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Lecture 3
Chromosome Map-1
(Linkage map, Genetic map or Crossing-over map)
MS Semester 1 2022
Genetics & Mol. Biol.
Course: ZoolM131
Prepared and presented by
Prof. Dr. M. S. Islam
Department of Zoology, RU
25 September 2023
Chromosome Map-1
Contents
Definition
How is the chromosome map constructed ?
Principles underlying chromosome mapping
Construction of a chromosome map in Drosophila
 Determination of the sequence of genes
 Calculation of the map distance
 Layout of the chromosome map
 Calculation of the co-efficient of coincidence (k) and
interference (I)
References
Definition
Chromosome map is a line on which the position of genes and their
sequences
are shown by points, separated by the distances proportional to the
percentage of crossing-over.
The first chromosome map in Drosophila melanogaster was prepared by A.
H.
Sturtevant in 1911.
Chromosome mapping in this fly was further developed by T. H. Morgan,
C. B.
Bridges and their colleagues in the USA.
Chromosome maps of Drosophila melanogaster
Chromosome maps of the major genes in D.
melanogaster
Chromosome maps of some genes on X and Y
chromosomes in man
Chromosome maps of some of the genes associated with sex-linked disorders in
man
1. A chromosome map is constructed entirely from the results of breeding
experiment.
2. The most commonly used method is the trihybrid or three-point test
cross.
P: AABBCC × aabbcc
F1: AaBbCc
Test cross: AaBbCc × aabbcc
F2: ABC/abc
ABc/abc
AbC/abc
Abc/abc
aBC/abc
aBc/abc
abC/abc
abc/abc
How is the chromosome map constructed?
Principles underlying the construction of
chromosome map
There are four basic assumptions that are made during the
construction of
a chromosome map in D. melanogaster:
1. The genes are arranged in a linear order on the chromosomes;
2. The percentage of crossing-over between the genes is an index
to their distance;
3. Parentals or non cross-over types are the highest number of
progenies in F2 generation; and
4. The double cross-overs represent the lowest number of progenies
in the F2.
Construction of a chromosome map
in D. melanogaster
A Drosophila melanogaster male having three genes for dominant characters
(ABC)
was crossed with a female having three recessive characters (abc). The F1
heterozygotes were test-crossed and the following F2 progenies were recorded:
1. aBc = 2
2. abc = 904
3. abC = 34
4. aBC = 76
5. ABC = 860
6. AbC = 2
7. Abc = 90
8. ABc = 32
Determine the sequence of the genes on the chromosome; and construct a
chromosome map from the breeding experiment results (F2 data) shown above.
Construction of a chromosome map
in D. melanogaster
According to the given problem, the F2 progenies may be arranged as to
their
phenotypes as follows:
Phenotypes No. of Progenies in F2 % F2 Progenies
Parentals or non-
cross-overs
ABC + abc
860 + 904=1764
88.20
Crossing-over
region I (COR-I)
Abc + aBC
90 + 76= 166
8.30
Crossing-over
region II (COR-II)
ABc + abC
32 + 34= 66
3.30
Double cross-
overs (DCO)
AbC + aBc
2 + 2 = 4
0.20
Total 2000 100.00
Determination of the sequence of genes
on the chromosome
The most common practice for determining the sequence of the three genes on the
chromosome is to assume one gene after another as follows:
Let us assume first that, A gene be located in the middle. If so, the double cross-
over progenies would have been: aBC and Abc, which are 76 and 90,
respectively in numbers among the F2 progenies. Since the numbers do not
correspond to the lowest numbers, obviously the assumption could not be
correct.
Let then, B gene be in the middle. In this situation, the double cross-over
progenies would have been: AbC and aBc, which are 2 and 2, respective. This
assumption might be correct because the numbers of progenies are the least
number in F2.
Finally, let C gene be in the middle. If so, the double cross-over progenies would
have been ABc and abC, which are 32 and 34, respectively. Since these do not
conform to the lowest number of progenies among F2, this assumption also
appears to be not correct.
So, the sequence of the genes on the chromosome is ABC or CBA, depending on
the actual position of the centromere of the chromosome.
Calculation of map distance
According to the basic assumptions made during the construction of
chromosome map-
The actual distance between A and B genes= % crossing-over region I
(COR-I)
+ % double cross-overs (DCO) = 8.30 + 0.20 = 8. 50 map units or 8.50
Centimorgans.
Again, the actual distance between B and C genes = % crossing-over
region II
(COR-I) + % double cross-overs (DCO) = 3.30 + 0.20 = 3. 50 map units or
3.50 Centimorgans.
So, the total distance between A and C genes = 8.50 + 3.50 = 12.0 map
units or 12.0 Centimorgans.
Layout of the ABC genes on the chromosome
The chromosome map of three genes under consideration is shown
below:
A.........................8.5..................B......3.5........C
a b c
→ → 12.0 ← ←
The distance between A and B genes is 8.5 map units, and that between
B
and C genes is 3.5 map units. Therefore, the total distance between A
and
C genes is 12.0 map units.
Calculation of interference (I) and
coefficient of coincidence (k)
Interference (I)
The tendency of a crossing-over is to prevent the occurrence of another
crossing-over near its vicinity. This phenomenon is known as interference
(I),
which could be shown by the equation, I = 1-k, where k is the coefficient of
coincidence.
Coefficient of coincidence (k)
The strength of interference is measured by the coefficient of coincidence
(k),
which is given by the formula:
k = Actual frequency of the double cross-overs (DCO) ÷ Calculated
frequency of DCO
In the given problem, we have,
Actual frequency of the double cross-overs (DCO) = 4 ÷ 2000 = 0.002; and
Calculated frequency of DCO = 166 + 4 ÷ 2000 × 66 + 4 ÷ 2000
= 0.085 × 0.035
= 0.003
So, k = 0.002 ÷ 0.003 = 0.67 or 67%; and I = 1- 0.67 = 0.33 or 33%.
References
• Ayala FJ. & Kiger Jr. JA. 1980. Modern Genetics.
• Burns, GW. 1980. The Science of Genetics.
• Gardner et al. 1991. Principles of Genetics (8th edn).
• Islam, MS 2018. Selected Lectures on Genetics.
• Islam MS et al. 2017. Genetics: The Science of Heredity and
Variation.
• Sinnott et al. 1973. Principles of Genetics (5th edn).
• Stansfield, WD. 1991. Theory and Problems of Genetics (3rd edn).
• Strickberger, MW. 1976. Introduction to Genetics.
• Winchester, AM. 1966. Genetics.
• Wikipedia. 2023. www.wikipedia.com.
-***-
3. Chromosome Mapping in Drosophila flies

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3. Chromosome Mapping in Drosophila flies

  • 1. Lecture 3 Chromosome Map-1 (Linkage map, Genetic map or Crossing-over map) MS Semester 1 2022 Genetics & Mol. Biol. Course: ZoolM131 Prepared and presented by Prof. Dr. M. S. Islam Department of Zoology, RU 25 September 2023
  • 2. Chromosome Map-1 Contents Definition How is the chromosome map constructed ? Principles underlying chromosome mapping Construction of a chromosome map in Drosophila  Determination of the sequence of genes  Calculation of the map distance  Layout of the chromosome map  Calculation of the co-efficient of coincidence (k) and interference (I) References
  • 3. Definition Chromosome map is a line on which the position of genes and their sequences are shown by points, separated by the distances proportional to the percentage of crossing-over. The first chromosome map in Drosophila melanogaster was prepared by A. H. Sturtevant in 1911. Chromosome mapping in this fly was further developed by T. H. Morgan, C. B. Bridges and their colleagues in the USA.
  • 4. Chromosome maps of Drosophila melanogaster Chromosome maps of the major genes in D. melanogaster
  • 5. Chromosome maps of some genes on X and Y chromosomes in man Chromosome maps of some of the genes associated with sex-linked disorders in man
  • 6. 1. A chromosome map is constructed entirely from the results of breeding experiment. 2. The most commonly used method is the trihybrid or three-point test cross. P: AABBCC × aabbcc F1: AaBbCc Test cross: AaBbCc × aabbcc F2: ABC/abc ABc/abc AbC/abc Abc/abc aBC/abc aBc/abc abC/abc abc/abc How is the chromosome map constructed?
  • 7. Principles underlying the construction of chromosome map There are four basic assumptions that are made during the construction of a chromosome map in D. melanogaster: 1. The genes are arranged in a linear order on the chromosomes; 2. The percentage of crossing-over between the genes is an index to their distance; 3. Parentals or non cross-over types are the highest number of progenies in F2 generation; and 4. The double cross-overs represent the lowest number of progenies in the F2.
  • 8. Construction of a chromosome map in D. melanogaster A Drosophila melanogaster male having three genes for dominant characters (ABC) was crossed with a female having three recessive characters (abc). The F1 heterozygotes were test-crossed and the following F2 progenies were recorded: 1. aBc = 2 2. abc = 904 3. abC = 34 4. aBC = 76 5. ABC = 860 6. AbC = 2 7. Abc = 90 8. ABc = 32 Determine the sequence of the genes on the chromosome; and construct a chromosome map from the breeding experiment results (F2 data) shown above.
  • 9. Construction of a chromosome map in D. melanogaster According to the given problem, the F2 progenies may be arranged as to their phenotypes as follows: Phenotypes No. of Progenies in F2 % F2 Progenies Parentals or non- cross-overs ABC + abc 860 + 904=1764 88.20 Crossing-over region I (COR-I) Abc + aBC 90 + 76= 166 8.30 Crossing-over region II (COR-II) ABc + abC 32 + 34= 66 3.30 Double cross- overs (DCO) AbC + aBc 2 + 2 = 4 0.20 Total 2000 100.00
  • 10. Determination of the sequence of genes on the chromosome The most common practice for determining the sequence of the three genes on the chromosome is to assume one gene after another as follows: Let us assume first that, A gene be located in the middle. If so, the double cross- over progenies would have been: aBC and Abc, which are 76 and 90, respectively in numbers among the F2 progenies. Since the numbers do not correspond to the lowest numbers, obviously the assumption could not be correct. Let then, B gene be in the middle. In this situation, the double cross-over progenies would have been: AbC and aBc, which are 2 and 2, respective. This assumption might be correct because the numbers of progenies are the least number in F2. Finally, let C gene be in the middle. If so, the double cross-over progenies would have been ABc and abC, which are 32 and 34, respectively. Since these do not conform to the lowest number of progenies among F2, this assumption also appears to be not correct. So, the sequence of the genes on the chromosome is ABC or CBA, depending on the actual position of the centromere of the chromosome.
  • 11. Calculation of map distance According to the basic assumptions made during the construction of chromosome map- The actual distance between A and B genes= % crossing-over region I (COR-I) + % double cross-overs (DCO) = 8.30 + 0.20 = 8. 50 map units or 8.50 Centimorgans. Again, the actual distance between B and C genes = % crossing-over region II (COR-I) + % double cross-overs (DCO) = 3.30 + 0.20 = 3. 50 map units or 3.50 Centimorgans. So, the total distance between A and C genes = 8.50 + 3.50 = 12.0 map units or 12.0 Centimorgans.
  • 12. Layout of the ABC genes on the chromosome The chromosome map of three genes under consideration is shown below: A.........................8.5..................B......3.5........C a b c → → 12.0 ← ← The distance between A and B genes is 8.5 map units, and that between B and C genes is 3.5 map units. Therefore, the total distance between A and C genes is 12.0 map units.
  • 13. Calculation of interference (I) and coefficient of coincidence (k) Interference (I) The tendency of a crossing-over is to prevent the occurrence of another crossing-over near its vicinity. This phenomenon is known as interference (I), which could be shown by the equation, I = 1-k, where k is the coefficient of coincidence. Coefficient of coincidence (k) The strength of interference is measured by the coefficient of coincidence (k), which is given by the formula: k = Actual frequency of the double cross-overs (DCO) ÷ Calculated frequency of DCO In the given problem, we have, Actual frequency of the double cross-overs (DCO) = 4 ÷ 2000 = 0.002; and Calculated frequency of DCO = 166 + 4 ÷ 2000 × 66 + 4 ÷ 2000 = 0.085 × 0.035 = 0.003 So, k = 0.002 ÷ 0.003 = 0.67 or 67%; and I = 1- 0.67 = 0.33 or 33%.
  • 14. References • Ayala FJ. & Kiger Jr. JA. 1980. Modern Genetics. • Burns, GW. 1980. The Science of Genetics. • Gardner et al. 1991. Principles of Genetics (8th edn). • Islam, MS 2018. Selected Lectures on Genetics. • Islam MS et al. 2017. Genetics: The Science of Heredity and Variation. • Sinnott et al. 1973. Principles of Genetics (5th edn). • Stansfield, WD. 1991. Theory and Problems of Genetics (3rd edn). • Strickberger, MW. 1976. Introduction to Genetics. • Winchester, AM. 1966. Genetics. • Wikipedia. 2023. www.wikipedia.com. -***-