1. 1 Lecture 7 Three point cross

2. 2 A geneticist has two mutations: ｷ A = tall ｷ a = short ｷ H = hairy ｷ h = no hair and constructs the following pure-breeding stocks: AAhh and aaHH Tallshort No hairhairy Linkage

3. 3 These individuals are mated and the F1 progeny are mated to the double recessive. The following results are obtained in the F2: indep assortmentlinked loci Tall, no hair Short, hairy Tall, hairy Short, no hair total Do these genes reside on the same or different chromosomes? Answer- If on the same chromosome, what is the distance between them? We simply identify the parental and recombinant classes and determine the recombinant frequency #Recombinants/Total progeny Which class is the recombinant? 997 1007 410 415 2827 700 710 690 700 2800 Test cross

4. 4 Mapping PAAhhxaaHH A hxa H A ha H F1AaHhxaahh A hxa h a Ha h ah A h a H A H a h Aahh aaHh AaHh aahh Parental Recomb 410 997 1007 415 Which of these are parental and which are recombinants? Tall, No hair short, hairy

5. 5 Distance Which are the parental and which are the recombinant classes? What is the recombination frequency? So the map distance between the A and H genes is 410+415825 =29% 410+1007+997+4152829

6. Orientation 6 AH 29 Is This Correct? AH 29

7. 7 Distance dependent accuracy Another mutation C (crinkled) is isolated and recombination frequencies between this gene and the A and H genes are determined % recombinants A to H29 A to C15 H to C16 AHC 29 1516 15+16=31 31 is close to 29! What is going on? The map is not very accurate There is a small error in all our results

8. 8 What is going on? The map is not internally consistent? ---A-------h------------------ ---a-------H------------------ A-hparental A-HRecomb a-hRecomb a-Hparental ---A--------------------h----- ---a--------------------H----- A-hparental a-Hparental Single cross-over What if we get two crossovers between A and H A double cross-over Now the parental class is over counted for progeny with 2 crossovers The recombinant class is under counted for the progeny with two crossovers Over large distance there will be a significant number of double crossovers that go undetected - the genetic distances are underestimated

9. 9 Double crossovers The double crossovers go undetected and therefore over large distances the genetic distances are underestimated The solution is to include additional markers between A and H to greatly reduce the probability of undetected doubles: For instance with the intervening C marker the double crossovers can be separated: ---A--------C------------H----- ---a--------c------------h----- A-C-Hparental A-c-Hdb Recomb a-C-hdb Recomb a-c-hparental A15C16H 29+ undetected double (2)

10. 10 Three point cross Because of the problem of undetected double crossovers, geneticists try to map unknown genes to marker genes that are closely linked (LESS than 10 m.u.) when constructing a detailed map. sc Scute Bristlev vermilion eyecv Crossveinless wingct Cut wingec Echinus eye g Garnet eyes f forked bristles 9.110.59.215.911.210.9

11. 11 This is one of the reasons behind a mapping technique known as The Three-Point Testcross To map three genes with respect to one another, we can use a series of pair-wise matings between double heterozygotes OR A more efficient method is to perform a single cross using individuals heterozygous for the three genes

12. 12 Three point crosses Here is a example involving three linked genes: v - vermilion eyes cv - crossveinless ct - cut wings To determine linkage, gene order and distance, we examine the data in pair-wise combinations When doing this, you must first identify the Parental and recombinant classes! P F1 F2

13. 13 Three point crosses PV+ cv ct x v Cv+ Ct+ V+ cv ctv Cv+ Ct+ F1v Cv+ Ct+xv cv ct V+ cv ct v cv ct F2 v cv ct v Cv+ Ct+580 V+ cv ct592 v cv Ct+45 V+ Cv+ ct40 v cv ct89 V+ Cv+ Ct+94 v Cv+ ct3 V+ cv Ct+5 P P R R R R R R v - vermilion eyes cv - crossveinless ct - cut wings Vermilion, norm vein, norm wing Norm eye, crossvein, cutwing Vermilion, crossvein, norm wing Norm eye, norm vein, cutwing Vermilion, crossvein, cutwing Norm eye, norm vein, norm wing Vermilion, norm vein, cutwing Norm eye, crossvein, norm wing

14. 14 Distance between v and cv v to cv v cv ct v Cv+ Ct+v Cv+ 580 V+ cv ctV+ cv 592 v cv Ct+v cv 45 V+ Cv+ ctV+ Cv+ 40 v cv ctv cv 89 V+ Cv+ Ct+V+ Cv+ 94 v Cv+ ctv Cv+ 3 V+ cv Ct+V+ cv 5 Parental V Cv+583 V+ cv597 Recombinant V+ Cv+134 v cv134 268/1448 = 18.5% The genes are linked!

15. 15 Distance between ct and cv ct to cv v cv ct v Cv+ Ct+Cv+ Ct+580 V+ cv ctcv ct 592 v cv Ct+cv Ct+45 V+ Cv+ ctCv+ ct40 v cv ctcv ct89 V+ Cv+ Ct+Cv+ Ct+94 v Cv+ ctCv+ ct3 V+ cv Ct+cv Ct+5 Parental Cv+ Ct+674 cv ct681 Recombinant Cv+ ct43 cv Ct+50 93/1448 = 6.4% The genes are linked!

16. 16 Distance between v and ct v to ct v cv ct v Cv+ Ct+v Ct+580 V+ cv ctV+ ct 592 v cv Ct+v Ct+45 V+ Cv+ ctV+ ct40 v cv ctv ct89 V+ Cv+ Ct+V+ Ct+94 v Cv+ ctv ct3 V+ cv ct+V+ Ct+5 Parental v Ct+625 V+ ct632 Recombinant V+ Ct+99 v ct92 191/1448 = 13.2% The genes are linked

17. Arranging the three genes vcv 18.5 vct 13.2 ct cv 6.4 vcv 18.5 ct 13.2 6.4 The accurate map is: vcv ct 13.2 6.4

18. 18 DCO Notice if one focuses on the v and cv markers, they will be scored as non-recombinant (parental). However if one also scores v-ct and ct-cv the double recombination event from which they arose can be detected. In fact, when scored, a number of recombinations occur between v and cv. These classes should be counted. By including these double recombinants the map is internally consistent. Parental chromosomes v----Ct+-----cv+&V+----ct----cv v----Ct+-----Cv+V+----ct----cv vCt+Cv+ V+ctcv The parental homologs will pair in meiosisI. Crossing over will occur and a Double crossover produces: vCt+Cv+ vctCv+ V+Ct+cv V+ctcv

19. 19 Another method to solve a three point cross Solving three-point crosses 1. Identify the two parental combinations of alleles 2. The two most rare classes represent the product of double crossover. v cv ct v Cv+ Ct+580 V+ cv ct592 v cv Ct+45 V+ Cv+ ct40 v cv ct89 V+ Cv+ Ct+94 v Cv+ ct3 V+ cv Ct+5 Parent DCO

20. 20 Solving three-point crosses 1.Identify the two parental combinations of alleles 2. The two most rare classes represent the product of double crossover. Parentv Cv+ Ct+ &V+ cv ct DCOv Cv+ ct &V+ cv Ct+ 3. With this knowledge, you can establish a gene order in which a double cross produces the allelic combination observed in the most rare class. There are three possible relative order of the three genes in the parent:

21. 21 There are three possible gene orders for the parental combination **basically we want to know which of the three is in the middle** Parentv Cv+ Ct+ &V+ cv ct vermillionred normal veincrossveinless normal wingcut wing DCOv Cv+ ct &V+ cv Ct+ vermillionred normal veincrossveinless cut wingnormal wing You are driving along Rte1 and you are told that there are three towns along this route- San Francisco, Half moon bay and Santa Cruz. You have no idea which town you will encounter first, second and last. How many possible orders are there? San Francisco----Santa Cruz----Half moon bay San Francisco----Half moon bay----Santa Cruz Half moon bay----San Francisco----Santa Cruz

22. 22 Parentv Cv+ Ct+ &V+ cv ct vermillionred normal veincrossveinless normal wingcut wing Observed DCOv Cv+ ct &V+ cv Ct+ There are three possible gene orders for the parental combination **basically we want to know which of the three is in the middle** Each relative order in the parent gives a different combination of the rarest class (DCO) v----Cv+----Ct+ V+---cv-----ct OR v----Ct+----Cv+ V+---ct-----cv OR Ct+----v----Cv+ ct-----V+---cv predicted DCO v----cv----Ct+ V+---Cv+---ct v----ct----Cv+* V+---Ct+---cv Ct+----V+---Cv+ ct-----v----cv

23. 23 Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified v cv ct v Cv+ Ct+580 V+ cv ct 592 v cv Ct+45 V+ Cv+ ct40 v cv ct89 V+ Cv+ Ct+94 v Cv+ ct3 V+ cv Ct+5 Gene Order v----ct----cv REWRITE THE COMBINATION IN THE PARENTS v---Ct+---Cv+ and V+---ct---cv

24. 24 3. Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified v..Ct+..Cv+ V+..ct..cv v..Ct+..cv V+..ct..Cv+ v..ct..cv V+..Ct+..Cv+ v..ct..Cv+ V+..Ct+..cv Gene Order v----ct----cv REWRITE THE COMBINATION IN THE PARENTS v---Ct+---Cv+ and V+---ct---cv vCt+Cv+ V+ctcv P SCO-II SCO-I DCO v cv ct v Cv+ Ct+580 V+ cv ct 592 v cv Ct+45 V+ Cv+ ct40 v cv ct89 V+ Cv+ Ct+94 v Cv+ ct3 V+ cv Ct+5 CO1CO2

25. 25 Now the non-recombinants, single recombinants, and double recombinants are readily identified Recombination freq in region I = 89+94 + 3+5 SCOIDCO Recombination freq in region II = 45+40 + 3+5 SCOIIDCO Now the DCO are not ignored. With this information one can easily determine the map distance between any of the three genes v--------13.2 m.u.--------ct--------6.4m.u.-------cv

26. 26 Now the non-recombinants, single recombinants, and double recombinants are readily identified Parental input: (As a check that you have not made a mistake, reciprocal classes should be equally frequent) With this information one can easily determine the map distance between any of the three genes:

27. 27 Interference Interference: this is a phenomenon in which the occurrence of one crossover in a region influences the probability of another crossover occurring in that region. Interference is readily detected genetically. For example, we determined the following map for the genes v ct and cv. v--------13.2 m.u.--------ct--------6.4m.u.-------cv Expected double crossovers = product of single crossovers The expected frequency of a double crossover is the product of the two frequencies of single crossovers: DCO= 0.132 x 0.064= 0.0084 Total progeny = 1448 Expected number of DCO is 0.0084 x 1448 = 12 Observed number of DCO = 8 The coefficient of coincidence is calculated by dividing the actual frequency of double recombinants by this expected frequency: c.o.c. = actual double recombinant frequency / expected double recombinant frequency Reduction is because of interference

28. 28 Interference is often quantified by the following formula: I= 1- observed frequency of doubles/ expected frequency of Doubles I= 1- 8/12 = 4/12 = 33% If actual frequency is the same as expected frequency then Interference is 1-1=0

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30. 30 Linked or unlinked? P sc ec vg x Sc+ Ec+ Vg+ sc ec vgSc+ Ec+ Vg+ F1 Sc+ Ec+ Vg+xsc ec vgsc ec vg F2 If these genes were on separate chromosomes, they should be assorting independently. Are all three assorting independently, are two assorting independently or are none assorting independently sc ec vgsc ec vgsc ec vg sc ec vg Sc+ Ec+ Vg+ sc ec Vg+ Sc+ Ec+ vg sc Ec+ vg Sc+ ec Vg+ sc Ec+ Vg+ Sc+ ec vg Sc= scutellar bristle Ec= echinus rough eye Vg= vestigial wing 23589130 24194145 24345132 23340149 123145 145132 14580133 16590145