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Modern Genetic Analysis Chapter 6. Gene Interaction
Solved Problem

2.

The leaves of pineapples can be classified into three types: spiny (S), spine tip (ST), and piping (nonspiny) (P). In crosses between pure strains followed by intercrosses of the F1,the following results were obtained:

a. Assign gene symbols. Explain these results in terms of the genotypes produced and their ratios.
b. Using the model from part a, give the phenotypic ratios you would expect if you crossed (1) the F1 progeny from piping × spiny with the spiny parental stock, and (2) the F1 progeny of piping × spiny with the F1 progeny of spiny × spiny tip.
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Solution
a. First, let's look at the F2 ratios. We have clear 3:1 ratios in crosses 1 and 2, indicating single-gene segregations. Cross 3, however, shows a ratio that approximates a 12:3:1 ratio. This would be a modification of a 9:3:3:1 dihybrid ratio. Therefore it looks as if we are dealing with a two-gene interaction. This seems the most promising place to start; we can go back to crosses 1 and 2 and try to fit them in later. Any dihybrid self-cross involving two independently assorting genes (with complete dominance at both) is based on the proportions 9:3:3:1. Since we are not told which phenotype is wild-type, we cannot use the wild-type symbolism, so we just use the generic A/a and B/b symbolism to represent the two interacting loci. The genotypes must be grouped as follows to obtain a 12:3:1 ratio:

So without worrying about the name of the type of gene interaction (we are not asked to supply this anyway), we can already define our three pineapple leaf phenotypes in terms of the proposed allelic pairs A, a and B, b:

(Note: we could have designated these, reversing the letters A and B, as follows: B/ (A, a irrelevant), b/b ; A/, and b/b ; a/ait makes no difference.)

What about the parents of cross 3? The spiny parent must be a/a ; b/b, and because the B gene is needed to produce F2 spiny-tip individuals, the piping parent must be A/A ; B/B. (Note that we are told that all parents are pure, or homozygous.) The F1 must therefore be A/a ; B/b. Without further thought, we can write out cross 1 as follows:

Cross 2 can also be written out partially without further thought, using our arbitrary gene symbols:

We know that the F2 of cross 2 shows single-gene segregation, and it seems certain now that the A/a allelic pair is involved. But the B allele is needed to produce the spiny-tip phenotype, so all individuals must be homozygous B/B:

Notice that the two single-gene segregations in crosses 1 and 2 do not show that the genes are not interacting. What is shown is that the two-gene interaction is not revealed by these crossesonly by 3, in which the F1 is heterozygous for both genes.
b. Now it is simply a matter of using the laws of segregation and independent assortment to predict cross outcomes:

See question

© 1999 by W. H. Freeman and Company. All rights reserved.