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Construction of Linkage Map or Genetic Mapping |
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Construction of Linkage Map or Genetic Mapping
The method of construction of genetic maps of different chromosomes is called genetic mapping. The genetic mapping includes following processes:
1. Determination of Linkage Groups
Before starting the genetic mapping of the chromosomes of a species, one has to know the exact number of chromosomes of that species and then, he has to determine the total number of genes of that species by undergoing hybridization experiments in between wild and mutant strains. By the same hybridization techniques, it can also be easily determined that how many phenotypic traits remain always together or linked and consequently their determiners or genes during the course of inheritance. And thus, the different linkage groups of species can be worked out.
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2. Determination of Map Distance
After knowing total number of genes in each linkage group of a species, the relative distances between each linked gene have to be determined. The distance between two given genes is calculated according to the percentage of crossing over, because, cross over frequency is directly proportional to distance between the genes. For example, if the percentage of crossing over between two linked genes is 1 per cent it means that the map distance between two linked genes is one unit of map distance, which is known as map unit, morgan or centimorgan.
Examples
1. If a F1 hybrid having the genotype Ab/aB produces 8% of cross over gametes AB and ab, then the distance between A and B is estimated to be 16 map units or centimorgan.
2. If the map distance between the gene loci B and C is 12 centimorgan, then 12% of gametes of genotype BC/bc should be cross over types, i.e., 6% Bc and 6%bC.
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Because, each chiasma products 50% crossover products, 50% crossing over is equivalent to 50 map units or centimorgans. If the mean number of chiasmata is known for a chromosome pair, the total length of the map for that linkage group may be predicted:
Total length = mean number of chiasmata X 50
Two Point Test Cross
The percentage of crossing over between two linked genes is calculated by test crosses in which a F1 dihybrid is crossed with a double recessive parent. Such crosses because involve crossing over at two points, so called two point test crosses. For example, a dihybrid having the genotype AC/ac is test crossed with a double recessive parent (ac/ac), then among F2 test cross hybrids we may get 37% dominant genes at both gene loci (AC/ac), 37% recessive genes at both gene loci (ac/ac), 13% dominant gene at first gene locus and recessive at the second gene locus (aC/ac), and 13% recessive gene at first gene locus and dominant gene at second gene locus (aC/ac).
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The last two groups (i.e., 13% Ac/ac and 13% aC/ac) were produced by crossover gametes. (13 + 13) from the dihybrid parent. Thus, 26% of all gametes (13+ 13) were of cross over types and the distance between the loci A and C is estimated to be 26 centimorgans. Because, double crossovers usually do not occur between genes less than 5 centimorgans apart, so for genes further apart, the three point rest crosses are used.
Three Point Test Cross
A three point test cross or trihybrid test cross (involving three genes) gives us information regarding relative distances between these genes, and also shows us the linear order in which these genes should be present on chromosome. Such a three point test cross may be carried out if three points Or gene loci on chromosome pair can be identified by marker genes. If, in addition to genes A and C indicated above, a third marker gene B is located in fairly closed proximity in the same linkage group, all three markers may be used together in conducting a more precise analysis of the map distance and the relative position of three points.
Suppose that we testcross trihybrid individuals of genotype ABC/abc and find in the progeny the following:
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| 36% ABC/abc |
9% Abc/abc |
4% ABc/abc |
1% AbC/abc |
| 36% abc/abc |
9% aBC/abc |
4% abC/abc |
1% aBc/abc |
| ---------------------------------------------------------- |
72% Parental:
type |
18% Single: crossovers between A and B. (region I) |
8% Single: crossovers between B and C. (region II) |
2% Double crossover |
To find the distance A-B we must count all crossovers (both singles and doubles) that occurred in region 1= 18% + 2% or= 20% or 20 map units between the loci A and B. To find the distance B- C we must again count all crossovers (both singles and doubles) that occurred in region 11=8%+2%= 10% or 10 map units between the loci Band C.
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The A--C distance is therefore 30 map units when double crossovers are detected in a three point linkage experiment and 26 map units when double crossovers are undetected in the two-point linkage experiment above.
Without the middle marker (B), double crossovers would appear as parental types and hence we underestimate the true map distance (crossover percentage). In this case the 2% double cross overs would appear with the 72% parental types, making a total of 74% parental types and 26% recombinant types.Therefore, for any three linked genes whose distances are known, the amount of detectable crossovers between the two outer markers A and C when the middle marker B is missing; (A-B crossover percentage) plus (B-C crossover percentage) minus (2xdouble crossover percentage).
3. Determination of Gene Order
After determining the relative distances between the genes of a linkage group, it becomes easy to place genes in their proper linear order. For example, if the linear order of three genes ABC is to be determined, then these three genes may be in anyone of three different orders depending upon that which gene is in the middle. For the time being we may ignore left and right end alternatives. If double crossovers do not occur, map distances may be treated as completely addititive units. Now, if we suppose that the distance between the gene A -B=12, B -C=7 and A-C=5, we can determine the order of genes correctly in the following manner.
Case I. Let us assume that gene A is the middle (e.g., B-A-C):
 In this case, because the distances between B-C are not equitable, therefore, gene B cannot be in the middle.
Case II. Let us assume that gene B is in the middle (e.g., A-B-C):
In this case, because the distances between A-C are not equitable, therefore, gene B cannot be in the middle.
Case III.
Let us assume that gene C is in the middle (e.g., A-C-B):
In this case, because the distances between A-B are equitable, therefore, gene C must be in the middle.
Thus, the relative distances and ordering of genes in a linkage group are determined in separate segments by two point test crosses or three point test crosses, as the case may be.
4. Combining Map Segments
Finally, the different segments of maps of a complete chromosome are combined to form a complete genetic map of 100 centimorgans long for a chromosome.
Example: For example, suppose we have to combine following three map segments.
We can superimpose each of these segments by aligning the genes shared in common.
Then finally we may combine the three segments into one map=
The a to d distance = (d to b)-(a to b)=22-8= 14
The a to e distance= (a to d)-(d to e)=14-2=12
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