Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 221 / 2 / ii

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- Solution ii

Solution

The common marginal trait variance is

V_{Y} = V_{A} + V_{C} + V_{E} .

For monozygotic pairs, shared genes and shared environment give

Cov (Y_{i 1}, Y_{i 2}) = V_{A} + V_{C}, r_{MZ} = \frac{V _{A} + V _{C}}{V _{Y}} .

For dizygotic pairs, the genetic covariance is halved while the common environment is unchanged:

Cov (Y_{i 1}, Y_{i 2}) = \frac{1}{2} V_{A} + V_{C}, r_{DZ} = \frac{V _{A} /2 + V _{C}}{V _{Y}} .

Subtracting cancels the common-environment variance and proves Falconer's formula

2 (r_{MZ} - r_{DZ}) = \frac{V _{A}}{V _{Y}} = h^{2},

the heritability under the ACE model.

 Ancestors (10)

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