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Crosses Involving Two or More Genes Help (page 2)

By — McGraw-Hill Professional
Updated on Aug 21, 2011

Testcross with Two Traits

A testcross of a homozygote (BB) results in offspring of all one phenotype, while a testcross of a heterozygote (Bb) gives a 1 : 1 phenotypic ratio, indicating that one pair of factors is segregating (see section on Monohybrid Crosses). A dihybrid testcross with a dihybrid (i.e., heterozygote; BbLl) gives 1 : 1 : 1 : 1 genotypic and phenotypic ratios, indicating that two pairs of factors are segregating and assorting independently. Testcrosses with individuals that are homozygous for one trait and heterozygous for the second trait give a 1 : 1 phenotypic ratio.

EXAMPLE 2.35 A testcross of a black-coated, short-haired individual with an incompletely known genotype (BL –) is performed with an individual whose genotype is homozygous recessive at all of the loci under consideration (bbll).

Testcross with Two Traits

Modified Dihybrid Ratios

The classical phenotypic ratio resulting from the mating of dihybrid genotypes is 9 : 3 : 3 : 1. This ratio appears whenever the alleles at both loci display complete dominant and recessive relationships. The classical dihybrid ratio may be modified if one or both loci have codominant alleles or lethal alleles. A summary of these modified phenotypic ratios in adult progeny is shown in Table 2-2.

Higher Combinations

The methods for solving two-factor (dihybrid) crosses may easily be extended to solve problems involving three (trihybrid) or more pairs of independently assorting autosomal factors. Given any number of heterozygous pairs of factors (n) in the F1, the following general formulas apply:

Higher Combinations

EXAMPLE 2.36 In the following cross, AaBb × AaBb, n = 2 (both loci are heterozygous). Then, 2n or 4 equals the number of different gametes that are possible from each parent, 3n or 9 equals the number of different genotypes possible, and 4n or 16 equals the possible gametic combinations.

For a conventional trihybrid F2 cross (AaBbCc × AaBbCc), eight phenotypic classes result and the phenotypic ratio is 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1. The branch diagram below helps illustrate this point, again using the law of products to combine probabilities.

Higher Combinations

Practice problems for these concepts can be found at: 

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