I. Genes found on the same chromosome = linked genes

Transcription

1 Genetic recombination in Eukaryotes: crossing over, part 1 I. Genes found on the same chromosome = linked genes II. III. Linkage and crossing over Crossing over & chromosome mapping I. Genes found on the same chromosome = linked genes Conflicting cytological evidence, only a few dozen chromosomes/individual so must be several genes per chromosome Testcross experiments revealed Conclusion: Genes assort independently if they are on different chromosomes but show linkage if they are on the same chromosome. 1

2 If a testcross is done and the genes are on separate chromosomes: Aa/Bb x aa/bb 2 genes, located on different chromosomes, will segregate independently. 2

3 Chromosome is the unit of transmission, not the gene Linkage = two or more genes located on the same chromosome Linked genes are not free to undergo independent assortment Instead, the alleles at all loci of one chromosome, should in theory, be transmitted as a unit during gamete formation. When two genes are compeletely linked, no crossing over occurs therefore, 3

4 II. Linkage and crossing over A. Crossing over breakage and rejoining process between Crossing over produces The % of recombinant gametes varies, dependent upon location of the loci. The closer the genes are, C.O. Breakage and rejoining process between two homologous non-sister chromatids, keep in mind: 4

5 Recombination Frequency (RF) = the # of recombinants/total progeny B. Recombination Frequency (RF), unlinked genes v. linked genes 1). In the case of unlinked genes, independent assortment holds true: Testcross: Heterozygous x homozygous mutant AaBb x aabb Offspring: So from a cross resulting in 100 progeny we would see 25 individuals from each genotype. the # of recombinants 5

6 2). In the case of linked genes, there is no independent assortment Testcross: Heterozygous x homozygous mutant AaBb x aabb Offspring: AaBb, aabb, Aabb, aabb So from a cross resulting in 100 progeny we would see a lot more of these two genotypes when compared to the recombinants: Crossing between adjacent non sister chromatids generates recombinants The two chromatids not involved in the exchange result in non-parental gametes 6

7 We can compare the RF to what one would expect with independent assortment RF Range Recombination by Crossing Over points to keep in mind: 1. CO s can occur between any two nonsister chromatids 2. If there is NO crossing over, only parental types will be observed 3. If there IS crossing over, RF will increase up to 50% 4. when the loci of two linked genes are very far apart, the RF approaches 50%, 1:1:1:1 ratio observed, thus transmission of the linked genes is indistinguishable from that of two unlinked genes 7

8 Morgan noted the proportion of recombinant progeny varied depending on which linked genes were being examined Testcross F1 results: pr + pr vg + vg x pr pr vg vg pr + vg pr vg 1195 pr + vg 151 pr vg y + y w + w x yy ww y w 43 y + w 2146 y w y + w + 22 As Morgan studied more linked genes, he saw that the proportion of recombinant progeny varied considerably. III. Chromosome mapping determined by analyzing Drosophila crosses Morgan hypothesized that variations in RF might indicate the actual distances separating genes on the chromosomes. Sturtevant (Morgan s student) compiled data on recombination between genes in Drosophila test crosses He found that the closer the two linked genes, the lower the recombination frequency- thus RF may be correlated with the map distance between two loci on a chromosome Alfred Sturtevant 8

9 A. Linkage Maps derived by Sturtevant Linkage of genes can be represented in the form of a genetic map, which shows the linear order of genes along a chromosome. The % recombinant offspring is correlated w/the distance between the two genes, thus the degree of crossing over between any two loci on a single chromosome is proportional to the distance between them, known as the interlocus distance Variations in recombination frequency indicate B. Map Units Map Unit (m.u.) = the distance between genes for which one product of meiosis out of 100 is recombinant [RF of 1% = 1 m.u. or 1 cm] e.g. if RF 12% between A & B, and 28% between B & C: A B C 9

10 Linkage map of Drosophila 4 linkage groups identified 10

11 F1: F2 males females F1 F2 males females 11

12 A plant of genotype: A B a b Is test-crossed to a b a b If the two loci are 10 m.u. apart, what proportion of progeny will be A B / a b? In the garden pea, orange pods (orp) are recessive to normal pods (Orp), and sensitivity to pea mosaic virus (mo) is recessive to resistance to the virus (Mo). A plant with orange pods and sensitivity is crossed to a true-breeding plant with normal pods and resistance. The F1 plants were then test-crossed to plants with orange pods and sensitivity. The following results were obtained: 160 orange pods/sensitive 165 normal pods/resistant 36 orange pods/resistant 39 normal pods/sensitive calculate the map distance between the two genes 12

13 C. Mapping multiple genes Threepoint mapping & Alfred s research Hypothesis = when multiple genes are located on the same chromosome, the distance between the genes can be estimated from the proportion of recombinant offspring. A. Sturtevant s First Genetic Map The linear order of these genes can be determined using testcross data Examined 5 different genes: y, w, v, m, r All alleles were found to be recessive and X linked. Crossed the double heterozygote female with hemizygous male recessive for the same alleles. Example: y+y w+w x yw y+w+ yw y+w yw+ RF = 214/21,736 = w+w r+r x wr w+r+ wr w+r wr+ RF = 2,062/6116 =

14 genes are arranged on the chromosome in a linear order- which can be determined The Complete Data: Alleles y and w y and v y and r y and m w and v w and r w and m v and r w and m # R./total# 214/21,736 1,464/4, / / /1,584 2,062/ /898 17/ /405 RF 1% 32.2% 35.5% 37.5% 29.7% 33.7% 45.2% 3% 26.9% 14

15 y-w = 1 m.u. v-r = 3 m.u. y-m = 37.5 m.u. w-r = 33.7 m.u. w-v = 29.7 m.u. Eukaryotic linkage, part 2 I. Three-point mapping to determine genetic maps A. Multiple cross-overs B. How to: analyzing the 3 pt testcross C. Mapping the results D. The accuracy of mapping E. Mitotic recombination and Sister Chromatid Exhanges II. Genetic mapping in haploid eukaryotes A. Ordered tetrad analysis B. Unordered tetrad analysis 15

16 I. Three-point mapping in Drosophila to determine genetic maps We can map 3 or more linked genes in a single cross, provided the following are true: The genotype of the organism producing crossover gametes must be heterozygous for all loci under consideration Offspring sample size must be high enough to recover a representative sample of all crossover classes DCOs (double crossovers) double exchanges of genetic material in two regions (RI & RII) RI RII Probability of a single crossover occurring between two loci is directly related to the distance separating the loci in the case of a DCO, two separate and independent crossovers must occur simultaneously. 16

17 three-point testcross for mapping Cross two true-breeding strains that differ with regard to three alleles to obtain F1 individuals that are heterozygous for all three alleles: y + y, w + w, ec + ec Perform a testcross by mating the F1 female to males that are homozygous recessive for all three alleles Results in 8 phenotypic classes (2 3 ) FI female: (heterozygous for 3 genes) her possible gametes: y w ec y+ w+ ec+ y w ec+ y w+ ec y w+ ec+ y+ w ec y+ w+ ec y+ w ec+ Because the F 2 phenotypes complement each other (i.e., one is wild type and the other is mutant for all three genes), they are called reciprocal classes of phenotypes. The distance between two genes in a three-point cross is equal to the percentage of all detectable exchanges occurring between them and includes all single and double crossovers. 17

18 How to: analyzing the 3 pt testcross Collect data from the F2 generation. Parental types (usually the two highest # s) Non-Parentals (recombinants) Double Crossovers (two lowest # s) Single Crossovers (two are RI, two are RII) Determine the gene order based upon the DCOs Calculate the RF for each region to determine the map distance between genes (#recombinants/total x 100) P: v + v +, cvcv, ctct x vv, cv + cv +, ct + ct + Testcross: v + v, cv + cv, ct + ct x vv, cvcv, ctct Phenotype v cv + ct + v + cv ct 592 v cv ct + 45 v + cv + ct 40 v cv ct 89 v + cv + ct + 94 v cv + ct 3 v + cv ct + 5 # offspring

19 Parental input Possible output t Only the first possibility is compatible with the data. Example: bb, prpr, vgvg x b + b, + pr + pr, + vg + vg + F1: b + b pr + pr vg + vg testcross: b + b pr + pr vg + vg x bb prpr vgvg Phenotype b + pr + vg + b pr vg b + pr vg 30 b pr + vg + 28 b + pr + vg 61 b pr vg + 60 b + pr vg + 2 b pr + vg # observed Distance between b & pr = Distance between pr & vg = 19

20 Mapping the results: The eye color gene must be in the middle. This order of genes is confirmed by the pattern of traits found in the double crossovers. Double crossover data b + pr vg + 2 b pr + vg 1 D. The accuracy of mapping: Interference Interference = When a crossover in one region affects the likelihood of there being a crossover in an adjacent region Expected frequency of DCOs DCOs rare between segments that are very short If crossovers in the 2 regions are independent, then: frequency of double recombinants = product of the recombinant frequencies in the adjacent regions Expected DCOs = x = x 1,005 = 7.5 Coefficient of coincidence (c.o.c) Observed/expected double recombinants 1 3/7.5 = 0.4 I = 1-c.o.c 20

21 Step by step summary: 1. Calculate the RF for each pair of genes 2. Draw the linkage map 3. Determine the double recombinants 4. Calculate the Frequency & # of double recombinants expected if there is no interference 5. Calculate Interference Problem: Vermilion eyes are recessive to normal, miniature wings are recessive to long wings, and sable body is recessive to gray body. A cross was made between a heterozygous female for all three genes and a homozygous recessive male. Data: 1,320 vermilion eyes, miniature wings, sable body 1,346 red eyes, long wings, gray body 102 vermilion eyes, miniature wings, gray body 90 red eyes, long wings, sable body 42 vermilion eyes, long wings, gray body 48 red eyes, miniature wings, sable body 2 vermilion eyes, long wings, sable body 1 red eyes, miniature wings, gray body A. Determine the gene order & calculate the map distance between the three genes B. Calculate interference 21

22 E. Mitotic recombination and Sister Chromatid Exhanges Mitotic recombination = crossing over that occurs during mitosis (it does happen, in often in Drosophila & fungi also in humans & mice) It is likely that the recombinational repair of DNA lesions occurs preferentially by sister chromatid exchanges that have no genetic consequences Those between non-sister chromatids producemutant patches in female flies = twin spot SCE s don t produce new allelic combinations, but may be involved in repairing DNA lesions II. Genetic mapping in haploid eukaryotes Fungi haploid (n) multicellular organisms that can reproduce asexually to create a diploid zygote Diploid zygote proceeds through meiosis to produce four haploid spores = ascospores 22

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24 Group of four spores is known as a tetrad In some species meiosis is followed by mitosis to produce eight cells known as an octad. Ascus = sac that contains the tetrad/octad Can do an ordered tetrad analysis because the 8 cells reflect the sequence of formation what s the fun in fungi? They are haploid They produce large # s of progeny They have short life cycles Can make direct observations on the behavior of genes during meiosis, can examine cross-overs, can map centromeres Chlamydomonas Neurospora 24

25 A. Ordered tetrad analysis: Linear tetrad analysis can be used to map distance between a gene and the centromere. 1). FDS PATTERN: 2) SDS PATTERN: a) 2:2:2:2 b) 2:4:2 segregate until the 2 nd meitotic division is complete! 25

26 Calculating map distance w/ ordered tetrads mapping the centromere - % of SDS or M2 asci can be used to calculate the map distance between the centromere & the gene of interest Map distance = B. Unordered tetrad analysis can be used to map genes in dihybrid crosses, Spores are randomly arranged Haploid cell AB x ab Haploid cell Aa Bb Diploid zygote meiosis AB AB AB Ab Ab Ab ab ab ab ab ab ab Parental ditype Tetratype Nonparental ditype 26

27 When two genes are located on different chromosomes: 27

28 When two genes are located on the same chromosome: Map distance = (1/2) ([TT] + 3[NPD]) + 4 [NPD] / total x 100 If the # of parentals = nonparentals, the two are unlinked If there are TT s, CO s have occurred, if there are NPDs then DCO s occurred. The following spore arrangements were obtained from tetrads in a cross between Neurospora strain com val (c v) and a wild type strain (+ +). Only 1 member of each pair of spores is shown. Spore pair Ascus composition 1-2 cv c+ cv +v cv 3-4 cv c+ c+ c+ +v v +v cv c v Number: What can you conclude about linkage? 35 PD 36 NPD 30 TT PD = NPD, genes not linked 28