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Unformatted text preview: 0 Linkage, Recombination, and the Mapping of Genes on Chromosomes Chapter 5 I. Gene linkage and recombination 0 A. Some genes on the same chromosome assort together more often than not 1. In dihybrid crosses, departures from 1:1:1:1 ratio of F1 gametes indicate that the 2 genes are on same chromosome 2. A preponderance of parental genotypes in the F2 generation defines linkage 3. % of parental and recombinant classes vary with the gene pair 4. Autosomal traits also exhibit linkage 5. The Chi-Square test pinpoints the probability that observed percentages are evidence for linkage 6. Applying the Chi-Square test B. Recombination results when crossing over during meiosis separates linked genes 1. Reciprocal exchanges between homologous chromosomes are the physical basis of recombination 2. Through the light microscope: chiasmata mark the sites of recombination 3. Recombination frequencies for pairs of genes reflect the distances between them along a chromosome 4. Experimental recombination frequencies between 2 genes are never greater than 50% II. Mapping: locating genes along a chromosome A. 2 point crosses: comparisons help establish relative gene positions 0 1. recombination mapping supports idea that genes arranged in a line 2. two-point crosses have their limitations B. Three-point crosses: a faster, more accurate way to map genes 1. 3-point crosses allow correction for double crossovers 2. Interference: the number of double crossovers may be less than expected 3. The arrangement of alleles in double recombinants indicates the relative order of 3 genes 4. comprehensive example: using a 3-point cross to reanalyze Sturtevant's map of the Drosophilia X chromosome C. How close is the correlation between a genetic map and physical reality? D. Multiple factor crosses help establish linkage groups by interference E. Tetrad analysis in fungi: a powerful tool for mapping and for 0 understanding the mechanisms of recombination 1. tetrads can be analyzed by the # of parental and recombinant spores they contain 2. when PD (parental ditype) = NPD (nonparental ditype), two genes are unlinked 3. when PDs significantly outnumber the NPDs, it is a sign of 2-way linkage 4. calculating the recombination frequency 5. tetrad analysis confirms that recombinantion occurs at the 4-strand stage 6. tetrad analysis also demonstrates that recombination is usually reciprocal 7. ordered tetrads help locate genes in relation to the centromere 8. a numerical example of ordered-tetrad analysis 0 Linkage, Recombination, and the Mapping of Genes on Chromosomes Linkage and meiotic recombination Genes linked together on the same chromosome usually assort together. Linked genes may become separated through recombination. The frequency with which genes become separated reflects the physical distance between them. Rarely, recombination occurs during meiosis. In eukaryotes mitotic recombination produces genetic mosaics. Mapping Mitotic recombination Independent assortment Genes on different chromosomes Homologous pair #10 A A a a B B b b Homologous pair #15 0 4 Gametes A B a B A b a b Linkage: Two genes on same chromosome segregate together The closer 2 genes are on a chromosome, the tighter they are linked A A a a B B b b A B A B a b a b "Crossing over" and "linkage" lead to separation of linked genes A A a a 0 x B B b b Gametes Parental Recombinant A B B a A b a b A. Some genes on the same chromosome assort together more often than not 0 1. In dihybrid crosses, departures from a 1:1:1:1 ratio of F1 gametes indicate that the two genes are on the same chromosome. 2. A bias in parental genotypes in the F2 generation defines linkage 3. The % parental and recombinant classes vary with the gene pair in question. For example, examine sex linkage... 3. Linkage at a sex-linked gene 0 Deviation from 1:1:1:1 ratio of phenotypes for males Draw traits on chromosomes and work through the cross 4. Linkage in an autosomal gene 0 Genotypes of F1 female revealed by test cross Parental class outnumbers recombinant class demonstrating linkage. 5. Chi square test pinpoints the probability that ratios are evidence of linkage 0 Transmission of gametes is based on chance events. Deviations from 1:1:1:1 ratios can represent chance events OR linkage. Ratios alone will never allow you to determine if observed data are significantly different from predicted values. The larger your sample, the closer your observed values are expected to match the predicted values. Chi square test measures "goodness of fit" between observed and expected (predicted) results. Accounts for sample size, or the size of the experimental population. Degree of Freedom (DF): # of independently varying parameters minus 1 6. Applying the chi square test 0 Framing a hypothesis Null hypothesis observed values are not different from the expected values For linkage studies no linkage is null hypothesis Expect a 1:1:1:1 ratio of gametes. Alternative hypothesis observed values are different from expected values For linkage studies genes are linked. Expect significant deviation from 1:1:1:1 ratio. 6. Applying the chi square test to a linkage study Genotype AB ab Ab AB Total Experiment 1 17 14 8 11 50 Experiment 2 34 28 16 22 100 0 Observed/Expected Observed/Expected Class Parentals 31 25 62 50 Recombination 19 25 38 50 0 Chi Square Experiment 1 & 2 2 = (observed expected)2 number expected Experiment 1 2 = (31 25)2 + (19 25)2 25 25 2 = (62 50)2 + (38 50)2 50 50 = 2.88 Experiment 2 = 5.76 0 Chi square table of critical values Experiment 1 2 = 2.88 Can NOT reject Null Hypothesis WHAT is DF? 4 - 1 = 3 Experiment 2 2 = 5.76 Can NOT reject Null Hypothesis So what does rejecting the Null Hypothesis versus NOT rejecting the Null Hypothesis really mean??? 0 2 > critical value at p = 0.05 REJECT NULL HYPOTHESIS Data not significant YES LINKAGE! Deviations NOT due to chance alone, something else is going on 2 < critical value at p = 0.05 ACCEPT NULL HYPOTHESIS Data are significant NO LINKAGE! Deviations just due to chance alone Experiment 1 2 = 2.88 Experiment 2 C.V. = 7.81 0 2 C.V. 2 C.V. Deviations just due to chance alone for both experiments Observed/Expected Observed/Expected Class Parentals 31 25 62 50 Recombination 19 25 38 50 2.88 Be careful not to "eyeball" values and think that it is a big difference in what was seen versus what was expected. Go strictly by what the numbers tell you. How can we apply the Chi-Square to data we examined already? Summer squash color patterns 1. 2. 3. white yellow green 4. 0 F2 phenotype White Yellow Green Expected ratio 12/16 3/16 1/16 Expected # 90.0 22.5 7.5 Observed # 82.0 29.0 9.0 (O E) -8 6.5 1.5 (O E)2 64.00 42.25 2.25 (O E)2 E 0.711 1.878 0.300 =2.889 82 white 29yellow 9 green 120 TOTAL WHAT is DF? WHAT is C.V. at 0.05? 31=2 5.99 vs. C.V. 0 = 2.889 2 C.V. Can NOT reject Null Hypothesis Deviations just due to chance alone, true 12:3:1 model happening B. Recombination results when crossing-over during meiosis separates linked genes 1909 Frans Janssens observed chiasmata, regions in which non-sister chromatids of homologous chromosomes cross over each other. Thomas Hunt Morgan suggested these were sites of chromosome breakage and change resulting in genetic recombination. 0 1. Reciprocal exchanges between homologous chromosomes are the physical basis of recombination 0 1931 Genetic recombination depends on the reciprocal exchange of parts between maternal and paternal chromosomes. Harriet Creighton and Barbara McClintock studied corn. Curtis Stern studied fruit flies. Physical markers to keep track of specific chromosome parts Genetic markers were points of reference to determine if particular progeny were result of recombination. Genetic recombination between car and Bar genes on the Drosophila X chromosome 0 2. Chiasmata mark the sites of recombination 0 0 Chiasmata mark the sites of recombination 3. Recombination frequencies for pairs of genes reflect distance between them 0 Alfred H. Sturtevant Percentage of recombination, or recombination frequency (RF) reflects the physical distance separating two genes. 1 RF = 1 map unit (or 1 centiMorgan) 4. Unlinked genes show a recombination frequency of 50% Genes on different chromosomes (use Chi-square test to verify this...) 4. Unlinked genes show a recombination frequency of 50% Genes so far apart on the same chromosomes, genetically unlinked (Use Chi square to verify this Summary of linkage and recombination Genes close together on the same chromosomes are linked and do not segregate independently Linked genes lead to a larger number of parental class than expected in double heterozygotes Mechanism of recombination is crossing over Chiasmata are the visible signs of crossing over Farther away genes are the greater the opportunity for chiasmata to form Recombination frequencies reflect physical distance between genes Recombination frequencies between two genes vary from 0% to 50% II. Mapping: Locating genes along a chromosome A. 2 point crosses: 1. Comparisons help establish relative gene positions. 2. Genes are arranged in a line along a chromosome. "signature" is 4 BIG classes and 4 small classes Mapping: Locating genes along a chromosome Genes are arranged in a line along a chromosome 3. Limitations of two point crosses Difficult to determine gene order if two genes are close together Actual distances between genes do not always add up. Pair-wise crosses are time and labor consuming. B. Three Point Crosses: A faster more accurate method to map genes "signature" appearance is... 2 BIG = NCO 4 Med = SCO 2 small= DCO 1. Analyzing the results of a three point cross Look at two genes at a time and compare to parental 1. Analyzing the results of a three point cross vg b distance 252 + 241 + 131 + 118 X 100 = 17.7 m.u. (map units) 4197 vg pr distance 252 + 241 + 13 + 9 4197 b pr distance 131 + 118 + 13 + 9 4197 X 100 = 12.3 m.u. X 100 = 6.4 m.u. 0 Correction for Double Crossovers vg b distance 131 + 118 + 13 + 9 4197 X 100 = 6.4 m.u. More accurate to sum the two smaller map distances versus doing the first total distance calculation 2. Interference: The number of double crossovers may be less than expected Sometimes the number of observable double crossovers is less than expected if the two exchanges are independent. Occurrence of one crossover reduces likelihood that another crossover will occur in adjacent parts of the chromosome. Chromosomal interference crossovers do not occur independently. Interference is not uniform among chromosomes or even within a chromosome. Measuring interference Coefficient of coincidence = ratio between actual frequency of DCO and expected frequency of DCO. Interference = 1 coefficient of coincidence. If interference = 0, observed and expected frequencies are equal. If interference = 1, no double crossovers can occur. 3. Double recombinants indicate order of three genes Summary of three point cross analysis Cross true breeding mutant with wild-type Analyze F2 individuals (males if sex linked) Parental class most frequent Double crossovers least frequent Determine order of genes based on parentals and recombinants Determine genetic distance between each pair of recombinants Calculate coefficient of coincidence and interference C. Do Genetic and Physical maps correspond? Order of genes is correctly predicted by physical maps. Distance between genes is not always similar on physical maps. Double, triple, and more crossovers Only 50% recombination frequency observable in a cross Variation across chromosome in rate of recombination Mapping functions compensate for inaccuracies, but are not often imprecise. D. Genes chained together by linkage relationships are known as linkage groups 4. Calculating recombination frequency RF = NPD + T 100 Total tetrads We will be using the formula = all SCO + 2 (DCO) Total # x 100 Mitotic recombination can produce genetic mosaics Mitotic recombination is rare. Initiated by Mistakes in chromosome replication Chance exposure to radiation Curt Stern observed "twin spots" in Drosophila a form of genetic mosaicism. Animals contained tissues with different genotypes. Mitotic crossing over between sn and centromere in Drosophila Crossing over between sn and y gene Twin spots in Drosophila ...
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This note was uploaded on 08/04/2009 for the course BIOL 2153 taught by Professor Larkin during the Fall '03 term at LSU.

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