1. LECTURE 6: LINKAGE

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 recessive
AaBb x aabb ab AB
AaBb
25% aB
aaBb
25%
Aabb Ab
25% ab
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 b b B AB ab Ab 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
AaBb A a B b A a b B OR
Test Cross
AaBb x aabb
AaBb x aabb a b a b A B
AaBb
25% A B
AaBb
25% a B
aaBb
aaBb a B A b
Aabb A b
Aabb a b
aabb a b
aabb

5. 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 Tall, white x short, Cream AACC aacc 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 the same chromosome:

7. Cross I:
A-C
Tall cream
a-c
Short white P X
A-C
a-c
a-c F1 X
a-c
A-C
a-c
Tall cream
A-C
a-c
Short white
a-c
A= Tall a= short
C= Cream c = white

8. Cross II:
A-c
Tall white
a-C
Short cream P X
A-c
a-C
a-c F1 X
a-c
A-c
a-c
Tall white
A-c
a-C
a-c
Short cream
a-C
A= Tall a= short
C= Cream c = white

9. AABB x aabb
AACC x aacc
AaBb
AaCc
AaBb x aabb
Test Cross
Test Cross
AaCc x aacc A a B b a A c C A a b B OR a b a c A B AC ac AB ab
25%
50% A C a b ab ac a c A b Ab ab aC ac Ac a C A c 0% a B aB ab

10. Purple vestigial
Morgan performed the following experiments in Drosophila to investigate the independent assortment of the genes pr and vg
PR+ = normal red eyes pr = purple eyes
VG+ = normal wings vg = vestigial wings
P PR+VG+/PR+VG+ x prvg/prvg
F1 PR+VG+/pr vg x prvg/prvg
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+
25%
PR+ VG+
~50%
~0%
pr vg
pr vg
pr vg
pr vg
pr VG+
pr vg
pr VG+
pr vg
pr VG+
pr VG+
PR+ vg
pr vg
PR+ vg
pr vg
PR+ vg
PR+ vg

11. When Morgan performed this cross, he obtained the following result:
P PR+VG+/PR+VG+ x prvg/prvg
F1 PR+VG+/pr vg x prvg/prvg
pr vg
PR+ VG+
pr vg
PR+ VG+
PR+ vg
Pr VG+
pr vg
1005
~44%
~6%
pr vg
968
Pr VG+
pr vg
143
PR+ vg
pr vg
153
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?

12. Morgan suggested that when homologous chromosomes pair during meiosis I, the chromosomes occasionally exchange parts
PR+ VG+
pr vg P
pr vg
pr vg F1
PR+ VG+
PR+ VG+
pr VG+
pr vg
Crossover
chromosome
PR+ vg
PR+ 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 structure is known as a chiasmata.

13. Crossing-over through the microscope
Duplicated homologous chromosomes
Synapsis in meiosis I
Crossing over between
Non-sister chromatids
Anaphase in meiosisI
Segregation of homologous
chromosomes
Haploid gametes in meiosis II

14. 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
Recombinant gametes are those with different allelic combinations than those gametes of the previous generation.

15. 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

16. 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)

17. If genes A and B are on different chromosomes:P A B a b
Gamete A a B b a b
F1 diploid
(tester) A B b
25% Parental a a b
25% Parental
Test cross
progeny A b a
25% Recombinant a B b
25% Recombinant

18. Genes A and B are linked on the same chromosome
A-B
a-b P
A-B
a-b
Gamete
A----B
a----b
A----b
a----b
F1 diploid x
(tester)
A----B
a----b
> 25% Parental
a----b
> 25% Parental
Test cross
progeny
A----b
a----b
< 25% Recombinant
A----B
a----b
< 25% Recombinant

19. Recombination frequency is always less than 50%
parental A B A b
recombinant a b a B
recombinant a b 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.

20. 100 75 50
% Recombination 25 25 50 75
Distance (MU)

21. 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
(Recombination ALWAYS occurs between paired homologs but it might not occur between the two genes YOU are studying)
Consequently the overall frequency of recombinants is significantly
reduced from 50% A B a b A B a b
parental

22. 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 B a b A B a b A B a

23. 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.

24. 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
Santa Cruz to San Francisco is 80 miles= 8 breakdowns
Santa Cruz to Monterrey is 40 miles = 4 breakdown
The distance between San Francisco and Santa Cruz is 8 units and the distance between Santa Cruz and Monterrey is 4 units
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.

25. Mapping using recombination frequency
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
Cen
Tel w cv sn
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

26. 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 F2
w sn y
Red eye
Normal bristle
102
Red eye
Normal bristle 96
W+ SN+
Parental
White eye
Singed bristle 88
White eye
Singed bristle 92
w sn
White eye
Normal bristle 24
White eye
Normal bristle 23
w SN+
Recomb
Red eye
Singed bristle 24
Red eye
Singed bristle 23
W+ sn
Fill out the phenotypes- recombinants can be determined by phenotype analysis

27. Recombination frequency equals the number of recombinants over total number of progeny
white eye red eye
# recombinant progeny = normal bristle + singed bristle
# total progeny # total progeny
= =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)

28. To determine the distance between the cv gene and the sn gene
P cv sn/cv sn x CV+ SN+/Y
F1 cv sn/CV+ SN+ x cv sn/Y F2
cv sn y
Normal vein
Normal bristle 90
Normal vein
Normal bristle 90
CV+ SN+
Parental
crossvein
Singed bristle 90
crossvein
Singed bristle 90
cv sn
crossvein
Normal bristle 6
crossvein
Normal bristle 8
cv SN+
Recomb
Normal vein
Singed bristle 7
Normal vein
Singed bristle 7
CV+ sn
Fill out the phenotypes- recombinants can be determined by phenotype analysis
Distance between CV and SN is 7MU

29. The next issue is where does cv map with respect to w and sn:
We find that there are 7 m.u. between cv and sn
This means cv can map to either one of two positions:- left of Sn and right of Sn
w______________20________________ sn_________7____ cv OR
w _________13?__________cv_______7___ sn
These models can be distinguished by determining the map distance between w and cv. Is the distance 27 or 13? 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?
Small Internal Inconsistency- 14+7=21 not 20
Map gives you order of genes but not PRECISE distance

30. Genetic Map
cv Crossveinless wing
g Garnet eyes
f forked bristles
sc Scute Bristle
ec Echinus eye
ct Cut wing
v vermilion eye
9.1
10.5
9.2
15.9
11.2
10.9
= 66.8 sc f 50