Calculation of recombination from F2 data.

 

•It is not always possible to calculate linkage using  testcross data. 

For example, the snail Biomphalaria glabrata is of great interest since it is a primary vector for the parasitic disease schistosomiasis in many tropical countries.  This snail is a functional hermaphrodite and can donate sperm (act as a male), receive sperm (act as a female) or internally fertilize its own eggs. 

 

•
•Although it is possible to detect the offspring of an F1 cross, it is nearly impossible to tell backcross progeny from internally crossed progeny (the ultimate F2).  It is simpler to map recombination distances using only F2 data.  This is also often true of the nematode C. elegans.
•
 
Example: Consider two new mutants in Drosophila, green eye (g) and feather (f) aristae.  These are both autosomal and recessive to wildtype red eye and normal arista.  The following crosses were made:

 

P green eye, feather arista x red eye; normal arista
 
F1 all red eye, normal arista
 
F2 Phenotype        Genotype                                Number
---------------------------------------------------------------------------------------------
red eye; normal arista  G_F_ 1044
red eye, feather                                        G_ff 144
green eye, normal                                    gF_   151
green eye, feather                                    ggff 261

The table below summarizes the Chi-square goodness of fit tests for the joint F2 cross, for segregation at the eye color locus, for segregation at the aristal shape locus and for independence of the eye color and shape loci. The tests all suggest that the deviation from the 9:3:3:1 ratio is due to linkage. For description of these tests Click here.

 

 
   
c2
Reject/Do not Reject
1.
Fits 9:3:3:1 ratio
480.82
Reject
2.
Fits 3:1 ratio eyes
.44 with 1 df
Do not Reject
3.
Fits 3:1 ratio arista
.067 with 1 df
Do not Reject
4.
Independence
421.94 with 1 df
 
Reject
 

The first step is to decide if the genes are in coupling and repulsion.

 
Since I set up  the parental cross as (GF/GF x gf/gf)  i already know that the parental gametes are in coupling
(gf , GF).

 

Hint:
•If the parental cross is unknown, you can always tell if the system is in coupling or repulsion by looking at the frequency of the double homozygote recessive (aabb).

 

•
•If the system is in independent assortment the aabb genotype (ab phenotype) has a frequency of 1/16. 
•

If the system is in coupling,  the aabb genotype (ab phenotype) will have a frequency greater than 1/16;

if the system is in repulsion the aabb genotype (ab phenotype) will have a frequency less than 1/16.

 
 

The phenotypic frequencies must be summarized as shown below

  G g
F GF (a) gF(c)
f Gf   (b) gf (d)

 

  G g
F a = 1044 c= 151
f b =   144 d = 262
For repulsion, calculate the value of ad/bc and for coupling, calculate the value bc/ad.

In our example the genes are in coupling so :

 

 
        bc 144 * 151
    ---------         = ------------------  =  .080
        ad 1044 * 261
 
 
Look now at the table provided.  The value of .08 for coupling places the recombination value between .20 and .205. Thus, the map distance between feather and green eyes is about 20 map units.
 
For Repulsion the Calculation is:
 
       ad  
    ---------         
       bc
•
 
 

Quick Links F2 Cross
1 Basic F2 cross - autosomal genes
2 Basic  F2 cross- X-linked gene
       Expected Phenotypes of F2 Cross - Single Autosomal Traits
        Expected Phenotypes of F2 Cross - X-linked gene
4 F2 cross - two unlinked autosomal genes
5 F2 cross - one autosomal - one X-linked gene
6 Linkage - Coupling and Repulsion
7  F2 cross - two linked autosomal genes -coupling
8  F2 cross - two linked autosomal genes - repulsion
9 Measuring the recombination rate from F2 data
          Recombination from F2 data - Table of Values
   
10 Basic  F2 cross- Z-linked gene (inactive)
12 Expected Phenotypes of F2 Cross - Z-linked gene (inactive)
13 F2 cross - one autosomal - one Z-linked gene (inactive)
 

Also See
  Detection of Linkage - Basic Info - Two Autosomal Traits - Complete Dominance
  Detection of Linkage -  More complex examples
  Detection of Linkage -  One More Example
  Detection of Linkage -  Two Autosomal Traits - Heterozygote has Unique Phenotype
   
  Home page
  Detailed List for Mendelian Genetics