Crosses Involving Two or More Genes Help (page 2)
The Dihybrid Cross
The simultaneous inheritance of two or more traits, each specified by a different pair of independently assorting autosomal genes (i.e., genes on different chromosomes other than the sex chromosomes) can be analyzed using a dihybrid cross. This type of cross demonstrates Mendel's second principle of independent assortment (see Mendel's Law). In the conventional dihybrid cross, two true-breeding parents are mated (cross #1 in Example 2.30) to yield an F1 generation. The F1 hybrids are then crossed to yield an F2 generation (cross #2).
EXAMPLE 2.30 In addition to the coat-color locus of guinea pigs introduced earlier in this chapter (B– = black, bb = white), another locus on a different chromosome (independently assorting) is known to govern length of hair, such that L– = short hair and ll = long hair. Any of four different genotypes exist for the black, short-haired phenotype: BBLL, BBLl, BbLL, BbLl. Two different genotypes produce a black, long-haired pig: BBll or Bbll; likewise two genotypes for a white, short-haired pig: bbLL or bbLl; and only one genotype specifies a white, longhaired pig: bbll. A dihybrid genotype is heterozygous at two loci. Dihybrids form four genetically different gametes with approximately equal frequencies because of the random orientation of nonhomologous chromosome pairs on the first meiotic metaphase plate (Chapter 1).
These gametes then randomly combine in the F1 cross. The use of a Punnett square helps demonstrate all 16 possible combinations of these gametes.
The four dark squares are homozygous BB genotypes, the eight light squares are heterozygous Bb, and the four open squares are homozygous bb in a 1 : 2 : 1 ratio, respectively. The same ratio occurs for the LL, Ll, and ll genotypes.
Summary of F1 cross results:
The phenotypes in this type of conventional dihybrid cross give a 9 : 3 : 3 : 1 ratio. This ratio is derived from grouping the phenotypes in the table above as follows:
9/16 Black coat, short hair (B – L–)
3/16 Black coat, long hair (B – II )
3/16 White coat, short hair (bbL–)
1/16 White coat, long hair (bbll)
The following examples demonstrate several other methods for determining the ratios in a dihybrid cross.
EXAMPLE 2.31 Determination of genotypic ratios and frequencies.
Considering only the B locus, Bb × Bb produces 1/4 BB, 1/2 Bb, and 1/4 bb. Likewise for the L locus, Ll × Ll produces 1/4 LL, 1/2 Ll, and 1/4 ll. Place these genotypic probabilities in a Punnett square and combine independent probabilities by multiplication (law of products).
Another way to arrive at the frequencies seen in the above Punnett square is to use the branch diagram. Again, the final frequencies are determined by using the law of products because traits are being combined (e.g., BB and LL).
EXAMPLE 2.32 Determination of phenotypic ratios and frequencies.
Considering the B locus, Bb × Bb produces 3/4 black and 1/4 white. Likewise at the L locus, Ll × Ll produces 3/4 short and 1/4 long. Place these independent phenotypic probabilities in a Punnett square and combine them by multiplication.
Another way to arrive at the frequencies seen in the above Punnett square is to use the branch diagram. Again, the final frequencies are determined by using the law of products because traits are being combined (e.g., black coat and long hair).
If only one of the genotypic frequencies or phenotypic frequencies is required, there is no need to be concerned with any other genotypes or phenotypes. A mathematical solution can be readily obtained by combining independent probabilities
EXAMPLE 2.33 Determination of the frequency of only one genotype, BBLl, in the offspring of dihybrid parents
First consider each locus separately: Bb × Bb = 1/4 BB; Ll × Ll = 1/2 Ll. Combining these independent probabilities, 1/4 × 1/2 = 1/8 BBLl.
EXAMPLE 2.34 To find the frequency of only one phenotype, white coat, short hair, in the offspring of dihybrid parents.
First consider each trait separately: Bb × Bb = 1/4 white (bb); Ll × Ll = 3/4 short (L–). Combining these independent probabilities, 1/4 × 3/4 = 3/16 white, short.
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 (B – L –) is performed with an individual whose genotype is homozygous recessive at all of the loci under consideration (bbll).
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.
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:
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.
Practice problems for these concepts can be found at:
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