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The example below assesses another double-heterozygote cross using RrYy x RrYy. As stated above, the phenotypic ratio is expected to be 9:3:3:1 if crossing unlinked genes from two double-heterozygotes. The genotypic ratio was obtained in the diagram below, this diagram will have more branches than if only analyzing for phenotypic ratio.
All the haploid sperm and eggs produced by meiosis received one chromosome. All the zygotes received one R allele (from the round seed parent) and one r allele (from the wrinkled seed parent). Because the R allele is dominant to the r allele, the phenotype of all the seeds was round. The phenotypic ratio in this case of Monohybrid cross is 1.
The traits observed in this cross are the same traits that Mendel was observing for his experiments. This cross results in the expected phenotypic ratio of 9:3:3:1. Another example is listed in the table below and illustrates the process of a dihybrid cross between pea plants with multiple traits and their phenotypic ratio patterns.
When conducting a dihybrid test cross, two dominant phenotypic characteristics are selected and crossed with parents displaying double recessive traits. The phenotypic characteristics of the F1 generation are then analyzed. In such a test cross, if the individual being tested is heterozygous, a phenotypic ratio of 1:1:1:1 is typically observed. [7]
The genotypic ratio is 1: 2 : 1, and the phenotypic ratio is 3: 1. In the pea plant example, the capital "B" represents the dominant allele for purple blossom and lowercase "b" represents the recessive allele for white blossom. The pistil plant and the pollen plant are both F 1-hybrids with genotype "B b". Each has one allele for purple and one ...
For example, in co-dominance, a red homozygous flower and a white homozygous flower will produce offspring that have red and white spots. When plants of the F1 generation are self-pollinated, the phenotypic and genotypic ratio of the F2 generation will be 1:2:1 (Red:Spotted:White). These ratios are the same as those for incomplete dominance.
An example in dog coat genetics is the homozygosity with the allele "e e" on the Extension-locus making it impossible to produce any other pigment than pheomelanin. Although the allele "e" is a recessive allele on the extension-locus itself, the presence of two copies leverages the dominance of other coat colour genes.
When there is only additive gene action, this sibling phenotypic correlation is an index of familiarity – the sum of half the additive genetic variance plus full effect of the common environment. It thus places an upper limit on additive heritability of twice the full-Sib phenotypic correlation.