Genotype of siblings given identity by descent status I have a question regarding genotype of siblings (specifically I am referring to the table on slide 19 here:
http://ibgwww.colorado.edu/workshop2005/cdrom/ScriptsA/evans/IBDestimation/IBD--2005.pdf  )

Say the prevalence of an allele, say $A_1$, is $p_1$ and the prevalence of allele $A_2$ is $p_2$ in the population. Is there anyway we can determine the conditional probability of the observed genotype of two siblings given the identity by descent (IBD) status (i.e. $P(\text{sibling }1=A_1A_1, \text{sibling }2=A_1A_1) = ?$, $P(\text{sibling }1=A_1A_2, \text{sibling }2= A_2A_2)=?)$?

I am asking the question because I am not sure how the author determines the probability in the table based on just knowing the IBD of the two siblings.
Can someone explain column of probability when $k=1$?  I don't really understand how one can get for example $p_1$ to the $2$ times $p_2$ in the table when $k=1$ when the sibling genotype are sibling $1=A_1A_1$ and sibling $2=A_1A_2$.
 A: I gather that you understand the columns $k=0$ and $k=2$, which correspond to the siblings having no identity by descent and complete identity by descent, respectively.
For $k=1$, the siblings share one allele. Thus there are three alleles in play, the one they share and the ones they each have that they don't share, and these three alleles are modelled as independent.
To get genotypes $A_1A_1$ and $A_1A_2$, the shared allele must be $A_1$ and the other two must be $A_1$ and $A_2$. The probability for three alleles to be $A_1$, $A_1$, $A_2$ is $p_1^2p_2$.
The more interesting case is where both genotypes are $A_1A_2$, since in this case both alleles could be the shared one. We can calculate this as the complement of the remaining probabilities, which are easier to determine: 
$$1-\left(p_1^3+2p_1^2p_2+2p_1p_2^2+p_2^3\right)=1-(p_1+p_2)^3+p_1^2p_2+p_2^2p_1=p_1p_2(p_1+p_2)=p_1p_2\;.$$
Or we can add the two cases: The probability for the shared allele to be $A_2$ and the other two to be $A_1$ is $p_1^2p_2$, and the probability for the shared allele to be $A_1$ and the other two to be $A_2$ is $p_1p_2^2$, for a total of $p_1^2p_2+p_1p_2^2=p_1p_2(p_1+p_2)=p_1p_2$.
