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 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


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


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
Determination
of linkage group
Determination
of map distance
Determination
of gene order
 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


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
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 %)
 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
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
 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
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
 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
 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)].
 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
 Finally

different segments of maps of a
complete chromosome are combined to
form a complete map of 100 cM long for a
chromosome
 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
 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)
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
 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
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.
 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
 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
 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
Chromosomal mappin highlighted

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Chromosomal mappin highlighted

  • 1.
  • 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
  • 12.
  • 13.
  • 14.
  • 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)].
  • 18.
  • 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)
  • 24.
  • 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
  • 29.
  • 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

Hinweis der Redaktion

  1. If yellow were in the middle, yellow phenotype would show up in DCO.If echinus were in the middle, echinus phenotype would show up in DCO.white is actually in the middle since white phenotype shows up in actual DCO data