linkage and crossing over - jnkvv
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Linkage and
Crossing over
Dhirendra khare Plant Breeding and Genetics
JNKVV, Jabalpur (India)
We consider two traits to test their inheritanceIndividually inheritance of each trait follows 3:1 ratio in F2
Therefore considering both the traits simultaneously should follow 9:3:3:1 ratio
In F2
four classes are formed as formed in case of simple dihybird but
the data did not follow 9:3:3:1 ratioit means
it is not the case of gene interaction(Generally In gene interaction four classes are not formed)
To follow 9:3:3:1 ratio both the genes under study should be located on two different chromosomes
7 3
Location of first gene on chromosome seventh
Second gene on chromosome third
When genes are located on two different chromosomes they follow
law of segregation and
Independent assortmentAs dihybrid case
i.e., study of both the genes simultaneously
A
B
a
b
A
B
Chromosome 7 Chromosome 3 genes under study are located on two different
chromosomes
Chromosome 7 Chromosome 3
Gamete formation
Chromosome 7 Chromosome 3
Law of segregation
Gamete formation
Chromosome 7 Chromosome 3
Law of segregation
Gamete formation
Law of independent assortment
3
A
B
If the genes are located on the same chromosome
a
b
A
B
a
b
A a B b
Genes are located on the same chromosome Considering individual gene separately
Gamete formation
3
A
B
a
b
A a B b
Allele of individual gene segregates as per law of segregation
A a
Genes are located on the same chromosome Considering individual gene separately
F2 for first gene
3
A
B
a
b
Law of segregation
Gamete formation
A a
A a
A AA Aa
a Aa aa
3:1 Law of independent assortment
3
A
B
a
b
Law of segregation
Gamete formation
3:1 Law of independent assortment
B b
B b
B b
B BB Bb
b Bb bb
3:1
A a
A a
A a
A AA Aa
a Aa aa
Genes are located on the same chromosome Considering individual gene separately
F2 for second gene
It shows that when genes are located on different chromosomes or in the same chromosome they follow
Law of Segregation &
Independent assortmentWhen considered individually
B b
Gamete formation
A a
B b
A a
Allele of individual gene segregates but
when we observed both the genes simultaneously then they do not segregate from each other
because both are located on the same chromosome
Consider both the genes present on the same chromosome simultaneously
A
B
a
b
Gamete formation
It shows that A will always go with B and a with b
A
B
a
b
Following combination will not be formed
A a
b B
This type of situation when two alleles of different genes inherited
with each other due to their location on the same
chromosome is known as LINKAGE
It means it will follow law of segregation
but not law of independent assortment
thereforenew combinations in next generation will not formed
A
B
a
b
In this case allele A of first gene and allele B of second gene are considered
linkedSimilarly
allele a of first gene and allele b of second gene are considered
linked
A
B
a
b
Linkage checks the Independent assortment therefore
A is not able to combine with b and
a not with B
Hence in dihybrid study it will not provide 9:3:3:1 ratio
Sometimes even in the presence of linkage four classes are formed
that do not follow simple 9:3:3:1 ratio Why
The new combination i.e., not possible because of linkage appear in F2
Again we consider the same case
These two are the homologous chromosomes that pair at stage of prophase during meiosis division
A
B
a
b
A
B
a
b
A
B
a
b
Homologous chromosomespair at ZYGOTENE stage of prophase
A
B
a
b
A
B
a
b
Four copies of each gene and two copies of each allele are formed during Pachytene stage
A
B
a
b
A
B
a
b
After pairing they form their copy
A
B
a
b
A
B
a
b
Sister chormatid
A
B
a
b
Sister chormatid
A
B
a
b
Now four copies of each gene and two copies of each allele are formed during Pachytene stage
After pairing they form their copy
A
B
a
b
Non sister chormatid
A
B
a
b
During pairing Non sister chromatids cross over each other
It means breakage and reunion of only two of the four strand at any given point on the chromosome
A
B
a
b
A
B
a
b
No cross over take place between sister chromatids
During pairing Non sister chromatids cross over each other
It means breakage and reunion of only two of the four strand at any given point on the chromosome
A
B
a
b
3
A aA
B
a
bbB
A
B
a
b
A a
bB
3
A
B
a
bbB
A
B
a
b
A a
bB
a A
New combination is due to crossing over it is not due to law of independent assortment
The exchange of genetic martial between sister chromatids of homologues chromosomes is known as
CROSSING OVER
It is of two types
Two pint test cross Three point test cross
Two point test crossWhen two genes on the same chromosomes are considered for
crossing over
a b+ +
a b
+ +a
+ b
+
a b+ +
+ +
a b
Three point test crossWhen three genes on the same chromosomes are considered for
crossing over
a b c
+ + +
Three point test crossWhen three genes on the same chromosomes are considered for
crossing over
a b c
+ + +
a b c
+ + +
a b c
+ + +
Four strand stage
a b c
+ + +
a b c
+ + +
Parental type
Parental type
Chromatid first and four are exactly parental type No crossing over takes place in these chromatids
therefore producing parental type
a b c
+ + +Non sister chromatids
Crossing over may take place between non sister chromatids only and they may produce other than parental type combination
a b c
+ + +
a b c
+ + +
In a three point test cross two types of crossing over may take place
Single Cross Over
Double Cross Over
+ + +
Non sister chromatids
a b c
Crossing over between any two genes in a chromosome
Single Cross Over
+ + +
a b c
+ + +
a b c
+ b c
a + +
+ + +
a b c
+ + c
a b +SCO I
SCO II
Non sister chromatids
Crossing over among three genes in a chromosome
Double Cross Over
+ + +
a b c
+ + +
a b c
+ + +
a b c
+ b +
a +c
In a three point test cross following combinations may appear
a b c
+ + +
In a three point test cross following combinations may appear
a b c
+ + +
a b c
+ + +
Parental type
In a three point test cross following combinations may appear
a b c
+ + +
a b c
+ + +
+ b c
a + +
Parental type
SCO I
In a three point test cross following combinations may appear
a b c
+ + +
a b c
+ + +
+ b c
a + +
+ + c
a b +
Parental type
SCO I
SCO II
In a three point test cross following combinations may appear
a b c
+ + +
+ b +
a +c
a b c
+ + +
+ b c
a + +
+ + c
a b +
Parental type
SCO I
SCO II
DCO II
In a three point test cross following combinations may appear
+ b +
a +c
a b c
+ + +
+ b c
a + +
+ + c
a b +
Parental type
SCO ISCO I
SCO II
DCO II
Ranking based on frequencyFirst (Highest)
SecondThird
FourthFifth (Lowest)
the two combinations with maximum number of frequency are considered as
parents
the two combinations with minimum number of frequency are considered as Double Cross Over
the two combinations between maximum and minimum number of frequency are considered as
Single Cross Over
In a population
In a three point test cross following combinations may appear
+ b +
a +c
a b c
+ + +
+ b c
a + +
+ + c
a b +
Parental type
SCO ISCO I
SCO II
DCO II
Ranking based on frequency
Class
First (Highest) ParentSecond Parent
Third SCOFourth SCOFifth DCO
Sixth (Lowest) DCO
Gene Order
The gene are positioned in a linear order on a chromosomeSequential order of genes under study on the chromosome is
known as gene order
It is determined by making double cross over in DCOthe produce should be same as parent
& the distance between the genes under consideration
In a three point test cross gene order may be
a b c
a c b
b a c
b c a
c a b
c b a
n(n-1)3 (2)=6
It is the determination of relative distance between genes
The unit of map distance is cetimorgan = one unit of map distanceIt is equivalent to 1% crossing over
Map distance is used in predicting the probability of crossing over between genes
Map distance
Coefficient of coincidence
An experimental value equal to the observed number of double cross over divided by the expected double cross over
Coefficient of coincidence =
% expected double cross over
CoincidenceChances of happening both the events at a time In this case it means single cross over between both the combination at a time i.e., double cross over
% observed double cross over
Interference
Crossing over at one place in a chromosome may change the actual probability of another crossing
over at the adjacent region This change is known as interference
It is of two types
Positive interference The interaction between cross over is such that the occurrence of one cross over reduces the chance of cross over of another.In this case coefficient of coincidence is less than 1
Negative interference The interaction between cross over is such that the occurrence of one cross over enhances the chance of cross over of another.In this case coefficient of coincidence is greater than 1
Coincidence is the complement of interference
Coincidence + interference = 1
When interference is complete (1.0), no double cross over will be observed and coincidence becomes zero
When all the expected cross over occur actually than interference becomes zero
Problems on Linkage
In maize following allelic pairs have been identified on chromosome number 3.+/b= plant color booster vs. non booster+/lg+ Liguled vs. ligule-less+/v = Green plant vs. Virescent ( turning green)A tri-hybrid test cross produces following data
+ v lg 305 b v + 66b + lg 128 + + + 22v b lg 18 + v + 112+ + lg 74 b + + 275
DetermineA. Gene OrderB. Map distanceC. Coefficient of coincidenceD. InterferenceE. Gene map
Parents The off springs with maximum
frequency are considered as parents The maximum frequency is of
+ v lg 305 b v + 66b + lg 128 + + + 22v b lg 18 + v + 112+ + lg 74 b + + 275
Parents The off springs with maximum
frequency are considered as parents The maximum frequency is of
+ v lg
b + +
= 305
= 275
+ v lg 305 b v + 66b + lg 128 + + + 22v b lg 18 + v + 112+ + lg 74 b + + 275
Double Cross Over (DCO)The off springs with minimum frequency are considered as DCO
The minimum frequency is of
+ v lg 305 b v + 66b + lg 128 + + + 22v b lg 18 + v + 112+ + lg 74 b + + 275
Double Cross Over (DCO)The off springs with minimum frequency are considered as DCO
The minimum frequency is of
+ + +
v b lg
= 22
= 18
+ v lg 305 b v + 66b + lg 128 + + + 22v b lg 18 + v + 112+ + lg 74 b + + 275
Double Cross Over (DCO)Application of double cross over in selected DCO
should form offspring of parental type
+ + +
v b lg
Double Cross Over (DCO)Application of double cross over in selected DCO
should form offspring of parental type
+ b +
v + lg
+ + +
v b lg
DOC type Parental type
+ b +
v + lg
+ + +
v b lg
DOC type Parental type
+ v lg 305
b + + 275
Remaining Off springs are considered as SCO
+ v lg 305 b v + 66b + lg 128 + + + 22v b lg 18 + v + 112+ + lg 74 b + + 275
Single Cross Over (SCO)SCO ISCO II
b v +
+ + lg
= 66
= 74
b + lg
+ v +
= 128
= 112
Correct gene sequence
The correct gene sequence means the sequence of genes on parents should be such that after double cross over it produces
correct DCO
b + +
+ v lg
Parental type DCO type
v b lg
+ + +
b + +
+ v lg
Parental type DCO type
v b lg
+ + +
Correct DCO
b + +
+ v lg
Parental type
+ + lg
b v +
DCO type
v b lg
+ + +
Correct DCO
b + +
+ v lg
Parental type
+ + lg
b v +
DCO type
v b lg
+ + +
These two are not correct DCO type therefore
sequence of gene on parental type has to be changed
Correct DCO
+ b +
v + lg
Parental type DCO type
v b lg
+ + +
Correct DCO
+ b +
v + lg
Parental type DCO type
v b lg
+ + +
Correct DCO
+ b +
v + lg
Parental type
v b lg
+ + +
DCO type
v b lg
+ + +
This is the correct DCO type therefore
it is the correct gene sequence
Correct DCO
+ b +
v + lg
Correct gene sequence of the Parental type
v b lg
+ + +
DCO type
+ b +
v + lg
22
18
128305
Correct gene sequence of the Parental type
v b +
+ + lg
Single Cross Over I
+ b +
v + lg
74
66
+ b +
v + lg
128305
Correct gene sequence of the Parental type
v + +
+ b lg
Single Cross Over II
+ b +
v + lg
128
112
+ b +
v + lg
128305
Map distanceFor this calculate percentage of parental type, DCO and SCO
Frequency Total PercentageParental type
DCO
SCO I (v-b)
SCO II(b-lg)Total
Map distance
Frequency Total PercentageParental type 275
305
DCO
SCO I (v-b)
SCO II(b-lg)Total
Map distance
Frequency Total PercentageParental type 275
305
DCO 2218
SCO I (v-b)
SCO II(b-lg)Total
Map distance
Frequency Total PercentageParental type 275
305
DCO 2218
SCO I (v-b)
7466
SCO II(b-lg)Total
Map distance
Frequency Total PercentageParental type 275
305
DCO 2218
SCO I (v-b)
7466
SCO II(b-lg)
128112
Total
Map distance
Frequency Total PercentageParental type 275
305
DCO 2218
SCO I (v-b)
7466
SCO II(b-lg)
128112
Total 1000
Map distance
Frequency Total PercentageParental type 275 580
305
DCO 2218
SCO I (v-b)
7466
SCO II(b-lg)
128112
Total 1000
Map distance
Frequency Total PercentageParental type 275 580
305
DCO 22 4018
SCO I (v-b)
7466
SCO II(b-lg)
128112
Total 1000
Map distance
Frequency Total PercentageParental type 275 580
305
DCO 22 4018
SCO I (v-b)
74 14066
SCO II(b-lg)
128112
Total 1000
Map distance
Frequency Total PercentageParental type 275 580
305
DCO 22 4018
SCO I (v-b)
74 14066
SCO II(b-lg)
128 240112
Total 1000
Map distance
Frequency Total PercentageParental type 275 580
305
DCO 22 4018
SCO I (v-b)
74 14066
SCO II(b-lg)
128 240112
Total 1000 1000
Map distance
Frequency Total PercentageParental type 275 580 58%
305
DCO 22 4018
SCO I (v-b)
74 14066
SCO II(b-lg)
128 240112
Total 1000 1000
Map distance
Frequency Total PercentageParental type 275 580 58%
305
DCO 22 40 4%18
SCO I (v-b)
74 14066
SCO II(b-lg)
128 240112
Total 1000 1000
Map distance
Frequency Total PercentageParental type 275 580 58%
305
DCO 22 40 4%18
SCO I (v-b)
74 140 14%66
SCO II(b-lg)
128 240112
Total 1000 1000
Map distance
Frequency Total PercentageParental type 275 580 58%
305
DCO 22 40 4%18
SCO I (v-b)
74 140 14%66
SCO II(b-lg)
128 240 24%112
Total 1000 1000
Map distance
Frequency Total PercentageParental type 275 580 58%
305
DCO 22 40 4%18
SCO I (v-b)
74 140 14%66
SCO II(b-lg)
128 240 24%112
Total 1000 1000 100
Distance between genes
v b lg
= % SCO I + DCO %= 14 + 4= 18%
Distance between gene v-b
Distance between genes
v b lg
= % SCO I + DCO %= 14 + 4= 18%
Distance between gene v-b
= % SCO II + DCO %= 24 + 4= 28%
Distance between gene b-lg
Distance between genes
v b lg
= % SCO I + DCO %= 14 + 4= 18%
Distance between gene v-b
= % SCO II + DCO %= 24 + 4= 28%
Distance between gene b-lg
Distance between gene v-lg
Distance between gene v-b
18
Distance between gene b-lg
28
+ = 46 map unit
Coefficient of coincidence
Percent observed DCO frequencyPercent expected DCO frequency=
Expected DCO Frequency at SCO I X Expected DCO Frequency at SCO II
SCO I + DCOExpected DCO Frequency at SCO I =
100
14+4 18 = = = 0.18%
100 100
SCOII + DCOExpected DCO Frequency at SCO II =
100
24+4 28 = = = 0.28%
100 100
Percent expected DCO frequency =
Percent observed DCO frequencyPercent expected DCO frequency
Coefficient of coincidence
Percent observed DCO frequency = 0.04%
Percent expected DCO frequency = 0.18X 0.28
0.04Coefficient of coincidence = = 0.7936
0.18X 0.28
Coefficient of coincidence = 0.7936 X 100 = 79.36 %
Interference
= 1- Coefficient of coincidence= 1- 0.7936= 0.2064= 20.64%
Gene map
v b lg
Gene map
v b lg
46map unit
Gene map
Gene map
v b lg
46map unit
18 map unit
Gene map
Gene map
v b lg
46map unit
18 map unit 28 map unit
Gene map
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