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Recombination & Genetic Analysis -The maximum recombination frequency -Quiz section 5 revisited -Genetic vs. Physical maps

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Test your understanding Dihybrid female X Predict the number of progeny for each type of offspring that result from the following cross. Assume 100 total offspring. b + sp + b sp Testcross male tan no speck black speck tan speck black no speck cM PR 25 Recombinants never exceed 50%

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vg + vg Are b and sp linked? b and sp appear to be unlinked even though they are on the same chromosome! These genes are so far apart, that they assort independently from one another. How do we know that b and sp are on the same chromosome? b is linked to vg and vg is linked to sp. No! b sp+ b+ sp+ b sp b+ sp All four gamete types are equally frequent sp + b+b+ b+b+ spb b

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In-class experiment… Mark two crossovers anywhere between the homologues: 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: After you are given the locations of loci A, B, and D… Why do we observe IA?

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In-class experiment… Mark two crossovers anywhere between the homologues: 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: After you are given the locations of loci A, B, and D… Why do we observe IA?

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In-class experiment… Mark two crossovers anywhere between the homologues: A B and a b A D and a 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: After you are given the locations of loci A, B, and D… A A a a B B b b D D d d Why do we observe IA?

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In-class experiment… Mark two crossovers anywhere between the homologues: A B and a b A D and a 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: After you are given the locations of loci A, B, and D… A A a a B B b b D D d d Why do we observe IA?

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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? Then look at just loci A/a and D/d… What are the genotypes of the products from your meiosis? Are these gametes all parental? All recombinant? 2 of each?

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Class aggregate data: 2 P, 2 R AB, ab, Ab, aB 4 RAb, aB 4 PAB, ab Number R gametesNumber P gametesNumber?P or R?Genotype A-B # Recombinants % Rec? 2 43 X Total # Gametes X X = What must have happened to create these gametes? In-class experiment (cont’d)

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Class aggregate data: 2 P, 2 R AD, ad, Ad, aD 4 RAd, aD 4 PAD, ad Number R gametesNumber P gametesNumber?P or R?Genotype A-D # Recombinants % Rec? X Total # Gametes = In-class experiment (cont’d)

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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. Why do we observe IA? A A a a B B b b D D d d Need to look closer at meiosis itself to see why.

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What is the maximum recombination frequency in any interval? Parental Recombinant Reminder… one crossover gives 2 parental, 2 recombinant gametes 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 out of 400 50% The range of possibilities: tightly linked independent assort. Maximum recombination frequency = 50% for single recombination events

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The effect of multiple crossovers: # xoversresulting gametes 4 parental 2 (2 strands) 2 parental 2 non-par. 2 (3 strands) 4 non-par. 2 (4 strands) 6 parental 6 non-par. = 50% recombinant and 50% parental. Also true for triple Xover, quadruple Xover, etc. What is the maximum recombination frequency in any interval?

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Human X- chromosome map… how could we get 180 cM?

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Now consider independent assortment …the “ultimate” in non-linkage Refer to one of Mendel’s F 1 x F 1 dihybrid cross (round yellow X round yellow): What were the parental types for the F 1 ? What were the parental and recombinant gametes made by the F 1 plants? What was the % recombinant gametes? RY and ry 1/4 RY 1/4 ry 1/4 Ry 1/4 rY 1/4 + 1/4 = 50%! So, even for independently assorting genes, the % recombinant products is only 50% R Y r y R Y r y X

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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 Conclusion 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. For widely separated genes

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Practice question In a certain plant species… flower fragrance (F) is dominant over unscented (f) blue flower color (B) is dominant over white (b) rounded leaves (R) is dominant over pointy (r); and thorny stems (T) is dominant over smooth stems (t). From the following crosses, can you determine whether the fragrance gene is linked to any of the other genes; if so, at what map distance? Bb Ff x bb ff 270 blue, fragrant 281 blue, non-fragrant 268 white, fragrant 275 white, non-fragrant Rr ff x rr Ff 219 rounded, fragrant 222 rounded, non-fragrant 209 pointy, fragrant 216 pointy, non-fragrant Tt Ff x tt ff 333 thorny, fragrant 36 thorny, non-fragrant 39 smooth, fragrant 342 smooth, non-fragrant F not linked to B Can’t tell! F and T linked at 10 cM R f r f r F r f X The parental and recombinant types are the same! Need to be heterozygous at both loci

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QS7 revisited What were the main points of QS5? etc. …were designed to help set up specific predictions -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…

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Setting up predictions for meiosis outcomes… 2 genes are linked they are independently assorting but each close to a centromere they are unlinked and at least one is distant from a centromere If… then we expect… a parental ditype (PD) PD > T >> NPD either PD or non- parental ditype (NPD), with a 50:50 chance of each PD = NPD > T But if we can’t see all of the products from a single meiosis we expect… mostly parental types an equal proportion of parental and recombinant types Can’t distinguishCan distinguish! mostly tetratypes (T), but also some PD, and NPD

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the 2 genes are linked they are independently assorting but each close to a centromere they are unlinked and at least one is distant from a centromere If we see… then we conclude that… only or mostly PD PD and NPD in equal proportions mostly tetratypes Conversely:

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What did the data from quiz section tell you? Mat haploid parent = ade his Mata haploid parent = ADE HIS Spore phenotype # of spores ADE HIS9 ADE his11 ade HIS11 ade his9 Total =40 Mat haploid parent = his LEU Mata haploid parent = HIS leu Spore phenotype # of spores HIS LEU3 HIS leu17 his LEU17 his leu3 Total =40 Mat haploid parent = LEU TS Mata haploid parent = leu ts Spore phenotype # of spores LEU TS11 Leu ts9 leu TS9 leu ts11 Total =40 Conclusion? Probably not linked Probably linked Probably not linked

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Looking at the 10 tetrads in terms of LEU & TS How many PD? How many NPD? How many T? So what do you conclude about the LEU and TS genes? they are independently assorting but each close to a centromere!

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Our completed map Diploid genotype: MATa ADE HIS leu ts URA1 ura2 MAT ade his LEU TS ura1 URA2 leu HIS LEU his ts TS ADE ade

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ADE2 HIS4 LEU2 CDC7 (TS) How well did we do? Let’s look in SacchDB… The actual gene names… ADE2CDC7 (TS) HIS4LEU2 Not bad!

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What about URA? Know the parental typesLook at spore phenotypes Parental types? U1 u2 & u1 U2 What spore genotypes would you expect in a PD tetrad? U1 u2 u1 U2 Phenotype on -ura plate? no growth So, given these parental types… 0/4 spores growing is diagnostic of PD

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What spore genotypes would you expect in a NPD tetrad? Growth phenotype on -ura?Genotype? What about URA? Parental types? U1 u2 & u1 U2 U1 U2 u1 u2 GROWTH no growth

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What spore genotypes would you expect in a T tetrad? Growth phenotype on -ura?Genotype? What about URA? Parental types? U1 u2 & u1 U2 U1 u2 U1 U2 u1 U2 u1 u2 no growth GROWTH no growth

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Looking at the 10 tetrads… How many PD? How many NPD? How many T? So what do you conclude about the ura genes?

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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 = BeBe bEbE Practice question

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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. Recombinant gametes = B E and b e, 20% total = 10% each Parental type gametes = B e and b E, 80% total = 40% each Heterozygote genotype = BeBe bEbE Practice question

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Be/Be 0.16 bE/Be 0.16 BE/Be 0.04 be/Be 0.04 Progeny phenotypes: Be/bE 0.16 bE/bE 0.16 BE/bE 0.04 be/bE 0.04 Be/BE 0.04 bE/BE 0.04 BE/BE 0.01 be/BE 0.01 Be/be 0.04 bE/be 0.04 BE/be 0.01 be/be 0.01 BE 0.51 Be 0.24 bE 0.24 be Be 0.4 bE 0.1 BE 0.1 be parental non- parental 0.4 Be0.4 bE0.1 BE0.1 be gamete genotypes and frequencies Practice question

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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)

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3-point testcross—predicting progeny from a known map Predict the progeny phenotypes and numbers from this cross: Parent 1: Parent 2: ++a++a bc+bc+ bcabca bcabca + = wild type, dominant Map: bca 3 cM7 cM Step 1.Determine the number of DCO products Probability of recombinant product in (b-c) = Probability of recombinant product in (c-a) = Probability of recombinant product in both = 3% = 0.03 Count 10,000 progeny 7% = x 0.07 =

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Predicting progeny from a known map (cont’d) Heterozygous parent: only one chromatid of each homologue shown on next slide

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DCO: both together = x = 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 = (SCO + DCO) = 9021 Predicting progeny from a known map (cont’d)

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SCO DCO NCO 3-point testcross—constructing a linkage map Construct a linkage map (gene order and map distance) for the following genes in Drosophila: Genespr/+ (purple or red eyes) v/+ (vestigial or long wings) b/+ (black or tan body) Parents Female: pr/+v/+b/+ Male:pr/prv/vb/b Progeny phenotypes b32 v4125 pr266 vb272 prb4137 prv30 prvb574 Step 1.Expand the shorthand pr + v + b + pr + b+v+b+v+ pr + v+v+ b+b+ Step 2.Identify the NCO and DCO classes Step 3.Which gene is in the middle? Total = ASK: Which order allows us to go from the NCO genotype to the DCO genotype.

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Constructing a linkage map… Step 3 (cont’d) We know: (b + vpr + ) (bv + pr) DCO (bv + pr + ) (b + vpr) 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. order unknown b + vpr + bv + pr XX b + v + pr + bvpr incorrect b + pr + v bprv + XX b + prv bpr + v + correct!

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Constructing a linkage map (cont’d) Step 4.Calculate % recombinant products b+prv+ bpr+v NCO: = 8262 SCO b-pr : = 538 b+pr+v+ bprv SCO pr-v : = 1138 b+prv bprv+ DCO: = 62 % recombinants in b-pr interval= (538+62)/10000 =600/10000 =6% % recombinants in pr-v interval= ( )/10000 =1200/10000 =12%

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Step 5.Draw the map bprv 6 cM12 cM bprv 6 cM12 cM or Constructing a linkage map (cont’d)

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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 = 72 Observed DCO = 62 COC = 62/72 = 0.86 Interference = = 0.14

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alleles! Genetic vs. physical maps Genetic maps… based on recombinant frequencies between markers variation at location #1 variation at location #2 pr + vg vg + pr Alleles are detected as associated phenotypes New combination of phenotypes new combination of alleles recombination recombination: how frequent?

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Physical maps… based on DNA sequence or landmarks in sequence The number of chromosomal bands separating the known locations of genes. For example: AB The number of bp of DNA. The pattern of restriction sites in a DNA sequence. site1site2site4site5site3 Genetic vs. physical maps (cont’d)

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Which chromosome? Relation to centromere and telomere? pr + vg+ vgpr How do we know where to place these genes relative to the centromere and telomeres? Number of bp? Questions for Thought How do we know which physical entity (chromosome) our linkage group describes? What is the relationship between crossover frequency and gene order/distance (in bp of DNA) along the chromosomes? “Linkage group” = chromosome

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white is X-linked Linking the Physical and Genetic Maps Color is on chromosome IX knob extra DNA C Wx Novel Strain Cri du chat is on chromosome V From lecture 6 From lecture 8

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Metaphase chromosomes *** Fluorescently labeled single-stranded DNA -Fluorescence in situ hybridization (FISH) Partially denature DNA * * * Linking the Physical and Genetic Maps Double-stranded DNA segment from a particular location in the human genome

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Fragile XR-G colorblindness 6 cM genetic map (recombination)units = cM Haemophilia 10 cM What does the genetic map position tell us? physical map (DNA) ~6Xbp ~10Xbp units = bp Fragile X Haemophilia R-G colorblindness Physical map (FISH) units = chrom. bands -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). telomerecentromere

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