Recurrence Risk from Pedigrees — Compute recurrence risk from pedigree information - Genetics

SECTION 1

Historical Context & Motivation

For centuries, people noticed that certain conditions — like hemophilia or colorblindness — seemed to "run in families." But nobody had a reliable way to predict whether a child would inherit a particular trait. Doctors and families were left guessing. The idea of drawing a pedigree (a diagram of a family tree that tracks a trait across generations) changed everything. Pedigrees gave scientists a visual tool to spot patterns, and probability gave them the math to turn those patterns into actual predictions.

1865 • Mendel's Laws Published

Gregor Mendel studied pea plants and discovered that traits are passed from parent to offspring in predictable ratios. His work laid the foundation for understanding inheritance patterns.

1905 • Pedigree Charts Formalized

Geneticists began using standardized pedigree symbols — circles for females, squares for males, and shading for affected individuals — to track traits through human families.

1953 • DNA Structure Discovered

Watson and Crick revealed the double-helix structure of DNA. This discovery explained how genetic information is stored, copied, and passed on.

1970s–Today • Genetic Counseling Emerges

Medical professionals now combine pedigree analysis with probability to counsel families about the chances of passing on genetic conditions. Recurrence risk calculations are a key part of this process.

The big question that drove all of this work is simple: if a genetic condition has appeared in a family, what is the chance it will appear again in a future child? That chance is called the recurrence risk, and learning to compute it from a pedigree is one of the most useful skills in genetics.

SECTION 2

Core Principles & Definitions

Before you can calculate recurrence risk, you need to understand a handful of foundational ideas. These concepts are the building blocks for every pedigree problem you will encounter.

Alleles & Genotype

Every gene comes in different versions called alleles. Your genotype is the combination of two alleles you carry for a gene — one from each parent.

Dominant vs. Recessive

A dominant allele (written as a capital letter, like A) shows its effect even with just one copy. A recessive allele (lowercase, like a) only shows up when both copies are recessive (aa).

Carrier

A carrier is someone with one dominant and one recessive allele (Aa). They don't show the recessive trait, but they can pass the recessive allele to their children.

Autosomal vs. X-Linked

A gene on chromosomes 1–22 is autosomal. A gene on the X chromosome is X-linked. Males (XY) have only one X, so one recessive X-linked allele is enough to show the trait.

Recurrence Risk

The recurrence risk is the probability that a trait or condition will appear in a future child, calculated by combining the parents' likely genotypes with Mendel's laws.

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KEY TAKEAWAY

Think of alleles like playing cards. Each parent holds two cards for a gene and randomly deals one card to each child. A pedigree lets you figure out which cards each parent is probably holding. Once you know the cards, you can calculate the odds of any particular hand being dealt — that's the recurrence risk.

SECTION 3

Reading a Pedigree — Visual Explanation

A pedigree uses standard symbols so that any geneticist in the world can read it. Let's look at a simple three-generation pedigree for an autosomal recessive condition. Study the diagram below, then read the explanation that follows.

This pedigree shows three generations. Squares represent males and circles represent females. A filled (solid-colored) symbol means the person is affected by the condition. The horizontal line connecting a male and a female is a mating line, and the vertical lines dropping down lead to their offspring. Genotypes (like Aa or aa) are written beneath each individual.

Notice a few important clues in the diagram above. In Generation I, neither parent is affected, yet they produce an affected daughter in Generation II. This tells us the trait is recessive — both parents must be carriers (Aa). The affected daughter has the genotype aa, meaning she received one recessive allele from each parent. The question marks in Generation III represent children whose genotypes we want to predict. That's where recurrence risk comes in.

SECTION 4

The Mathematical Framework

Computing recurrence risk boils down to two steps. First, figure out the most likely genotypes of the parents by reading the pedigree. Second, use a Punnett square to find the probability of each possible offspring genotype. Let's look at the formulas and rules that make this work.

PUNNETT SQUARE PROBABILITY

P(genotype) = (number of boxes with that genotype) ÷ (total boxes)

A standard Punnett square has 4 boxes. If 1 out of 4 boxes shows genotype aa, then P(aa) = 1/4 = 0.25, or 25%.

MULTIPLICATION RULE (AND)

P(A and B) = P(A) × P(B)

When two events are independent, multiply their probabilities. For example, if the chance a parent is a carrier is 2/3 and the chance of passing the recessive allele is 1/2, the combined probability is 2/3 × 1/2 = 1/3.

BAYESIAN ADJUSTMENT (CONDITIONAL PROBABILITY)

P(carrier | unaffected) = (# carrier outcomes) \div (# unaffected outcomes)

When you know a person is unaffected, you can rule out the aa genotype. Among the remaining outcomes from a Punnett square of two carriers (Aa × Aa), there are 3 unaffected outcomes (1 AA + 2 Aa). Two of those 3 are carriers. So the probability that an unaffected child of two carriers is actually a carrier is 2/3.

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Why 2/3 and Not 1/2?

This trips up a lot of students! When a Punnett square for Aa × Aa gives 1 AA : 2 Aa : 1 aa, and we already know a person is not affected (not aa), we remove that possibility. Of the 3 remaining outcomes, 2 are carriers. So the adjusted probability is 2/3, not 1/2. This is called conditional probability — you're updating the odds based on new information.

SECTION 5

Inheritance Patterns & Their Recurrence Risks

The recurrence risk depends on the inheritance pattern of the trait. Different patterns produce different probabilities. The table below summarizes the most common patterns you'll see in pedigree problems.

Inheritance Pattern	Parent Genotypes	Recurrence Risk	Key Pedigree Clues
Autosomal Recessive	Aa × Aa (both carriers)	1/4 (25%) for each child	Trait skips generations; unaffected parents can have affected children
Autosomal Dominant	Aa × aa (one affected parent)	1/2 (50%) for each child	Trait appears every generation; affected child always has at least one affected parent
X-Linked Recessive	X^AX^a × X^AY	1/4 overall; 1/2 of sons affected	Mostly males affected; carrier mothers pass to sons
X-Linked Dominant	X^AX^a × X^aY	1/2 for all children	Affected fathers pass trait to ALL daughters; never to sons

Common inheritance patterns and their associated recurrence risks

The Punnett square for two carrier parents (Aa × Aa) produces a 1 in 4 chance (25%) that any given child will be affected (aa). This is the fundamental tool for computing recurrence risk.

The Punnett square above is the workhorse of recurrence risk calculation. For autosomal recessive conditions, when both parents are carriers, each pregnancy has an independent 25% chance of producing an affected child. That word independent is crucial — previous children's outcomes don't change the odds for the next child. It's like flipping a coin: getting heads three times in a row doesn't make tails more likely on the fourth flip.

SECTION 6

Worked Example — Computing Recurrence Risk

Let's work through a full problem step by step. Imagine a couple comes to a genetic counselor. The husband's brother has cystic fibrosis (an autosomal recessive condition). Neither the husband nor his wife is affected. The wife has no family history of cystic fibrosis. What is the probability that their first child will have cystic fibrosis?

Recurrence Risk for Cystic Fibrosis

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Step 1 — Determine the Inheritance Pattern

Cystic fibrosis (CF) is autosomal recessive. This means an affected person has the genotype ff (two copies of the recessive allele). A carrier has genotype Ff, and an unaffected non-carrier has genotype FF.

Pattern: Autosomal Recessive
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Step 2 — Assign Genotypes to Known Individuals

The husband's brother has CF, so his genotype is ff. For the brother to be ff, both of the husband's parents must be carriers (Ff). The husband himself is unaffected, so he is either FF or Ff.

Husband's parents: Ff × Ff
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Step 3 — Compute the Husband's Probability of Being a Carrier

From the cross Ff × Ff, the expected genotype ratio is 1 FF : 2 Ff : 1 ff. Since the husband is unaffected, we rule out ff. Of the 3 remaining outcomes, 2 are Ff (carrier). So the probability the husband is a carrier = 2/3.

P(husband is Ff) = 2/3
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Step 4 — Estimate the Wife's Carrier Probability

The wife has no family history of CF. For a common autosomal recessive disease in the general population, the carrier frequency is often given in the problem or can be looked up. For CF, approximately 1 in 25 people of European descent are carriers. So P(wife is Ff) = 1/25.

P(wife is Ff) = 1/25
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Step 5 — Use the Multiplication Rule

For their child to be affected (ff), three independent events must all happen: (1) the husband must be a carrier (2/3), (2) the wife must be a carrier (1/25), and (3) both parents must each pass on the recessive allele (1/4 from a Ff × Ff cross). Multiply all three probabilities together:

P(child has CF) = 2/3 × 1/25 × 1/4 = 2/300 = 1/150 ≈ 0.67%

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Check Your Logic

Always ask: "Which events need to happen for the child to be affected?" List them, find each probability from the pedigree or given data, and then multiply. This is the core strategy for every recurrence risk problem.

SECTION 7

Strengths & Limitations of Pedigree-Based Risk

Pedigree analysis is a powerful tool, but like any method it has strengths and weaknesses. Understanding both helps you know when the recurrence risk you calculate is reliable and when you should be cautious.

Strengths	Limitations
Works without DNA testing — you only need a family history	Small families make it hard to identify the correct inheritance pattern
Identifies carriers who don't show the trait	Assumes complete penetrance (the genotype always causes the trait), which isn't always true
Easy to communicate risk as a simple fraction or percentage	Cannot account for new (de novo) mutations that arise spontaneously
Applicable to any single-gene (Mendelian) disorder	Does not work well for traits controlled by multiple genes (polygenic traits)

Strengths and limitations of pedigree-based recurrence risk analysis

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KEY TAKEAWAY

A pedigree-based recurrence risk is like a weather forecast — it gives you the best prediction based on available information. Just as a forecast improves with more data (satellite images, temperature readings), your recurrence risk estimate gets more accurate with a larger, more detailed family history. And just as weather can surprise us, biology can too — so the number you calculate is a probability, not a guarantee.

SECTION 8

Connection to Advanced Genetics

The recurrence risk calculations you've learned here are the foundation of genetic counseling. In more advanced courses, you'll encounter situations where the math gets more complex — but the same core logic applies.

Basic Concept (This Lesson)	Advanced Extension
Complete dominance (A is fully dominant over a)	Incomplete dominance & codominance — heterozygotes show a blended or dual phenotype
Single-gene (Mendelian) traits	Polygenic traits — many genes contribute, so risk is calculated with statistical models, not simple Punnett squares
Carrier probability from pedigree alone	Genetic testing — DNA tests can confirm carrier status, updating the risk from a probability to a certainty
100% penetrance assumed	Reduced penetrance — not everyone with the genotype shows the trait, adding another probability factor to the calculation

How basic recurrence risk concepts connect to advanced genetics

As you advance in genetics, you'll also learn about Bayesian analysis, which lets you update probabilities as new information comes in (for example, if a person has three unaffected children, what does that tell us about whether they are a carrier?). The conditional probability you learned in Section 4 — the 2/3 carrier probability — is actually a simple form of Bayesian reasoning. So you've already started thinking this way!

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two unaffected parents have a child with an autosomal recessive condition. What must the genotype of each parent be? Explain your reasoning.

Show answer

Both parents must be carriers (Aa). The affected child has genotype aa, meaning it received one recessive allele from each parent. Since both parents are unaffected, they cannot be aa. Since each parent contributed an 'a' allele, they must each carry at least one. Therefore, each parent's genotype is Aa — one dominant allele (which explains why they are unaffected) and one recessive allele (which they passed to the affected child).

Question 2: BASIC CALCULATION

Both parents are carriers (Aa) for an autosomal recessive trait. Using a Punnett square, calculate the probability that their next child will be (a) affected, (b) a carrier, and (c) homozygous dominant.

Show answer

The Punnett square for Aa × Aa gives: 1 AA, 2 Aa, and 1 aa. So (a) the probability of being affected (aa) is 1/4 = 25%; (b) the probability of being a carrier (Aa) is 2/4 = 1/2 = 50%; and (c) the probability of being homozygous dominant (AA) is 1/4 = 25%. Notice that these three probabilities add up to 100%, as they should, since they cover all possible outcomes.

Question 3: INTERMEDIATE

A man whose sister has phenylketonuria (PKU, autosomal recessive) marries a woman who is a known carrier. The man is unaffected. What is the probability that their first child will have PKU?

Show answer

Since the man's sister has PKU (pp), both of his parents must be carriers (Pp × Pp). The man is unaffected, so he is either PP or Pp. Using conditional probability: of the 3 non-affected outcomes from Pp × Pp (1 PP and 2 Pp), 2 are carriers. So P(man is Pp) = 2/3. The wife is a confirmed carrier (Pp). If both are carriers, P(child has PKU) = 1/4. So the overall risk = P(man is carrier) × P(affected child if both carriers) = 2/3 × 1/4 = 2/12 = 1/6 ≈ 16.7%.

Question 4: APPLIED

A woman is a carrier for hemophilia A (X-linked recessive). She marries an unaffected man. What is the probability that (a) their son will have hemophilia? (b) their daughter will be a carrier? (c) any given child will be affected?

Show answer

Let X^H = normal allele and X^h = hemophilia allele. The mother is X^HX^h and the father is X^HY. The possible offspring are: X^HX^H (normal daughter), X^HX^h (carrier daughter), X^HY (normal son), and X^hY (affected son). (a) A son has a 1/2 (50%) chance of being affected. (b) A daughter has a 1/2 (50%) chance of being a carrier. (c) Looking at all four equally likely outcomes, 1 out of 4 is affected, so the overall probability for any child is 1/4 = 25%.

Question 5: CRITICAL THINKING

Two parents are both carriers (Aa) for an autosomal recessive condition. They already have three unaffected children. A friend tells them: "You've been lucky — the next child is more likely to be affected because you're 'due' for one." Is the friend correct? Explain using probability concepts, and calculate the actual recurrence risk for the fourth child.

Show answer

The friend is incorrect. This is an example of the gambler's fallacy — the mistaken belief that past outcomes influence future independent events. Each pregnancy is an independent event, just like each coin flip is independent. The Punnett square for Aa × Aa doesn't change based on what happened with previous children. The recurrence risk for the fourth child is exactly the same as it was for the first, second, and third: 1/4 = 25%. The fact that the first three children are unaffected does not increase or decrease the probability for child number four. In fact, from a Bayesian perspective, having three unaffected children in a row might even slightly lower our confidence that both parents truly are carriers, but as long as we know they are Aa, the risk remains 25% per pregnancy.

SUMMARY

Lesson Summary

Summary

Recurrence risk is the probability that a genetic trait will appear again in a future child, and it can be computed from a pedigree — a standardized family tree diagram that uses squares for males, circles for females, and shading for affected individuals. The process starts by identifying the inheritance pattern (autosomal dominant, autosomal recessive, X-linked recessive, or X-linked dominant), then assigning likely genotypes to individuals in the pedigree, and finally using a Punnett square to calculate offspring probabilities.

Key mathematical tools include the multiplication rule for combining independent probabilities and conditional probability for adjusting carrier odds when a person is known to be unaffected (for example, an unaffected sibling of someone with an autosomal recessive condition has a 2/3 chance of being a carrier, not 1/2). Each pregnancy is an independent event, so the recurrence risk stays the same regardless of previous children's outcomes. While pedigree analysis is a powerful and widely used tool, it works best for single-gene (Mendelian) traits and can be enhanced further with modern genetic testing.