Lecture 6
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Transcript of Lecture 6
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Lecture 6
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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?
Answer-
Linkage
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P
F1
Parental
Recomb
Tall, No hair short, hairy
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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
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Another mutation C (crinkled) is isolated and recombination frequencies between this gene and the A and H genes are determined
% recombinants
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What is going on? The map is not internally consistent?
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The double crossovers go undetected and therefore over large distances the genetic distances are underestimated
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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
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First example
P
F1
F2
If these genes were on separate chromosomes, they should be assorting independently and all the classes should be equally frequent.
sc ec 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
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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 vgsc vg
sc ec vg 235sc+ ec+ vg+ 241sc ec vg+ 243sc+ ec+ vg 233sc ec+ vg 12sc+ ec vg+ 14sc ec+ vg+ 14sc+ ec vg 16
247
255
257
249
# recombinant/total progeny =
Therefore sc and vg are
Next: What about sc and ec?
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sc and ec are ???
sc vgsc ec vg 235sc+ ec+ vg+ 241sc ec vg+ 243sc+ ec+ vg 233sc ec+ vg 12sc+ ec vg+ 14sc ec+ vg+ 14sc+ ec vg 16
478
474
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What about sc and ec?
# recombinant/total progeny =
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ec and vg are not linked
sc vgsc ec vg ec vg 235sc+ ec+ vg+ ec+ vg+ 241sc ec vg+ ec vg+ 243sc+ ec+ vg ec+ vg 233sc ec+ vg ec+ vg 12sc+ ec vg+ ec vg+ 14sc ec+ vg+ ec+ vg+ 14sc+ ec vg ec vg 16
251
255
257
245
# recombinant/total progeny = 502/1008 = 50%
Therefore ec and vg are NOT LINKED!
From these observations what is the map distance between ec and vg?
scec
vg
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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
F2v cv ct
v cv+ ct+v+ cv ctv cv ct+v+ cv+ ct+v cv ctv+ cv+ ct+v cv+ ctv+ cv ct+
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v and cv
v to cv
v cv ctv cv+ ct+ v cv+ 580v+ cv ct v+ cv 592v cv ct+ v cv 45v+ cv+ ct+ v+ cv+ 40v cv ct v cv 89v+ cv+ ct+ v+ cv+ 94v cv+ ct v cv+ 3v+ cv ct+ v+ cv 5
Parental
v cv+ 583
v+ cv 597
Recombinant
v+ cv+ 134
v cv 134268/1448 = 18.5%
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ct and cv
ct to cv
v cv ctv cv+ ct+ cv+ ct+ 580v+ cv ct cv ct 592v cv ct+ cv ct+ 45v+ cv+ ct cv+ ct 40v cv ct cv ct 89v+ cv+ ct+ cv+ ct+ 94v cv+ ct cv+ ct 3v+ cv ct+ cv ct+ 5
Parental
cv+ ct+ 674
cv ct 681
Recombinant
cv+ ct 43
cv ct+ 50
93/1448 = 6.4%
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v and ct
v to ct
v cv ctv cv+ ct+ v ct+ 580v+ cv ct v+ ct 592v cv ct+ v ct+ 45v+ cv+ ct v+ ct 40v cv ct v ct 89v+ cv+ ct+ v+ ct+ 94v cv+ ct v ct 3v+ cv ct+ v+ ct+ 5
Parental
v ct+ 625
v+ ct 632
Recombinant
v+ ct+ 99
v ct 92
191/1448 = 13.2%
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Three possible relative orders
v cv
v ct
v cv mapI
v cvct
18.5
13.2
mapII
ct
cv ct6.4
mapIIIvcvct
18.5
13.2 6.4
18.5
18.5
13.2
13.2
6.4
6.4
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The map
v cv
18.5
ct13.2 6.4
The map is not very accurate
It is internally inconsistent!!!!
Undetected DCO
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DCO
Parental chromosomes
v----ct+-----cv+ & v+----ct----cv
The parental homologs will pair in meiosisI. Crossing over will occur and….
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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 ctv cv+ ct+ 580v+ cv ct 592v cv ct+ 45v+ cv+ ct 40v cv ct 89v+ cv+ ct+ 94v cv+ ct 3v+ cv ct+ 5
Parent
DCO
3. Establish the gene order
There are three possible relative order of the three genes in the parent.
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There are three possible gene orders for the parental combination
**basically we want to know which of the three is in the middle**
Parent v cv+ ct+ & v+ cv ctvermillion rednormal vein crossveinlessnormal wing cut wing
DCO v cv+ ct & v+ cv ct+vermillion rednormal vein crossveinlesscut wing normal wing
Each relative order in the parent gives a different combination of the rarest class (DCO)
predicted DCO
OR
OR
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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+ ct 40
v cv ct 89
v+ cv+ ct+ 94
v cv+ ct 3
v+ cv ct+ 5
Gene Order v----ct----cv
REWRITE THE COMBINATION IN THE PARENTS
v---ct+---cv+ and v+---ct---cv
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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
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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: