Lecture 7 Three point cross

Test cross A geneticist has two mutations on the X-chromosome: Cv+ = Normal vein cv = crossveinless V+ = Norm red eye v =Vermillion eye Construct pure-breeding stocks: Cv+ v/Cv+ v and cv V+/Y Normal vein crossveinless Vermillion norm red eye These are mated: F1= females = Cv+ v/ cv V+ males= Cv+ v/Y F1 females are mated to a double recessive male (cv v/Y) The following results are obtained in the F2: indep assortment linked loci Norm vein/Vermillion Crossveinless/Red eye Norm vein/norm eye Crossveinless/Vermillion total 380 120 1000 240 260 255 250 1000 Do these genes reside on the same or different chromosomes?

If on the same chromosome, what is the distance between them? We first identify the parental and recombinant classes Then Recombinant frequency = #Recombinants/Total progeny F2: indep assortment linked loci Norm vein/Vermillion Crossveinless/Red eye Norm vein/norm eye Crossveinless/Vermillion total 380 120 1000

Mapping P Cv+--v x cv--V+ Cv+--v Y F1 Cv+--v x cv--v cv--V+ Y cv--v Cv+--v cv--v Cv+--v 390 Parental cv—V+ cv--v cv—V+ 370 Cv+--V+ cv--v Cv+--V+ 125 Recomb cv--v cv--v 115 Which of these are parental and which are recombinants? Map distance between the Cv+ and V+ genes is 125+115 = 240 =24% 390+370+125+115 1000

Orientation Cv+ V+ 24 Is This Correct? Cv+ V+ 24

Mapping Another mutation Ec+ (echinus eye) is isolated and recombination frequencies between this gene and V+ are determined P Ec+--v x ec--V+ Ec+--v Y F1 Ec+--v x ec--v ec--V+ Y ec--v Ec+--v ec--v Ec+--v 325 Parental ec--V+ ec--v 335 ec--V+ Ec+--V+ ec--v 165 Ec+--V+ Recomb 175 ec--v ec--v MU = 165+175/325+335+165+175 = 340/1000 = 34MU

Orientation Cv to V is 24MU V to Ec is 34 MU Cv+ V+ 24 Ec+ 34 34 Ec+ Cv+ V+ 24 Prediction1= Cv+ to Ec+ = 58 Prediction2 = Cv+ to Ec+ = 10

Mapping recombination frequencies between Ec+ (echinus eye) and V+ are: P Ec+--cv x ec--CV+ Ec+--cv Y F1 Ec+--cv x ec--cv ec--CV+ Y ec--v Ec+--v ec--v Ec+--cv 447 Parental ec--V+ ec--v 448 ec--CV+ Ec+--V+ ec--v 52 Ec+--CV+ Recomb 53 ec--cv ec--v MU = 52+53/1000 = 10.5 MU

Which is correct? Cv to V is 24MU V to Ec is 34 MU V to Ec is 10.5 MU Cv+ V+ 24 Ec+ 34 Cv+ V+ 24 34 Ec+ 10.5

Distance dependent accuracy % recombinants Cv+---V+ 24 Ec+---V+ 34 Ec+---Cv+ 10.5 Cv+ Ec+ 10.5 34 V+ 24 Cv+ V+ 24 34 Ec+ 10.5 10.5+24=34.5 10.5 is close to 10 The map is not very accurate There is a small error in our results

What is going on? The map is not internally consistent? Single cross-over ---Ec+-------v------------------ Ec+--v parental Ec+--V+ Recomb ---ec-------V+------------------ ec--v Recomb ec--V+ parental What if we get two crossovers between Ec+ and V+ A double cross-over ---Ec+--------------------v----- Ec+--v parental ---ec--------------------V+----- ec--V+ parental Now the parental class is over counted because progeny with 2 crossovers are counted as parental class instead of recombinant class The recombinant class is under counted for the progeny with two crossovers

Double crossovers Over large distance there will be a significant number of double crossovers that go undetected - the genetic distances are underestimated The solution is to include additional markers to greatly reduce the probability of undetected doubles: For instance with the intervening Cv+ marker the double crossovers can be separated: ---Ec+------Cv+----------v----- Ec+--Cv+--v parental Ec+--cv--v db Recomb ---ec------cv------------V+----- ec--Cv+--V+ db Recomb ec--cv--V+ parental ---Ec+------Cv+----------v----- Ec+--Cv+--v parental Ec+--cv--V+ s Recomb ---ec------cv------------V+----- ec--Cv+--v s Recomb ec--cv--V+ parental ---Ec+------Cv+----------v----- Ec+--Cv+--v parental Ec+--CV+--V+ s Recomb ---ec------cv------------V+----- ec--cv--v s Recomb ec--cv--V+ parental

Three point cross Because of the problem of undetected double crossovers, geneticists try to map unknown genes to marker genes that are closely linked when constructing a detailed map. sc Scute Bristle v vermilion eye cv Crossveinless wing ec Echinus eye g Garnet eyes f forked bristles 9.1 10.5 24 11.2 10.9

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 three separate pair-wise crosses between heterozygotes OR A more efficient method is to perform a single cross using individuals heterozygous for the three genes

Three point crosses Here is a example involving three linked genes: Q: Is cut wings linked to vermilion and cross veinless and if so what is the gene order and distance between these three 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

Vermilion, norm vein, norm wing Norm eye, crossvein, cutwing Three point crosses v - vermilion eyes cv - crossveinless ct - cut wings P V+ cv ct x v Cv+ Ct+ V+ cv ct Y (normal eye, crossvein, cut) (vermilion, norm vein, norm wing) F1 v Cv+ Ct+ x v cv ct V+ cv ct Y F2 v cv ct v Cv+ Ct+ V+ cv ct v cv Ct+ V+ Cv+ ct V+ Cv+ Ct+ v Cv+ ct V+ cv Ct+ Phenotype 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 377 43 75 5 P R

Distance between ct and cv 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 norm vein, norm wing crossvein, cutwing crossvein, norm wing norm vein, cutwing v cv ct v Cv+ Ct+ Cv+ Ct+ 377 V+ cv ct cv ct 377 v cv Ct+ cv Ct+ 43 V+ Cv+ ct Cv+ ct 43 v cv ct cv ct 75 V+ Cv+ Ct+ Cv+ Ct+ 75 v Cv+ ct Cv+ ct 5 V+ cv Ct+ cv Ct+ 5 Parental Cv+ Ct+ 377+75 cv ct 377+75 Recombinant Cv+ ct 43+5 cv Ct+ 43+5 96/1000 = 9.6% The genes are linked!

Distance between v and ct v to ct Vermilion, norm wing Norm eye, cutwing Vermilion, norm wing Norm eye, cutwing Vermilion, cutwing Norm eye, norm wing Vermilion, cutwing Norm eye, norm wing 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 v cv ct v Cv+ Ct+ v Ct+ 377 V+ cv ct V+ ct 377 v cv Ct+ v Ct+ 43 V+ Cv+ ct V+ ct 43 v cv ct v ct 75 V+ Cv+ Ct+ V+ Ct+ 75 v Cv+ ct v ct 5 V+ cv ct+ V+ Ct+ 5 Parental v Ct+ 377+43 V+ ct 377+43 Recombinant V+ Ct+ 75+5 v ct 75+5 160/1000= 16% The genes are linked

Distance between v and cv Vermilion, norm vein Norm eye, crossvein Vermilion, crossvein Norm eye, norm vein 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 v cv ct v Cv+ Ct+ 377 V+ cv ct 377 v cv Ct+ 43 V+ Cv+ ct 43 v cv ct 75 V+ Cv+ Ct+ 75 v Cv+ ct 5 V+ cv Ct+ 5 v Cv+ V+ cv v cv V+ Cv+ v cv V+ Cv+ v Cv+ Parental v Cv+ 377+5 V+ cv 377+5 Recombinant V+ Cv+ 43+75 v cv 43+75 Recombinants/Total = 236/1000 = 23.6% The genes are linked!

Arranging the three genes v cv 23.6 v ct 16 ct cv 9.6 v cv 23.6 ct 16 9.6 (25.6) The accurate map is: v cv ct 16 9.6

DCO Parental chromosomes v----Ct+-----cv+ & V+----ct----cv v----Ct+-----Cv+ V+----ct----cv v Ct+ Cv+ The parental homologs will pair in meiosisI. Crossing over will occur and a Double crossover produces: v Ct+ Cv+ V+ ct cv V+ ct cv v Ct+ Cv+ v ct Cv+ V+ Ct+ cv V+ ct cv 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. By including these double recombinants the map is internally consistent.

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 Bristle v vermilion eye cv Crossveinless wing ct Cut wing ec Echinus eye g Garnet eyes f forked bristles 9.1 10.5 9.6 16 11.2 10.9

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--------16 m.u.--------ct--------9.6 m.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= 0.16 x 0.096= 0.0154 Total progeny = 1000 Expected number of DCO is 0.0154 x 1000 = 15 Observed number of DCO = 10 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

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P sc ec vg x Sc+ Ec+ Vg+ F1 Sc+ Ec+ Vg+ x sc ec vg F2 Another example Sc= scutellar bristle Ec= echinus rough eye Vg= vestigial wing P sc ec vg x Sc+ Ec+ Vg+ sc ec vg Sc+ Ec+ Vg+ F1 Sc+ Ec+ Vg+ x sc ec vg sc ec vg sc ec vg F2 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+ ec vg 235 241 243 233 12 14 16 If these genes were on separate chromosomes, they should be assorting independently and all the classes should be equally frequent. ARE ALL THREE GENES ON THE SAME CHROMOSOME?

What about sc and ec? Are they linked not linked? sc vg sc ec vg 235 Sc+ Ec+ Vg+ 241 sc ec Vg+ 243 Sc+ Ec+ vg 233 sc Ec+ vg 12 Sc+ ec Vg+ 14 sc Ec+ Vg+ 14 Sc+ ec vg 16 sc ec Sc+ Ec+ sc Ec+ Sc+ ec 478 474 26 30 # recombinant/total progeny = 56/1008 =5.5% Sc and Ec are Linked sc ec What about Sc and Vg?

Are sc and vg linked/not linked??? To map them, we simply examine the pair-wise combinations and identify the parental and recombinant classes: To determine the distance between sc vg we remove ec sc vg sc ec vg 235 Sc+ Ec+ Vg+ 241 sc ec Vg+ 243 Sc+ Ec+ vg 233 sc Ec+ vg 12 Sc+ ec Vg+ 14 sc Ec+ Vg+ 14 Sc+ ec vg 16 sc vg Sc+ Vg+ sc Vg+ Sc+ vg 247 255 257 249 # recombinant/total progeny = 506/1008 = 50% Therefore sc and vg are NOT LINKED! sc vg Sc and Ec are linked Sc and Vg are Not linked Prediction: Ec and Vg should not be linked

Are ec and vg linked? In theory they should not be. From these observations what is the map distance between ec and vg? sc vg sc ec vg 235 Sc+ Ec+ Vg+ 241 sc ec Vg+ 243 Sc+ Ec+ vg 233 sc Ec+ vg 12 Sc+ ec Vg+ 14 sc Ec+ Vg+ 14 Sc+ ec vg 16 ec vg Ec+ Vg+ ec Vg+ Ec+ vg 251 255 257 245 # recombinant/total progeny = 502/1008 = 50% Therefore ec and vg are NOT LINKED! sc ec vg

P sc ec vg x Sc+ Ec+ Vg+ F1 Sc+ Ec+ Vg+ x sc ec vg F2 Linked or unlinked? Sc= scutellar bristle Ec= echinus rough eye Vg= vestigial wing P sc ec vg x Sc+ Ec+ Vg+ sc ec vg Sc+ Ec+ Vg+ F1 Sc+ Ec+ Vg+ x sc ec vg sc ec vg sc ec vg F2 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+ ec Vg+ sc Ec+ Vg+ Sc+ ec vg 235 580 130 241 590 145 243 45 132 232 40 149 10 3 145 12 5 132 12 89 133 16 94 145 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

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