Lecture 6

A geneticist isolates two mutations: ｷ A = tall ｷ a = short Linkage A geneticist isolates 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 Tall short No hair hairy These individuals are mated and the F1 progeny are mated to the double recessive. The following results are obtained in the F2: Indep assortment Linked loci Tall, hairy Tall, no hair Short, hairy Short, no hair total Do these genes reside on the same or different chromosomes? Answer- If they reside on the same chromosome, what is the distance between them?

P Tall, No hair short, hairy F1 Parental Recomb

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

Another mutation C (crinkled) is isolated and recombination frequencies between this gene and the A and H genes are determined % recombinants

What is going on? The map is not internally consistent?

The double crossovers go undetected and therefore over large distances the genetic distances are underestimated

Three point cross Because of the problem of undetected double crossovers, geneticists try to use closely linked markers (less than 10 m.u.) when constructing a map. 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 have used a series of pair-wise matings between double heterozygotes A more efficient method is to perform a single cross using individuals triply heterozygous for the three genes

P F1 F2 First example sc ec vg sc ec vg 235 sc+ ec+ vg+ 241 If these genes were on separate chromosomes, they should be assorting independently and all the classes should be equally frequent.

sc and vg are ??? To map them, we simply examine the pair-wise combinations and identify the parental and recombinant classes: For example to determine the distance between sc 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 16 247 255 257 249 # recombinant/total progeny = Therefore sc and vg are Next: What about sc and ec?

sc and ec are ??? What about sc and 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 16 478 474 26 30 # recombinant/total progeny =

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

More three point crosses Here is another 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 v cv ct v cv+ ct+ v+ cv ct v cv ct+ v+ cv+ ct+ v cv+ ct v+ cv ct+

v and cv v to cv v cv ct v cv+ ct+ v cv+ 580 v+ cv ct v+ cv 592 v cv ct+ v cv 45 v+ cv+ ct+ v+ cv+ 40 v cv ct v cv 89 v+ cv+ ct+ v+ cv+ 94 v cv+ ct v cv+ 3 v+ cv ct+ v+ cv 5 Parental v cv+ 583 v+ cv 597 Recombinant v+ cv+ 134 v cv 134 268/1448 = 18.5%

ct and cv ct to cv v cv ct v cv+ ct+ cv+ ct+ 580 v+ cv ct cv ct 592 v cv ct+ cv ct+ 45 v+ cv+ ct cv+ ct 40 v cv ct cv ct 89 v+ cv+ ct+ cv+ ct+ 94 v cv+ ct cv+ ct 3 v+ cv ct+ cv ct+ 5 Parental cv+ ct+ 674 cv ct 681 Recombinant cv+ ct 43 cv ct+ 50 93/1448 = 6.4%

v and ct v to ct v cv ct v cv+ ct+ v ct+ 580 v+ cv ct v+ ct 592 v cv ct+ v ct+ 45 v+ cv+ ct v+ ct 40 v cv ct v ct 89 v+ cv+ ct+ v+ ct+ 94 v cv+ ct v ct 3 v+ cv ct+ v+ ct+ 5 Parental v ct+ 625 v+ ct 632 Recombinant v+ ct+ 99 v ct 92 191/1448 = 13.2%

Three possible relative orders cv 18.5 v ct 13.2 cv ct 6.4 18.5 v ct cv mapI 13.2 6.4 6.4 ct v cv mapII 18.5 13.2 13.2 ct cv v mapIII 6.4 18.5

The map 18.5 v cv ct 13.2 6.4 The map is not very accurate It is internally inconsistent!!!! Undetected DCO

DCO Parental chromosomes v----ct+-----cv+ & v+----ct----cv The parental homologs will pair in meiosisI. Crossing over will occur and….

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 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 Parent DCO 3. Establish the gene order There are three possible relative order of the three genes in the parent.

Parent v cv+ ct+ & v+ cv ct vermillion red normal vein crossveinless normal wing cut wing DCO v cv+ ct & v+ cv ct+ cut wing normal wing There are three possible gene orders for the parental combination **basically we want to know which of the three is in the middle** predicted DCO OR Each relative order in the parent gives a different combination of the rarest class (DCO)

Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified Gene Order v----ct----cv REWRITE THE COMBINATION IN THE PARENTS v---ct+---cv+ and v+---ct---cv v cv ct v cv+ ct+ 580 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

Now the non-recombinants, single recombinants, and double recombinants are readily identified Recombination freq in region I = SCOI DCO Recombination freq in region II = SCOII DCO Now the DCO are not ignored. With this information one can easily determine the map distance between any of the three genes

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: