# Pedigree symbols

Males are represented by squares.

Females are represented by circles.

Filled symbols show individuals who exhibit the trait in question.

A horizontal line between two symbols represents a mating.

A vertical line descends from parents to a horizontal line shared by all their offspring.

A diagonal line through a symbol indicates that person is deceased.

Roman numerals (I, II, III,…) represent generations.

Arabic numerals (1, 2, 3,…) represent the birth order of siblings.

# Pedigree Analysis

## Pedigrees reflecting a family affected by an Autosomal recessive disease typically show the following:

– Healthy couples have affected children (characteristic of a recessive trait)

– number of affected males = number of affected females (the pedigree should be big enough to show a large number of descendants or a statistical study with a large sample size should be performed (this is a characteristic of autosomal traits, whether dominant or recessive)

– Affected couples have ALL their children affected.

– Consanguineous matings increase odds for recessive trait (relatives may both be carriers of recessive allele)

# Genetic Risk calculation

A three-generation pedigree for a particular human genetic disease is shown in the following figure:

a. What is the mechanism of inheritance for the trait?

b. Which persons in the pedigree are known to be heterozygous for the trait?

c. What is the probability that III-2 is a carrier (heterozygous)?

d. If III-3 and III-4 marry, what is the probability that their first child will have the trait?

(source: iGenetics a molecular approach, Peter J Russel)

Answers:

a) Ok, let’s consider the mechanism of inheritance of this trait:

-Healthy couples I1 and I2 have affected children (II3 and III1 respectively), this shows that the allele of this disease was masked in parents and appeared in their affected children, hence the morbid allele is recessive.

– The gene of this disease is not carried by the non-homologous fraction of Y since there are females affected with this disease and females do not have a Y chromosome.

– The gene of this disease is not carried by the non-homologous fraction of the X chromosome either since, if it were there, the genotype of the affected daughter III1 would’ve been X^{d}X^{d} and she would’ve received X^{d} from her father II1 whose genotype would’ve been X^{d}Y and he would’ve been affected, which is obviously not the case.

– The gene is thus autosomal (or pseudo-autosomal)

b) I1 and I2 as well as II1 and II2 as they have affected children whose genotype is dd and who surely received the morbid allele d from each of their healthy parents. Also, III4 and III5 are heterozygotes as their mother is affected and her genotype is dd, she must have transmitted the d allele to each of her healthy children.

c) Since III2 is phenotypically normal, her genotype is either Nd or NN, the ratio of these genotypes in the Punnet square is 2:1 and hence the probability of III2 being a carrier is 2/3.

d) III3 is just like III2 in that he has 2/3 as probability of being heterozygous, III4 is however different, he is surely heterozygous as she must have received the morbid allele from her affected mother II3, therefore the risk for this couple to have an affected child is :

Risk of the father being heterozygous (2/3) x risk of the mother being heterozygous (1/1) x risk of each of them transmitting the morbid allele to the child (1/4)= 1/6