33. Genetic and phenotypic components

Covariance in genetic and environmental components of weight variability of twins.

• $\eta$ is the phenotype (weight)
• $\xi$ is the genetic component of the phenotype
• $\epsilon$ is the environmental effect

(a) Calculate the covariance and correlation coefficient between ${\eta }_1$ and ${\eta }_2$ in identical twins.

Let's individually look at $E({\eta }_i)$:

Given that the mean of ${\epsilon }_i$ has been scaled to zero, its $E({\epsilon }_i)=0$. And $c$ is a constant baseline weight, so $E(c)=c$. Therefore,

This also means that $E({\eta }_1)=E({\eta }_2)=E(\eta )$, and the same for variance.

Plugging into our Covariance alternative formula

Now we can apply the expectation to all of these terms. Since the $E(\epsilon )=0$, all the terms containing $\epsilon$ as a factor will be zero.

Since $E(c^2)=c^2$ and $E(2c\xi )=2cE(\xi )$ (expectation with a coefficient), these get subtracted and we end up with just:

This is same as the equation for the True Variance, therefore:

Correlation coefficient is therefore just (since $V({\eta }_1)=V({\eta }_2)$):

(b) The quotient $V(\xi )/V(\eta )$ is called the broad sense heritability, which tells which fraction of the observable phenotypic variance $V(\eta )$ is due to genetic effects. Suggest a practical way to measure the heritability of body weight and comment on if this measurement is correct for the entire population.

With identical twins who share the same genetics, we can determine the genetic component of variance by measuring their weights, except we would only have two data points. Also, they would have shared environmental effects so they would not have a mean of zero.

36. ANOVA (Maths) with groups of animals

In our example:
$n=10$
$k=4$
$\overline{\xi }=4.025$

"Full variance":

$var=2.625$

"Average of the sample variance":

average of sample variance = $1.2875$

F-statistic

$F=2.625/1.2875=2.039$
Therefore, the F-statistic is less than the threshold and we fail to reject the null hypothesis. It is possible this result is because of random chance.

41. Truncation selection in animal breeding

Continuous Distributions
This is a normal distribution $N(10,\sqrt{5})$.
(a) When $T=13.67$, there is a 5% chance of all the animals being bigger than this weight.
(b) When $T=9.43$, there is a 60% chance of all the animals being bigger than this weight.

42. Bayes' Theorem with continuous distributions

(a)
Probability of being this height if you are a Finnish man: P(data|man) = 0.0682
Probability of being this height if you are a Finnish woman: P(data|woman) = 0.1535

From Bayes' Theorem, we calculate the posterior:

So there is a 31% chance the suspect was a man given the height data.

If the prior probability of being a man $P(man)$ was instead 0.80 (and thus $P(woman)=0.2$):

(b) See solution text.

Extra exercise (an example that came up during the lecture):

Suppose we take two measurements, $\xi$ and $\eta$, on every individual of a large population. There are only four combinations of numbers we get:

• (i) xi=1, eta=0;
• (ii) xi=-1, eta=0;
• (iii) xi=0, eta=1;
• (iv) xi=0, eta=-1
  It is clearer if you sketch these as four points in a coordinate system. Each combination occurs with probability 1/4. xi and eta are obviously not independent: if xi=0, we know that eta is not zero and if eta=0, we know that xi is not zero.

Show that, nevertheless, Cov(xi,eta) is zero

Sketch as four points in coordinate system:

The average values $\overline{\xi }$ and $\overline{\eta }$ are:

So our sample covariance is:

In every case given above, either $\xi$ or $\eta$ is 0. So the numerator will be zero in each part of the sum.