Reading Pedigrees: Human Inheritance

A woman’s brother has cystic fibrosis. Her parents do not, and neither does she. Asked the chance that she carries a cystic fibrosis allele, most students say 1/2. Under the standard model the answer is 2/3.

You cannot cross people, so human genetics works backwards from the families that already exist. The data set is a pedigree, a family tree drawn to a code. This lesson teaches you to read the code, to test the five simple modes of inheritance against a chart, and to calculate the probabilities a pedigree supports, including that 2/3. Then it shows why real pedigrees are less certain than they look.

Reading the chart: a family tree drawn to a code

A pedigree uses a small set of symbols. In the 2022 revision of the standard from the National Society of Genetic Counselors, a square is a man or boy, a circle is a woman or girl, and a diamond is a nonbinary person or someone whose sex or gender is not specified. The revision ties the shape to gender and records sex assigned at birth separately. A filled symbol means affected, for the condition named in the chart’s key. A diagonal slash through a symbol marks someone who has died.

Lines carry the relationships. A horizontal line joins two partners. A vertical line drops from it to a horizontal sibship line, from which each child hangs. A double partner line marks consanguinity: the partners are related by descent, as first cousins are. Generations are numbered in roman numerals (I, II, III) and people within a generation from left to right, so II-3 is a single person. An arrow marks the proband, the person through whom the family came to attention.

A carrier is a person heterozygous for a recessive allele, who carries it without showing the trait. Many textbook charts put a dot in the centre of a circle for a woman known to carry an X-linked recessive allele. The 2022 revision changes how carrier status is shown, so always read the key.

Five patterns, and the one observation that rules each out

Reading a pedigree resembles checking alibis. You rarely prove that a mode is right, but one observation can show that a mode is wrong. Each rule below assumes full penetrance (everyone with the genotype shows the trait), no new mutation and correct family relationships.

1. Autosomal dominant. One copy of the allele, in a heterozygote, is enough. The trait usually appears in every generation, in both sexes, and can pass from father to son. A heterozygous parent with an unaffected partner passes the allele to each child with probability 1/2. Ruled out by: an affected child of two unaffected parents.
2. Autosomal recessive. Affected people have two pathogenic alleles at the locus, the same one twice or two different ones. They are usually the children of two unaffected carriers, so the trait appears among brothers and sisters of either sex, with few affected relatives elsewhere. Partners related by descent are more often seen, because they are more likely to carry the same rare allele. Ruled out by: two affected parents with an unaffected child.
3. X-linked recessive. A male has one X, so one copy of the allele affects him. Affected men are much more common than affected women, and the allele passes from an affected man through his carrier daughters to half of their sons. Heterozygous women can show some signs, partly through X-inactivation. Ruled out by: father-to-son transmission, since a father gives his son a Y.
4. X-linked dominant. An affected father passes it to all his daughters and none of his sons; an affected heterozygous mother passes it to half her children of either sex. Ruled out by: an affected father with an unaffected daughter, or with a son who inherited it from him.
5. Mitochondrial. Human mitochondrial DNA normally comes from the mother, so an affected mother can have affected children of both sexes and an affected father passes the trait to none. Mitochondrial DNA encodes only a handful of respiratory-chain subunits (see oxidative phosphorylation); most mitochondrial proteins are encoded in the nucleus, so many mitochondrial diseases follow Mendelian patterns instead. Ruled out by: transmission from an affected father.

Two traps follow. Consanguinity raises the chance of a recessive disorder but does not prove that a trait is recessive. And “skips a generation” is no proof of recessive inheritance, because a dominant allele with reduced penetrance does the same. Rett syndrome and fragile X syndrome, which you meet in synapse disorders, are X-linked examples whose pedigrees are complicated by new mutations and repeat expansion.

Conditional probability: the 2/3 rule

Worked problem, part 1. A woman’s brother has cystic fibrosis, an autosomal recessive disorder caused by two pathogenic alleles of the CFTR gene. Her parents are unaffected, and so is she. Her partner has no family history. What is the probability that their first child is affected?

Assume full penetrance, no new mutations, correct relationships, and a carrier frequency of 1/25 in the partner’s population. That figure is often quoted for people of northern European ancestry and is lower in many other populations; it is an assumption of the exercise, not a universal number.

1. Her parents. Two unaffected parents have an affected child, so both are carriers, Aa.
2. Her genotype before her phenotype is known. From Aa x Aa, four equally likely outcomes: AA, Aa (A from the mother), aA (A from the father) and aa, each 1/4.
3. Condition on what you know. She is unaffected, so aa is impossible. Three equal outcomes remain, two of them carriers:

P(carrier | unaffected) = (1/2) / (1/4 + 1/2) = (1/2) / (3/4) = 2/3

4. Her partner. P(carrier) = 1/25.
5. The child. The child is affected only if both parents are carriers and both pass a, which has probability 1/4 for two carriers:

P(affected child) = 2/3 x 1/25 x 1/4 = 2/300 = 1/150

Check it another way: 1/150 is about 0.0067, and 0.667 x 0.04 x 0.25 = 0.0067.

The answer 1/2 is the carrier probability before you learn that she is unaffected. Knowing her phenotype removes one of four outcomes, and the denominator shrinks from 4 to 3. These numbers are exercises in the logic, not risk figures for a real family.

Updating with evidence: Bayes and the unaffected sons

Conditioning can use more than one fact. Bayes’ rule combines what you believed before (the prior) with how likely the new evidence is under each hypothesis (the likelihood).

Worked problem, part 2. A woman’s mother is a known carrier of an X-linked recessive allele that is fully penetrant in males. The woman has not been tested and has two sons, both unaffected. What is the probability that she is a carrier?

1. Prior. She received one of her mother’s two X chromosomes, so P(carrier) = 1/2 and P(not carrier) = 1/2.
2. Likelihood. If she is a carrier, each son is unaffected with probability 1/2, so two unaffected sons have probability (1/2)^2 = 1/4. If she is not, the probability is 1.
3. Joint. Carrier: 1/2 x 1/4 = 1/8. Not carrier: 1/2 x 1 = 1/2.
4. Posterior.

P(carrier | two unaffected sons) = (1/8) / (1/8 + 1/2) = (1/8) / (5/8) = 1/5

Check with odds. Prior odds are 1 : 1, the evidence is four times likelier if she is not a carrier, so posterior odds are 1 : 4, which is 1/5. Under the same model, the chance that her next son is affected is 1/5 x 1/2 = 1/10.

Her sons did not change her genotype. They changed what you know about it.

Why real pedigrees mislead

Every rule above assumed a tidy family. Real families break the assumptions in five common ways.

Reduced and age-dependent penetrance. In Huntington disease, inherited as a dominant, symptoms usually begin in adulthood, so a young unaffected child of an affected parent may carry the allele. Penetrance also depends on the length of a CAG repeat in HTT: about 36 to 39 repeats is usually described as reduced penetrance and 40 or more as full penetrance. Variable expressivity, the same genotype giving different clinical pictures in different people, blurs the chart further.

New mutations. A new pathogenic variant can cause a dominant disorder in a child with no affected parent, which looks like the exclusion rule for dominance.

Germline mosaicism. A parent can carry a variant in only some cells, including some sperm or eggs. In one study of 58 families with one affected child and an apparently new variant, sequencing several parental tissues estimated a recurrence risk below 0.1% for 35 families, while 5 had parental mosaicism associated with higher risk. In three fathers the variant was in 5.6 to 12.1% of sperm DNA. In two further families with repeated affected pregnancies, routine testing of parental blood had not identified the parent of origin.

Heteroplasmy. A cell can hold a mixture of normal and mutant mitochondrial DNA. A heteroplasmic mother can pass different proportions to different children, so severity varies and some children may be unaffected.

Small families. Two carriers with two children have no affected child with probability (3/4)^2 = 9/16, more than half the time. Such families look as if nothing is inherited at all.

Apparent exceptions need molecular checks. When 11,035 trios were searched for mitochondrial DNA from fathers, the seven candidate signals were each explained by a mitochondrial-derived insert in a nuclear chromosome shared by father and child.

Design problem

An affected woman has five children, all affected. Her affected son has three children, none affected. Her affected daughter has two children, both affected. Severity varies among the affected. Decide between mitochondrial and X-linked dominant inheritance, and say what data would settle it.