1. Lecture 6

2. 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?

3. P
Tall, No hair short, hairy F1
Parental
Recomb

4. 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

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

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

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

8. 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

9. 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.

10. 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
sc+ ec+ vg
sc ec vg
sc+ ec+ vg
sc ec+ vg
sc+ ec vg
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?

11. sc and ec are ???
What about sc and ec?
sc vg
sc ec vg
sc+ ec+ vg
sc ec vg
sc+ ec+ vg 233
sc ec+ vg 12
sc+ ec vg
sc ec+ vg+ 14
sc+ ec vg 16
478
474 26 30
# recombinant/total progeny =

12. 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
sc ec vg+ ec vg
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

13. 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+

14. v and cv
v to cv
v cv ct
v cv+ ct+ v cv
v+ cv ct v+ cv
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%

15. 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%

16. 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%

17. Three possible relative orderscv
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

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

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

20. 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.

21. 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)

22. 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
v+ cv ct
v cv ct
v+ cv+ ct 40
v cv ct
v+ cv+ ct
v cv+ ct 3
v+ cv ct+ 5

23. 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

24. 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: