GGAZ
,Wl
L2 -
Meiosis and Genetic Recombination
recombination process that combination transmitted
generates alleles
from parent to
gospring
~
of
new
1
why
t
important ? ↳
this transmitted
1) produces genetic diversity keeps populations going genetic info .
1) out later
is
packaged gametes
into
Find
3) ( must have 1
copy of each oh .
diploid This
so
zygote is
=
.
due to meiosis
)
For recombination to have the
an
affect in
next there must be
generation a
heterozygous
individual for at least 2
genes .
÷siÉdiÉt
( would not be evident with
homozygous) due to independent assort .
This requires the parental to be AABB aabb A Abb aaBB
generation Parental
✗ or ✗
generation
:
t
Aaa Bbb F1
parental
combination
parental
combination of alleles
alleles
of
AABB aabb
IF f- If f-
I : p
✗
ABF f-
5 Gametes
y
f- §
AD ! F1
AaDb
, Genes A and B could be on
shame or on
different chromosomes (;)
" ^^ ""
{
←
•
it att Independent assortment at Mt
b leads to inheritance
tf Tt of
new
•
combinations alleles
of of genes
on
different chromosomes
-
equally probable so 1 1
:
recombination between different
f. of genes
on
chromosomes is 50%
For A Abb ✗ aaBB parental generation :
This time parental combination =
Ab ,
AB
and recombinant combination =
AB ab
,
So which combination alleles
of is
and which is
parental recombinant
depends on
genotype of parents of F1
, Genes
" "
chromosome (1)
}
on sand
It I 11=8
it
each combo will.
away remain
together
as on same oh .
→
independent assortment has not
recombination
given us for alleles
chromosome
of genes on same
/
so what recombination alleles
gives of
chromosome ?
of genes on same
I
chiasma at Mt
(need bivalent)
how do we know that a
new combination alleles
n.is
→
of
experiment
genetic
ab →
genetically
is due to a new
AB
combination chromosomal
Ab
of
material ?
aka how do we know there
If :#I µ Aff f f
" " "
> is
exchange among non sister
-
, B- B- b
b -
?
chromatids
-
P R R P
, Genetic experiment to
find whether recombination due to
genetic is
crossing
over :
Need an individual
heterozygous for 2
genes @ wxgenes) +
aid laid
this oh Wx
only genetically distinct
is not oh
.
from c. ,
a
D ④ but has acquired a chromatin blob (LHS) and translocation @Hs)
PARENTAL
D d
Aid / Aid
a
*
A d
F1 A.cl/aiD
÷
µ÷,&, {
Aid /Aid And crossing over
)
A D does not happen
AD / aid chromatids A. D ineachmeiocyk
p a d i.
f. recombination
Aid must be -50%
a
D
I
test -
cross = one between Ai AID
gemsareiinned
Hetero F1 Homo
atestcwss :
a hetero + homo Aide /
Dad
a.
/ ad AD / ad
FL
a.m.mn, ,
between non -
sister chromatids independent so
since new different phenotype > 50%1-8
/
"" " "" "
So two processes can result in recombination :
1) independent assortment
for genes on
doggonest chromosomes
a) chiasma
formation / crossing over
forgeries on same chromosome
-
\ > involves
Paf
AND recombination
crossing
over
, Recombination Genetic
↳
frequency linkage
-
or
/
→ 50%
stays far
2 + crossovers
between A- +B
So recombination between chromosome is 50%
ma_✗mu f. of genes on same
I
50% for both doggonest
i. since
genes
+ same
1
Max f- of recombination between two is 50%
any genes
.
Genetic linkage
show
'
recombination chromosome
if genes they are LINKED and same
- -
on
Genes on DIFFERENT chromosomes
will show 50% recombination and are said to be unlinked .
c÷÷¥)
1-
Genes show 50% recombination be unlinked
on SAME chromosome
may
and
genetically
born "" " h
different oh
-
.
I
if
501 .
or more
p of
. recombination not linked
( as
crossing
over d)
less than 50% linked
game eh
-
. if
, Crossing over
gives reciprocal products # AB # ab
: =
# Ab =
#aB
recombination
Measuring f. of
Outcome 2 test
of different crosses
-
:
reciprocal pairs equal (as should be)
x
but the four classes are not
↳ I
A and C linked
genes
:
.
are
all ✗ same #
✓
recombination
so
f. of
must be v50 %
what the recombination
is
f. of
① between A and C ?
A not
gene
i.
linked to B V
# recombinant
f. recombination
=
✗ no
total
I progeny
f. of recombination gene distance
if 1 genes
dude >
are the
phenotypes belonging 93+107
=
✗ no
less chance to
together #412 # 388
and 1000
over must be the PARENTAL 20%
of crossing
=
as most frequent =
20 map units or 20cm
chance
if far more
If f- Ef f-
E &
recombination PARENTAL AC /ac aclac
so
f. of
: :
✗
can use
✗
-
guideoh
as a to their distance
apart on .
mhimap> 1% recombination = 1 map unit = 1cM
, Making genetic a map
Using
and so
a 3-point test
distance values
cross
:
as more
informative that 2- point
for recombination f.
ABC combo alleles received
of
=
from one parent
parental combos as most
V9
" "t
?
aye
it from other patent
other combo
{
=
.
i. recombinant
i.
ABC/ abc ✗ abc /abc test cross
""
to
?•÷; dont have
Map distance =
% Recombination =
Total A B recombinant
-
✗ too f. 4 combos)
*
%,
=
145 + 155 + 77 + 73
✗ too
1000
=
45 map units
-
",mµ
Total A -
C recombinant =
combos not with
" "" → • " " %
AC or ac
f- 4)
"
"
=
1451-155+33+37
A C B
37p
=
"5
{ 37
> (
22 }
Y ,
,
=
,, ,,
22maPF
, ,, , ,,
whyadddo they not
up ?
↳ 2 crossovers
,W 2
Lk -
Genetic o
Physical maps
4- 1
Effect of a double crossover
✓ = recombinant pair
✗ = Parental pain
Acr cB✓ AB ✗
A- Bc
ac
✓ Cb ✓ ab ✗
abc
u
both double crossovers have
non recombinant
given gene
-
in meiotic products
pairing
-
when
calculating the map distance between A B
-
¢15 ) mu this was underestimated
=
as the double recombinant
cancelled out and were not counted
Double recombination reduces recombinant between outside
f. of genes .
✗
} products of double crossover
( note the
K-astf-due.to double crossing over
)
-
✗
So how can we use identified recombinants to correct map distance A B ? -
!
remember ✗ 2 as double crossovers
happening
Add them in recombinant f. equation y
:
, 4- 2
Coefficient of coincidence
Are crossover events independent in same meiosis ?
Coefficient coincidence observed
f. of doubles 1 independent each other
of
=
if
=
of
expected f. of doubles
usually observed e
expected
From ex .
T¥ .
of
doubles =
?%_3 =
0.07
=
assume
independence
as
expected f. of doubles 0.37
¥ 022
0.08¥
= =
t
: .
Coc =
0--07 37% recombination between A -
C
0.0814
between A 0.37
f. of crossing C
i. -
over
- =
=
0.8£ not independent
am
observed expected
suggesting that
-
occurrence crossover
of one
makes it less
likely of another
happening in same meiosis .
So
they interfere with one another .
1
Interference =
coefficient of coincidence
-
=
1 -
0.86
0¥
=
,Wl
L2 -
Meiosis and Genetic Recombination
recombination process that combination transmitted
generates alleles
from parent to
gospring
~
of
new
1
why
t
important ? ↳
this transmitted
1) produces genetic diversity keeps populations going genetic info .
1) out later
is
packaged gametes
into
Find
3) ( must have 1
copy of each oh .
diploid This
so
zygote is
=
.
due to meiosis
)
For recombination to have the
an
affect in
next there must be
generation a
heterozygous
individual for at least 2
genes .
÷siÉdiÉt
( would not be evident with
homozygous) due to independent assort .
This requires the parental to be AABB aabb A Abb aaBB
generation Parental
✗ or ✗
generation
:
t
Aaa Bbb F1
parental
combination
parental
combination of alleles
alleles
of
AABB aabb
IF f- If f-
I : p
✗
ABF f-
5 Gametes
y
f- §
AD ! F1
AaDb
, Genes A and B could be on
shame or on
different chromosomes (;)
" ^^ ""
{
←
•
it att Independent assortment at Mt
b leads to inheritance
tf Tt of
new
•
combinations alleles
of of genes
on
different chromosomes
-
equally probable so 1 1
:
recombination between different
f. of genes
on
chromosomes is 50%
For A Abb ✗ aaBB parental generation :
This time parental combination =
Ab ,
AB
and recombinant combination =
AB ab
,
So which combination alleles
of is
and which is
parental recombinant
depends on
genotype of parents of F1
, Genes
" "
chromosome (1)
}
on sand
It I 11=8
it
each combo will.
away remain
together
as on same oh .
→
independent assortment has not
recombination
given us for alleles
chromosome
of genes on same
/
so what recombination alleles
gives of
chromosome ?
of genes on same
I
chiasma at Mt
(need bivalent)
how do we know that a
new combination alleles
n.is
→
of
experiment
genetic
ab →
genetically
is due to a new
AB
combination chromosomal
Ab
of
material ?
aka how do we know there
If :#I µ Aff f f
" " "
> is
exchange among non sister
-
, B- B- b
b -
?
chromatids
-
P R R P
, Genetic experiment to
find whether recombination due to
genetic is
crossing
over :
Need an individual
heterozygous for 2
genes @ wxgenes) +
aid laid
this oh Wx
only genetically distinct
is not oh
.
from c. ,
a
D ④ but has acquired a chromatin blob (LHS) and translocation @Hs)
PARENTAL
D d
Aid / Aid
a
*
A d
F1 A.cl/aiD
÷
µ÷,&, {
Aid /Aid And crossing over
)
A D does not happen
AD / aid chromatids A. D ineachmeiocyk
p a d i.
f. recombination
Aid must be -50%
a
D
I
test -
cross = one between Ai AID
gemsareiinned
Hetero F1 Homo
atestcwss :
a hetero + homo Aide /
Dad
a.
/ ad AD / ad
FL
a.m.mn, ,
between non -
sister chromatids independent so
since new different phenotype > 50%1-8
/
"" " "" "
So two processes can result in recombination :
1) independent assortment
for genes on
doggonest chromosomes
a) chiasma
formation / crossing over
forgeries on same chromosome
-
\ > involves
Paf
AND recombination
crossing
over
, Recombination Genetic
↳
frequency linkage
-
or
/
→ 50%
stays far
2 + crossovers
between A- +B
So recombination between chromosome is 50%
ma_✗mu f. of genes on same
I
50% for both doggonest
i. since
genes
+ same
1
Max f- of recombination between two is 50%
any genes
.
Genetic linkage
show
'
recombination chromosome
if genes they are LINKED and same
- -
on
Genes on DIFFERENT chromosomes
will show 50% recombination and are said to be unlinked .
c÷÷¥)
1-
Genes show 50% recombination be unlinked
on SAME chromosome
may
and
genetically
born "" " h
different oh
-
.
I
if
501 .
or more
p of
. recombination not linked
( as
crossing
over d)
less than 50% linked
game eh
-
. if
, Crossing over
gives reciprocal products # AB # ab
: =
# Ab =
#aB
recombination
Measuring f. of
Outcome 2 test
of different crosses
-
:
reciprocal pairs equal (as should be)
x
but the four classes are not
↳ I
A and C linked
genes
:
.
are
all ✗ same #
✓
recombination
so
f. of
must be v50 %
what the recombination
is
f. of
① between A and C ?
A not
gene
i.
linked to B V
# recombinant
f. recombination
=
✗ no
total
I progeny
f. of recombination gene distance
if 1 genes
dude >
are the
phenotypes belonging 93+107
=
✗ no
less chance to
together #412 # 388
and 1000
over must be the PARENTAL 20%
of crossing
=
as most frequent =
20 map units or 20cm
chance
if far more
If f- Ef f-
E &
recombination PARENTAL AC /ac aclac
so
f. of
: :
✗
can use
✗
-
guideoh
as a to their distance
apart on .
mhimap> 1% recombination = 1 map unit = 1cM
, Making genetic a map
Using
and so
a 3-point test
distance values
cross
:
as more
informative that 2- point
for recombination f.
ABC combo alleles received
of
=
from one parent
parental combos as most
V9
" "t
?
aye
it from other patent
other combo
{
=
.
i. recombinant
i.
ABC/ abc ✗ abc /abc test cross
""
to
?•÷; dont have
Map distance =
% Recombination =
Total A B recombinant
-
✗ too f. 4 combos)
*
%,
=
145 + 155 + 77 + 73
✗ too
1000
=
45 map units
-
",mµ
Total A -
C recombinant =
combos not with
" "" → • " " %
AC or ac
f- 4)
"
"
=
1451-155+33+37
A C B
37p
=
"5
{ 37
> (
22 }
Y ,
,
=
,, ,,
22maPF
, ,, , ,,
whyadddo they not
up ?
↳ 2 crossovers
,W 2
Lk -
Genetic o
Physical maps
4- 1
Effect of a double crossover
✓ = recombinant pair
✗ = Parental pain
Acr cB✓ AB ✗
A- Bc
ac
✓ Cb ✓ ab ✗
abc
u
both double crossovers have
non recombinant
given gene
-
in meiotic products
pairing
-
when
calculating the map distance between A B
-
¢15 ) mu this was underestimated
=
as the double recombinant
cancelled out and were not counted
Double recombination reduces recombinant between outside
f. of genes .
✗
} products of double crossover
( note the
K-astf-due.to double crossing over
)
-
✗
So how can we use identified recombinants to correct map distance A B ? -
!
remember ✗ 2 as double crossovers
happening
Add them in recombinant f. equation y
:
, 4- 2
Coefficient of coincidence
Are crossover events independent in same meiosis ?
Coefficient coincidence observed
f. of doubles 1 independent each other
of
=
if
=
of
expected f. of doubles
usually observed e
expected
From ex .
T¥ .
of
doubles =
?%_3 =
0.07
=
assume
independence
as
expected f. of doubles 0.37
¥ 022
0.08¥
= =
t
: .
Coc =
0--07 37% recombination between A -
C
0.0814
between A 0.37
f. of crossing C
i. -
over
- =
=
0.8£ not independent
am
observed expected
suggesting that
-
occurrence crossover
of one
makes it less
likely of another
happening in same meiosis .
So
they interfere with one another .
1
Interference =
coefficient of coincidence
-
=
1 -
0.86
0¥
=
