4. Ch5_Oct_23_posted - 10/25/09 Linkage
Maps
 y 0 w 1.1...

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Unformatted text preview: 10/25/09 Linkage
Maps
 y 0 w 1.1 1.1 32.1 v 33.2 4 m 37.2 17.8 r 55 •  NB:
In
any
given
cross,
recombina4on
frac4on
can’t
be
>50%
 (=unlinked)
 •  Gene4c
maps
can
have
>50
m.u.
b/c
cumula4ve
across
a
 chromosome.
 Oct 23rd A Describing
Linkage
 + B + •  Cis
arrangement:
mutant
alleles
are
on
same
homologue
 •  Trans
arrangement:
mutant
alleles
are
on
different
homologues
 Cis
heterozygote
 Trans
heterozygote
 1 10/25/09 Genes
on
the
X
chromosome
can
be
mapped
without
a
 testcross.
 1.  True
 2.  False
 Determining
Gene
Loca4on
 Two
Genes
 •  2
point
cross
=
cross
with
2
loci
 •  2
point
testcross
=
cross
b/w
dihybrid
parent
&
homozygous
recessive
 
 
 
 
 




individual
 Three
Genes
 •  3
point
testcross
(3
point
crosses
too)
‐
advantages
over
2
point
crosses
 •  Can
establish
rela4ve
order
of
all
3
genes
–
beXer
image
 •  Can
analyze
effects
of
mul4ple
X.O.
events
on
map
distances

 Represen-ng
Genes
 1.  bb
prpr
cc
 2.  b/b;
pr/pr;
c/c
 3.  b
pr
c/b
pr
c
 2 10/25/09 Gene4c
Mapping
with
>
3
markers…
 Phenotypes
of
offspring
 correspond
with
genotype
 of
homolog
inherited
 from
mom.
 •  Test
individual
(here
F1
female)
has
to
be
heterozygous
for
the
genes
of
 interest.
 Gene4c
Mapping
with
>
3
markers…
 1.
Determine
which
genotypes
are
recombinant
&
which
are
parental
for
each
 PAIR
of
genes.
 2.
Determine
map
distances
for
each
pair.
 How?
Look
at
recombina4on
frequency
b/w
pairs.
 3 10/25/09 Gene4c
Mapping
with
>
3
markers…
 •  Recombina4on
b/w
vg
&
b
 Rf
=
100
(252
+
241
+
131
+
118)/4197
 Rf
=
17.7%
=
17.7
m.u.
(vg

b)
 •  Recombina4on
b/w
vg
&
pr
 Rf
=
100
(252
+
241
+
13
+
9)/4197
 Rf
=
12.3%
=
12.3
m.u.
(vg

pr)
 •  Recombina4on
b/w
b
&
pr
 Rf
=
100
(131
+
118
+
13
+
9)/4197
 Rf
=
6.4%
=
6.4
m.u.
(b

pr)
 Gene4c
Mapping
with
>
3
markers…
 Recombination frequency (Rf) = # Offspring with recombinant phenotypes Total # of offspring 4 10/25/09 Gene4c
Mapping
with
>
3
markers…
 1.
Determine
which
genotypes
are
recombinant
&
which
are
parental
for
each
 PAIR
of
genes.
 2.
Determine
map
distances
for
each
pair.
(use
Rf)
 3.
Determine
gene‐order
based
on
map
distance.
 •  vg

b

 
17.7
m.u.
 •  vg

pr 
12.
3
m.u.
 •  b

pr 
6.4
m.u.
 18.7 = 17.7!! Why
discrepancies
in
distance?
 •  Order
is
consistent,
but
numbers
don’t
add
up!
 5 10/25/09 Why
discrepancies
in
distance?
 w 1.1 m.u. y 32.8 m.u. w y 1.1 w m 32.8 m •  1.
Random
Chance:
 •  Recombina4on
is
stochas4c
 •  2.
Double
crossovers
 34.3 Single
&
Double
Crossovers
 6 10/25/09 •  Double
X.O.

smallest
group
 •  A
double
X.O.
=
2
crossover
events

need
to
include
 •  2
X.O.

needs
to
be
counted
twice
 •  Recombina4on
b/w
vg
&
b
 Correc4ng
for
Double
Crossovers
 Rf
=
100
(252
+
241
+
131
+
118)/4197
 Rf
=
17.7%
=
17.7
m.u.
(vg

b)
 * * * * •  Correc4on
for
RF
(vg
&
b)

 
RF
(vg
&
b)
=
100
(252
+
241
+
131
+
118
+
2[13
+9])/4197
 

 




=
18.7
m.u.


 Correc4ng
for
Double
Crossovers
 •  Correc4on
for
RF
(vg
&
b)

 
RF
(vg
&
b)
=
100
(252
+
241
+
131
+
118
+
2[13
+9])/4197
 

 




=
18.7
m.u.


 
RF
(vg
&
b)
=
18.7
m.u.
=
12.3
m.u.+
6.4
m.u.


 7 10/25/09 Correc4ng
for
Double
Crossovers
 •  RF
(vg
&
b)
=
17.7
m.u.
 •  RF
(vg
&
pr)
=
12.3
m.u.
 •  RF
(b
&
pr)
=
6.4
m.u.
 •  We
know
the
order
of
the
genes

can
see
where
double
X.O.
occurs.
 •  vg
b
pr+
and
vg+b+pr

due
to
2
crossover
events
 •  Double
X.O.
switches
gene
in
middle
wrt
gene4c
markers
on
either
side
 Short
Cuts
with
Double
Recombinants
 •  We
know
that
double
recombinants
are
the
smallest
class
of
progeny
 •  F1
female
was

vg
pr
b
/
vg+
pr+
b+ 







these
combos
are
thus
parental
 •  Smallest
group
was:
 •  13
vg

b

pr+
 •  9
vg+
b+
pr
 •  Look
parental
with
excep4on
of
pr

recombined

 •  progeny
that
show
double
X.O.
 •  Gene
with
recombined
alleles
rela4ve
to
parental
 configura4ons
of
other
2
genes

must
be
in
the
middle.
 56
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This note was uploaded on 01/10/2010 for the course BIOLOGY biol2040 taught by Professor Tamarakelly during the Fall '09 term at York University.

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