This document discusses genetic mapping and linkage analysis. It explains that genetic mapping involves determining the order and distance between genes on chromosomes using crossover frequencies. Linkage groups containing linked genes are identified through test crosses. Map distances between genes are measured in centimorgans based on crossover rates. Gene order is determined using two-point and three-point test crosses. Interference affects double crossover rates. Cytological mapping can verify gene order by correlating genetic maps to chromosome banding patterns.

2.  Number
of genes exceeds the number of
chromosome in different species ,many genes
located on same chromosome
 Number
of inklage groups directly proportional to
bivalents of species

Sometimes crossing over occurs ,closely linked
genes have less chance of departure

3. 
Each gene has definite order and location in a
linkage group or chromosome, as the crossing
over frequency has found to be constant

Eg. in Drosophilia,3 genes(white eyes, yellow
body and cut wings)white and yellow always 1%
cross over frequency

Thus %crossing over appears to be closely
related to physical distance between genes

4. 
All genes, linkage groups and their number is
known then by using crossing over as the
tool, relative distance between the genes in a
linkage group and their order could be determined

This may give diagrammatic representation of
chromosome showing the genes as points
separated by distances proportional to amount of
crossing over

This is said to be genetic map, cross over map or
linkage map

5. Determination
of linkage group
Determination
of map distance
Determination
of gene order

6.  By
hybridization experiments b/n wild and mutant
strains
 Helps
to determine how many phenotypes remains
completely linked and consequently their genes
during the course of inheritance
 Different
linkage groups of a species could be
worked out

7. 
It is measured using map unit or Morgan unit

One map unit or 1 cM = 1 % cross over(recombinants)

can be calculated using the formula,

Total no of recombinants produced in coupling phase or
repulsive phase/total number of gametes formed in
coupling phase or repulsive phase *100

Each chiasma=50% cross over .if mean chiasma is known
for a chromosome pair ,then the map distance would
be =mean number of chiasma*50

8. Even
if every meiocyte had
a crossing over between two
linked genes only 2 /4
chromatids in a bivalent was
involved in crossing
over(2/4=50 %)

9.  In
which f1 hybrid crossed with a double recessive
parent
 Crossing
over at two points ,so called two point
test cross
 Acac*acac
 F2
hybrid 37 % ACac 37 % aCac of which 26 %
recombinants aCac and ACac.
 26
Cm

10. Underestimate map distance when >10cM
(double crossovers cancel each other)


Provide no info of relative position of two
linked genes

Do not allow detection of double cross over

11.  It
involves three gene
 It
helps to know gene order or gene
sequence
 Two
single cross over and double crossing
over occurs between different genes

15. Genotype
Observed
Type of Gamete
ABC
390
Parental
abc
374
Parental
AbC
27
Single-crossover between
genesC and B
aBc
30
Single-crossover between
genesC and B
ABc
5
Double-crossover
abC
8
Double-crossover
Abc
81
Single-crossover between
genesA and C
aBC
85
Single-crossover between
genesA and C
Total
1000

16.  Determine
 Determine
the parental genotype
the gene order:
The double-crossover gametes are always in
the lowest frequency. From the table
the Abc and abC genotypes are in the lowest
frequency. a double-crossover event moves
the middle allele from one sister chromatid
to the other

17.  ACB
order
 .linkage
distance between AC and CB is
measured as usual two point test cross but here
double crossover is also included
 So
the distance between genes A and C is 17.9
cM [100*((81+85+5+8)/1000)], and the distance
between C and B is 7.0 cM
[100*((27+30+5+8)/1000)].

19.  After
determining the relative distance between
the genes of a linkage group ,it becomes easy to
place genes in linear order
 Eg
:ABC linear order any one could be in middle
 If suppose distance b/n A-B =12,B-C=7, A-C=5
 B-A-C,
in this case distance between BC is not
equitable n so A cannot be in middle
 A-B-C ,in this case distance between AC is not
equitable n so B cannot be in middle
 A-C-B ,this is correct order

20.  Finally
different segments of maps of a
complete chromosome are combined to
form a complete map of 100 cM long for a
chromosome

21.  In
most higher organisms it has been found that
one chiasma formation will reduce the formation of
other in an immediately adjacent region on the
chromosome because of physical inability of
chromosome to bend back upon them within
minimum distance
 This
tendency of one cross over interfering on
other cross over is called interference.
 Thus
interference reduces the double crossover
frequencies than excepted through map distance

22.  The
strength of interference varies in different
segments of the chromosome and is usually
expressed in terms of a coefficient of coincidence
 Coefficient
of coincidence=% observed double cross
over /%expected double cross over
 Coincidence
n interference are inversely
proportional to each other
 Eg
study made in 3 genes of corn ,c (colorless
aleurone), sh( shrunken grains) and wx (waxy
endosperm)

23. regions
genes
Percentage
crossover
Map
distance
1
C-sh
3.4
11
Sh-wx
18.3
3.4+0.1
=3.5
18.3+0.
1=18.4
Double cross
over
C-sh-wx 0.1

25.  Linkage
map distances between genes are not
necessarily proportional to physical linear
measurements
 Special
cytological techniques have been used to
determine location of gene on the chromosome
 Polytene
chromosomes of salivary glands of
Drosophilia were very useful-T-H Painter first
geneticist
 Bridges
did extensive and detailed investigation

26. A
cytological map is a graphic
representation of the location of genes
on a chromosome, based on correlating
the genetic recombination results of
testcrosses with the structural analysis of
chromosomes that have undergone
changes, such as deletions or
translocations, as detected by banding
techniques.

27.  5000
single cross bands-4 pairs of salivary gland
chromosome in D.melanogaster
 Some
genes have been associated with individual
bands

bridges sys of designating parts of chromosomes
with numbers ,sub division with letters and
numbering bands within subdivision had made it
possible for investigators to discuss precise location

28.  The
gene (w) for white eyes is in bands 3C2.
 In
linkage units this gene is located at 1.5 in the
X chromosome
 Linkage
data do not correspond exactly with
cytological locations, but the linear sequence of
genes can be verified from salivary preparations

30.  The
chromosome maps display the exact
location , arrangement and combination of
genes in a linkage group or chromosomes
 They
are useful in predicting results of dihybrid
and trihybrid crosses