# Is heritability just slope?

Is broad-sense heritability just the slope of the line that passes through the coordinates:

1. (1, The phenotypic correlation among monozygotic twins)
2. (0.5, The phenotypic correlation among dizygotic twins
3. [Optionally] (0, The phenotypic correlation among unrelated individuals)?

Here, the X axis is the relatedness (0 for unrelated, 0.5 for dizygotic, and 1 for monozygotic) and the Y axis is the correlation coefficient r. Is this the same as twin-based broad-sense heritability?

(This would also mean you can estimate heritability by taking the slope of the line through only dizygotic siblings, as they don't always share 50/50 of their DNA and there will be some spread in the X dimension.)

• Context here is important, slope of what? Wht kind of plot are you looking at.
– John
Jul 13 at 15:12
• @John Slope of the line which passes through dizygotic twin correlation and monozygotic twin correlation, where the x-axis is relatedness (0 to 1) and the y-axis is correlation (0 to 1). Jul 13 at 22:13
• typo: bounty should not say "twin A," it should just say "A." Jul 18 at 6:57
• In a population without genetic variation, the slope would be 0 while the offsprings are essential clones. Wording a meaningful definition of what it is, that you are calculating is not so easy. Maybe it would be easier the other way round: You start with a definition of what you want to calculate. Jul 18 at 13:02
• The broad-sense heritability is the ratio of total genetic variance to total phenotypic variance. H2 = VG/VP (as defined here), so no, unless in a very limited sense of one trait with binary values for phenotype.
– bob1
Jul 19 at 3:56

Given genetic relatedness (1 for mz identical twin, 0.5 for dz fraternal twin) on the $$x$$ axis and twin correlation for the phenotype on the $$y$$ axis, we can define these two points: \begin{align} (x_1, y_1) &= (0.5, r_{dz})\\ (x_2, y_2) &= (1, r_{mz}) \end{align}
Then, we can write the line between them in the two-point form: $$(y - y_1) = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)$$ $$(y - r_{dz}) = \frac{r_{mz} - r_{dz}}{1 - 0.5} (x - 0.5)$$ Therefore, the slope is $$\frac{r_{mz} - r_{dz}}{1 - 0.5} = 2(r_{mz}-r_{dz})$$, which is equal to the broad-sense heritability.