CHAPTER 6 GENETIC RECOMBINATION IN EUKARYOTES + CHAP[TER 14, PAGES )

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1 CHAPTER 6 GENETIC RECOMBINATION IN EUKARYOTES + CHAP[TER 14, PAGES ) Questions to be addressed: 1. How can we predict the inheritance patterns of more than one gene? 2. How does the position of genes within the genome affect the inheritance patterns? 3. How can we determine the distance between genes on the chromosome using the recombination frequency? Terminology (see also Glossary pages and web site recombinant gamete is any gamete that has a genotype that differs from the genotypes of the two haploid parents that fused to form the diploid meiocyte parental gamete: any gamete that has the same genotype as the two haploid parents that contributed to the diploid meiocyte dihybrid: an individual heterozygous at two genes linkage the association of two genes on the same chromosome coupling configuration arrangement of linked alleles in a dihybrid such that the two dominant alleles are on one chromosome, the two recessive alleles are on the other chromosome (AB/ab) repulsion configuration arrangement of linked alleles in a dihybrid such that the a dominant and a recessive allele are together on each chromosome (Ab/aB) cis dihybrid a dihybrid in which the alleles are coupled (AB/ab) trans dihybrid a dihybrid in which the alleles are in repulsion (Ab/aB) map unit (centimorgan, cm): the distance between two genes where one product of meiosis out of 100 is recombinant gene locus: the position of a gene within the genome

2 1. How can we predict the inheritance patterns of more than one gene? Question: If we consider 2 genes controlling two different characteristics, is their segregation independent or dependent of one another? Prediction: consider the behaviour of chromosomes if genes are on different chromosomes, we expect alleles of the genes to segregate independently of one another if genes are on the same chromsome, we expect that the alleles that came from one parent will remain together unless a crossover occurs between them a recombinant gamete is any gamete that has a genotype that differs from the genotypes of the two haploid parents that contributed to the diploid meiocyte (Figure 6-2) a parental gamete is any gamete that has the same genotype as the two haploid parents that contributed to the diploid meiocyte e.g. genotypes of haploid parents Ab x ab genotype of diploid meiocyte AaBb (= a dihybrid) genotypes of meiotic products Ab parental ab parental AB recombinant ab recombinant Analysing inheritance of genes in organisms with a diploid life cycle PART 1: INHERITANCE OF GENES THAT ARE ON DIFFERENT CHROMOSOMES (INDEPENDENT ASSORTMENT)

3 Question: what proportion of each meiotic product will be generated, if A and B genes are on different chromosomes? Mendel choose two characteristics, seed colour (yellow = Y or green = y) and seed shape ( round = R or wrinkled = r), and analysed their segregation in 1) individuals differing in one characteristics and 2) individuals differing in two characteristics First, he determined that the two genes segregate normally in monohybrid crosses: 1) Monohybrid Crosses monohybrid formed in F1 generation - heterozygous at one gene F2 generation shows 3:1 phenotypic segregation (i.e. normal segregation) 2) Dihybrid Cross (Figure 6-7 Next, he crossed parents that were different at two genes to determine how the genes segregated relative to one another Seed Colour and Seed Shape Parental cross: YY rr X yyrr (yellow, wrinkled X green and round) gametes: F1 generation: all YyRr yellow ( & round)

4 dihybrid formed in F1 generation - heterozygous at two genes Expected results: if alleles are segregating independently: 1) what gametes do we expect to be produced by the F1 generation? (same logic as what products will be produced by the diploid meiocyte described above) product rule: the probability of two independent events occurring together is equal to the product of their individual probabilities (Figure 6-4) p(yr) = p(y) x p(r) = (recombinant gamete) p(yr) = p(y) x p(r) = (parental gamete) p(yr) = p(y) x p(r) = (recombinant gamete) p(yr) = p(y) x p(r) = (parental gamete) 2) what F2 progeny will be produced? product rule: sum rule: the probability of either of two mutually exclusive events occurring is equal to the sum of their individual probabilities using these two laws and the expected gamete frequency, we can determine the expected genotypic frequencies (Figure 6-7): p(yyrr) = p(yyrr)= p(yyrr) = but, phenotypically YYRR = YyRr how to figure out expected phenotypic ratios?

5 For any monohybrid cross, the expected F2 phenotypic frequency is 3:1 Rr x Rr = 3 R_ : 1rr (_ represents either R or r ) Yy x Yy = 3 Y_ : 1yy (_ represents either Y or y) if genes controlling two phenotypes are assorting independently, can apply the product rule to individual phenotypic probabilities to determine the frequency of phenotypic combinations: p(r_y_) = p(r_) x p(y_) = p(rry_) = p(rr) x p(y_) = p(r_yy) = p(r_) x p(yy) = p(rryy) = p(rr) x p(yy) = therefore, if genes are assorting independent, expected phenotypic ratio = Mendel s F2 data (Figures 6-7, 6-8): 315 yellow round 106 yellow wrinkled 101 green round 32 green wrinkled = 9.8 : 3.3 : 3.1 : 1 independent assortment of colour and shape characteristics Mendel s Second Law: During gamete formation, the segregation of alleles of one gene is independent of segregation of the alleles of another gene Forked Diagrams

6 organized way of writing down product rule, such that all possible outcomes at one event are multiplied by all possible outcomes at another independent simultaneously occurring event deriving phenotypic classes e.g. what is the probability of (R_Y_) in the F2 generation, assuming the phenotype at R and phenotype at Y are two independent events? EVENT 1 R phenotype 3/4 R_ EVENT 2 Y phenotype 3/4 Y_ 1/4 yy progeny 3/4 X 3/4 = 9/16 R_ Y_ 3/4 X 1/4 = 3/16 R_yy 1/4 rr 3/4 Y_ 1/4 yy 1/4 X 3/4 = 3/16 rry_ 1/4 X 1/4-1/16 rr yy Test Cross (Figures 6-5, 6-6) Crossing an individual showing the dominant phenotype to a homozygous recessive individual phenotype of progeny reveals genotypes of meiocytes YyRr X yyrr event 1 event 2 What ratio do you expect? - find possible gamete combinations for each event: 1) segregation of alleles at gene Y in YyRr 2) segregation of alleles at gene R in YyRr

7 Forked diagram expected ratio = MESSAGE: IF A DIHYBRID INDIVIDUAL IS TESTCROSSED OR SELFED, A 1:1:1:1 OR A 9:3:3:1 RATIO RESPECIVELY INDICATES THAT THE TWO GENES ARE ASSORTING INDEPENDENTLY Multiple genes AaBbccddEeFF x AaBbCCDdeeff p(aabbccddeeff) =? if independent: = = PART 2 INHERITANCE OF GENES ON THE SAME CHROM0SOME Bateson and Punnett (early 1900s) -studying inheritance in Sweet Pea

8 P - controls flower colour L - controls seed shape hypothesis: genes control different characteristics (seed shape and flower colour), therefore the genes will show independent segregation. experiment: cross true-breeding parents, allow F1 plants to self-fertilize. expected results: The F2 progeny will show a phenotypic ratio of 9:3:3:1 (see figures 6-4 and 6-7) CROSS: purple, long X red, round PARENTAL: PPLL X ppll F1 PROGENY: PpLl (purple, long) F2 PROGENY: Expect Observe 9 P_L_ (215) P_ll (71) 21 3 ppl_ (71) 21 1 ppll (24) 55 conclusion: this is not a 9:3:3:1 ratio. additional observations: P_ll and ppl_ groupings are smaller ppll, P_L_ groupings are bigger larger classes are those which have P and L or p and l - P and L are coupled - p and l are coupled - (dominant alleles are together, recessive alleles are together) Drosophila - Thomas Morgan - similar results with two genes in Drosophila, one regulating wing development and the other eye colour

9 pr = purple eyes pr + = red eyes vg = vestigal wings vg + = normal wings hypothesis: pr and vg are coupled, because they are linked on the same chromosome; expect that linked alleles will stay together during meiosis, and be present in the same gamete experiment: a test cross (recall that phenotypes resulting from a test cross reveal the genotypes of the gametes) (figure 6-5) CROSS: red eyed, normal wings X purple eyed, vestigal wings PARENTAL: pr + pr + vg + vg + X prprvgvg GAMETES: pr + vg + pr vg F1 PROGENY: pr + prvg + vg pr + vg + pr vg TEST CROSS: pr + prvg + vg X prprvgvg (assuming independent assortment, see figures 6-5, 6-6) Morgan s results: expected observed 1/4 pr + vg /4 pr + vg /4 prvg /4 prvg TOTAL again, largest classes are those in which wild type alleles are coupled or recessive alleles are coupled

10 = arrangement in the parents = parental arrangement Hypothesis: the coupling of alleles is based on their arrangement in the parents, and not on their dominant/recessive relationships experiment: progeny expected observed 1/4 pr + vg /4 pr + vg /4 prvg /4 prvg TOTAL largest classes are pr + vg and prvg + - this time, the dominant allele pair is in repulsion - parental alleles are linked conclusion: the alleles which are together in the parents stay together through subsequent generations, independent of their dominance/recessive relationship - these genes are on the same chromosome = linked 2 crosses pr + vg + pr vg

11 note: - cross 1 - alleles are in coupling (dominant together, recessive together) = cis dihybrid (Figure 6-11) pr + vg + pr vg - cross 2 - alleles are in repulsion (recessive with dominant) = trans dihybrid pr vg + pr + vg Question: if genes are linked on the same chromosomes, how do any gametes arise that alter the original arrangement? - during Prophase I, see chiasmata - evidence of chromosome exchange = RECOMBINATION or CROSSING OVER (figures 6-9, 6-10) question: how do we know if an unexpected ratio indicates linkage? - consider a test cross involving 2 linked genes, A and B:

12 Two genes, A and B are linked PARENTAL CROSS: AAbb X aabb A b A b a a B B F1 PROGENY: A a b B (test cross X aabb) parental gametes = Ab, ab recombinant gametes = AB, ab expected results: test cross, therefore the simplest hypothesis is that the genes are independently assorting (unlinked) and we expect 1:1:1:1 ratio (see figure 6-5)] observed results: PROGENY genotype phenotype numbers AaBb AB 50 aabb ab 150 AaBB Ab 150 aabb ab 50 Use a statistical test to determine if the observed results are significantly different from the expected (pages ) Chi-Square (χ2) Test 1) develop null hypothesis The genes, A and B, are independently assorting (i.e. they are not linked) and the test cross progeny do not deviate from the expected 1:1:1:1 ratio

13 2) calculate the χ2!2 = "(observed - expected)2 expected = "(O - E) 2 E phenotype expected observed AB ab Ab ab degrees of freedom (df) = number of classes (n) -1 = compare the calculated chi-square value to the critical value from the chisquare distribution table (Figure 6-1) - if χ2 is < critical value, P is > than 0.05, accept null hypothesis - if χ2 is > critical value, P is < than 0.05, reject null hypothesis 3. How can we determine the distance between genes on the chromosome using the recombination frequency? Alfred Sturtevant (Morgan s student - Box 6-2) hypothesis: the frequency of recombinant products of meiosis is related to distance between genes - the further apart genes are, the more likely a cross-over will occur between them increased distance = increased recombinant meiotic products

14 - created a unit (map unit, later called a centimorgan, cm) which was equivalent to the frequency of recombination: - 1 genetic map unit = 1% recombinant products of meiosis genetic map unit (mu) = centimorgan (cm) conclusion: using the recombination frequency between linked genes, one can determine position of genes relative to one another (Figure 6-12) genetic LOCUS = position of a gene Consider Morgan's Cross PARENTAL: pr + pr + vgvg X prprvg + vg + TEST CROSS: pr + prvg + vg X prprvgvg expected observed 1/4 pr + vg /4 pr + vg /4 prvg /4 prvg TOTAL recombinant parental parental recombinant parental = pr + vg, prvg + - no recombination, therefore LARGEST classes recombinant = pr + vg +, prvg - products of recombination, therefore SMALLEST classes

15 - map distance = percent recombinant progeny = number of recombinants x 100 total Note: the frequency of recombinate products never exceeds 50%, therefore, the farthest distance that can be calculated between 2 genes is 50 m.u. Genes on X-chromosome - NO crossing over in male (XY) - X-linked genes in Drosophila = w + (allele for eye color; red / white), f + (allele for bristle type; unforked / forked) experiment: because the male is hemizygous, crossing to a mutant male F1 fly has the same effect as crossing to a tester (homozygous recessive fly) -therefore, instead of a test cross, carry through to F2

16 CROSS white, forked X red, unforked (wild type) PARENTAL: wf X w + f + wf Y F1: w + f + X wf wf Y (wild type) (white, forked) F2: w + f + parental 435 w + f recombinant 42 wf + recombinant 58 wf parental 465 w-f distance = Predicting progeny frequencies from known map distance Hypothesis: if we know the map distance between two genes, we should be able to predict the frequency of recombinants and parental phenotypes resulting from a given cross. consider 2 cases:1) genes A and B assort independently, and 2) genes C and D are linked

17 1) if A and B assort independently PARENTAL: aabb X AAbb F1: AaBb test cross: expected AB 25% recombinant Ab 25% parental ab 25% parental ab 25% recombinant (of 200 progeny, 25% should be ab, therefore 50 will be ab) Example 2: Genes C and D are known to be 15 map units apart Parental ccdd x CCdd Three Point Test Cross (see pages ) Question: how to map three genes relative to one another Arabidopsis chromosome 5 contains 3 genes lfy + - flowers/ lfy - leaves cer3 + - epidermal wax/ cer3 - no wax yi + - green inflorescence/ yi - yellow inflorescence

18 PARENTAL CROSS: lfy cer3 + yi + X lfy + cer3 yi lfy cer3 + yi + lfy + cer3 yi F1: lfy cer3 + yi + / lfy + cer3 yi Test Cross Progeny: rules: 1) there should be 8 phenotypic classes 2) progeny classes are grouped in pairs = reciprocal cross over events 3) largest classes are the parentals 4) smallest classes are the double cross-over classes (DCO) There are two methods for doing a 3-point test cross 1) Longer method - consider genes pairwise 1. distance between lfy and cer3 (ignore yi) PARENTAL lfy cer lfy + cer RECOMBINANT lfy + cer lfy cer distance = % recombinants = = 3.0 map units X distance between cer3 and yi 3. distance between lfy and yi

19 Determine orientation of genes most likely orientation = note discrepancy in outer distance WHY? draw the F1 chromosome configuration cer3 lfy yi cer3 + lfy + yi + we considered all cer3 yi and cer3 + yi + as parentals in fact, some of these are the result of a double crossover (one crossover in each interval) which classes? the reciprocal pair in which the middle gene switches places relative to the outside genes (see Figures 6-14, 6-15)

20 cer3 lfy yi cer3 + lfy + yi + cer3 lfy + yi = cer3 lfy yi and cer3 + lfy + yi + cer3 + yi + lfy note - these are the smallest classes to calculate the true distance between cer3 and yi distance = % recombinants = (! SCO classes) + 2(DCO classes) X 100 total = ( ) + 2(1 + 1) = 9.0 map units 1000 distance between outer genes = sum of distance between two pairs OR X 100 = (sum of SCO classes) + 2(DCO classes) X 100 total 2) method 2 - Short cut to 3 point test cross (determining gene order by comparing double crossover class to parentals) 1) perform chi square (to determine whether genes are independent or unlinked) 2) identify parental classes (the largest classes) 3) identify double crossover class (the smallest class)

21 4) compare parentals to DCO (to determine the gene in middle) (figure 6-14) 5) find distance between 2 pairs of genes one outer gene - middle gene middle gene - other outer gene 6) find distance between outer genes (by adding the distance of the 2 pairs calculated above) 7) draw a map (showing the relative gene positions and distances) 8) calculate interference value INTERFERENCE -what is the probability of double crossovers? p(dco) = p(scoi) x p (SCOII) -this assumes that the crossovers are independent of one another, that having a crossover in one region does not inhibit/promote a crossover in another region interference = a measure of how much one crossover has interfered with another positive (+) interference - fewer DCO than expected negative (-) interference - more DCO than expected INTERFERENCE = 1 - coefficient of coincidence = 1 - = 1 - observed DCO expected DCO observed DCO p(scoi) X p(scoii) X 1000

22 in our 3 point test cross example: observed DCO interference = 1 - expected DCO observed DCO = 2 (lfycer3 yi + lfy + cer3 + yi + ) expected DCO = (0.06) X (0.03) X 1000 = 1.8 (this value calculated using the actual gene distances) interference = = one more example: TEST CROSS = TEST CROSS PROGENY: AABBCC X aabbcc ABC 450 abc 440 Abc 4 abc 6 ABc 25 abc 30 abc 25 AbC 20 A B C a b c 1) calculate chi-square statistic (to determine if this is a 1:1:1:1:1:1:1:1 ratio?; are genes linked or unlinked?) 2) what are the parental classes? 3) what are the DCO classes? 4) which gene is in the middle? 5) distance between one outer gene and middle gene 6) distance between other outer gene and middle gene

23 7) distance between two outer genes 8) map Mendel s findings: one gene, two alleles each gene controls a single different character one allele is dominant to the other ratios = 3 : 1 = 1 : 2 : 1 (monohybrid) heterozygotes 9 : 3 : 3 : 1 (dihybrid) = 1 : 1 (monohybrid) test cross of heterozygotes 1 : 1 : 1 : 1 (dihybrid) Complexity arises (see Chapter 14, ): a gene may have multiple alleles: each gene consists of many nucleotides, which when mutated result in different alleles different alleles may have different phenotypes a gene may be pleiotrophic: a single gene acts to control more than one phenotypic characteristic e.g. gene controlling pollen grain germination also controls root hair growth suggests a common underlying physiological or developmental mechanism

24 more than one gene may control a single characterisic since many biological pathways consist of a series of steps, each controlled by a different protein (coded by a gene), a single characteristic may be the result of multiple gene activity 1. Each gene can have multiple alleles How do you know if 2 mutant lines with the same phenotype are the result of: a) mutations in 2 genes in the same pathway b) 2 alleles of the same gene example: consider the trp biosynthetic pathway trpa + X trpb+ trpc + Y tryptophan experiment: two new auxotrophic mutants for tryptophan are found in the plant Arabidopsis thaliana hypothesis 1: the two mutants are alleles of same gene, A (e.g. mutant 1 = a1, mutant 2 = a2) hypothesis 2: the two mutants are alleles of different genes A and B (e.g. mutant 1 = a1, mutant 2 = b1) experiment: cross two mutants together and the results will differ depending on which hypothesis is correct expected results for hypothesis 1

25 conclusion: F1 progeny is mutant, therefore mutations do not complement one another, and the mutations are alleles of the same gene expected results for hypothesis 2 conclusion: F1 progeny is wild type, therefore two mutants complement one another, and must be mutations in different genes RULES FOR A COMPLEMENTATION TEST (see Figure 14-3) can only be done with recessive mutations If the mutations are in different genes, the two mutations will complement one another (progeny will be wild type) If the mutations are alleles of the same gene, the two mutations will not complement one another (progeny will be mutant)

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