This is a true story.   My stepson was doing some Drosophila crosses for his AP biology class.  I had a recently isolated mutation called "pinhead" that I had discovered in my lab. Pinhead flies had no compound eyes.   For extra credit, my stepson decided to cross "pinhead" flies to vestigial-winged flies.  We knew that both mutations were autosomal and recessive to wildtype.  We expected the F₁ flies to be wildtype and to recover pinhead: vestigial flies in the F₂.  He wanted to get flies which would neither see nor fly.  Earlier work showed that we could expect 3:1 ratios of wildtype: mutation in the F₂. That is, there was no known lethality or penetrance problems.  We expected a 9:3:3:1 ratio in the  F₂. 

He crossed a homozygous normal wing, pinhead fly to a homozygous vestigial, eyes present fly. In the F₂ he obtained the following results:

The critical value for 3 df and a .05  level of significance is 7.82. Obviously, the observed data do not fit an expected 9:3:3:1 ratio.

Analysis: It was pretty obvious to me that this was a case of epistatic lethality.  That is, the combination of pinhead and vestigial killed the flies.  I already  knew that pinhead alone gave a 3:1 ratio in the F₂ of a monohybrid cross. The same was true of vestigial wing.   Furthermore, I knew from previous work that pinhead was on chromosome 2 and vestigial was on chromosome 3,  so linkage could not explain the results.  Nevertheless, his teacher insisted (in that annoying way that she had) that the results were obviously due to extremely tight linkage and nothing else!  We set out to test her hypothesis. To do so we again carried out the standard set of procedures.  

a.    Test for Chance (already done)

b.     Test for Problems with segregation at one or both loci.

c.     Test for Problems with independent assortment of the two loci.

d.     Decide whether the results are due to problems with segregation at one or both loci and additional problems with independent assortment of the two loci.

First we collapsed the data for wildtype eyes vs pinhead and tested for goodness of fit.

The critical value for 1 df and .05 level of significance is 3.85. We rejected  the null hypothesis of no problem with segregation at the pinhead locus (as long as vestigial is present).

Next we collapsed the data for wildtype vs vestigial wing, and tested for goodness of fit.

The critical value for 1 df and .05 level of significance is 3.85. We reject the null hypothesis of no problem with segregation at the vestigial locus (as long as pinhead is present).

Finally we did the contingency Chi-square test.

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

           [ |3170*0-1066*1059| - 0.5*5295]2*5295
c2 =   ------------------------------------------------------------------
                    (4229)(4236)(1066)(1059)

c2 = 332.1

The degrees of freedom is the (rows -1) * (columns -1) = 1. The critical value for 1 df and a .05 level of significance is 3.84. The null hypothesis of independence is rejected. That is, the deviation from the 9:3:3:1 ratio is due to problems with segregation at the both loci and some interaction between loci. Since the genes are on different chromosomes, the best explanation is lethal epistasis.

One more statistical test to drive the point home to the teacher.

The teacher was still not satisfied with our answer. She was sure that the two loci were completely linked. She was sure that I had been incorrect when I mapped pinhead to chromosome 2.   We decided to test her model. even though it was not consistent with our analyses. 

Let A be wildtype wing and a be vestigial wing. Similarly, let B be wildtype eyes-present and b be pinhead.   The original cross would have been AAbb (pinhead) x aaBB (vestigial).  

Under the idea of complete linkage there would be only two gametes produced by the double heterozygote: Ab and aB. These gametes would occur with equal frequencies. The expected genotypes and phenotypes produced in a cross of Ab/aB x Ab/aB would be:

Ab/Ab : Ab/aB : aB/aB  (pinhead: wildtype: vestigial) in a 1:2:1 ratio.  The observed and expected distributions for complete linkage are given below.

The critical value for 2 df and a .05 level of significance is 5.99. The teacher's explanation for the observed results does not fit.  We strongly rejected the null hypothesis of complete linkage!!

Detection of Linkage

More complex examples

Still one more example

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