LECTURE 5: LINKAGE
1
Linked genes, recombination, and chromosomal mapping
Mendel's Law of Independent Assortment is a consequence of the
fact that chromosomes segregate independently in meiosis
Take two individuals and two genes
One heterozygous and one homozygous
AaBb
x
aabb
2
Linked genes, recombination, and chromosomal mapping
Mendel's Law of Independent Assortment is a consequence of the
fact that chromosomes segregate independently in meiosis
Take two individuals
One heterozygous and one homozygous
AaBb
x
aabb
ab
AB
AaBb
aB
aaBb
Ab
ab
25%
25%
Aabb
25%
aabb
25%
3
Chromosome alignment in MeiosisI
These results are readily explained by the two alternative
ways the chromosomes can line up on the metaphase plate
during meiosis I:
A
a
A
a
OR
B
AB
b
b
ab
Ab
B
aB
Because the A and B genes assort independently, AaBb
dihybrids constructed from different parental genotypes will
behave the same.
AABBxaabb
AAbbxaaBB
4
AABB
x
AAbb
aabb
x
aaBB
B
a
b
a
AaBb
b
A
B
A
OR
Test Cross
AaBb
x
aabb
AaBb
ab
AB
AaBb
aB
aaBb
Ab
ab
x
aabb
ab
25%
25%
AB
AaBb
aB
aaBb
Aabb
25%
Ab
aabb
25%
ab
25%
25%
Aabb
25%
aabb
5
25%
Why 50:50
Why not 25:25:25:25
A hypothetical dihybrid cross involving the genes A and C
produced the following results:
·
A= Tall
a= short
·
C= Cream
c = white
Cross I:
Cross II:
Tall, Cream x short, white
AACC
aacc
Tall, white x short, Cream
AAcc
aaCC
Tall, Cream AaCc
Tall, Cream AaCc
X
X
short, white aacc
short, white aacc
50% Tall, Cream
50% Tall, white
50% short, white
50% short, Cream
Why 50:50?
Why not 25:25:25:25?
6
In these crosses, independent assortment is not occurring.
For example in the first cross, the alleles Tall and Cream
behave as if they are linked to one another.
Similarly in the second cross the alleles Tall and white appear
as if they are linked to one another.
These results are readily explained if the genes A and C lie
next to one another on a chromosome:
7
Cross I:
A-C
a-c
X
a-c
Short white
A-C
Tall cream
a-c
A-C
X
a-c
a-c
a-c
A-C
a-c
A-C
a-c
a-c
a-c
Tall cream
Short white
A= Tall
a= short
C= Cream
c = white8
Cross II:
A-c
a-C
X
a-C
Short cream
A-c
Tall white
a-c
A-c
X
a-C
a-c
a-c
A-c
A-c
a-C
a-c
a-C
a-c
Tall white
Short cream
A= Tall
a= short
C= Cream
c = white9
Purple vestigial
Morgan performed the following experiments in Drosophila to
determine if the genes pr and vg were linked.
PR+ = normal red eyes
pr = purple eyes
VG+ = normal wings
vg = vestigial wings
P
PR+PR+ VG+VG+ x
prpr vgvg
F1
PR+pr VG+vg x
prpr vgvg
If they are on different
chromosomes they should
assort independently
If they are next to one another
on the same chromosome they
should not assort independently
pr vg
PR+ VG+
PR+ vg
pr VG+
pr vg
pr vg
PR+ VG+
pr vg
25%
PR+ VG+
PR+ vg
pr vg
25%
PR+ vg
pr VG+
pr vg
25%
pr VG+
pr vg
pr vg
25%
pr vg
PR+ VG+
pr vg
PR+ vg
pr vg
pr VG+
pr vg
pr vg
pr vg
~50%
~0%
~0%
10 ~50%
When Morgan performed this cross, he obtained the following
result:
pr vg
PR+ VG+
PR+ vg
Pr VG+
pr vg
PR+ VG+
pr vg
1005
PR+ vg
pr vg
153
~6%
Pr VG+
pr vg
143
~6%
pr vg
pr vg
968
~44%
~44%
Although the non- parental classes are present, their
frequencies are dramatically reduced from that expected from
independent assortment.
The two loci are linked !!!
How do we explain the presence of non-parental classes?
11
Morgan suggested that when homologous chromosomes pair during
meiosis I, the chromosomes occasionally exchange parts
P
PR+
VG+
pr
vg
PR+
VG+
pr
vg
pr
F1
vg
pr
vg
PR+ VG+
PR+
VG+
pr
vg
pr
VG+
PR+
vg
PR+
Crossover
chromosome
VG+
The chromosomes that have gone through this crossover are
known as crossover products or recombinants. The original
chromosomes and those that have not undergone a crossover
are known as parental.
Evidence for the model that chromosomes physically exchange
during meiosis is found in meiotic structures known as
chiasmata.
During meiosisI when homologs pair, non-sister chromatids
appear to cross with each other. The resulting cross-shaped
12
structure is known as a chiasmata.
Crossing-over through the microscope
Duplicated homologous chromosomes
Synapsis
Crossing over between
Non-sister chromatids
AnaphaseI
Segregation of homologous
chromosomes
Haploid gametes
13
Answer:
How does one determine whether two genes reside on
different chromosomes or reside on the same chromosome as
linked genes?
To explain this we need to define the terms parental and
recombinant:
Parents: AB/AB x ab/ab
Gametes:
AB
ab
F1:
AB/ab
Meiosis produces the following gametes:
AB
Ab
aB
ab
Recombinant gametes are those with different allelic
combinations than those gametes of the previous generation.
14
Coupling/repulsion
Genes located on the same pair of homologous chromosomes are
called LINKED GENES
Therefore when the A and C alleles are introduced from one
parent they are physically located on the same chromosome and
they do not assort independently. We say that they are linked.
In the above cross we say that the A and C genes are linked.
Therefore when we write the genotype of a dihybrid for two
linked genes, there are two possible conformations:
AaCc
15
Coupling/repulsion- PHASE
Results like these led Morgan to suggest that the A and C
genes are located on the same pair of homologous chromosomes.
Therefore when the A and C alleles are introduced from one
parent they are physically located on the same chromosome and
they do not assort independently. We say that they are linked.
In the above cross we say that the A and C genes are linked.
Therefore when we write the genotype of a dihybrid for two
linked genes, there are two possible conformations:
AaCc
----A----C------o--------a----c------o-----
Coupling conformation
(linkage of two dominant
or two recessive alleles)
---A----c-------o-------a----C-------o-----
Repulsion conformation
(linkage of a dominant and
recessive allele)
16
If genes A and B are on different chromosomes:
A
A
A
F1 diploid
Test cross
progeny
B
B
a
b
a
b
a
b
B
A
B
a
b
P
Gamete
b
a
a
A
B
a
b
a
b
a
b
A
b
a
b
(tester)
b
25% Parental
25% Parental
25% Recombinant
a
B
a
b
25% Recombinant
17
Genes A and B are linked on the same chromosome
A-B
a-b
A-B
a-b
A-B
F1 diploid
a-b
Gamete
A-B
a-b
a-b
a-b
(tester)
A-B
a-b
a-b
Test cross
progeny
P
a-b
A-b
> 25% Parental
> 25% Parental
< 25% Recombinant
a-b
a-B
a-b
< 25% Recombinant
18
Recombination frequency
A
B
A
B
parental
A
a
B
b
A
b
recombinant
a
b
a
B
recombinant
a
b
parental
IF a crossover occurred between linked genes each time
homologs paired, the recombinant frequency would be 50%
This is because crossing-over involves only two of the four
chromatids on the metaphase pair (each of the paired
homologs consists of two sister chromatids).
For example, the frequency of recombinant gametes between
linked genes A and B is 50% if crossing-over occurred each
time the homologs paired.
19
However there are many instances in which the homologs pair and
crossing over does not occur between genes A and B.
It occurs somewhere else
Consequently the overall frequency of recombinants is significantly
reduced from 50%
A
B
A
B
A
a
B
b
A
B
a
b
a
b
a
b
parental
parental
parental
parental
20
A
B
A
b
a
B
a
b
A B
A
B
A B
a b
A
B
a
b
a
b
A
B
A
a
B
b
a
b
a
b
A
B
A
a
B
b
a
b
A
B
A
B
a
b
a
b
21
Distance
The larger the distance between two genes residing on the
same chromosome, the higher the probability there is that a
crossover event will occur between them.
That is for any chromosome, there is a fixed probability per
given distance on the chromosome that a crossover event will
event.
Sturtevant realized that this property could be used to map
genes with respect to one another. For each pair of genes on a
chromosome a recombination frequency can be determined.
By determining the recombination frequency between many
pairs of genes on a chromosome, the relative distance between
genes and the order of the genes on the chromosome can be
determined.
22
Distance
The larger the distance between two genes residing on the
same chromosome, the higher the probability there is that a
crossover event will occur between them.
That is for any chromosome, there is a fixed probability per a
given distance on the chromosome that a crossover event will
event.
Sturtevant realized that this property could be used to map
genes with respect to one another. For each pair of genes on a
chromosome a recombination frequency can be determined.
By determining the recombination frequency between many
pairs of genes on a chromosome, the relative distance between
genes and the relative order of the genes on the chromosome
can be determined.
On average a car breaks down every 40 miles
Some break down after 30 some after 60 etc…
Santa Cruz to San Francisco is 80 miles
2 breakdowns
Santa Cruz to Monterrey is 45 miles
1 breakdown
There is greater probability that your car will break down
between Santa Cruz and San Francisco.
If crossing-over occurs once every 50 kb of DNA, then there
is greater probability of a crossover between two genes 100 kb
apart than two genes 50 kb apart.
23
For example Sturtevant identified three recessive mutations
that reside on the X chromosome of Drosophila
W+ red eyes
w- white eyes
CV+ normal wings
cv- crossveinless
SN+ normal bristle
sn- singed bristle
w
cv
Cen
sn
Tel
X chromosome
By calculating recombination frequencies between each pair of
genes we can begin to establish where these three genes reside
on the X chromosome with respect to one another
24
To determine the distance between the w gene and the sn gene
P
w sn/w sn
x
W+ SN+/Y
F1
w sn/W+ SN+
x
w sn/Y
w sn
F2
W+ SN+
Red eye
Normal bristle
102
y
Red eye
Normal bristle
96
Parental
w sn
White eye
Singed bristle
88
White eye
Singed bristle
92
w SN+
White eye
Normal bristle
24
White eye
Normal bristle
23
Recomb
W+ sn
Red eye
Singed bristle
24
Red eye
Singed bristle
23
25
Fill out the phenotypes- recombinants can be determined by phenotype analysis
Recombination frequency equals the number of recombinants over
total number of progeny
white eye
# recombinant progeny =
# total progeny
normal bristle
+
red eye
singed bristle
# total progeny
= 24+24+23+23=94/472
1 map unit (m.u.) = 1% recombination frequency
Therefore _19.92% or ~20cM or ~20 m.u. separate the W+ and
SN+ genes.
This is a relative distance- depends upon recombination between
two genes. Not an absolute distance like bp
In the above cross, we could have determined recombination
frequency by counting only males (or only females)
26
w -------20--------sn
The next issue is where does cv map with respect to w and sn:
By crosses similar to those described above, we find that there
are 7 m.u. between cv and sn
This means cv can map to either one of two positions:
A)
w ____________14?______________cv_________7___ sn
OR
B)
w______________20____________ sn_________7____ cv
These models can be distinguished by determining the map
distance between w and cv. Recombination analysis indicates
14 m.u. between w and cv.
By determining the map distance between w and cv and the
map distance between cv and sn, we can determine the
distances and order of all three genes.
Which map is consistent with this distance?
Internal inconsistency- 14+7=21 not 20
27
Map gives you order of genes but not PRECISE distance