Incomplete Dominance and Codominance
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You should be comfortable with basic Mendelian monohybrid crosses and the concept of alleles. Familiarity with the chi-squared test for goodness-of-fit is helpful, as we will apply it to non-Mendelian ratios.
Why this matters
Real-world inheritance often defies the simple "dominant/recessive" binary taught in introductory genetics.
While Mendel’s peas provided a clean model of discrete inheritance, most biological traits—especially in humans and animals—exhibit more complex patterns of gene expression. Incomplete dominance and codominance are two fundamental mechanisms where the heterozygous genotype produces a phenotype that is distinct from both homozygous parents. Understanding these patterns is critical for predicting inheritance in medical genetics, agricultural breeding, and evolutionary biology. This lecture moves beyond the binary view of alleles to explore how gene products interact at the molecular level to produce intermediate or combined phenotypes.
The Spectrum of Allelic Interaction
Imagine two paint buckets, one containing red paint and the other white. If you mix them, you do not get a result that is simply "red" or "white"; you get pink. This is the intuitive basis of incomplete dominance. In classical Mendelian genetics, we often assume that one allele completely masks the other. For example, in pea plants, the allele for purple flowers is dominant over the allele for white flowers. A plant with one purple and one white allele looks exactly like a plant with two purple alleles. However, this "all-or-nothing" model is a simplification. In many organisms, the presence of one functional allele does not produce enough protein to fully mask the effect of the other, or the two proteins interact to create a new, visible trait.
Incomplete dominance occurs when the heterozygous phenotype is an intermediate blend of the two homozygous phenotypes. A classic example is the snapdragon flower, where red and white parents produce pink offspring. The pink is not a separate "pink allele"; it is the result of half the normal amount of red pigment being produced. Codominance, on the other hand, is different. Here, both alleles are fully expressed simultaneously. The phenotype shows both traits distinctly, rather than blending them. A prime example is the ABO blood group system, where the A and B alleles are codominant. A person with both alleles expresses both A and B antigens on their red blood cells. They do not have "AB" blood that is a blend of A and B; they have cells that carry both distinct markers. This distinction is crucial: blending implies a quantitative reduction of one trait, while codominance implies the qualitative presence of both.
Molecular Basis of Incomplete Dominance
To understand why incomplete dominance occurs, we must look at the dosage of gene products. In many cases, the dominant allele in a Mendelian system is "haplosufficient," meaning one copy produces enough protein to achieve the full phenotype. In incomplete dominance, the gene product is often haploinsufficient. This means that one copy of the gene produces only half the amount of functional protein (such as an enzyme or pigment), and this half-dose is not sufficient to produce the full wild-type phenotype. Instead, the phenotype correlates directly with the amount of protein present.
Consider the flower colour in Mirabilis jalapa (four o'clock plant). The red allele produces a red pigment, while the white allele produces no pigment. A homozygous red plant has two copies of the pigment-producing gene and thus high pigment levels. A homozygous white plant has zero pigment. A heterozygote has one copy, resulting in intermediate pigment levels, which appear pink. This is a linear relationship between genotype dosage and phenotype intensity. Mathematically, if we define the pigment concentration \( P \) as proportional to the number of functional alleles \( n \), then \( P \propto n \). In a standard dominant system, the phenotype \( \Phi \) is a step function: \( \Phi = 1 \) if \( n \ge 1 \), and \( \Phi = 0 \) if \( n = 0 \). In incomplete dominance, \( \Phi \) is continuous and proportional to \( n \). This dosage sensitivity is common in traits controlled by enzymes where the reaction rate depends on enzyme concentration, or in structural proteins where the amount of material affects the physical property.
Codominance and Multiple Alleles
Codominance is frequently confused with incomplete dominance, but the molecular mechanisms are distinct. In codominance, both alleles in a heterozygote are transcribed and translated into functional proteins that are stable and detectable. The phenotype reflects the presence of both gene products. The most clinically significant example is the ABO blood group system in humans. The \( I^A \) and \( I^B \) alleles code for different glycosyltransferase enzymes that add specific sugar residues to the H antigen on red blood cells. The \( I^A \) enzyme adds N-acetylgalactosamine, while the \( I^B \) enzyme adds galactose. In a person with genotype \( I^A I^B \), both enzymes are produced. Since the enzymes act on the same substrate (the H antigen) but add different sugars, the red blood cells display both A and B antigens. This is not a blend; it is a simultaneous expression. An AB individual makes neither anti-A nor anti-B antibodies, so their plasma does not attack their own cells.
Another example is the coat colour in cattle. The Shorthorn cattle have alleles for red (\( C^R \)) and white (\( C^W \)) coat colours. Heterozygotes (\( C^R C^W \)) are roan, meaning they have a mixture of red and white hairs. This is often cited as codominance because individual hairs are either red or white, not pink. The phenotype is a mosaic of the two homozygous phenotypes. This contrasts with the snapdragon, where the entire flower is a uniform pink. The distinction lies in the cellular level: in codominance, different cells or structures may express different alleles, or the same cell expresses both distinct products. In incomplete dominance, the same cell expresses a reduced level of a single product.
Genetic Ratios and Chi-Squared Testing
When performing genetic crosses, the expected phenotypic ratios change depending on the mode of inheritance. In a standard Mendelian monohybrid cross (dominant/recessive), the F2 generation shows a 3:1 ratio of dominant to recessive phenotypes. However, in incomplete dominance or codominance, the heterozygote is phenotypically distinct. Therefore, the F2 generation from a cross between two heterozygotes (e.g., \( Aa \times Aa \)) yields a 1:2:1 phenotypic ratio, corresponding to the genotypic ratio of \( AA : Aa : aa \). This is a critical point for statistical analysis. If you observe a 1:2:1 ratio in your data, you cannot simply test it against a 3:1 Mendelian ratio. You must test it against the 1:2:1 expectation.
Suppose you cross two pink snapdragons and obtain 120 red, 240 pink, and 120 white offspring. The total is 480. The expected numbers for a 1:2:1 ratio are 120 red, 240 pink, and 120 white. The chi-squared statistic \( \chi^2 \) is calculated as \( \sum \frac{(O - E)^2}{E} \). Here, \( O \) and \( E \) are identical for all classes, so \( \chi^2 = 0 \). This indicates a perfect fit. However, if you had observed 150 red, 180 pink, and 150 white, the total is still 480. The expected values remain 120, 240, 120. The calculation would be: for red, \( \frac{(150-120)^2}{120} = \frac{900}{120} = 7.5 \); for pink, \( \frac{(180-240)^2}{240} = \frac{3600}{240} = 15 \); for white, \( \frac{(150-120)^2}{120} = 7.5 \). The total \( \chi^2 = 30 \). With 2 degrees of freedom (3 classes - 1), the critical value at \( p = 0.05 \) is 5.99. Since 30 > 5.99, we reject the null hypothesis that the data fits a 1:2:1 ratio. This suggests that the inheritance pattern is not simple incomplete dominance, or that there is another factor at play, such as linked genes or environmental effects.
A common misconception is that codominance and incomplete dominance are the same phenomenon described differently. They are mechanistically distinct. In incomplete dominance, the heterozygote has a reduced quantity of a single gene product, leading to an intermediate phenotype. In codominance, the heterozygote has two distinct gene products, both fully functional and detectable. Confusing these can lead to incorrect predictions in molecular diagnostics, where detecting the presence of both proteins (codominance) is different from measuring the total level of a single protein (incomplete dominance). C
Worked example
Problem. In a population of rabbits, coat colour is determined by a single gene with two alleles, \( C^B \) (black) and \( C^W \) (white). The alleles exhibit codominance, such that \( C^B C^B \) rabbits are black, \( C^W C^W \) rabbits are white, and \( C^B C^W \) rabbits are spotted (black and white patches). A breeder crosses two spotted rabbits and obtains 100 offspring. The observed counts are 25 black, 50 spotted, and 25 white. Calculate the chi-squared value to test if this data fits the expected codominant ratio.
- Determine the expected phenotypic ratio for a cross between two heterozygotes (\( C^B C^W \times C^B C^W \)). The genotypic ratio is 1 \( C^B C^B \) : 2 \( C^B C^W \) : 1 \( C^W C^W \). Since the phenotypes correspond directly to the genotypes in codominance, the expected phenotypic ratio is 1 black : 2 spotted : 1 white.
- Calculate the expected numbers (\( E \)) for each phenotype based on the total offspring count of 100. Total \( N = 100 \). Expected black = \( \frac{1}{4} \times 100 = 25 \). Expected spotted = \( \frac{2}{4} \times 100 = 50 \). Expected white = \( \frac{1}{4} \times 100 = 25 \).
- List the observed numbers (\( O \)) for each phenotype. Observed black = 25. Observed spotted = 50. Observed white = 25.
- Calculate the chi-squared contribution for each phenotype using the formula \( \frac{(O - E)^2}{E} \). For black: \( \frac{(25 - 25)^2}{25} = 0 \). For spotted: \( \frac{(50 - 50)^2}{50} = 0 \). For white: \( \frac{(25 - 25)^2}{25} = 0 \).
- Sum the contributions to find the total chi-squared value. \( \chi^2 = 0 + 0 + 0 = 0 \). Compare this to the critical value for 2 degrees of freedom at \( p = 0.05 \), which is 5.99. Since 0 < 5.99, we fail to reject the null hypothesis. The data fits the expected codominant ratio perfectly.
Units check. Chi-squared is a dimensionless statistic, so no units are required, but the input counts must be in the same units (individuals).
Three quick tests
- If a cross between two heterozygotes yields a 1:2:1 phenotypic ratio, what mode of inheritance is most likely?
Answer
Incomplete dominance or codominance, where the heterozygote is phenotypically distinct from both homozygotes. - What is the key molecular difference between incomplete dominance and codominance?
Answer
In incomplete dominance, the heterozygote has a reduced quantity of a single gene product. In codominance, the heterozygote has two distinct, fully functional gene products. - Why is the ABO blood group system considered an example of codominance?
Answer
Because individuals with the \( I^A I^B \) genotype express both A and B antigens on their red blood cells, rather than a blended intermediate antigen.
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
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