Recombination frequency is used directly as an estimate of map units. –The measure is more accurate when alleles are close together. –Scoring large numbers of progeny increases accuracy.

18 GENERATING A LINKAGE MAP Genetic map is generated from estimating the crossover rate in a particular segment of a the chromosome. It may not exactly match the physical map because crossover is not equally probable at all sites on the chromosome. 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+).

19 LINKED or NON-LINKED? A recombination frequency of 50% means that genes are unlinked. There are two ways in which genes maybe unlinked: –They may be on separate chromosomes. –They may be far apart on the same chromosome.

20 MULTIPLE CROSSOVERS 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.

21 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Demonstration that the recombination frequency between two genes located far apart on the same chromosome cannot exceed 50 percent

22 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Demonstration that the recombination frequency between two genes located far apart on the same chromosome cannot exceed 50 percent

23 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Demonstration that the recombination frequency between two genes located far apart on the same chromosome cannot exceed 50 percent

24 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Three-point mapping, showing the testcross used and the resultant progeny

25 Mapping using three-point testcrosses 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 –In the progeny, each gene has two possible phenotypes. For three genes there are (2)^3=8 expected phenotypic classes in the progeny.

26 Establishing the order of genes The order of genes on the chromosome can be deduced from results of the cross. Of the eight expected progeny phenotypes: –Two classes are parental (p+ r+ j+/ p r j and p r j / p r j) and will be the most abundant. –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.

27 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Consequences of a double crossover in a triple heterozygote for three linked genes

28 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Rearrangement of the three genes in Figure 13.9 to p j r

29 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Rewritten form of the testcross and testcross progeny in Figure 13.9, based on the actual gene order p j r

30 Calculating the recombination frequencies Cross data is organized to reflect the gene order, and this example the region between genes p and j is called region I, and that between j and r is region II.

31 Calculating recombination frequencies 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. Recombination frequencies are used to position genes on the genetic map (each 1% recombination frequency = 1 map unit) for the chromosomal region. Recombination frequencies are not identical to crossover frequencies, and typically underestimate the true map distance.

32 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Genetic map of the p-j-r region of the chromosome computed from the recombination data in Figure 13.12

33 Interference and Coincidence 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 crossover occuring in the region.

34 Coefficient of coincidence express the extent of interference Interference = 1-coefficient of coincidence. The values are inversely related. 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. A value of 0 means that none of the expected crossovers occurred, and interference is total.

35 Calculating accurate map distances Recombination frequency generally underestimates the true map distance: –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 take place. –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. –Map distances for genes that are less than 7 mu apart are very accurate. As distance increases, accuracy declines because more crosses go uncounted.

36 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Progeny of single and double crossovers

37 Mapping functions 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. Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.

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45 Mitotic recombination Crossing over during mitosis was first observed by Stern (1936) in Drosophila. –The alleles involved are sex-linked and recessive to the wild type: Y produces yellow body color instead of wild type grey. Sn produces short, twisty bristles (“signed”) rather than the wild-type long, curved ones. Bristles follow body color (y+/- are black, and y/y are yellow.

46 Mitotic recombination Female progeny from the cross y+ sn / y+ sn x y sn+ /Y Generally have wild type phenotype of grey bodies and normal bristles, corresponding to their genetoype (y+ sn / y sn+). But exceptions: –Some flies had patches of yellow and/or signed bristles. This could be explained by nondisjunction or chromosomal loss. –Other flies had twin spots, adjacent regions of bristles, one yellow and the other signed, a mosaic phenotype. The spots are reciprocal products of the same genetic event, a mitotic crossing over. –Mitotic crossover occurred either between the centromere and the sn locus or between the sn and the y locus.

47 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Body surface phenotype segregation in a Drosophila strain

48 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Production of the twin spot and single yellow spot shown in Figure by mitotic crossing-over

49 Mechanism of Mitotic Crossing over 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. If the starting genotype is d+ e / d e+, the two possible orientations of the resulting chromatids are: –One cell with d+ e+ / d+ e+, and one with the 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. –Reversal of the alleles, d e+ / d+ e. Phenotypically indistinguishable from non-recombinant cells, there are not useful for mapping, but are nonetheless derived from a crossover event.

50 Retinoblastoma Most common childhood eye cancer. –Non hereditary (sporadic) form occurs in an individual with no family history of the disease, and affects only one eye (unilateral). –Heteditary form affects both eyes (bilateral) and usually occurs at an earlier age than sporadic. –A single gene (Rb) on chromosome 13q14 involved. 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. 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.

51 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Normal mitotic segregation of genes in a theoretical diploid cell with one homologous pair of chromosomes

52 Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. Fig Result of a mitosis of the same cell type as the cell in Figure but in which a rare mitotic crossing-over occurs


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