Three point mapping II Gene Order II Gene Distance II Genetics Problem Linkage

Let us learn how to solve three point linkage cross using 4 simple steps
With 3 genes There are 3 possible cross over type or recombination that can take place
Lets take the example of two parent type ABC and abc, A single crossing over can happen between genes A and B, or between genes B and C, or double cross over can happen between genes AB and BC.
The recombinants by a single cross over between A and B will be Abc and aBC
Similarly, a single crossover between genes B and C will give Abc and abC recombinants
A double crossover between genes AB and BC will give
Similarly in the second condition, the recombinants will be
In every question we have 8 types of Gametes, two Parental, two SCO between genes 1-2, 2 sCO between genes 2-3 and 2 DCO.
It is very easy to identify the parental and DCO gametes in most of the questions
The gametes showing the highest number of progenies are parental types, and the gametes showing the lowest number of progenies are the DCO types.
Let's look at this question where we need to find the distance between the genes and we provided with the genotype and number of observed progeny.
To solve this we must know the type of gamete
Our first step is to find the parental and DCO types. As we know the highest number of progenies are parental types and the lowest number of progenies come from DCO
The question mentions the genes as ABC, but the actual gene order can be different
To find the gene order, we will compare the parental and the DCO type
On comparison, we notice that C gene does not match in the parental and DCO type, thus C gene is present in the center of gene A and B
The third step is to determine SCO between genes. Considering this gene order ACB, we will do a single cross of the two parental type between genes A and C
The recombinants obtained by this cross will be Acb and aCB
Now we need to compare this recombinant with the table in the question.
If we rearrange this, we have Capital A and small bc and small a and capital BC recombinants which are our single crossovers between genes A-C
So naturally, the remaining two genotypes are SCO between genes C and B. You can do a crossover yourself and check.
The 4th step is to determine the distance between the genes. To determine the distance between genes A and C we need to add the sum of SCO between AC and DCO progenies, divide by total progenies * 100
Which is 17.9 cM apart
Similarly, the distance between genes C and B will be the sum of the SCO between genes CB and DCO progenies divided by total progenies *100 which is 7cM
The distance from A-C is 17 cM, and the distance between C-B is 7 cM
For some problems like this one, we must use an alternative approach. Since the values are not letting us decide the parental and DCO types we cannot follow our 4 steps here
When crossing over occurred between two genes, the two gene forms will not be together in the recombinants
Like crossing over occurred between capital A and capital B, but in the recombinants, Capital A and Capital B will not be together.
Similarly, when crossing over occurred between capital B and Capital C, the recombinants will not have both capital forms together
Capital A and small a are two forms of the gene
Similarly, in this problem + and or no plus are two forms.
So another way to identify Recombinants is by finding combinations where the two forms are not together.
Like in p q r and P+Q+R+…both are in same form so we won't consider these
Here p and r are in different forms
Similarly, we select all combinations where they p and r are in different forms




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