A geneticist isolates two mutations:

1
Lecture 6
2
Linkage
A geneticist isolates two mutations ? A tall
? a short ? H hairy ? 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-
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?
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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
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First example
P F1 F2
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
If these genes were on separate chromosomes, they
should be assorting independently and all the
classes should be equally frequent.
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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 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?
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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
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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
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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 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 v cv ct
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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
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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 c
v ct 43 cv ct 50
93/1448 6.4
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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
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Three possible relative orders
v
cv
18.5
v
ct
13.2
cv
ct
6.4
18.5
v
cv
mapI
ct
13.2
6.4
6.4
ct
mapII
v
cv
18.5
13.2
13.2
ct
cv
v
mapIII
18.5
6.4
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The map
18.5
v
cv
ct
13.2
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 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.
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Parent v cv ct v cv ct vermillion red
normal vein crossveinless normal wing cut
wing DCO v cv ct v cv ct vermillion r
ed normal vein crossveinless 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 OR
Each relative order in the parent gives a
different combination of the rarest class (DCO)
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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
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
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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