Structure

• Define families with genders and relationships.
• Define easy and intermediate relationship predicates.
• Specify pedigree collapse determination using the Coefficient of Inbreeding (Wright, 1922) (see SOURCES [4]) by using helper predicates and bundling them into advanced logic.
• Define helper functions for the COI calculation.
• Sources.

Goal of This Program

This program aims to determine the degree of pedigree collapse (incest) in families. The Coefficient of Inbreeding (COI) is measured with Wright's 1922 formula (see SOURCES [4]).

Note: The pedigree collapse coefficient is usually not used in science due to misleading approximations. For scientific purposes, COI is normally used.

Usage: coi

?- coi(<COI>, <Person>, <BaseRiskRate>).

• <COI> should not be set with a value.
• <Person> can be any person in the knowledge base (e.g., charles_II_kingOfSpain) or a variable.
• <BaseRiskRate> is a float in [0, 1].
• Recommended base risk rates: 0, 0.03, 0.05.

Example:

coi(COI, charles_II_kingOfSpain, 0.03).

Usage: cois_to_json

?- cois_to_json(<BaseRiskRate>, <JSON>), write_jsoned_cois(<JSON>).

• Set <BaseRiskRate> to a value like 0.03.
• Use JSON as the variable for <JSON>.
• This writes JSON-formatted COIs to the console.

Formula

FP = SUM[ 0.5 ^ (N1 + N2 + 1) * (1 + FCA) ]

• FP: Coefficient of Inbreeding of person P.
• SUM: Sum of all results for each common ancestor of P.
• N1: Number of generations between P and a found common ancestor of P's father (Parent1).
• N2: Same as N1, for the mother (Parent2).
• FCA: Coefficient of Inbreeding for the found common ancestor.

Example Equation

Example: see the simple animal relationship example (see SOURCES [3], page 150):

• animal_P has the same grandfather animal_c.
• The equation is:

FP = 0.5 ^ (1 + 1 + 1) * (1 + 0) = 0.125

• N1 is 1, as one generation difference between grandfather and father.
• N2 is 1, as one generation difference between grandfather and mother.
• FCA is 0, as no information about the grandfather's ancestors was given.
• Result: 0.125 → 12.5% of the genes are duplicates.

Possible Further Steps

• Plot the solution for a given person, or the average of many people, with different base rates.
  • Ask teammates if anyone does Graphviz in Prolog (visualization).
  • One person as a function of base rate [0, 10].
  • All people in a population, averaged, with base rates [0, 10].
• Check other methods of evaluating pedigree collapse.