Can you marry your half-sister's daughter?
Gary Trudeau poses an interesting question:
Let's lend Zipper and Jeff a hand. Here is a pedigree of the family with Jeff, Alex (Jeff's half-niece), J.J. (Jeff's half-sister and Alex's mom), and Joanie (Jeff and J.J.'s mom) labeled. Jeff and Alex are the filled in symbols. Circles indicate females and squares indicate males. Horizontal lines are matings and the vertical lines lead downward toward the progeny of that mating.
We will define the inbreeding coefficient (f) as the probability that both alleles in a single individual are identical by descent (IBD). IBD simply means that both alleles came from the same allele in one of the ancestors -- in this case, Joanie. We are trying to figure out what is the probability that Jeff and Alex's hypothetical child would get the same allele from both of his/her parents. To do this, we will redraw the pedigree so that we only focus on the individuals in question. Each passing of gametes (birth) is represented by an arrow. Only one of the two parents is shown for each mating. Jeff and Alex's hypothetical child is represented by a diamond.
Joanie has two alleles for the gene in question -- let's call it the "A" gene -- A1 and A2. She could have passed either allele on to either of her children, J.J. and
- The probability Jeff gets the A1 allele from Joanie is 0.5. The probability he passes on the A1 allele to his child assuming he has the allele is 0.5. The probability he passes on the A1 allele to his child is the product of these two probabilities: 0.25.
- The probability Alex gets the A1 allele from Joanie is the probability J.J. gets the allele from Joanie times the probability J.J. passes it along to Alex. The product of those two events (0.5 x 0.5) is 0.25. The probability Alex passes on the A1 allele to her child assuming she gets the allele from Joanie is 0.5. Therefore, the probability Alex passes along the A1 allele to her child is the probability she gets it times the probability she passes it along, or 0.125 (0.25 x 0.5).
- Now that we have the probability Jeff passes the A1 allele to his child (0.25) and the probability Alex passes the allele to her child (0.125) we can calculate the probability they both pass the allele on to their hypothetical child if they were to mate. This just the product of the previous two probabilities (0.25 x 0.125): 0.03125, or 1/32.
Up to this point, we have only dealt with the A1 allele. Jeff and Alex's hypothetical child could also receive the A2 allele from both parents. Repeating the steps above for the A2 allele gives the same answer as for the A1 allele, 1/32. Because the child can either have the genotype A1A1 or A2A2, the two events are mutually exclusive. This means we can add the probability that the child will be A1A1 and the probability the child will be A2A2 to get the probability the child will have two alleles that are IBD for a single gene. This gives us a final answer of 1/16.
For any given gene, Jeff and Alex's child has a 6.25% chance (1/16) of being IBD. For most genes this should not be a problem, but if the child ended up being homozygous for a recessive deleterious allele, this could be catastrophic. As the probability of being IBD at a given locus increases, so does the probability of having two copies of a recessive deleterious allele. As a point of comparison, the probability of being IBD for a child from a mating between first cousins is
1/32 1/16. Matings between cousins is generally frowned upon because of the risks associated with heritable diseases. Jeff and Alex's child would have twice the same risk of a heritable disease due to recessive deleterious alleles than as a child from a mating between first cousins.
In conclusion, Jeff can have his half-niece Alex stay with him, but only Zipper is allowed to fool around with her.
Hedrick, PW. 1999. Genetics of Populations. Jones and Bartlett Publishers, Sudbury, MA, USA.