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So, I thought I understood how to calculate the probability for a pedigree, until I came across this answer in my textbook. The question is: what is the probability the offspring of IV1 x IV2 will show the recessive trait? Assume individuals outside the family are not carriers of r unless there is evidence.
White = Dominant phenotype, Black = recessive phenotype.
What I did is:
III3 = 1/1 Rr, III4 = 1/2 Rr, IV1 = (1/2*1/1*2/3=1/3)
III7=1/2 Rr, III8= 1/2 Rr, IV=(1/2*1/2*2/3=1/6)
IV1 x IV2 = 1/3*1/6*1/4 = 1/72.
The answer in the book is 2/35. Can somebody please explain where I went wrong? enter image description here

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Here is my attempt!

Generation III

  • III3 is rR

  • II3 is rR and therefore III4 is rR with probability $\frac{1}{2}$ and RR with probability $\frac{1}{2}$.

  • Just like III4, III7 and III8 are rR with probability $\frac{1}{2}$.

Generation IV

Let's start with IV1.

  • IV1 is rR if ...

    • III3 transmits the r allele (probability $\frac{1}{2}$)
    • III4 is rR (probability $\frac{1}{2}$) and transmits the r allele (probability $\frac{1}{2}$). The total probability is $\frac{1}{4}$.
  • However, we have to remove the probability that both parent transmit the r allele as we know IV1 is not rr. This probability is $\frac{1}{2} \frac{1}{4} = \frac{1}{8}$. So, in total, the probability of IV1 to be rR is $\frac{1}{2} + \frac{1}{4} - \frac{1}{8} = \frac{5}{8}$.

Now, IV2

  • IV2 is rR if ...
    • III7 is rR (probability $\frac{1}{2}$) and transmits the r allele (probability $\frac{1}{2}$) or if III8 is rR (probability $\frac{1}{2}$) and transmits the r allele (probability $\frac{1}{2}$).
  • However, we have to remove the probability that IV2 is rr as we know this is not the case. The total probability is therefore $\frac{1}{4} \frac{1}{4} - \frac{1}{16} = \frac{7}{16}$.

Generation V

Finally, we get to your question

[W]hat is the probability the offspring of IV1 x IV2 will show the recessive trait?

Both parents need to be rR which happen at probability $\frac{5}{8} \frac{7}{16} = \frac{35}{128}$ and both need to transmit the $r$ allele which happen with probability $\frac{1}{4}$. Hence the probability is $\frac{35}{512}$.

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