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 X tan no speck P 21 29 25 Recombinants never exceed 50% black speck R tan speck black no speck

Are b and sp linked? No! 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! b sp+ b+ sp+ b sp b+ sp All four gamete types are equally frequent vg+ sp + b+ sp b 1 2 3 4 vg 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 D 1 A B 2 D a b 3 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 D 1 A B 2 D a b 3 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. 3. 2. 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. 3. 2. 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 4 P 81 324 Ab, aB 4 R 2 8 AB, ab, Ab, aB 2 P, 2 R 43 86 86 # Recombinants X = % Rec? X Total # Gametes X

In-class experiment (cont’d) Class aggregate data: A-D Genotype P or R? Number? Number P gametes Number R gametes AD, ad 4 P 40 160 Ad, aD 4 R 19 76 AD, ad, Ad, aD 2 P, 2 R 61 122 122 # Recombinants 76+122 = % Rec? X Total # Gametes 76+122 +160 +122

Why do we observe IA? A B D A B D a b 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.

The range of possibilities: tightly linked  independent assort. What is the maximum recombination frequency in any interval? Parental Reminder… one crossover gives 2 parental, 2 recombinant gametes Recombinant Recombinant Parental 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  There are four ways the double crossover event can happen; together they result in a 50% recombinant and 50% parental. Look at page 179 in the latest edition of griffiths. 20 recombinant out of 400  5.0% if all cells have crossovers  200 recombinant out of 400  50% Maximum recombination frequency = 50% for single recombination events

The effect of multiple crossovers: What is the maximum recombination frequency in any interval? The effect of multiple crossovers: # xovers resulting gametes 2 (2 strands) 4 parental 2 (3 strands) 2 parental 2 non-par. Save this for later when you talk about maximum recombination frequency; also tally up the totals at the bottom (an equal number of recombinant and non recombinant. 2 (4 strands) 4 non-par. 6 parental 6 non-par. = 50% recombinant and 50% parental. Also true for triple Xover, quadruple Xover, etc.

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

Now consider independent assortment R Y r y X …the “ultimate” in non-linkage Refer to one of Mendel’s F1 x F1 dihybrid cross (round yellow X round yellow): What were the parental types for the F1? What were the parental and recombinant gametes made by the F1 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%

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 R f r f r F X 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? R f r f r F X The parental and recombinant types are the same! Need to be heterozygous at both loci 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-fragrant F and T linked at 10 cM F not linked to B Can’t tell!

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 they are independently assorting but each close to a centromere they are unlinked and at least one is distant from a centromere a parental ditype (PD) PD > T >> NPD mostly parental types either PD or non-parental ditype (NPD), with a 50:50 chance of each PD = NPD > T an equal proportion of parental and recombinant types Can’t distinguish Can distinguish! Fine-but emphasize that being able to see all four meiotic products. Compare what would be expected if you could or could not see mostly tetratypes (T), but also some PD, and NPD an equal proportion of parental and recombinant types

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

What did the data from quiz section tell you? Mat haploid parent = ade his Mata haploid parent = ADE HIS Spore phenotype # of spores ADE HIS 9 ADE his 11 ade HIS ade his Total = 40 Mat haploid parent = his LEU Mata haploid parent = HIS leu Spore phenotype # of spores HIS LEU 3 HIS leu 17 his LEU his leu Total = 40 Mat haploid parent = LEU TS Mata haploid parent = leu ts Spore phenotype # of spores LEU TS 11 Leu ts 9 leu TS leu ts Total = 40 Conclusion? No. Instead, summarize data from quiz section and what it told us (include linked and unlinked genes-> then go to database and point out that His is also linked to the centromere-> that a close look at the data also demonstrates this. What about Ura? The two genes are unlinked but affect different steps in the pathway for adenine biosynthesis-need to think about how the alleles would segregate. Conclusion? Conclusion? Probably not linked Probably linked Probably not linked

Looking at the 10 tetrads in terms of LEU & TS How many PD? How many NPD? How many T? 4 5 1 So what do you conclude about the LEU and TS genes? No. Instead, summarize data from quiz section and what it told us (include linked and unlinked genes-> then go to database and point out that His is also linked to the centromere-> that a close look at the data also demonstrates this. What about Ura? The two genes are unlinked but affect different steps in the pathway for adenine biosynthesis-need to think about how the alleles would segregate. they are independently assorting but each close to a centromere!

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

How well did we do? Let’s look in SacchDB… Not bad! The actual gene names… ADE2 CDC7 (TS) HIS4 LEU2 ADE2 HIS4 LEU2 Side point here is that since leu2 is linked to the centromere, and is also linked to his4, we might expect that his4 would provide evidence of centromere linkage, but the evidence would be less compelling than for leu2 (based on the database). CDC7 (TS) Not bad!

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

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

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

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

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

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: + + a b c + + = wild type, dominant Parent 1: b c a 3 cM 7 cM Map: b c a Parent 2: 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

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 = Both together refers to both products of the recombination event. (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/+ v/+ b/+ Male: pr/pr v/v b/b Progeny phenotypes + + + 564 b 32 v 4125 pr 266 v b 272 pr b 4137 pr v 30 pr v b 574 Step 1. Expand the shorthand SCO pr+ v+ b+ pr+ b+ v+ pr+ v+ b+ DCO Step 2. Identify the NCO and DCO classes NCO SCO Step 3. Which gene is in the middle? ASK: Which order allows us to go from the NCO genotype to the DCO genotype. Total = 10000

incorrect correct! Constructing a linkage map… Step 3 (cont’d) We know: (b+ v pr+) (b v+ pr) DCO (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+ b v+ pr b+ v+ pr+ b v pr incorrect X b+ pr+ v b pr v+ X b+ pr v b pr+ v+ correct!

Constructing a linkage map (cont’d) Step 4. Calculate % recombinant products NCO: 4125+4137 = 8262 % recombinants in b-pr interval= b+ pr v+ b pr+ v SCOb-pr: 266+272 = 538 (538+62)/10000 =600/10000 =6% b+ pr+ v+ b pr v SCOpr-v: 564+574 = 1138 % recombinants in pr-v interval= b+ pr v b pr v+ DCO: 30+32 = 62 (1138+62)/10000 =1200/10000 =12%

Constructing a linkage map (cont’d) Step 5. Draw the map b pr v 6 cM 12 cM b pr v 6 cM 12 cM or

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 observed DCO COC = Interference = 1 - COC expected DCO 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 variation at location #1 variation at location #2 alleles! pr+ vg pr vg+ Next year, use your genetic vs. physical map. recombination: how frequent? Alleles are detected as associated phenotypes New combination of phenotypes  new combination of alleles  recombination

Genetic vs. physical maps (cont’d) 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? Relation to centromere and telomere? pr+ 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 knob extra DNA C Wx Novel Strain From lecture 8 Cri du chat is on chromosome V

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

What does the genetic map position tell us? Fragile X R-G colorblindness 6 cM genetic map (recombination) units = cM Haemophilia 10 cM telomere centromere Physical map (FISH) units = chrom. bands physical map (DNA) ~6Xbp ~10Xbp units = bp Fragile X Haemophilia R-G colorblindness X~1,000,000 bp in humans (how many centimorgans in humans?) X~600,000bp in flies X~3.5kb in yeast. -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).