Why Separate These Two Comparisons?
For a biallelic locus with alleles \(A\) and \(a\), the three genotypes are \(aa\), \(Aa\), and \(AA\). A quantitative trait can have a different mean in each genotype group. Two distinct questions can then be asked: how far apart are the two homozygote means, and how far is the heterozygote mean from the midpoint between them? The first describes a homozygote contrast; the second describes a dominance deviation.
These quantities are easy to confuse because both describe differences among genotype means. They answer different questions, however. The homozygote contrast uses only the \(aa\) and \(AA\) means. The dominance deviation also uses the \(Aa\) mean, comparing it with the midpoint fixed by the homozygotes. Keeping the comparisons separate makes the three genotype means easier to describe without implying that a particular genetic mechanism has been established.
All three quantities are measured in the trait’s units. For example, if the trait is height in centimeters, then \(m_0\), \(a\), and \(d\) are also in centimeters. The sign of \(a\) depends on which allele is labelled \(A\): exchanging the labels \(A\) and \(a\) reverses the sign of \(a\). The dominance deviation \(d\) does not change under that relabelling, because the two homozygote means still define the same midpoint.
Parameterizing the Three Genotype Means
The definitions can be rearranged to express each genotype mean using the midpoint, average homozygote contrast, and dominance deviation. By construction, the homozygote means lie equally far below and above their midpoint; the heterozygote mean is the midpoint plus its deviation.
This is a re-expression of the three means, not an additional biological assumption. To verify it, substitute the definitions: \(m_0-a=(M_{aa}+M_{AA})/2-(M_{AA}-M_{aa})/2=M_{aa}\), and \(m_0+a=(M_{aa}+M_{AA})/2+(M_{AA}-M_{aa})/2=M_{AA}\). Also, \(m_0+d=m_0+(M_{Aa}-m_0)=M_{Aa}\). Thus, any three genotype means on the same scale have exactly this representation.
If \(d=0\), the heterozygote mean equals the midpoint. The steps from \(aa\) to \(Aa\) and from \(Aa\) to \(AA\) are then equal, each of size \(a\) when \(A\) is the allele with the larger homozygote mean. If \(d\ne 0\), those two genotype steps are not equal. The deviation describes the heterozygote’s position relative to the midpoint; it does not by itself state which homozygote has the larger mean.
These identities follow by subtracting the parameterized means. They show why the two quantities are distinct: \(a\) describes the homozygote separation, while \(d\) shifts the heterozygote away from the midpoint and changes the two adjacent genotype steps in opposite directions.
Worked Example: Heterozygote Below the Midpoint
Suppose a hypothetical study measures a trait in millimeters and estimates the genotype means as \(M_{aa}=10\) mm, \(M_{Aa}=14\) mm, and \(M_{AA}=22\) mm. Calculate the additive midpoint, average homozygote contrast, and dominance deviation.
Additive midpoint. Average the homozygote means:
Average homozygote contrast. Take half the difference between the \(AA\) and \(aa\) means:
Dominance deviation. Subtract the midpoint from the heterozygote mean:
The heterozygote mean is 2 mm below the midpoint of 16 mm. The homozygote means are 12 mm apart, and their average contrast is 6 mm. These statements describe different features of the same three means; the value \(-2\) mm is not the difference between the two homozygote means.
As a check, the parameterization gives \(M_{aa}=16-6=10\) mm, \(M_{Aa}=16-2=14\) mm, and \(M_{AA}=16+6=22\) mm, recovering all three supplied means.
Estimating the Quantities From Genotype Groups
When genotype-specific means are estimated from data, the same formulas apply to the estimated means. Let \(\bar{x}_{aa}\), \(\bar{x}_{Aa}\), and \(\bar{x}_{AA}\) be the sample means of the trait among individuals with each genotype. The corresponding sample estimates are the midpoint of the two homozygote sample means, half their difference, and the heterozygote sample mean minus that midpoint.
These are descriptive calculations from the three group means. Their interpretation requires that the means refer to the same trait, measured on the same scale, and that the genotype labels have been assigned consistently. If the values are sample estimates, they may differ from the underlying genotype-specific means because of sampling variation. The formulas alone do not quantify that uncertainty or establish that genotype differences are causal.
Worked Example: Calculating Deviations From Sample Means
Suppose a hypothetical dataset records trait values in centimeters. The sample means are \(\bar{x}_{aa}=8\) cm, \(\bar{x}_{Aa}=13\) cm, and \(\bar{x}_{AA}=16\) cm. Calculate the sample estimates of \(m_0\), \(a\), and \(d\).
First, the homozygote midpoint is:
Next, the average homozygote contrast is:
Finally, the estimated dominance deviation is:
The heterozygote sample mean is 1 cm above the homozygote midpoint. The homozygote means differ by 8 cm, not 4 cm; 4 cm is half that difference. The reconstructed means are \(12-4=8\) cm, \(12+1=13\) cm, and \(12+4=16\) cm.
Worked Example: A Midpoint Does Not Have to Be the Overall Mean
Consider another hypothetical dataset, with sample means \(\bar{x}_{aa}=10\) units, \(\bar{x}_{Aa}=15\) units, and \(\bar{x}_{AA}=22\) units. Suppose there are 20 individuals with genotype \(aa\), 60 with genotype \(Aa\), and 20 with genotype \(AA\). Find the additive midpoint and dominance deviation, and compare the midpoint with the overall sample mean.
The additive midpoint uses only the two homozygote means:
Therefore, the estimated dominance deviation is:
The overall sample mean weights each genotype group by its sample size. Here it is:
The overall sample mean, 15.4 units, is not the additive midpoint, 16 units. The overall mean depends on how many observations are in each genotype group; the additive midpoint is defined by the two homozygote means alone. Unequal genotype group sizes do not change the definition of the midpoint or the dominance deviation.
Interpreting the Sign and Size of the Deviation
A positive \(d\) means the heterozygote mean is above the homozygote midpoint; a negative \(d\) means it is below; and \(d=0\) means it equals the midpoint. The size is in the original trait units, so the interpretation should name both the direction and the scale. For example, a deviation of \(-2\) mm means the heterozygote mean is 2 mm below the midpoint, not that it is 2 mm below either homozygote mean.
The deviation is not, by itself, a percentage, a probability, or a statement about the molecular action of an allele. It is a contrast among trait means. In particular, describing a heterozygote as “between” the homozygotes is not enough to establish \(d=0\): a mean can lie between the homozygote means but still be above or below their midpoint.
A short algebraic check makes the distinction explicit. If the heterozygote mean equals the midpoint, then \(M_{Aa}=m_0\), so \(d=M_{Aa}-m_0=0\). Conversely, if \(d=0\), then \(M_{Aa}=m_0+d=m_0\). Thus, equality to the midpoint and a zero dominance deviation are equivalent.
Use means for \(aa\), \(Aa\), and \(AA\) measured on the same trait scale.
Set \(m_0=(M_{aa}+M_{AA})/2\) and \(a=(M_{AA}-M_{aa})/2\).
Calculate \(d=M_{Aa}-m_0\), retaining its sign and the trait units.
Confirm that \(m_0-a\), \(m_0+d\), and \(m_0+a\) reproduce the three genotype means.
Common Mistake: Calling the Homozygote Difference a Dominance Deviation
A common error is to subtract the two homozygote means and call the result the dominance deviation. That calculation gives \(M_{AA}-M_{aa}\), the full homozygote difference. Half of it is \(a\), the average homozygote contrast. Neither quantity compares the heterozygote with the additive midpoint.
The correction is to calculate the midpoint first, using only the homozygotes, and then subtract that midpoint from the heterozygote mean. This order makes the reference explicit and prevents a difference between the wrong pair of means from being given the wrong name. Another related error is to call \(a\) the full homozygote difference: by definition, \(a\) is half that difference.
Check Your Understanding
Use the definitions and show the arithmetic where appropriate.
- If \(M_{aa}=5\), \(M_{Aa}=11\), and \(M_{AA}=17\) in the same units, what are \(m_0\), \(a\), and \(d\)?
- For \(M_{aa}=12\), \(M_{Aa}=18\), and \(M_{AA}=20\) centimeters, calculate the midpoint and explain the meaning of the sign of \(d\).
- Why is \((M_{AA}-M_{aa})/2\) not the dominance deviation?
- If the heterozygote mean lies between the two homozygote means, must \(d=0\)? Explain.
- Suppose the genotype group sample sizes change but the three genotype-specific sample means stay the same. Does the additive midpoint change? Does the overall sample mean necessarily stay the same? Explain.