In my previous post SAMIAM Introduction, we have succeeded in helping the bank modele a simple Credit-worthness Predictor! That’s fabulous! Do you have the feeling that we have found a new world! I really appreciate my teacher, Prof.Chen, who taught me the course - Probability and Stochastic Process. He inspired my interest in Probability and I was the top 3 student in his class. This lays a strong foundation for PGM.

Task Description:

Today, we will continue our exploration in PGM. This time, we will help Genetic Counselors, Milly Osworth(NPC in the figure below), advise couples with a family history of a genetic disease. Specifically, we need to help couples decide whether to have a biological child or to adopt by assessing the probability that their unborn child will have the disease.

Completion:

Upon completion of this task you will gain:

• 1000 experience
• 50 reputation with Biologists
• 100 skill proficiency in PGM

TASK GUIDE & WALKTHROUGH

• Problem Analysis
• Bayes Network
• Alternate Bayes Network
• Traits Controlled by Multiple Genes
• Reference

1. Problem Analysis

The question is so familiar for us, because it seems that this is just a common biological exercise in high school! Half and half. This is different from the high school exercise because we didn’t learn Probability Theory in high school. For a high school student, if “E” is the dominant allele for double eyelids and “e” is the recessive allele, then a person with genotype “EE” will have the physical trait, double eyelids, with one hundred percent probability. This is called Mendelian genetics. In Mendelian genetics, a dominant allele is always expressed, and a recessive allele is expressed only when the domiant allele is not present. A person with genotype “EE” or “ee” is called homozygous for the gene for double eyelids and “Ee” is called heterozygous for the gene for double eyelids.

In fact, inheritance of many physical traits is not Mendelian. That’s what we need to solve now. For example, the traits of polydactyly which is the trait of having an additional finger on one or both hands. Assuming there are two alleles for this gene, “P” and “p”, where “P” is dominant. But here “P” doesn’t cause the polydactyly with one hundred probability but is just a risk factor. While “p” might not prevent someone from having polydactyly but just make the person less likely to have polydactyly. For instance, “PP” might cause polydactyly with probability 0.8, while “pp” will cause polydactyly with probability 0.05. (It is not necessary that all the probability sum to 1)

Because Gene passed down from generation to generation and the relationship between genotype and phenotype is represented by a risk factor. If we use a graph to represent the whole relationship, it is convenient to use a directed graph. So we can use Bayes Network to model the mechanism of genetic inheritance.

Genetic inheritance patterns are generally consistent from generation to generation, so template model is a natural way to model them. Then parameters and structure are reused within all the Bayes Networks and across different Bayes Networks.

2. Bayes Network

Milly Osworth wants you to predict the Cystic Fibrosis in a family: Given the family tree, allele frequency in the population, and the risk factor for all the genotypes. Construct the Bayes Network and help the Genetic Counselors to give advice to the couples in this family.

2-1. Construct the Template

First, we need to construct a template which only contains a son(James), his mother(Ira) and father(Robin). Find the factors in this template, then construct Bayes Network for the whole family tree with those factors in the template. Template is as follows:

We find there are mainly three kinds of factors in this template:

P(person’s phenotype | person’s genotype)

P(person’s genotype | genotype of person’s first parent, genotype of person’s second parent)

P(person’s genotype) , its values are based on the allele frequencies in the population

Calculate the first kind of factor: consider a general case, a genotype may contains “n” allels, then the a person has C(2 in “n” )+n possible genotypes, (here C denotes combination). Given all the genotypes’ risk factor ,it is easy to complete it.

Calculate the second kind of factor: this is simple because we only need to use different combination and the probability is in 0, 1/2, 1.

Calculate the third kind of factor: given the alleles’ frequency, this factor is simply the product of frequencies of its constituent alleles in the population. That’s it!

2-2. Construct a Complete Bayesian Network

Having constructed all the necessary factors, you can now build a Bayesian Network for the inherantice of an autosomal trait with one gene. Put all the given information: pedigree, allele frequency, genotype risk factors. Show the Bayesian Network in SAMIAM.

2-3. Giving Advice

Congratulations! We have constructed a Bayesian Network for Genetic Inheritance. In practice, a genetic counselor often knows whether certain members of a family have a disease or have some genetic test results for the family members. So some of the information is obeserved such as the genotype and phenotype.

Now let’s try to give some advice to help the genetic counselor!

James and Rene are debating whether to have another child or adopt a child. They are concerned that, if they have a child, the child wil have cystic fibrosis because both of James’s parents have Ff genotype observed, and James’s cousin Eva has cystic fibrosis. No other information is provided. What’s the probability that their unborn child will have cystic fibrosis.

The following is the family tree and the probability when no information is provided. We find that the probability their unborn child will have cystic fibrosis is 19.70%

Then we add the information that James’ parents have one Ff genotype and his cousin Eva has cystic fibrosis. They probability their unborn child will have cystic fibrosis increase to 38.50% Also we notice that the probability didin’t change whether we have the information James’s cousin Eva’s information. Because James’s parents’ information is observed and the active trail is blocked. The influence didn’t transfer to his unborn child.