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Recombination & Genetic Analysis -The maximum recombination frequency -Quiz section 5 revisited -Genetic vs. Physical maps Test your understanding Predict the number of progeny for each type of offspring that result from the following cross. Assume 100 total offspring. Dihybrid female Testcross male 58.2 cM b+ sp+ b sp P R X tan no speck 21 25 black speck 21 25 tan speck 29 25 black no speck 29 25 Recombinants never exceed 50% No! Are b and sp linked? • These genes are so far apart, that they assort independently from one another. • b and sp appear to be unlinked even though they are on the same chromosome! sp + b+ 2 3 sp sp + b+ 4 sp b+ 1 b+ b b vg+ vg sp+ sp b sp+ b sp All four gamete types are equally frequent How do we know that b and sp are on the same chromosome? b is linked to vg and vg is linked to sp. Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: Write down the parental types with respect to A/a and D/d: Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: Write down the parental types with respect to A/a and D/d: Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: A B 1 D A a B b 2 3 D d a b 4 d After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: A B and a b Write down the parental types with respect to A/a and D/d: A D and a d Why do we observe IA? In-class experiment… Mark two crossovers anywhere between the homologues: A B 1 D A a B b 2 3 D d a b 4 d After you are given the locations of loci A, B, and D… Write down the parental types with respect to A/a and B/b: A B and a b Write down the parental types with respect to A/a and D/d: A D and a d In-class experiment (cont’d) Looking first at just loci A/a and B/b… What are the genotypes of the products from your meiosis? 1. 2. 3. 4. Are these gametes all parental? All recombinant? 2 of each? Then look at just loci A/a and D/d… What are the genotypes of the products from your meiosis? 1. 2. 3. 4. Are these gametes all parental? All recombinant? 2 of each? In-class experiment (cont’d) What must have happened to create these gametes? Class aggregate data: A-B Genotype P or R? Number? Number P gametes Number R gametes AB, ab 4P 81 324 0 Ab, aB 4R 2 0 8 AB, ab, Ab, aB 2 P, 2 R 43 86 86 # Recombinants Total # Gametes = X X % Rec? X In-class experiment (cont’d) Class aggregate data: A-D Genotype P or R? Number? Number P gametes Number R gametes AD, ad 4P 40 160 0 Ad, aD 4R 19 0 76 AD, ad, Ad, aD 2 P, 2 R 61 122 122 # Recombinants Total # Gametes 76+122 = % Rec? X 76+122 +160 +122 Why do we observe IA? A B D A a B b D d a b d Everyone in the class drew crossovers somewhere between A/a and D/d, yet the overall % recombinants for the class was only ~50%. If we look at a large enough sample, even genes that are very far apart on the same chromosome cannot show more than 50% recombinant products. Need to look closer at meiosis itself to see why. What is the maximum recombination frequency in any interval? Parental Recombinant Recombinant Parental Reminder… one crossover gives 2 parental, 2 recombinant gametes The range of possibilities: tightly linked independent assort. Consider 100 cells undergoing meiosis… if one cell has a crossover 2 recombinant out of 400 0.5% if 10 cells have crossovers 20 recombinant out of 400 5.0% if all cells have crossovers 200 recombinant 50% out of 400 Maximum recombination frequency = 50% for single recombination events What is the maximum recombination frequency in any interval? The effect of multiple crossovers: # xovers resulting gametes 2 (2 strands) 2 (3 strands) 2 (4 strands) = 50% recombinant and 50% parental. Also true for triple Xover, quadruple Xover, etc. 4 parental 2 parental 2 non-par. 4 non-par. 6 parental 6 non-par. Human Xchromosome map… how could we get 180 cM? Now consider independent assortment …the “ultimate” in non-linkage R r Y R X y r Y y Refer to one of Mendel’s F1 x F1 dihybrid cross (round yellow X round yellow): What were the parental types for the F1? RY and ry What were the parental and recombinant gametes made by the F1 plants? 1/4 1/4 1/4 1/4 RY ry Ry rY What was the % recombinant gametes? 1/4 + 1/4 = 50%! So, even for independently assorting genes, the % recombinant products is only 50% Conclusion For widely separated genes 1) An odd number of crossovers gives, on average, an equal number of parental and recombinant types. 2) An even number of crossovers gives, on average, an equal number of parental and recombinant types. 3) Alleles on two different chromosomes line up on the metaphase plate independently, giving on average equal numbers of parental and recombinant types. Thus, the maximum recombinant frequency = 50% Loci can appear to be unlinked because: • They are on separate chromosomes • They are so far apart on the same chromosome that recombination always occurs Practice question In a certain plant species… R f r F X flower fragrance (F) is dominant over unscented (f) r color (B) fis dominantr over white (b) f blue flower rounded leaves (R) is dominant over pointy (r); and The parental and recombinant thorny stems (T) is dominant over smooth stems (t). types are the same! Need to be From the following crosses, can you determine whether the fragrance heterozygous at both loci gene is linked to any of the other genes; if so, at what map distance? Bb Ff x bb ff Rr ff x rr Ff Tt Ff x tt ff 270 blue, fragrant 281 blue, non-fragrant 268 white, fragrant 275 white, non-fragrant 219 rounded, fragrant 222 rounded, non-fragrant 209 pointy, fragrant 216 pointy, non-fragrant 333 thorny, fragrant 36 thorny, non-fragrant 39 smooth, fragrant 342 smooth, non-fragra F not linked to B Can’t tell! F and T linked at 10 cM QS7 revisited What were the main points of QS5? -To give you an opportunity to see actual data from a meiosis and to draw conclusions from the data based on your knowledge of this process. -To show what can be learned from looking at all four products of a single meiosis. The diagrams used in quiz section… etc. …were designed to help set up specific predictions Setting up predictions for meiosis outcomes… But if we can’t see all of the products from a single meiosis we expect… If… then we expect… 2 genes are linked a parental ditype (PD) PD > T >> NPD they are independently assorting but each close to a centromere either PD or nonparental ditype (NPD), with a 50:50 chance of each PD = NPD > T an equal proportion of parental and recombinant types mostly tetratypes (T), but also some PD, and NPD an equal proportion of parental and recombinant types Can distinguish! they are unlinked and at least one is distant from a centromere mostly parental types Can’t distinguish What did the data from quiz section tell you? Mat haploid parent = ade his Mat haploid parent = his LEU Mat haploid parent = LEU TS Mata haploid parent = ADE HIS Mata haploid parent = HIS leu Mata haploid parent = leu ts Spore phenotype # of spores Spore phenotype # of spores Spore phenotype # of spores ADE HIS 9 HIS LEU 3 LEU TS 11 ADE his 11 HIS leu 17 Leu ts 9 ade HIS 11 his LEU 17 leu TS 9 ade his 9 his leu 3 leu ts 11 Total = 40 Total = 40 Total = 40 Conclusion? Conclusion? Conclusion? Probably not linked Probably linked Probably not linked Looking at the 10 tetrads in terms of LEU & TS How many PD? 4 How many NPD? 5 How many T? 1 So what do you conclude about the LEU and TS genes? they are independently assorting but each close to a centromere! Our completed map Diploid genotype: MATa ADE HIS MAT ade his leu ts URA1 ura2 LEU TS ura1 URA2 leu HIS LEU his ts ADE TS ade How well did we do? The actual gene names… Let’s look in SacchDB… ADE2 HIS4 CDC7 (TS) LEU2 ADE2 HIS4 LEU2 CDC7 (TS) Not bad! What about URA? Know the parental types Look at spore phenotypes Parental types? U1 u2 & u1 U2 What spore genotypes would you expect in a PD tetrad? Phenotype on -ura plate? U1 u2 no growth no growth u1 U2 no growth u1 U2 no growth U1 u2 So, given these parental types… 0/4 spores growing is diagnostic of PD What about URA? What spore genotypes would you expect in a NPD tetrad? Parental types? Genotype? U1 U2 U1 u2 & u1 U2 Growth phenotype on -ura? U1 U2 GROWTH GROWTH u1 u2 no growth u1 u2 no growth What about URA? What spore genotypes would you expect in a T tetrad? Parental types? Genotype? U1 u2 U1 u2 & u1 U2 Growth phenotype on -ura? U1 U2 no growth GROWTH u1 U2 no growth u1 u2 no growth Looking at the 10 tetrads… How many PD? How many NPD? How many T? So what do you conclude about the ura genes? Practice question Brown seed pods (B) in a plant species is dominant to green (b), and elongated pods (E) is dominant over squished (e). (a) A fully heterozygous plant has the dominant alleles linked in trans (i.e., dominant alleles not on the same homologue) at a map distance of 20 cM. What will be the genotypes of gametes produced by this plant, and in what frequencies (or percentages)? (b) If this plant is self-pollinated, what progeny phenotypes will you expect to see, and in what frequencies? Use a Punnett square to illustrate your answer. Heterozygote genotype = B e b E 3-point testcrosses The problem with using two markers (like a and d below)… double crossovers can go undetected underestimation of recombinant frequency Solution: include a third marker between the other two… more DCOs revealed Plus… gene order revealed (more later) 3-point testcross—predicting progeny from a known map Predict the progeny phenotypes and numbers from this cross: Parent 1: + b + a c + Parent 2: b b c a c a + = wild type, dominant Map: b 3 cM c 7 cM Count 10,000 progeny Step 1. Determine the number of DCO products Probability of recombinant product in (b-c) = 3% = 0.03 Probability of recombinant product in (c-a) = 7% = 0.07 Probability of recombinant product in both = 0.03 x 0.07 = 0.0021 a Predicting progeny from a known map (cont’d) Heterozygous parent: only one chromatid of each homologue shown on next slide Predicting progeny from a known map (cont’d) DCO: both together = 10000 x 0.0021 = 21 SCO in b-c interval: Both together = (10000 x 0.03)-21 = 279 SCO in c-a interval: Both together = (10000 x 0.07)-21 = 679 NCO (non-crossover): Both together = 10000-(SCO + DCO) = 9021 3-point testcross—constructing a linkage map Construct a linkage map (gene order and map distance) for the following genes in Drosophila: Genes pr/+ (purple or red eyes) v/+ (vestigial or long wings) b/+ (black or tan body) Parents Female: pr/+ Male: pr/pr Progeny phenotypes + + + 564 SCO b pr+ v+ 32 DCO v b+ pr+ 4125 NCO pr b+ v+ 266 SCO v b pr+ 272 pr b v+ 4137 NCO pr v b+ 30 DCO pr v b 574 SCO Total = 10000 Step 1.Expand the shorthand v/+ v/v b/+ b/b Step 2.Identify the NCO and DCO classes Step 3.Which gene is in the middle? ASK: Which order allows us to go from the NCO genotype to the DCO genotype. Constructing a linkage map… Step 3 (cont’d) We know: DCO (b+ v pr+) (b v+ pr) (b v+ pr+) (b+ v pr) order unknown The process: • Try out the parental genotypes in the 3 possible orders • Do a “virtual double crossover” to see which one would give the correct DCO genotype. b+ v pr+ X X b v+ pr b+ v+ pr+ b+ pr+ v X X b pr v+ b+ pr v b b v pr pr+ v+ Constructing a linkage map (cont’d) Step 4. Calculate % recombinant products b+ pr v+ b pr+ v b+ pr+ v+ b b+ pr pr v v NCO: 4125+4137 = 8262 % recombinants in b-pr interval= SCOb-pr: 266+272 = 538 (538+62)/10000 SCOpr-v: 564+574 = 1138 DCO: 30+32 = 62 b pr v+ =600/10000 =6% % recombinants in pr-v interval= (1138+62)/10000 =1200/10000 =12% Constructing a linkage map (cont’d) Step 5. Draw the map b 6 cM pr v 12 cM or v 12 cM pr 6 cM b Interference and coefficient of coincidence (COC) Interference: Lower-than-expected frequency of DCO products - Chiasma at one one location blocks other chiasmata from forming nearby COC = observed DCO expected DCO Interference = 1 - COC In our example… expected DCO = 0.06 x 0.12 x 10000 = 72 Observed DCO = 62 COC = 62/72 = 0.86 Interference = 1 - 0.86 = 0.14 Genetic vs. physical maps Genetic maps…based on recombinant frequencies between markers alleles! variation at location #2 variation at location #1 pr+ vg pr vg+ recombination: how frequent? alleles! Alleles are detected as associated phenotypes New combination of phenotypes new combination of alleles recombination Genetic vs. physical maps (cont’d) Physical maps… based on DNA sequence or landmarks in sequence For example: A B The number of chromosomal bands separating the known locations of genes. site1 site2 site3 site4 site5 The pattern of restriction sites in a DNA sequence. The number of bp of DNA. Questions for Thought Which chromosome? pr+ Relation to centromere and telomere? vg+ pr vg Number of bp? How do we know where to place these genes relative to the centromere and telomeres? How do we know which physical entity (chromosome) our linkage group describes? “Linkage group” = chromosome What is the relationship between crossover frequency and gene order/distance (in bp of DNA) along the chromosomes? Linking the Physical and Genetic Maps white is X-linked From lecture 6 Color is on chromosome IX Novel Strain C knob Wx extra DNA Cri du chat is on chromosome V From lecture 8 Linking the Physical and Genetic Maps -Fluorescence in situ hybridization (FISH) Double-stranded DNA segment from a particular location in the human genome ** * Metaphase chromosomes Fluorescently labeled single-stranded DNA Partially denature DNA * * * What does the genetic map position tell us? genetic map (recombination) units = cM R-G colorblindness R-G colorblindness 10 cM QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. telomere 6 cM Fragile X Fragile X Haemophilia centromere Physical units = map (FISH) chrom. bands Haemophilia ~10Xbp ~6Xbp physical map (DNA) units = bp -Order of genes is conserved in genetic and physical maps. -Distance separating markers in genetic and physical maps is ~proportional (but X varies in different organisms).