# Pedigree - Dealing With Unknown Parent Genotypes

I'm dealing with a pedigree problem, and I'm having some trouble dealing with problems of unknown parental genotypes (where there are multiple possibilities). This is not a homework question; it's an exam practice problem. The question asks: Suppose brother and sister 6 and 7 are mated. What is the probability that their first pup will be albino?

The answer choices are 1/4, 1/8, 1/16, 1/32, or Not Enough Info.

Here's my attempt at the problem:

Since #2 is black, and some of its offsprings ended up sepia, #2 must carry the sepia gene (since the male mate is cream, and sepia is dominant over cream). So, we know that #2 is $$Cc^k$$.

Now, the male mate can be either $$c^d c^d$$, or $$c^d c^a$$.

This means that in order for the offspring of #6 and #7 to be albino, we need:

1) Male parent needs to carry the albino gene (i.e. male needs to be $$c^d c^a$$).

2) Male parent needs to pass on the albino gene to both #6 and #7.

3) #6 and #7 both need to pass on the albino gene.

I think the probability that the male is $$c^d c^a$$ is $$\frac{1}{2}$$, since the male parent can be either $$c^d c^a$$ or $$c^d c^d$$.

If we draw out a punnett square, the probability that the albino gene gets passed onto 1 individual is $$\frac{1}{2}$$ (since we already know the offsprings are sepia, we only need to consider the genotypes that contain $$c^d$$).

The probability that #6 and #7 pass on the albino gene is $$\frac{1}{4}$$.

This led me to think that the final probability is then P(male parent has albino gene) * P(#6 inherits albino gene) * P(#7 inherits albino gene) * P(#6 and #7 pass on the albino gene), which is $$\frac{1}{2}$$ * $$\frac{1}{2}$$ * $$\frac{1}{2}$$ * $$\frac{1}{4}$$, which gives us $$\frac{1}{32}$$, but I do not feel confident in my answer. Specifically, I do not know how to account for the fact that the male parent can be either $$c^d c^d$$ or $$c^d c^a$$.

Is my thought process correct? If not, where in my logic did I go wrong? Thank you for your help.