Crossing-Over and Recombination

Similar documents
Recombination. The kinetochore ("spindle attachment ) always separates reductionally at anaphase I and equationally at anaphase II.

Genetics Lecture Notes Lectures 6 9

Chapter 4 Gene Linkage and Genetic Mapping

Midterm#1 comments#2. Overview- chapter 6. Crossing-over

Gene Linkage and Genetic. Mapping. Key Concepts. Key Terms. Concepts in Action

Genetics and Genomics in Medicine Chapter 1. Questions & Answers

Recombination Models. A. The Holliday Junction Model

Recombination Models. A. The Holliday Junction Model

Introduction to Plant Genetics Spring 2000

Human Anatomy & Physiology I Dr. Sullivan Unit IV Cellular Function Chapter 4, Chapter 27 (meiosis only)

Chapter 7. Linkage and Chromosome Mapping

FINDING THE PAIN GENE How do geneticists connect a specific gene with a specific phenotype?

Chapter 9. Gene Interactions. As we learned in Chapter 3, Mendel reported that the pairs of loci he observed segregated independently

Exam Plant Genetics CSS/Hort 430/530

6.2 Chromatin is divided into euchromatin and heterochromatin

Chapter 11 Homologous Recombination at the Molecular Level. 吳彰哲 (Chang-Jer Wu) Department of Food Science National Taiwan Ocean University

Concepts of Genetics Ninth Edition Klug, Cummings, Spencer, Palladino

GENES AND CHROMOSOMES-I. Lecture 3. Biology Department Concordia University. Dr. S. Azam BIOL 266/

C. Incorrect! Second Law: Law of Independent Assortment - Genes for different traits sort independently of one another in the formation of gametes.

GENETICS. Chapter 1: Cell cycle. Thème 1 : La Terre dans l Univers A. Expression, stabilité et variation du patrimoine génétique.

Chapter 3. DNA Replication & The Cell Cycle

Genetics or Genomics?

-Genes on the same chromosome are called linked. Human -23 pairs of chromosomes, ~35,000 different genes expressed.

FINDING THE PAIN GENE How do geneticists connect a specific gene with a specific phenotype?

Biology 105: Introduction to Genetics Page 1 Midterm Fall 2004

Vocab Word 1: Interphase

1. I can describe the stages of the cell cycle.

7.06 Problem Set Four, 2006

1. I can describe the stages of the cell cycle.

LINKAGE AND CHROMOSOME MAPPING IN EUKARYOTES

Problem Set 2B Name and Lab Section:

Identifying Genes Underlying QTLs

LECTURE 5: LINKAGE AND GENETIC MAPPING

UNIVERSITY OF KWAZULU-NATAL EXAMINATIONS: JUNE SUBJECT, COURSE AND CODE: GENE 240: Introductory Genetics DURATION: 3 HOURS TOTAL MARKS: 180

21.5cM. So probability of double recombinant in the absence of interference is:

CELLULAR PROCESSES; REPRODUCTION. Unit 5

A single meiotic gene, ZIP4, is responsible for the Ph1 locus effect on recombination. Azahara C. Martín John Innes Centre, Norwich, UK

CHAPTER 4 STURTEVANT: THE FIRST GENETIC MAP: DROSOPHILA X CHROMOSOME LINKED GENES MAY BE MAPPED BY THREE-FACTOR TEST CROSSES STURTEVANT S EXPERIMENT

Cell cycle. Chen Li. Department of cellular and genetic medicine

DNA is the genetic material found in cells Stands for: Deoxyribonucleic Acid Is made up of repeating nucleic acids It s the Unit of Heredity

Genetics and Genomics in Medicine Chapter 1 Questions

Human linkage analysis. fundamental concepts

1. What is DNA? 2. List anything you know about DNA. (from readings, class, TV?)

MUTANT: A mutant is a strain that has suffered a mutation and exhibits a different phenotype from the parental strain.

Concepts: What are RFLPs and how do they act like genetic marker loci?

Human linkage analysis. fundamental concepts

Map-Based Cloning of Qualitative Plant Genes

1. What is DNA? 2. List anything you know about DNA. (from readings, class, TV?)

1. What is DNA? 2. List anything you know about DNA. (from readings, class, TV?)

3I03 - Eukaryotic Genetics Repetitive DNA

The Synaptinemal Complex and Four-Strand Crossing Over

CHAPTER 5 Principle of Genetics Review

Molecular Cell Biology - Problem Drill 06: Genes and Chromosomes

How to use genetic methods for detecting linkage. David D. Perkins. Background

A) The constituent monomer of DNA and RNA. C) The basic structural unit of chromatin with "bead-on-a-string" morphology

Applicazioni biotecnologiche

Chapter 6 Linkage and Chromosome Mapping in Eukaryotes

four chromosomes ` four chromosomes correct markers (sister chromatids identical!)

Linkage & Genetic Mapping in Eukaryotes. Ch. 6

D) Gene Interaction Takes Place When Genes at Multiple Loci Determine a Single Phenotype.

Overview of Human Genetics

Requirements for the Genetic Material

Received November 4, 1964

Packing ratio the length of DNA divided by the length into which it is packaged

Chapter 1: Basic Genetics 2

Releasing Natural Variation in Bread Wheat by Modulating Meiotic Crossovers. James Higgins (WGIN trait coordinator for Recombination)

Genetics - Problem Drill 05: Genetic Mapping: Linkage and Recombination

Enzyme that uses RNA as a template to synthesize a complementary DNA

results may conflict yet NOT allelic by the SEGREGATION test (nonparental alleles recovered by meiotic recombination)

DESIGNER GENES SAMPLE TOURNAMENT

Lecture 6. Chromosome Structure and Function in Mitosis

B) You can conclude that A 1 is identical by descent. Notice that A2 had to come from the father (and therefore, A1 is maternal in both cases).

Plant Molecular and Cellular Biology Lecture 9: Nuclear Genome Organization: Chromosome Structure, Chromatin, DNA Packaging, Mitosis Gary Peter

4.1 CELL DIVISION AND GENETIC MATERIAL

environment (diffusion, etc.). High SA:V ratio is favorable. Ex. 6:1 is better than 6:5

Cell cycle: Cell growth and division (when the replication and segregation of chromosomes occurs). Interphase: Interphase Chromosomes

DNA Recombination Biochemistry 302. Bob Kelm February 2, 2005

Why do we need statistics to study genetics and evolution?

Appendix A DNA and PCR in detail DNA: A Detailed Look

Lateral DNA Transfer MECHANISMS AND CONSEQUENCES Frederic Bushman

Human Molecular Genetics Assignment 3 (Week 3)

Chapter 29. DNA as the Genetic Material. Recombination of DNA. BCH 4054 Fall 2000 Chapter 29 Lecture Notes. Slide 1. Slide 2. Slide 3.

Mutations during meiosis and germ line division lead to genetic variation between individuals

All Paired Up with No Place to Go: Pairing, Synapsis, and DSB Formation in a Balancer Heterozygote

DNA and GENETICS UNIT NOTES

THE CELLULAR AND MOLECULAR BASIS OF INHERITANCE

High-throughput genotyping with microarrays

COMPETITOR NAMES: TEAM NAME: TEAM NUMBER:

GENETICS. I. Review of DNA/RNA A. Basic Structure DNA 3 parts that make up a nucleotide chains wrap around each other to form a

Midterm Sample Questions Biology 1001A Page 1 of 9

Introduction to some aspects of molecular genetics

Phenotype analysis: biological-biochemical analysis. Genotype analysis: molecular and physical analysis

B. Incorrect! Centromeric DNA is largely heterochromatin, which is inactive DNA.

Gregor Mendel. Austrian Monk Worked with pea plants

Lecture 23: Causes and Consequences of Linkage Disequilibrium. November 16, 2012

Biological Sciences 50 Practice Exam 1

Chapter 4 The Transmission of DNA at Cell Division. DNA Replication Cell Division Mitosis Meiosis

Chapter 11 DNA Replication and Recombination

Genetics and Heredity. Mr. Gagnon

What is a chromosome and where is it located and what does it

Transcription:

Updated 2/20/06 Crossing-Over and Recombination Required Readings: Fu, H. et al., 2002. Recombination rates between adjacent genic and retrotransposon regions in maize vary by 2 orders of magnitude. PNAS 99:1082-1087. Yao, H. et al. 2002. Molecular characterization of meiotic recombination across the 140- kb mutigenic a1-sh2 interval of maize. PNAS 99:6157-6162. Sherman JD, Stack SM (1995) Two-dimensional spreads of synaptonemal complexes from solanaceous plants. VI. High-resolution recombination nodule map for tomato (Lycopersicon esculentum). Genetics 141:683-708. Additional Useful Reading: Dawe, R.K. 1998. Meiotic chromosome organization and segregation in plants. Ann Rev Plant Physiol Plant Mol Biol 49:371-95 Hamant, O. et al. 2006. Genetic of meiotic prophase I in plants. Ann. Rev. Plant Bio. Prepublication PDF file available online at Ann. Rev. Plant Bio. Website. Schnable, P. et al. 1998. Genetic recombination in plants. Curr. Op. Plant Bio. 1:123-129. How does genetic recombination occur? If genes are on different chromosomes, recombination occurs simply by the random assortment of chromosomes into daughter cells at the first meiotic division. But, if genes are on the same chromosome, recombination requires crossing-over of non-sister chromatids from homologous chromosomes. This occurs during the first meiotic division, when homologous chromosomes are paired. Outline of events involved in crossing-over My presentation of the events in meiosis at the chromosomal level follows Dawe (1998). See his excellent review article for more details. 1

Fig. 1 of Dawe (1998) Premeiotic interphase: DNA replication occurs, chromosomes are diffuse and homologs are not paired. Prophase of Meiosis I begins: Leptotene: Chromosomes condense so they are individually identifiable. Sister chromatids are tightly associated because they are bound to a common protein core called the Axial Element (AE). Zygotene: Presynaptic alignment occurs: homologs align but are not yet near enough to each other to synapse. The controlling mechanism of this is not known. Early Recombination Nodules (RN), small protein bodies, appear between homologs that are aligned but not yet synapsed. The RNs seem to be involved in the homology search that helps chromosomes find their homologs: sometimes RNs form between nonhomologous chromosomes, but then they disappear again, so synapsis does not occur in those cases. Fibers connecting RNs and the two homologous chromosomes have been observed, this may be how the homologs are brought into closer contact so that crossingover and synapsis can occur. See Figure 4 of Dawe (1998, below), and Figures in Sherman and Stack (1995) for pictures of recombination nodules. Crossing-over is probably initiated at this point in a subset of the RNs. We will return to the mechanism of crossing-over, but note that this is the point at which the double strand 2

breaks that initiate crossing-over likely occur. Homologs of two yeast enzymes that mediate recombination have been identified in plant RNs at zygotene. Figure 4 of Dawe (1998): Axial elements and early RNs Following the initiation of crossing-over, synapsis begins. Synapsis is accomplished by the synaptonemal complex (SC), a protein structure that binds the AEs of the two homologs together. The SC is initiated at several points between the homologs, and the SC grows, bringing the homologs into tight association. The precise mechanistic role of the SC is not known; it may act as a scaffold to stabilize those RNs in which crossing over occurs. SCs also seem to mediate recombination interference, which is a reduction in the probability of cross-overs occurring in adjacent chromosomal regions, relative to 3

what would be expected if cross-overs occur independently across chromosomes (more on this later). Pachytene: Homologs are completely synapsed. Most early RNs are degraded or dissociate from the SC by pachytene. Those RNs that remain in pachytene are called late RNs, and these are very likely the sites at which crossing-over has occurred (or is still occurring). The recombination-mediating proteins observed in early RNs are no longer associated with the late RNs. Thus, RNs are involved in bringing together homologs and initiating synapsis, and a small proportion of RNs represent sites of crossing-over; these are the late RNs. Diplotene: Homologs separate, but are held together at chiasmata. Chiasmata appear only at sites of crossing-over, and only where late RNs were located. Probably some proteins hold chiasmata together until anaphase. Diakinesis: The chromosomes condense further by spiraling along their axes. Metaphase I: Spindle forms, chromosomes are aligned on metaphase plate. Anaphase I: Spindles pull homologous centromeres to opposite poles. Chiasmata dissolve, finally releasing the homologous chromatids. 4

These events are also nicely summarized in Table 1 of Roder (1997): Table 1. Events in meiotic prophase Chromosome morphology Stage in meiotic prophase and SC morphogenesis Bouquet formation DSB repair Leptotene axial elements telomeres begin DSBs appear begin to to cluster develop Zygotene Pachytene Diplotene Diakinesis chromosome synapsis initiates chromosomes fully synapsed SC disassembled; chromosomes condense further chromosome telomeres tightly clustered telomeres disperse DSBs disappear double Holliday junctions mature recombinants Cytological signs of recombination early nodules early nodules late nodules chiasmata chiasmata compaction The precise sequence of events varies somewhat from one organism to another. Meiotic DSB repair has been studied exclusively in S. cerevisiae, whereas most observations of recombination nodules and all observations of chiasmata have been made in other organisms. (SC) Synaptonemal complex; (DSB) double-strand break. Mechanism of Crossing-Over The Double-strand break (DSB) repair model seems most likely based on evidence from yeast. See Figure 1 from Schnable et al. (1998) don t worry about Figures 2 and 3, they are models to explain nonhomologous recombination. Also, don t worry about part f of Figure 1, because that is not what happens during crossing-over. Finally, note that there is an error in Figure 1, the white chromatid orientation is mislabeled reverse the orientation of 5 and 3 ends of the white chromatid strands to get the correct orientation. A better picture is Figure 1 of Smith (2001): 5

Figure 1. Double-strand break repair model of recombination, after Szostak et al. (112 ). For explanation, see the text. Alternative resolutions at the last step are not shown. Arrowheads indicate 3' ends; solid dots, Spo11 protein linked to 5' ends at the DNA break (56 ); wavy arrows, newly synthesized DNA; thin lines, DNA from one parent; thick lines, DNA from the other parent; open and closed triangles, points of resolution by strand cutting, swapping, and ligation. Brackets with and without an asterisk indicate regions of symmetric and asymmetric hybrid DNA, respectively, which can give rise to aberrant 4:4, 5:3, or 6:2 segregations. 6

The double Holliday junction depicted in the figure above can be resolved by cutting the DNA strands in two places, but this can happen in two ways, with different consquences. First, if the same two strands are cut at the different positions (the white horizontal arrows in the figure above), the result is noncrossover products. Outside of the region defined by the two cut sites, there is no resulting recombination. In the relative small region between the two cut sites, the chromatids potentially have heterologous DNA duplexes (e.g., at some points a G may be paired with an A instead of a C because the two alleles differed in that region). The differences between the two heterologous strands can be repaired with DNA, but the repair can use either strand as a template, so the result could be an allelic change on in this region on one or both chromatids. The other resolution of the double Holliday junction occurs if all four DNA strands are cut, resulting in ligation of non-sister chromatids. This resolution causes a chiasmata to result at diplotene and outside of the cuts sites, genetic recombination results. Between the two cut sites, heterologous DNA duplexes can occur, and these can be repaired in the same way as the noncrossover products. If the repair of heterologous DNA duplexes results in the conversion of one allele to another between the two cut sites, this is referred to as gene conversion. To more explicitly demonstrate how gene conversion occurs with crossing-over, consider a sequence that is polymorphic between the two homologs at the site where recombination occurs (imagine that the sequence is otherwise quite similar outside of this region). This is a very schematic example, where I show the cuts and new DNA synthesis all happening in the space of four base pairs. In reality the size of these regions is at least an order of magnitude greater (and in plants sometimes greater than 1 kb): Homolog A Chromatid 1 3 ATCGATCGATCG 5 5 TAGCTAGCTAGC 3 Homolog A Chromatid 2 3 ATCGATCGATCG 5 5 TAGCTAGCTAGC 3 Homolog B Chromatid 1 3 ATCGGCTAATCG 5 5 TAGCCGATTAGC 3 Homolog B Chromatid 2 3 ATCGGCTAATCG 5 5 TAGCCGATTAGC 3 Assume the DSB occurs in Homolog A Chromatid 2: 5 3 exonuclease does its thing: 3 ATCGAT CGATCG 5 5 TAGCTA GCTAGC 3 3 AT CGATCG 5 5 TAGCTA GC 3 7

Homolog B Chromatid 1 serves as template for DNA repair of Homolog A Chromatid 2: Homolog A Chromatid 2 3 ATCGGCCGATCG 5 Homolog B Chromatid 1 5 TAGCCGATTAGC 3 Homolog A Chromatid 2 5 TAGCTA ATTA GC 3 Homolog B Chromatid 1 3 ATCGGCTAATCG 5 To resolve the resulting Holliday junctions, the four chromatids involved in the repair are broken and religated at two positions, which we will show to match Figure 1 of Schnable et al. (1998). Homolog A Chromatid 2 3 ATCG GCCGATCG 5 Homolog B Chromatid 1 5 TAGCCGATTA GC 3 Homolog A Chromatid 2 5 TAGCTAATTA GC 3 Homolog B Chromatid 1 3 ATCG GCTAATCG 5 Double helices are resolved (remember that the double helix has one 5 3 strand and one 3 5 strand) by swapping the first and second halves of this sequence between homologs: Homolog A Chromatid 2 3 ATCGGCTAATCG 5 Homolog A Chromatid 2 5 TAGCTAATTAGC 3 Homolog B Chromatid 1 3 ATCGGCCGATCG 5 Homolog B Chromatid 1 5 TAGCCGATTAGC 3 Notice that at the sequences that were removed by the nuclease and resynthesized during crossing over, the two DNA strands do NOT have complementary nucleotides, they are heteroduplexes. When this occurs, the cell can repair the faulty sequences by converting one or the other strand so they are complements (but maybe the repair will not happen, in which case, the first mitotic daughter cells of the heteroduplex chromatid, which may be in the zygote, will segregate!). If the top strand of chromatid is used as the correct strand and the bottom strand is repaired, the result is: Homolog A Chromatid 2 3 ATCGGCTAATCG 5 5 TAGCCGATTAGC 3 Or if the bottom strand is used as correct and the top strand is repaired: Homolog A Chromatid 2 3 ATCGATTAATCG 5 5 TAGCTAATTAGC 3 Correcting the bottom strand of the other chromatid gives: 8

Homolog B Chromatid 1 3 ATCGGCCGATCG 5 5 TAGCCGGCTAGC 3 or using its bottom strand as correct and fixing the top strand gives: Homolog B Chromatid 1 3 ATCGGCTAATCG 5 5 TAGCCGATTAGC 3 Then, remembering that there are two other chromatids that were not involved in the recombination, and considering only the four polymorphic bases, and considering only the 5 3 strand sequences, the four meiotic products are: ATCG ATCG ATCG ATCG CGAT CGAT TAAT TAAT CGGC or CGAT or CGGC or CGAT CTAG CTAG CTAG CTAG 9

Thus, if one could observe the products of a single meiosis, one could observe ratios of 3:1 instead of 2:2 among the gametes (as in the first bp in the first meiosis above, A:C:C:C). This is gene conversion, and it has been observed in fungi where tetrads can be studied. If you cannot observe the products of each meiosis as a group, you would not be able to observe this. At the level of an individual base pair, you could also recover three unique sequences among the four gametes (as in the 4 th bp in the first meiosis above, G:T:C:G)! I don t know if there is evidence for this or not. Notice that gene conversions will occur with every recombination, but the conversions will only occur in a small region around where the double strand break occurs (less than 1 kbp, (Smith 2001), so if one is studying a particular locus and looking for evidence of gene conversion, it will be found only with very small frequency at any one locus. Recombination Hot Spots Do crossovers occur with equal frequency at all regions in the genome? It should be obvious that the answer is no, because crossovers are almost entirely suppressed around the centromere. But, we now have good evidence that cross-overs occur more frequently within genes than in intergenic regions. Civardi et al. (1994) measured recombination within the sh1 gene of maize. To estimate intra-locus recombination, they used two different recessive alleles of the locus, which were null alleles because each contained a transposon insertion, but the insertions were at two different locations, about 1 kbp apart. They estimated the frequency of recombination between these two positions located to be 0.0046. Next, they estimated the frequency of recombination between the sh1 gene and the a1 gene to be 9%. Finally, they identified a single YAC clone with both loci in it, and used that clone to physically measure the distance between the two loci, which they reported to be 140 kbp. This allowed a measurement of the number of base pairs per percent of recombination within the sh1 gene and between the genes: Within a1 Between sh1 and a1 Physical Distance 1 kbp 140 kbp Recomb. Freq. 0.0046 0.09 Bp/ %recomb 217 bp 1,555 bp The frequency of recombination per physical size within this gene is 7 times greater than in the adjacent intergenic region! More recently, Yao et al. (2002) sequenced two alleles of the 140-kb region between a1 and sh2 and mapped precisely the position of 101 recombination events in this region using a very high density of DNA markers. They found that, as expected from the above experiment, most of the recombination events occurred in the genic rather than intergenic regions. But they also found that there was one gene in this interval that had a very low frequency of recombination and, conversely, there was a nongenic region that was a hotspot for recombination. So, not all genes are recombination hotspots and not all 10

hotspots are genes. They did find, however, that the regions that were primarily made up of repetitive retrotransposon-derived sequences (which other studies suggest make up about half of the maize genome!) were consistently very low in recombination activity. This last result was also observed by Fu et al. (2002), who found that the retrotransposondense region on one side of the bz1 locus had very few recombinations, whereas the gene-rich region on the other side of the locus had about two orders of magnitude greater recombination frequency. Why might suppression of crossing-over in regions containing repetitive DNA elements that compose a high proportion of a genome be important for maintaining fitness? Creation of Novel Alleles from the Crossing-Over Process Intragenic recombination A major implication of genes being recombination hotspots is that crossing-over may be an important force for producing novel alleles because intragenic cross-over events can generate novel alleles much more easily than cross-overs between genes. For example, if one allele has a gene sequence of: ACGTACGTACGT, and another allele of the same locus has sequence: TCGAACGTACGA, then a cross-over within this region could generate novel recombinant alleles: ACGTACGTACGA and TCGAACGTACGT. Richter et al. (1995) found examples progeny from heterozygotes at the complex Rp1 (resistance to common rust) locus in maize that possessed new Rp1 alleles conferring resistance to some rust strains that neither parent plant had resistance to. Each of these progeny had recombinant genotypes at marker loci flanking the locus, indicating that the new alleles were associated with recombination (not simple mutation). Novel alleles may be generated with relatively high frequency by crossing-over within complex loci like Rp1 that represent tandemly repeated genes with divergent functions. In such regions, mispairings may occur more frequently at meiosis, because there are numerous related sequences side by side. A recombination between mispaired copies of the replicated locus can easily produce new alleles (and new haplotypes, some of which can have even more copies of the replicated gene). Gene conversion Gene conversion resulting from crossing-over can also generate novel alleles, as already shown. Using the parental sequences in the previous example, gene conversion can generate novel alleles: ACGAACGTACGA and TCGTACGTACGT. Note than gene conversion can change alleles only within less than 1 kbp of the initial double strand breakpoint. 11

Double Crossovers and Crossover Interference Consider what happens when there is one crossover between two loci, A and B, in a heterozygote with parental allelic combinations AB and ab: A--------------- B A--------------- B Parental gamete A--------------- B A--------------- b Recombinant gamete a --------------- b a --------------- B Recombinant gamete a --------------- b a --------------- b Parental gamete With one crossover event, half of the resulting gametes are parental types and half are recombinant types. Recombination frequency is defined as the frequency of recombinant gametes formed between two loci. Thus, if exactly one crossover occurred between two loci at every meiosis, there would be 50% recombination frequency between the two genes. Now consider what happens when there are two crossovers between A and B. There are four equally likely ways for the double crossover to occur: A--------------- B A--------------- B Parental gamete A--------------- B A--------------- B Parental gamete a --------------- b a --------------- b Parental gamete a --------------- b a --------------- b Parental gamete A--------------- B A--------------- b Recombinant gamete A--------------- B A--------------- b Recombinant gamete a --------------- b a --------------- B Recombinant gamete a --------------- b a --------------- B Recombinant gamete A--------------- B A--------------- B Parental gamete A--------------- B A--------------- b Recombinant gamete a --------------- b a --------------- b Recombinant gamete a --------------- b a --------------- B Parental gamete A--------------- B A--------------- b Recombinant gamete A--------------- B A--------------- B Parental gamete a --------------- b a --------------- B Recombinant gamete a --------------- b a --------------- b Parental gamete If all four of these events are equally likely, then we still expect 50% recombinant gametes to occur! In fact, any number of crossovers greater than zero will produce 50% recombinant gametes. This is why recombination frequency is not a linear function of the average number of crossovers between two loci. If the A and B loci are widely separated on the 12

chromosome, several crossovers may occur between them regularly at each meiosis, but they will still have only a maximum recombination frequency of 50%. A third locus, C, that is even more distant from A will also still only have a recombination frequency of 50%, so just by comparing the two-point recombination frequencies, one cannot determine that B is physically closer to A and C is. Therefore, two-point recombination frequencies alone cannot be directly related to the average number of crossovers between two loci, and recombination frequencies are not additive between pairs of loci. To clarify the relationship between multiple crossovers and multiple recombination events, consider three loci linked in the order A B C and consider the different products that result from one crossover in the A B interval and one crossover in the B C interval: A B pair B C pair A C pair A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c a -------B-------- c Recomb Recomb Parental a -------b-------- c a -------b-------- c Parental Parental Parental a -------b-------- c a -------B--------C Recomb Parental Recomb a -------b-------- c a -------b-------- C Parental Recomb Recomb a -------b-------- c a -------B-------- c Recomb Recomb Parental a -------b-------- c a -------b-------- C Parental Recomb Recomb A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c a -------B--------C Recomb Parental Recomb a -------b-------- c a -------b-------- c Parental Parental Parental Sum r = 50% r = 50% r = 50% What about two crossovers between A B and one crossover between A C? The result is the same! (See Appendix 1 this requires enumeration of 16 possible arrangements of the three crossovers, and 48 possible products.) What about two crossovers between A B and zero crossovers between A C? In this case, the result is the same for the A B and A C intervals, but there is no recombination between B and C (Appendix 2). 13

In all cases (irrespective of the actual number of crossover events that generate a recombination), if the A B interval and the B C are both parental or both recombinant, then the A C interval is parental. In all cases, if the A B interval is parental and the B C interval is recombinant or the A B interval is recombinant and the B C interval is parental, then the A C interval is recombinant. If recombination events (having one or more crossovers in an interval) occur independently of each other on the chromosome, then the relationship between recombination frequencies between three loci is as follows: r AC = r AB (1 - r BC ) + r BC (1 - r AB ) = r AB + r BC 2r AB r BC. If a recombination event in one interval (say A B) hinders a simultaneous recombination even in an adjacent interval (say B C), then this relationship will not hold, and interference is said to occur. To account for interference, the equation is written as: r AC = r AB + r BC 2Cr AB r BC = r AB + r BC 2(1-i)r AB r BC, where C = coefficient of coincidence, and i = coefficient of interference = 1 - C. If there is no interference (pairs of recombination events occur in random positions), then i = 0, and we have the original equation. If a crossover in interval A B always completely suppresses recombination in interval B C and vice versa, then there is complete interference and i = 1, and r AC = r AB + r BC. The implications of this are that: 1) Unless there is complete interference (not likely), then recombination fractions are not additive, so generally, r AC r AB + r BC. As we collect data on many loci, and we wish to generate a complete genome map, we would prefer to use map distances that are additive. With additive map distances, one can then indicate that, for example, locus B is more closely physically linked to locus A than C is, even if A has 50% recombination frequency with both loci B and C. The way this is done is to use map functions that report linkage distances in map units which are in terms of the number of crossovers that occur on average between two loci, rather than the number of recombination events. So, you could have loci linked 150 map units apart, and these would have 50% recombination frequency. We will deal with mapping functions in a later lecture. 2) To derive an appropriate mapping function that relates observed number of recombination events to expected number of crossover events, one must know (or 14

assume) how important interference is. If the wrong assumption is made about interference, then the mapping function will be wrong. So, before attacking the issue of mapping functions, what do we know from plant biology about interference? There are at least two approaches that have been used to study interference: 1. Tetrad analysis Recovery of the four products of a single meiosis allows one to determine if recombinations occurred independently between linked markers or not. Copenhaver et al. used this approach in Arabidopsis (Copenhaver et al. 1998). Their conclusion: the number ofdouble COs in adjacent intervals was 65% lower than that expected from the genetic map. Thus, crossover interference seems to be reasonably strong. 2. Recombination nodule maps - by carefully studying the numbers and positions of late recombination nodules (RNs), Sherman and Stack (1995) were able to develop a recombination nodule map that corresponds to a survey of the positions and frequencies of crossovers and double crossovers of the tomato genome. This is a map of crossovers, which is not necessarily the same as a map of recombinations! Sherman and Stack (1995) reported the following results from a survey of 270 female meioses: 1. There is a 1:1 relationship between recombination nodules (RNs) and chiasmata. 2. Every synaptonemal complex (SC) has at least one RN. 3. There are no RNs at telomeres or kinetochores (centromeres) (Figs. 12 and 18). 4. RNs are more common in euchromatin than heterochromatin. 5. The number of RNs per chromosome pair is mainly related to the length of euchromatin in the chromosome pair. 6. Crossover interference is common: a RN in one chromosome arm seems to drive the second RN into the other chromosome arm. 7. Both positive and negative crossover interference was observed (Figure 17). Pairs of RNs tend not to occur in closely adjacent intervals as often as expected, nor did they tend to occur in very distant intervals, instead, pairs tended to occur more frequently than expected at intermediate distances (Figure 17). These results suggest that a proper mapping function should account for crossover interference, but the problem is that crossover interference is not constant throughout the genome! Additional References: Copenhaver GP, Browne WE, Preuss D (1998) Assaying genome-wide recombination and centromere functions with Arabidopsis tetrads. Proc Nat Acad Sci USA 95:247-252. 15

Dawe RK (1998) Meiotic chromosome organization and segregation in plants. Ann Rev Plant Physiol Plant Mol Biol 49:371-95 Fu H, Zheng A, Dooner HK (2002) Recombination rates between adjacent genic and retrotransposon regions in maize vary by 2 orders of magnitude. PNAS 99: 1082-1087. Richter TE, Pryor TJ, Bennetzen JL, Hulbert SH (1995) New rust resistance specificities associated with recombination in the Rp1 complex in maize. Genetics 141:373-381 Roeder GS (1997) Meiotic chromosomes: it takes two to tango. Genes Development 11:2600-2621 Smith GR (2001) Homologous recombination near and far from DNA breaks: Alternative Roles and Contrasting Views. Ann Rev Genet 35:243-274 Yao H, Zhou Q, Li J, Smith H, Yandeau M, Nikolau BJ, Schnable PS (2002) Molecular characterization of meiotic recombination across the 140-kb multigenic a1-sh2 interval of maize PNAS 99:6157-6162. 16

Appendix 1. Results of two crossovers between loci A and B plus one crossover between B and C. A B pair B C pair A C pair a -------b-------- c 6 a -------b-------- C Parental Recomb Recomb a -------b-------- c a -------b-------- c Parental Parental Parental A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------B-------- c Recomb Recomb Parental a -------b-------- c a -------B-------- C Recomb Parental Recomb A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------b-------- c Parental Parental Parental a -------b-------- c a -------B-------- c Recomb Recomb Parental A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------B-------- c Recomb Recomb Parental a -------b-------- c a -------b-------- c Parental Parental Parental a -------b-------- c 6 a -------b-------- c Parental Parental Parental a -------b-------- c a -------b-------- C Parental Recomb Recomb A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------B--------C Recomb Parental Recomb a -------b-------- c a -------B-------- c Recomb Recomb Parental a -------b-------- c 6 a -------B--------C Recomb Parental Recomb a -------b-------- c a -------b-------- C Parental Recomb Recomb a -------b-------- c 6 a -------b-------- C Parental Recomb Recomb a -------b-------- c a -------B--------C Recomb Parental Recomb 17

a -------b-------- c 6 a -------b-------- c Parental Parental Parental a -------b-------- c a -------b-------- C Parental Recomb Recomb A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------B-------- c Recomb Recomb Parental a -------b-------- c a -------B--------C Recomb Parental Recomb a -------b-------- c 6 a -------B-------- c Recomb Recomb Parental a -------b-------- c a -------b-------- C Parental Recomb Recomb a -------b-------- c 6 a -------b-------- C Parental Recomb Recomb a -------b-------- c a -------B-------- c Recomb Recomb Parental a -------b-------- c 6 a -------b-------- C Parental Recomb Recomb a -------b-------- c a -------b-------- c Parental Parental Parental A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------B--------C Recomb Parental Recomb a -------b-------- c a -------B-------- c Recomb Recomb Parental A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------B--------C Recomb Parental Recomb a -------b-------- c a -------b-------- c Parental Parental Parental A-------B-------- C A-------b-------- C Recomb Recomb Parental a -------b-------- c 6 a -------b-------- c Parental Parental Parental a -------b-------- c a -------B--------C Recomb Parental Recomb Sum r = 50% r = 50% r = 50% The result is the same as for one crossover between A and B plus one crossover between B and C! 18

Appendix 2. Results of two crossovers between A and B and zero crossovers between A and C A B pair B C pair A C pair a -------b-------- c 6 a -------b-------- c Parental Parental Parental a -------b-------- c a -------b-------- c Parental Parental Parental a -------b-------- c 6 a -------B-------- C Recomb Parental Recomb a -------b-------- c a -------B-------- C Recomb Parental Recomb a -------b-------- c 6 a -------b-------- c Parental Parental Parental a -------b-------- c a -------B-------- C Recomb Parental Recomb a -------b-------- c 6 a -------B-------- C Recomb Parental Recomb a -------b-------- c a -------b-------- c Parental Parental Parental Sum r = 50% r = 0% r = 50% The results for the A B and A C intervals are the same as for previous examples with one or two crossovers between A and B! Appendix 3. Rewording Lynch and Walsh (1998) p. 394 to make it correct. Each recombination frequency is half the probability that at least one crossover occurs between the markers, while 1 c is the probability that zero crossovers occur between the markers. There are two different ways to get a recombination in the interval A C: at least one crossover either in A B or in A C, or at least one crossover in both A B and A C. In either case, 50% of the meiotic products will be recombinant for A and C. In terms of recombination, this means that a recombination between A and C will occur if A recombines with B and B does not recombine with C, or if A does not recombine with B and B recombines with C. If A recombines with B and B recombines with C, then A does not recombine with C. If there is no interference, so that the presence of a crossover in one region has no effect on the frequency of crossovers in adjacent regions, these recombination frequencies can be related as: c AC = c AB (1 - c BC ) + c BC (1 - c AB ) = c AB + c BC -2c AB c BC 19

In terms of crossover probabilities, if crossovers in the two intervals are independent, then: c AC = (½)Prob(1 or more crossovers in A-B)[1 (1/2)Prob(1 or more crossovers in B C)] + (1/2)Prob(1 or more crossovers in B C)[1 (1/2)Prob(1 or more crossovers in A-B)] = (½)[Prob(1 or more crossovers in A-B) + Prob(1 or more crossovers in B C)] (1/2)[Prob(1 or more crossovers in A-B)Prob(1 or more crossovers in B C)] 20