Calculate heritability from an anova result QUESTION
Is it possible to extrapolate the heritability from an ANOVA result?
POSITIONING THE PROBLEM
Heritability is an indicator specific of a phenotype and a an experiment. It is usually calculate with mixed-model allowing us to estimate the $V_{G}$ (genetic variance) and the $V_{R}$ (residual variance)
There Broad sense heritability is given by the following formula:
$h^2_{bs} = V_{G}/(V_{G}+V_{R}/nrep)$
$nrep$ being the mean number of repetition for one genotype in the experiment.
My problem is that I don't have access to raw data for an experiment, only the result of an ANOVA like this one:  
           Df Sum Sq Mean Sq F value   Pr(>F)  
  data$clon  2 2437.1  1218.5  8.0915 0.001765 **  
  Residuals 27 4066.1   150.6   

PRACTICAL EXAMPLE DATA WITH R 
set.seed(15)
data <-data.frame(clon=as.factor(c(rep(1,10),rep(2,10),rep(3,10))),value=c(runif(n=10,min=30,max=70),runif(n=10,min=10,max=60),runif(n=10,min=40,max=90)))
res <- lm(data$value~data$clon)
res.lmer <- lmer(data$value~1|clon,data=data)
anova(res)  

Analysis of Variance Table

Response: data$value
      Df Sum Sq Mean Sq F value   Pr(>F)   
      data$clon  2 3064.4 1532.22  7.7408 0.002201 **
      Residuals 27 5344.4  197.94                    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

res.lmer
Linear mixed model fit by REML ['lmerMod']
Formula: data$value ~ 1 | clon
    Data: data
REML criterion at convergence: 243.1439
Random effects:
  Groups   Name        Std.Dev.
  clon     (Intercept) 11.55   # THIS is the squart of VG
  Residual             14.07   # THIS is the squart of VR
  Number of obs: 30, groups:  clon, 3
  Fixed Effects:
  (Intercept)  
        53.31  

I've tried to go empiricaly
Looking at the result, I found $V_{R}$ could be find in ANOVA with the $mean.sq$ Residuals value. But I couldn't find a good formula to calculate $V_{G}$. the best empirical result i've got was with the formula: $mean.sq/(nrep+1)$ using mean.sq of Clon value but it is not quite good.
I will be welcoming a more theorical point of view, or a better empirical way to do it. Even explaniation of why it is impossible (or assumptions needed to extrapolate such an Anova results)
 A: I estimate heritability (broadsense) as following.
H2 (broadsense) = Mean sq-group/(Mean sq-group+Mean sq-residual).
This is proportion of genetic varaiance out of total phenotypic variance, this is heritability in broadsense not in narrow sense. Heritability in narrow sense in the proportion of additive genetic variance out of total phenotypic variance.
A: $$H^2=\frac{\sigma^2_G}{\sigma^2_G+\sigma^2_e/r}$$




Term
Mean Sq
MS in terms of Variance




G
1532.22
$\sigma^2_e + r\sigma^2_G$


e
197.94
$\sigma^2_e$




This implies that  $\sigma^2_e = MS_e = 197.94 $ and $ \sigma^2_G = (MS_G -MS_e)/r = (1532.22 -197.94)/10 = 133.428 $ and therefore $$H^2=\frac{\sigma^2_G}{\sigma^2_G+\sigma^2_e/r}=\frac{133.428}{133.428+197.94/10}=0.870815 $$
A: The between-group variance is the genetic variance (VG) and the within-group variance is the "noise" or environmental variance (VE). Heritability ("narrow sense", h2) is the proprtion of VG out of the total variance (VG+VE), as you have described. Thus, using ANOVA:  
Sum Sq-group = (Mean Sq-group)/(df-group)) = V-between-groups = VG (genetic variance)
Sum Sq-residual = (Mean Sq-residual)/(df-residual) = V-within-groups (or V-residual) = VE (environmental variance)
h2 = VG/(VG+VE) 
