Detection of Linkage
Deconstructing the Phenotypic Ratios of Two Genes - Detecting Deviations from Independent Assortment

Consider a cross of two double heterozygotes. AaBb x AaBb. If both genes show complete dominance, no lethality or penetrance problems, and independent assortment we would expect to obtain four phenotypes in a 9:3:3:1 ratio. You have already seen that lethality and penetrance-less-than-100% can skew the expected ratios for a monohybrid cross.  If we observed deviation from ratio different from a 9:3:3:1 ration (when their is complete dominance) the difference might be due to:

a.    Chance

b.     Problems with segregation at one of both loci.

c.     Problems with independent assortment of the two loci.

d.     Problems with segregation at one of both loci and additional problems with independent assortment of the two loci.  Such deviation might be due to epistasis or linkage.

These four possibilities can be statistically tested.  Statistics will only tell us if there is a problem.  The exact nature of the problem might have to be elucidated by a series of experimental genetic crosses.

The primary statistic that is used in genetics is the Chi-square test for goodness of fit.  On order to carry out the c2 test one calculates an expected distribution of phenotypes based on a model or null hypothesis.   In genetics, the model is based on Mendelian laws.  For example, in the F2 of a cross between a homozygous gray, bicorn dust rhino and a homozygous white, unicorn dust rhino, with proper segregation at both loci, complete dominance and independent assortment,  we expect to get four genotypes in a 9/16:3/16:3/16:1/16 ratio.  If we have 32 progeny from this cross, we expect to see 18:6:6:2.

3 * 3 = 9 gray,  unicorn         Expected = 9/16 * 32 = 18

3 * 1 = 3 gray, bicorn             Expected = 3/16 * 32 = 6

1 * 3 = 3 white, unicorn          Expected = 3/16 * 32 = 6

1 * 1 = 1 white, bicorn         Expected = 1/16 * 32 = 2

 

We observe 32 dust rhino with the following results:

Phenotype Obs. Exp. Obs-Exp= d d2 d2/exp
gray, unicorn 21 18 3 9 2.0
gray, bicorn  5 6 -1 1 0.17
white, unicorn 5 6 -1 1 0.17
white, bicorn  1 2 1 1 0.5

  c2  = 2.84

Obviously the number of observed phenotypes does not equal the number of expected phenotypes.   The Chi-square test for goodness of fit indicates whether or not the differences small and  due to chance (refer to the book and my lecture notes) or large and probably due to a violation of the null hypothesis. 

                                        (obs-exp)2
General Chi- Square = S ----------
                                                                    exp

The Chi square value is 2.84.  The critical value for k-1 = 3 degrees of freedom and a  .05 level of significance is 7.82.  Since the calculated  c2  value is less than the critical value,  the differences between the observed and expected distributions are due to chance.  That is, we do not reject the null hypothesis.

Example 2:

Eye color in dust rhinos is either black or silver.   The genetics of this trait has not yet been worked out. Workly Drudge (a 21st  year graduate student and somewhat of legend around Rutgers) crosses a homozygous black-eyed, unicorn dust rhino to a homozygous silver-eyed, bicorn dust rhino (refer to Calculating Expected Ratios From a Cross Involving Two or More Genes).   He looks at a few of the F1 progeny and they appear to be black-eyed, unicorn. He intercrosses the F1 and obtains the following distributions of the 640 progeny:

Phenotype Obs. Exp.* Obs-Exp= d d2 d2/exp
black, unicorn 336 360 -24 576 1.6
black, bicorn  112 120 -8 64 0.53
silver, unicorn 144 120 24 576 4.8
silver, bicorn  48 40 8 64 1.6

  c2  = 8.53

(* Expected values based a null hypothesis of no problems with segregation at either locus, complete dominance, and independent assortment).

The  c2  value is 8.53.  The critical value for 3 degrees of freedom and a .05 level of significance is 7.82. Since the calculated  c2 value is greater than the critical value,  the differences between the observed and expected distributions are not  due to chance.  That is, we  reject the null hypothesis.  The differences between the observed and expected distributions are due to some thing other than chance.

Beyond Chance

The first step in the analysis the deviation from the 9:3:3:1 ratio is to check for problems with segregation at the eye color and horn number loci.  To do this we have to collapse the data. That is, we will  ignore the information on horn number and look only  at eye color.

(phenotypes in smaller font are being ignored)

black,unicorn black, bicorn silver, unicorn silver, bicorn
336 112 144 48

 Now add up the identical columns (e.g., columns one and two are both black).  This gives you the observed distributions. Calculate the expected distributions based on a null hypothesis of complete dominance and no problems with segregation. Check the model using the  special case Chi-square test for two phenotypes.

 

Phenotype black silver Total
obs 448 192 640
exp. 480 160 640
|obs-exp-.5| = d 31.5 31.5  
d2 992.25 992.25  
d2/exp 2.07 6.20 8.27

  c2  = 8.27

 The  c2  value is 8.27.  The critical value for 1 degrees of freedom and a .05 level of significance is 3.84. Since the calculated  c2 value is greater than the critical value,  the differences between the observed and expected distributions is not due to chance.  That is, we  reject the null hypothesis.  There is a problem with either dominance or segregation at the eye color locus.

Now we test the horn number locus:

First we collapse the data:

black, unicorn black, bicorn silver, unicorn silver, bicorn
336 112 144 48

 Now add up the identical columns. This gives you the observed distributions. Calculate the expected distributions based on a null hypothesis of complete dominance and no problems with segregation. Check the model using the special case Chi-square test for two phenotypes.

Phenotype unicorn bicorn Total
obs 480 160 640
exp. 480 160 640
|obs-exp-.5| = d .5 .5  
d2 .25 .25  
d2/exp 0 0 0

There is actually no reason to do a Chi-square test. This is an exact fit. We do not reject the null hypothesis. There is nothing wrong with segregation or dominance at the horn number locus.

Finally, we  test for independence of the horn number and eye color loci. 

These loci would not be independent if they somehow influenced each other. This occurs most often due to linkage. Certain forms of epistasis also cause the loci to deviate from independent assortment.

The statistic that we  use here is called the contingency chi-square or the chi-square test for independence. In this course we are only going to deal with  2 x 2 contingency tables.   Before proceeding, It is best to reformat your data in the following manner:

Dominant Phenotype 1 Recessive Phenotype 1 Total
Dominant Phenotype 2 Both dominant phenotypes Recessive 1 and Dominant 2 phenotypes Row Sum
Recessive Phenotype 2 Recessive 2 and Dominant 1 phenotypes Both recessive phenotypes Row Sum
Total Column Sum Column Sum Grand Sum

Proceeding with our example:
Black Silver Total
Unicorn 336 (a) 144 (b) 480 (a+b)
Bicorn 112 (c) 48 (d) 160 (c+d)
Total 448 (a+c) 192 (b+d) 640 N = (a+b+c+d)

 Each of these numbers has a letter value given in bold and italics.

The formula for the contigency Chi-square is:

        [ |ad-bc| - 0.5N]2N
c2 = --------------------------
        (a+b)(a+c)(c+d)(b+d)

         [ |336*48-112*192| - 0.5*640]2*640
c2 = ------------------------------------------
                (448)(480)(160)(192)

 

The degrees of freedom for a contingency Chi-square test is the (rows -1) * (columns -1) .   For a 2 x 2 table this will always be 1 df.   The critical value for 1 df and a .05 level of significance is 3.84.   Since the c2 value is less than the critical value the null hypothesis of independence is not rejected. That is, the deviation from the 9:3:3:1 ratio is due solely to problems with segregation at the eye color locus. Elucidation of the nature of that problem with segregation will take additional, experimental, work.

Example 3:

Long horns (F) is dominant to short horns (f).  A homozygous long horned, unicorn dust rhino is mated to a homozygous short-horned, bicorn dust rhino. All of the F1 progeny are long, unicorns. These F1 progeny are intermated to produce an F2 of 800 with the following results.

Phenotype Obs. Exp.* Obs-Exp= d d2 d2/exp
long, unicorn 500 450 50 2500 5.56
long, bicorn 100 150 -50 2500 16.7
short, unicorn 100 150 -50 2500 16.7
short, bicorn 100 50 50 2500 50.0

(* Expect 9:3:3:1. Expected values based on null hypothesis of  no problems with segregation at either locus, complete dominance, and independent assortment.)

  c2  = 88.9

The Chi-square value is 88.9. The critical value for 3 degrees of freedom and a .05 level of significance is 7.82. Since the Chi-square value exceeds the critical value we reject the null hypothesis.  The difference between the observed and expected distributions is due to something other than chance.

Now we are going to do the same procedure as before: 

1.     Check for segregation at the "A" locus (ie. the horn length locus)

2.     Check for segregation at the "B" locus (ie. the horn number locus)

3.     Check for independence of the two loci.

First, we collapse the data for horn length and check for an expected 3:1 ratio.

Phenotype long short Total
obs 600 200 800
exp. 600 200 800

No need to do the Chi-square for goodness of fit. The fit is exact. Segregation at the horn length locus is okay. We do not reject the null hypothesis.

Next, we collapse the data for horn number and check for an expected 3:1 ratio.

Phenotype unicorn bicorn Total
obs 600 200 800
exp. 600 200 800

No need to do the Chi-square for goodness of fit. The fit is exact. Segregation at the horn number locus is okay. We do not reject the null hypothesis.

Finally, we check for independence:

Phenotype Black Silver Total
Unicorn 500 (a) 100(b) 600 (a+b)
Bicorn 100 (c) 100 (d) 100 (c+d)
Total 600 (a+c) 100 (b+d) 800 N = (a+b+c+d)

        [ |ad-bc| - 0.5N]2N
c2 = --------------------------
        (a+b)(a+c)(c+d)(b+d)

         [ |500*100-100*100| - 0.5*800]2*800
c2 = ---------------------------------------
                (600)(600)(200)(200)

c2 = 87.2 

The critical value for 1 df and a .05 level of significance is 3.84. Since the c2  value is greater than the critical value, we reject the null hypothesis of independence.  The loci are either linked or the interact epistatically.

More complex examples

Still one more example

Calculating Expected Ratios From a Cross Involving Two or More Genes.

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