LECTURE 06: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION I CH4 key concepts linkage recombination linkage maps 3-point test cross 2 test for linkage CH4 problems... so far LECTURE 06: EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION I can new combinations of “linked genes” be inherited? by what mechanism? is the frequency of new combinations related to their distance apart on the chromosome? how do we know if genes are linked? MENDELIAN ANALYSIS: 1 n GENES ... a quick review from CHAPTER 2 3 methods of working out expected outcomes of controlled breeding experiments: 1. Punnet square 2. tree method (long) – genotypes 3. tree method (short) – phenotypes MENDELIAN ANALYSIS: PUNNET SQUARE all pairings of and gametes = probabilities of all pairings some pairings occur >1 different Ps for different genotypes & phenotypes 1 gene, 2 x 2 = 4 cells 2 genes, 4 x 4 = 16 cells 3 genes, 8 x 8 = 64 cells... too much work ! MENDELIAN ANALYSIS: LONG TREE 1 gene, alleles A, a GENOTYPE 1/4 A/A 1/4 A/A 1/2 A/a 1/2 A/a 1/4 a/a 1/4 a/a MENDELIAN ANALYSIS: LONG TREE 1 gene, alleles A, a GENOTYPE PHENOTYPE 3/4 A 1/4 A/A 1/4 A/A 1/2 A/a 1/2 A/a 1/4 a/a 1/4 a/a 1/4 a MENDELIAN ANALYSIS: LONG TREE 2 genes, alleles A, a; B, b... GENOTYPE 1/4 A/A 1/2 A/a 1/4 a/a 1/4 B/B 1/2 B/b 1/4 b/b 1/4 B/B 1/2 B/b 1/4 b/b 1/4 B/B 1/2 B/b 1/4 b/b 1/16 A/A B/B 1/8 A/A B/b 1/16 A/A b/b 1/8 A/a B/B 1/4 A/a B/b 1/8 A/a b/b 1/16 a/a B/B 1/8 a/a B/b 1/16 a/a b/b MENDELIAN ANALYSIS: LONG TREE 2 genes, alleles A, a; B, b... GENOTYPE PHENOTYPE 1/4 A/A 1/2 A/a 1/4 a/a 1/4 B/B 1/2 B/b 1/4 b/b 1/4 B/B 1/2 B/b 1/4 b/b 1/4 B/B 1/2 B/b 1/4 b/b 1/16 A/A B/B 1/8 A/A B/b 1/16 A/A b/b 1/8 A/a B/B 1/4 A/a B/b 1/8 A/a b/b 1/16 a/a B/B 1/8 a/a B/b 1/16 a/a b/b 9/16 AB 3/16 Ab 3/16 aB 1/16 ab MENDELIAN ANALYSIS: SHORT TREE 2 genes, alleles A, a; B, b... PHENOTYPE ¾B 9/16 AB ¼b 3/16 Ab ¾B 3/16 Ab ¼b 1/16 ab ¾A ¼a much easier... can extend to >2 genes CHAPTER 4: KEY CONCEPTS genes close together on a chromosome do not assort independently at meiosis recombination genotypes with new combinations of parental alleles homologous chromosomes can exchange segments by crossing-over recombination results from either independent assortment or crossing-over CHAPTER 4: MORE KEY CONCEPTS genes can be mapped by measuring frequencies of recombinants produced by crossing-over interlocus map distances based on recombination measurements are ~ additive one crossover can influence the occurrence of a second one in an adjacent region LINKAGE observe deviations from 9 : 3 : 3 : 1 ratios derived from dihybrid crosses 1 : 1 : 1 : 1 ratios derived from test crosses too many too few LINKAGE SYMBOLISM & TERMINOLOGY alleles on the same homologue have no punctuation between them... a b alleles on different homologues separated by a slash... A B / a b alleles written in same order on each homologue... A d B c / a D b C unlinked genes separated by semicolon... A/a ;B / b genes of unknown linkage separated by dot... A/a • B / b LINKAGE unlinked genes, different chromosomes write as ... Aa; Bb or A/a; B/b illustrate as ... A a ; B b or A a ; b B segregate and assort independently test cross to a/a; b/b 4 progeny types ... 1 AB : 1 Ab : 1 aB : 1 ab LINKAGE linked genes, same chromosomes write as ... Aa Bb or AB/ab or Ab/aB 2 possible dihybrids, illustrate as ... AB a b ‘cis’ or ‘coupling’ or Ab a B ‘trans’ or ‘repulsion’ segregate and assort dependently ~ distance test cross to ab/ab 4 progeny types ... in CIS: < 1 AB : > 0 Ab : > 0 aB : < 1 ab in TRANS: > 0 AB : < 1 Ab : < 1 aB : > 0 ab LINKAGE linked genes, same chromosomes LINKAGE chiasmata: visible manifestations of crossing over LINKAGE linked genes (in cis here), same chromosomes also called non-parental or recombinant chromosomes LINKAGE crossing over when in meiosis? (tetrad analysis) if if LINKAGE how many chromatids involved? (tetrad analysis) RECOMBINATION products of meiotic recombination linkage unknown ati RECOMBINATION detection of recombination using a test cross RECOMBINATION test cross unlinked genes independent assortment always “recombination” frequency of 50% RECOMBINATION recombinants arise when non-sister chromatids cross over between genes under study RECOMBINATION test cross linked genes independent assortment recombination frequency < 50% 2 TEST FOR LINKAGE A O a E O E B 140 / 127.5 115 / 127.5 255 b 110 / 122.5 135 / 122.5 245 250 250 500 2 TEST FOR LINKAGE A O a E O E B 140 / 127.5 115 / 127.5 255 b 110 / 122.5 135 / 122.5 245 250 250 500 2 TEST FOR LINKAGE 2 TEST FOR LINKAGE express this as: 0.05 > P(2c = 5.02) > 0.01 the data derived from a test cross deviate significantly from a 1:1:1:1 ratio we reject our H0 (alternatively we could not reject) “genes A and B are linked” the probability of deviation by chance from the linked genes model is between 1 and 5% (i.e., deviation from the 1:1:1:1 ration occurs because linkage is not supported by data) RECOMBINATION Autosome Linkage P F1 F2 cn bw+ x cn+ bw cn bw+ cn+ bw cn bw x cn bw+ cn bw cn+ bw score progeny no recombination in Drosophila s must use s only X-Chromosome X-Chromosome Linkage Linkage y w+ x y+ w Y y+ w y+ w x y w+ Y y+ w score progeny RECOMBINATION Autosome Linkage P F1 F2 X-Chromosome X-Chromosome Linkage Linkage cn bw+ x cn+ bw cn bw+ cn+ bw cn bw x cn bw+ cn bw cn+ bw score progeny tester chromosomes y w+ x y+ w Y y+ w y+ w x y w+ Y y+ w score progeny LINKAGE MAPS sites of recombination are ~ random probability of recombination increases ~ distance LINKAGE MAPS complications: RECOMBINATION FREQUENCY 1. recombination rates position on chromosome Telomere 2. 3. 4. 5. 6. centromere telomere linked genes >50 cM apart look like unlinked genes double recombination events are usually not detected multiple strand (>2) exchanges tester for multiple autosomal genes not always available interference – reduced probability of adjacent events (later) LINKAGE MAPS 2 possibilities for map add the shorter distances 3-POINT TEST CROSS P v+ cv+ cv ct x v v+ cv ct Y ct+ F1 F2 v+ cv ct x v cv+ ct+ v cv+ ct+ Y Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS double crossover involving 2 chromatids outer alleles look like parental types recombination cryptic without middle gene 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT with DCO gives expected 3. calculate distances: 2 genes at a time 4. construct map 5. calculate interference Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3. calculate distances: 2 genes at a time 4. construct map 5. calculate interference Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see your if PT alleles with on... Base decision DCOsimilar gives expected (1) numbers 3. calculate distances: (2) reciprocal genotypes 2 genes at a time 4. construct map 5. calculate interference Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3. calculate distances: 2 genes at a time 4. construct map 5. calculate interference Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3. calculate distances: linear Draw all 3 possible 2 genes at a(next), time what do arrangements 4. construct map double crossovers between 5. calculate interference give... both PT chromosomes Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS test all 3 possible orders of PT allele combinations ct+ 3 5 ct+ 94 89 ct+ 45 40 3-POINT TEST CROSS test all 3 possible orders of PT allele combinations 3 5 PT chromosomes with genes in the order v ct cv (NOT the order shown in the table... it usually won’t be) give you the two F2 groups with the smallest numbers with double crossovers between the genes. 3-POINT TEST CROSS REAL GENE ORDER DOES NOT HAVE TO BE AS LISTED IN THE DATA TABLE. Consider genes v & ct. PT are v ct+ or v+ ct in TRANS. All single crossovers between them yield reciprocal v ct & v+ ct+ CIS products as shown. Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS order of operations Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3. calculate distances: 2 genes at a time; e.g.: v ct = [ ( 89 + 94 + 3 + 5 ) / 1448 ] x 100 # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3. calculate distances: 2 genes at a time 4. construct map... 5. calculate interference Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS distances between outer genes is less accurate because double recombinants are cryptic v ct 13.2 cM cv 6.4 cM 18.5 cM or... 13.2 + 6.4 = 19.6 cM add distances for accurate map 3-POINT TEST CROSS order of operations 1. look for reciprocal groups: parental type (PT), single cross over (SCO), double cross over (DCO) 2. gene order: test all 3 to see if PT alleles with DCO gives expected 3. calculate distances: 2 genes at a time 4. construct map 5. calculate interference Genotype v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct v cv ct v+ cv+ ct+ v cv+ ct v+ cv ct+ total # 580 592 45 40 89 94 3 5 1448 3-POINT TEST CROSS interference = 1 – COEFFICIENT OF COINCIDENCE = 1 – [ OBSERVED # DCO / EXPECTED # DCO ] = 1 – [ 8 / ( 0.132 x 0.064 x 1448 ) ] = 1 – [ 8 / 12 ] = 1 / 3 = 33 % LINKAGE MAPS why is mapping important ? fundamental aspect of genetic analysis, concerns: gene function gene evolution gene isolation LINKAGE MAPS gene function ... location influences function... “position effect” relative to heterochromatin other genes genes of related function often clustered operons in Prokaryotes homeotic genes in Eukaryotes LINKAGE MAPS gene evolution ... relative positions of genes in related organisms infer history of change LINKAGE MAPS gene isolation ... mapping 1st step Drosophila ... LINKAGE MAPS gene isolation ... mapping 1st step tomato ... prophase nucleus chromosome numbering map (ca. 1952) LINKAGE MAPS - HUMANS why has this been difficult ? controlled breeding programs not possible (not ethical) sample sizes small (relatively) human genome is LARGE... EUKARYOTE CHROMOSOME MAPPING AND RECOMBINATION: PROBLEMS in Griffiths chapter 4, beginning on page 141, you should be able to do questions #1-30 begin with the solved problems on page 138 if you are having difficulty look at the way Schaum’s Outline discusses linkage and mapping for alternative explanations - especially tetrad analyses try Schaum’s Outline questions in chapter 4, beginning on page 208 TUTORIAL #2: T.9.26 or R.9.28 go to the TUTORIAL page on the genetics web site to download the file tut2-06F.pdf complete all six questions before attending tutorial #2 next week during tutorial, you will review solutions to these questions with your group and prepare to present solutions to all six questions one member of your group will be selected by the TA to present a solution to one of these questions presentations will be 10 min each + 5 min for questions and discussion