Any Phenotypic Pattern |
We have already discussed the 2 X 2 Contingency Table. The 2 x 2 table and its associated formula only works for two autosomal genes with complete dominance. However, it is possible to extend the Chi Square Test for Independence to any type of dominance system.
Consider the case where we have two loci A and B each with two alleles (A1;A2 and B1; B2 ). Locus A shows complete dominance and locus B shows incomplete dominance. The associate phenotypes can be represented as A1;= & for the A locus and as B1,F1 and B2 for the B locus (the heterozygote B1B2 has the phenotype called F1).
Example: Consider the following cross:
| Parental | A1A1B1B1 | x | A2A2B2B2 |
| F1 | A1A2B1B2 | A1A2B1B2 | |
In the F2 generation there are 3200 offspring divided among 6 phenotypes.
| F2 Phenotypes | Obs | Exp.* | |
| A1B1 | 750 | 600 | |
| A1F1 | 1339 | 1200 | |
| A1B2 | 311 | 600 | |
| A2B1 | 40 | 200 | |
| A2F1 | 268 | 400 | |
| A2B2 | 492 | 200 | |
|
Total |
3200 |
*expected assuming no problems with segregation and independent assortment
The observed results show a large deviation from the expected values (c2 = 906, 5df p >.01). Further work shows that there is not a problem with segregation at either the A or B loci.
The 3 x 2 general table is given below for the observed values.
| A Locus |
B locus |
||||
| Phenotypes | B1 | F1 | B2 |
|
|
| A1 |
|
|
|
|
|
| A2 |
|
|
|
|
|
| Column Totals |
|
|
|
|
|
Now we need to calculate the expected values for each cell in the table. This is done by multiplying the row totals times the column totals divided by the grand total (N). For example, the A1B1 phenotype would have an expected value of (a+b+c)(a+d)/N.
For the example above the observed values are:
| A Locus |
B locus |
||||
| Phenotypes | B1 | F1 | B2 |
|
|
| A1 |
|
1339 |
|
2400 | |
| A2 |
|
|
|
800 | |
| Column Totals |
|
|
|
3200 | |
The row totals divided by N are the phenotypic array for the A locus. Similarly, the column totals divided by N are the phenotypic array for the B locus.
For the example above the expected values are:
| A Locus |
B locus |
||||
| Phenotypes | B1 | F1 | B2 |
|
|
| A1 |
|
1200 |
|
2400 | |
| A2 |
|
|
|
800 | |
| Column Totals | 800 | 1600 | 800 | 3200 | |
Once the expected values have been calculated we can use the basic formula for a c2 test for goodness of fit.
| Sc2 = S(obs-exp)2/exp |
c2 = 906; Degrees of Freedom = (c - 1)(r - 1) = 2(1) = 2 ; p > .01. Reject hypothesis that A and B loci are independent.
Similar tables can be constructed for two loci where both loci show incomplete dominance or if there is lethality or penetrance. In genetics, the Chi Square Test of Independence is used to determine whether the phenotypic array for two loci considered jointly can be predicted by multiplying the phenotypic array of locus A by the phenotypic array of locus B. Linkage or epistasis can disrupt that relationship
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Quick Links F2 Cross |
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| 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) |
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Also See |
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| 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 | |