Lean mass plays a gender-specific role in familial resemblance for femoral neck bone mineral density in adult subjects

Abstract

Introduction

Whether the femoral neck bone mineral density (FN BMD) of children may be better predicted from that of their parents when taking into account the anthropometry of the children was assessed in a healthy adult sample consisting of 86 mother-daughter, 32 mother-son, 32 father-daughter, and 23 father-son pairs from 128 families. Heritability for FN BMD, which is considered to be a measurement of general resemblance, was defined as the regression coefficient of the mean of the parents’ BMD. Among the anthropometric factors, lean mass was the most strongly associated with FN BMD following the adjustment for age in women (r=0.52, p<0.0001) and men (r=0.25, p=0.02). After adjustment for age, calcium intake, physical activity, and menopause and hormonal replacement therapy if relevant, heritability estimates (h²) for FN BMD were 0.68±0.23 [95% credible interval (CI): 0.15–0.99] in father-daughter pairs, 0.40±0.17 (95% CI: 0.08–0.74) in mother-daughter pairs, and 0.19±0.15 (95% CI: 0.01–0.57) in father-son pairs. Adjustment for lean mass of children increased the h² for FN BMD in mother-son pairs [from 0.24±0.17 (95% CI: 0.01–0.57) to 0.66±0.18 (95% CI: 0.26–0.95)]. The present results show that FN BMD is heritable in adult father-daughter pairs (7.2% of a daughter’s FN BMD variance was explained by the father’s FN BMD) and that taking into account the lean mass of sons might improve the prediction of their FN BMD based on that of their mother’s (reduction of sons’ FN BMD residual variance by 5.1%). Taking the lean mass of children into account might improve the prediction of their FN BMD by 9.1% in daughters and by 18.1% in sons, irrespective of their parent’s FN BMD. These results, obtained using a Bayesian regression model, have to be confirmed in further studies involving a greater number of adult parent-offspring pairs of both genders before extrapolation to clinical practice.

Appendices

Appendix A. Unified formulation of heritability regression

Given the following equations

$$y_{i} \tilde{}N{\left( {0,\sigma ^{2}_{0} } \right)}$$

(1)

$$y_{i} = ay^{{{(f)}}}_{i} + \varepsilon _{{i1}} $$

(2)

$$y_{i} = h^{2} {\left( {\frac{{y^{{{(f)}}}_{i} + y^{{{\left( m \right)}}}_{i} }} {2}} \right)} + \varepsilon _{{i2}} $$

(3)

with $\varepsilon _{{i1}} \tilde{}{\text{N}}{\left( {0,\sigma ^{2}_{1} } \right)}$ and $\varepsilon _{{i2}} \tilde{}{\text{N}}{\left( {0,\sigma ^{2}_{2} } \right)}$the three equations describe the distribution of the centered BMD variable in the population, respectively, when (i) the parents’ measurements are unknown (Eq. 1), (ii) only those of the father is known (Eq. 2), and (iii) when the two measurements are available (Eq. 3).

Equation (2) holds symmetrically when the mother’s measurement is the only one available.

Expressing a from h²

From model 3 we can write

$$cor{\left( {y_{i} ,y^{{{(f)}}}_{i} } \right)} = cor{\left( {h^{2} {\left( {\frac{{y^{{{(f)}}}_{i} + y^{{{(m)}}}_{i} }} {2}} \right)} + \varepsilon _{{i2}} ,y^{{{(f)}}}_{i} } \right)}$$

(4)

Making the assumption that the residuals ɛ _I2 are independent of the BMD variable, it follows that

$$\eqalign {cor{\left( {y_{i} ,y^{(f)}_{i} } \right)} = \frac{h^{2} } {2}cor{\left( {y^{\left( f \right)}_{i} ,y^{(f)}_{i} } \right)} + \frac{h^{2} } {2}cor{\left( {y^{\left( m \right)}_{i} ,y^{(f)}_{i} } \right)} \cr = \frac{h^{2} } {2}\sigma ^{2}_{0} }$$

(5)

From model 2, it follows that

$$cor{\left( {y_{i} ,y^{{{(f)}}}_{i} } \right)} = cor{\left( {ay^{{{(f)}}}_{i} ,y^{{{(f)}}}_{i} } \right)} = a\sigma ^{2}_{0}$$

Finally, from 4 and 5, we get

$$a = \frac{{h^{2} }} {2}$$

Expressing $\sigma ^{2}_{1}$ and $\sigma ^{2}_{2}$ from $\sigma ^{2}_{0}$ and h²

Replacing a in Eq. (2) gives $y_{i} = \frac{{h^{2} }} {2}y^{{{\left( f \right)}}}_{i} + \varepsilon _{{i1}}$.

This result is rather counter-intuitive, as it means that if we do not consider the residuals of models 2 and 3, Eq. (3) resolves to Eq. (2) by setting the mother’s value to 0. In fact that loss of information results in an increase of the residual variance as seen below.

Indeed, from Eq. (2), $\operatorname{\rm var} {\left( {y_{i} } \right)} = \frac{{h^{4} }} {4}\operatorname{\rm var} {\left( {y^{{{(f)}}}_{i} } \right)} + \sigma ^{2}_{1}$.

And from Eq. (1) $\operatorname{\rm var} {\left( {y_{i} } \right)} = \operatorname{\rm var} {\left( {y^{{{(f)}}}_{i} } \right)} = \sigma ^{2}_{0}$.

It follows that

$$\sigma ^{2}_{1} = \sigma ^{2}_{0} {\left( {1 - \frac{{h^{4} }} {4}} \right)}$$

In the same way, considering Eq. (3) leads to

$$\sigma ^{2}_{2} = \sigma ^{2}_{0} {\left( {1 - \frac{{h^{4} }} {2}} \right)}$$

The relative reduction in residual variance based on knowledge of the value of a parent is $\frac{{h^{4} }} {4}$.

Finally, we can synthesize Eqs. (1), (2), and (3) in a unique formulation

$$y_{i} \tilde{}N{\left( {h^{2} {\left( {\frac{{y^{{{\left( f \right)}*}}_{i} + y^{{{\left( m \right)}*}}_{i} }}{2}} \right)},\sigma ^{2}_{0} {\left( {1 - I_{m} {\left( i \right)}\frac{{h^{4} }}{4} - I_{f} {\left( i \right)}\frac{{h^{4} }}{4}} \right)}} \right)}$$

(6)

where $\left\{ {\begin{array}{*{20}c} {I_{m} {\left( i \right)} = 0{\text{ if }}y^{{{\left( m \right)}}}_{i} {\text{ is missing,}}} \\ {I_{m} {\left( i \right)} = 1{\text{ otherwise}}} \\ \end{array} } \right.$ and $\left\{ {\begin{array}{*{20}c} {y^{{{\left( m \right)}*}}_{i} = 0{\text{ if }}y^{{{\left( m \right)}}}_{i} {\text{ is missing,}}} \\ {y^{{{\left( m \right)}*}}_{i} = y^{{{\left( m \right)}}}_{i} {\text{ otherwise}}} \\ \end{array} } \right.$ and $\left\{ {\begin{array}{*{20}c} {I_{f} {\left( i \right)} = 0{\text{ if }}y^{{{\left( f \right)}}}_{i} {\text{ is missing,}}} \\ {I_{f} {\left( i \right)} = 1{\text{ otherwise}}} \\ \end{array} } \right.$ and $\left\{ {\begin{array}{*{20}c} {y^{{{\left( f \right)}*}}_{i} = 0{\text{ if }}y^{{{\left( f \right)}}}_{i} {\text{ is missing,}}} \\ {y^{{{\left( f \right)}*}}_{i} = y^{{{\left( f \right)}}}_{i} {\text{ otherwise}}} \\ \end{array} } \right.$

Different heritability parameters depending on the gender of the parents and the sibs

Let us denote h₁₁, h₁₂, h₂₁, and h₂₂ heritability parameters of the father-son, father-daughter, mother-son, and mother-daughter pairs, respectively.

It can be shown, in a similar way as above, that model 6 becomes

$$y_{i} \tilde{}N{\left( {\frac{1}{2}h^{2}_{{1,s{\left( i \right)}}} y^{{{\left( f \right)}*}}_{i} + \frac{1}{2}h^{2}_{{2,s{\left( i \right)}}} y^{{{\left( m \right)}*}}_{i} ,\sigma ^{2}_{0} {\left( {1 - I_{f} {\left( i \right)}\frac{{h^{4}_{{1,s{\left( i \right)}}} }}{4} - I_{m} {\left( i \right)}\frac{{h^{4}_{{2,s{\left( i \right)}}} }}{4}} \right)}} \right)}$$

(7)

where s(i) takes the value of 1 if individual i is a man, 2 if i is a woman.

Appendix B

The Winbugs program to implement model 7 is given below:

model

{

for(i in 1:n)

{

y[i] ∼ dnorm(mu[i],tau.i[i])

mu[i]<− m + h2.f[sexe[i]]/2*(y.f[i]−m)*I.f[i] + h2.m[sexe[i]]/2*(y.m[i]−m)*I.m[i]

tau.i[i]<- tau/(

1−pow(h2.f[sexe[i]],2)*I.f[i]/4−pow(h2.m[sexe[i]],2)*I.m[i]/4)

res[i]<− y[i] − mu[i]

}

# priors

m ∼ dnorm(0.0,1.0E-6)

h2.f[1] ∼ dunif(0.0,1.0) # father-son

h2.f[2] ∼ dunif(0.0,1.0) # father-daughter

h2.m[1] ∼ dunif(0.0,1.0) # mother-son

h2.m[2] ∼ dunif(0.0,1.0) # mother-daughter

tau ∼ dgamma(0.001,0.001)

sigma<- 1 / sqrt(tau)

# signification levels

# ( =2*min(p,1−p) )

p.h2.f[1]<− step(h2.f[1])

p.h2.f[2]<− step(h2.f[2])

p.h2.m[1]<− step(h2.m[1])

p.h2.m[2]<− step(h2.m[2])

}

Note that all distributions in Winbugs are parameterized in precision (the inverse of the variance).

Data file extract:

list(

sexe=c(1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2,...

y=c(0.724, 0.637, 0.727, 0.893, 0.957, 1.101, 1.015, 1.073, 1.012, 0.767,...

y.p=c(0, 0, 0, 0, 0.893, 0.893, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,...

y.m=c(0, 0, 0, 0.727, 0.767, 0.767, 0.727, 1.015, 0.727, 0, 0, 0, 0.851, 0,...

I.m=c(0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0,...

I.p=c(0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0,...

n=301)