Pedigree Analysis Virtual Lab
Build fictional family pedigrees, test dominant, recessive and X-linked hypotheses, compare likelihoods and calculate carrier and recurrence risks with Bayesian working.
- Genetics
- Secondary to first-year university
- About 60 minutes
- Simulation, guided tasks, exam and notes
1Learn: what you will be able to do
A pedigree is a family tree that shows who has a trait. By the end of this laboratory you will read pedigrees, decide which modes of inheritance are possible, calculate how likely each one is, and work out the chance that a future child is affected. The trait is fictional.
Learning outcomes
- Draw and read pedigrees with the standard symbols: squares, circles, filled shapes, couples, sibships and the proband arrow.
- Recognise the patterns of autosomal dominant, autosomal recessive, X-linked recessive and X-linked dominant inheritance.
- Use exclusion rules to rule out modes, and explain why passing the rules does not prove a mode.
- Compare how likely a pedigree is under each mode and interpret a likelihood ratio.
- Calculate probabilities with Mendel's rules, the binomial formula and Bayes' theorem for carriers.
- Estimate a recurrence risk, with working and an idea of its uncertainty.
- Discuss the ethics and privacy of genetic information and the aims of non-directive counselling.
Equipment in this virtual laboratory
- A pedigree generator for four modes (or a hidden Mystery mode), with 3 or 4 generations, adjustable family size and population allele frequency.
- A canvas chart with clickable and keyboard-selectable individuals, a hypothesis test that outlines broken rules, and optional carrier dots.
- A data table with CSV download and four linked graphs.
- A likelihood and consistency check for all four modes, and a recurrence-risk calculator with a Bayesian table.
Where this topic appears in courses
Pedigree analysis is part of the inheritance and human genetics content of secondary and first-year university biology in the UK and Cambridge pathways, WAEC/WASSCE, the IB Diploma and North American high-school, AP and introductory university courses. The depth varies: some courses stop at reading a pedigree and drawing genetic diagrams, while others require probability calculations, Hardy-Weinberg and ethical discussion. The three level tags used in the notes (Foundation, Core, Extension) help you match the material to your course. This page supports, and does not replace, the specification you are following. Check it for the exact requirements.
2Prepare: ideas and predictions
Scientists predict before they measure. Commit to an answer for each question, then open it to check your reasoning.
Each child receives one allele from each parent at random. Everything in this laboratory follows from that statement, the rules for the X and Y chromosomes, and Bayes' theorem.
Prediction 1. Two unaffected parents have an affected child. Which kind of inheritance does this suggest?
Prediction 2. An X-linked recessive trait runs in a family. Who is more likely to be affected, males or females, and why?
Prediction 3. A man has an X-linked dominant trait. What happens to his sons and his daughters?
Prediction 4. An unaffected woman has an affected sibling with a recessive trait. Is her chance of being a carrier 1/2, 1/3 or 2/3?
Prediction 5. A small pedigree fits two modes of inheritance. Can you tell which is true?
3Experiment: the pedigree builder
Generate a family, click on individuals, choose a hypothesis and test it. When you are ready, record the pedigree: the simulator then reveals the true mode and the hidden carriers, and adds a row to the data table.
4Observe: guided investigations
Five short investigations, each checked against the pedigrees in your own table. Complete them in order, using the simulator above.
5Analyse: predict, test and calculate risk
Predict a cross with full working, test every mode of inheritance against your pedigree, calculate a recurrence risk with a Bayesian table, and check your data against Mendel's rules with a line of best fit.
6Examine: practical examination
Ten multiple-choice questions at three levels and five calculations with numbers generated for you each time. Marking is instant, and you are told which notes to review.
7Notes: the complete theory
Eighteen sections from principles to exam technique, with worked examples, diagrams and a section on the ethics of genetic information. Use the search box, or open everything and print.
A. Overview and learning outcomes
A pedigree is a family tree drawn with standard symbols that records who has a particular trait and who does not. Geneticists and genetic counsellors read pedigrees to work out how a trait is inherited, to find which relatives may carry a hidden allele, and to estimate the chance that a future child will be affected. Reading a pedigree is detective work: the chart shows the evidence, and the rules of inheritance tell you which explanations are possible.
Every trait in this laboratory is fictional. The "silver-streak trait" is an invented, harmless trait, used so that nobody is reminded of a real family or a real condition. The probabilities, the rules and the reasoning are the real ones. The notes are written in three layers. Foundation material needs fractions and the idea that children inherit one copy of each gene from each parent. Core material matches typical A-level, IB Diploma, AP Biology, WASSCE and Cambridge AS/A Level genetics, including conditional probability. Extension material goes towards IB HL and first-year university: likelihood, Bayes' theorem with several pieces of evidence, Hardy-Weinberg and consanguinity.
Use the simulator alongside these notes. It builds a random family for one of four modes of inheritance, draws it on a canvas, lets you select individuals and test hypotheses, and then calculates how likely the pedigree is under each mode. Always predict first, then test.
Learning outcomes
Foundation
- Name the pedigree symbols for male, female, affected, unaffected, mating, children and the proband.
- Explain that a person inherits one allele of each autosomal gene from each parent.
- Distinguish dominant from recessive inheritance in a simple family.
- State that a carrier of a recessive allele does not show the trait but can pass the allele on.
- Give the chance (1/2 or 1/4) that a child is affected in the simplest crosses.
Core (A-level, IB, AP, WAEC)
- Recognise the typical pedigree patterns of autosomal dominant, autosomal recessive, X-linked recessive and X-linked dominant inheritance.
- Use exclusion rules to rule a mode of inheritance out, and explain why passing the rules does not prove a mode.
- Work out the genotype of each individual that can be deduced from the pedigree, including obligate carriers.
- Calculate the probability that a child is affected, and the probability of several affected children, with the binomial formula.
- Use Bayes' theorem to update a carrier probability when an unaffected relative or child is observed.
- Explain why an unaffected person with an affected parent in a dominant pedigree needs careful interpretation (reduced penetrance).
Extension (HL, AP, university)
- Compute the likelihood of a pedigree under each mode and a likelihood ratio between two modes.
- Combine prior probability, conditional probability and a population carrier frequency from Hardy-Weinberg into a recurrence risk.
- Explain how consanguinity raises the risk of recessive disorders.
- Discuss the ethics and privacy of genetic information, and the aims of non-directive counselling.
B. Core principles
Genes, alleles and genotypes
Foundation
A gene is a stretch of DNA that influences a trait. Different versions of a gene are alleles. In this laboratory the normal allele is written N and the trait allele is written S. Your genotype is the pair of alleles you carry at a gene on an autosome (a non-sex chromosome), such as NN, NS or SS. Your phenotype is what can be observed, here whether you show the silver-streak trait. One allele of each pair comes from each parent, and which one is passed on is a matter of chance, each with probability 1/2.
Dominant and recessive
An allele is dominant if one copy is enough to show the trait: both NS and SS people are affected. An allele is recessive if two copies are needed: only SS people are affected, and NS people are carriers who look unaffected. These words describe the effect of an allele on the phenotype. They do not mean that an allele is common, or stronger, or more harmful.
Sex linkage
Females have two X chromosomes. Males have one X and one Y. A gene on the X chromosome is X-linked. A male has only one copy, so a single trait allele decides his phenotype whether it is dominant or recessive. A male cannot be a carrier in the usual sense, and he passes his X to every daughter and his Y to every son. A female has two copies and behaves like an autosomal case, but she receives one X from each parent and passes one of them, at random, to each child.
| Mode | Who is affected | Typical pedigree clues |
|---|---|---|
| Autosomal dominant (AD) | NS and SS | Appears in every generation. An affected person usually has an affected parent. Males and females equally. About half of the children of an affected heterozygote are affected. |
| Autosomal recessive (AR) | SS only | Can skip generations. Affected children often have two unaffected parents who are carriers. Males and females equally. More common in families with related parents. |
| X-linked recessive (XR) | XS Y males; XS XS females | Many more affected males than females. Affected males are linked through unaffected carrier women. No father-to-son transmission. An affected daughter needs an affected father. |
| X-linked dominant (XD) | XS Y males; XS XN and XS XS females | All daughters of an affected father are affected, and none of his sons. Affected mothers pass it to about half of sons and daughters. |
Pedigree symbols
The standard symbols are shown below. Squares are males, circles are females, filled shapes are affected, a horizontal line joins a couple, a vertical line descends to the children, and a bar joins siblings. Generations are labelled with Roman numerals, and individuals within a generation are numbered from the left, so III-4 is the fourth person from the left in the third generation. The proband (also called the propositus or index case) is the first affected person who brought the family to attention, marked with an arrow. A dot inside a shape marks a carrier who is known or presumed to carry the allele without showing the trait.
Why probability appears
Core
Each child receives alleles by chance, so the same parents can have an affected child and an unaffected child. Probabilities attach to individual children, and the children are independent: the result for one child does not change the chance for the next. This is the same idea as tossing a fair coin several times, and it is the reason the binomial formula appears in genetic counselling.
C. Equations and derivations
The basic cross probabilities
Foundation
Let each parent pass the allele S with probability a (father) and b (mother). For an autosomal gene a = 1/2 for an NS parent, 0 for NN and 1 for SS. The child's genotype probabilities are:
For a cross of two carriers (NS × NS) a = b = 1/2, giving 1/4 NN, 1/2 NS and 1/4 SS. For an autosomal recessive trait the chance of an affected child is 1/4. For an autosomal dominant trait with one heterozygous affected parent and one NN parent (NS × NN), a = 1/2 and b = 0, so the chance of an affected child is 1/2.
X-linked crosses
Core
A son receives the mother's X only. A daughter receives the father's X and one of the mother's two X chromosomes. If the mother passes the trait allele with probability b = 0, 1/2 or 1, then:
- Son is affected with probability b.
- Daughter has a trait allele from the mother with probability b, and from the father with probability 0 or 1 (1 if he has the allele).
For X-linked recessive inheritance with a carrier mother and an unaffected father: sons are affected with probability 1/2, daughters are never affected, and half of the daughters are carriers. With an affected father and a non-carrier mother: no son is affected, and every daughter is a carrier. For X-linked dominant inheritance with an affected father and an unaffected mother: every daughter is affected and no son is.
Several children: the binomial formula
If each child is affected with probability p, independently, then the probability that exactly r of n children are affected is:
The probability that none is affected is (1 − p)n, so at least one is affected with probability 1 − (1 − p)n.
Bayes' theorem for carrier probability
Core
Suppose a woman may be a carrier (hypothesis C) or not (hypothesis not-C). Before looking at her own children, her prior probabilities are P(C) and P(not-C). Then she has an observation E, for example three unaffected sons. For each hypothesis there is a conditional probability P(E | C) and P(E | not-C). Bayes' theorem gives the posterior:
A table is the clearest way to do it: list the hypotheses, write the prior and the conditional probability beside each, multiply to get the joint probability, add the joints, then divide each joint by the total. The Bayesian table in the Analyse section prints exactly this.
Hardy-Weinberg and the population carrier frequency
Extension
If the trait allele has frequency q in a large population with random mating, the genotype frequencies are NN: (1 − q)2, NS: 2q(1 − q), SS: q2. For a recessive trait the incidence is q2, so q = √incidence and the carrier frequency is 2q(1 − q). For a male X-linked trait the frequency of affected males is q itself. A partner who is known to be unaffected is a carrier with probability 2q(1 − q) / (1 − q2) for a recessive trait, because the unaffected people are those who are not SS.
Likelihood of a pedigree
Extension
The likelihood of a mode of inheritance is the probability of the observed phenotypes if that mode is true. Founders (people with no parents in the chart) are given genotypes from the Hardy-Weinberg frequencies. Each child's genotype then follows from the parents by the cross probabilities above. Hidden genotypes are summed over. If any observed pattern is impossible under the mode (for instance an affected child of two unaffected parents under a dominant mode), the likelihood is exactly 0. A likelihood ratio L1 / L2 compares two modes: values near 1 mean the data do not separate them.
D. Worked examples
All trait names are fictional. N is the normal allele and S the trait allele.
Example 1 (Foundation): an autosomal dominant cross
A man affected by the dominant silver-streak trait is heterozygous (NS). His partner is NN. They plan four children. Find the chance that a child is affected, that none of four is affected, and that exactly two of four are affected.
- Man passes S with probability 1/2, partner passes N always, so each child is affected with probability p = 1/2.
- None of four affected: (1/2)4 = 0.0625.
- Exactly two of four: C(4, 2) × (1/2)4 = 6 × 0.0625 = 0.375.
Each child 0.5. None affected 0.0625. Exactly two affected 0.375. At least one affected: 1 − 0.0625 = 0.9375.
Example 2 (Core): carriers of a recessive trait
Two unaffected parents have a child with a recessive trait. They have three children in all. Find (a) the chance that a further child is affected, (b) the probability that exactly one of three children is affected, (c) the probability that an unaffected child is a carrier.
- The affected child is SS, so each parent passed S. Both parents are NS carriers.
- (a) Each child is SS with probability 1/4.
- (b) C(3, 1) × (1/4)1 × (3/4)2 = 3 × 0.25 × 0.5625 = 0.421875.
- (c) An unaffected child is NN, NS or SS with probabilities 1/4, 1/2, 1/4 before conditioning. Given unaffected, NN and NS remain: P(NS) = (1/2) / (3/4) = 2/3.
(a) 0.25. (b) 0.4219. (c) 2/3, about 0.667.
Example 3 (Core): X-linked recessive crosses
A carrier woman and an unaffected man have children. Find the chance that a son is affected, a daughter is affected, a daughter is a carrier, and that three sons are all unaffected.
- Mother XN XS passes XS with probability 1/2 to each child.
- A son is affected if he receives XS: 1/2.
- A daughter receives the father's XN, so she is never affected. She is a carrier with probability 1/2.
- Three sons all unaffected: (1/2)3 = 0.125.
Son affected 0.5. Daughter affected 0. Daughter a carrier 0.5. Three unaffected sons 0.125.
Example 4 (Core): Bayes' theorem for an X-linked trait
A woman has a brother with an X-linked recessive trait. Their mother is unaffected and is assumed to be a carrier. The woman has three sons, all unaffected. What is the probability that she is a carrier?
- Prior: she inherited the mother's XS with probability 1/2, so carrier 1/2 and not carrier 1/2.
- Conditional: if she is a carrier, each son is unaffected with probability 1/2, so three unaffected sons has probability 1/8. If not a carrier, the probability is 1.
- Joint: carrier 1/2 × 1/8 = 0.0625. Not carrier 1/2 × 1 = 0.5.
- Posterior: 0.0625 / (0.0625 + 0.5) = 0.0625 / 0.5625.
The probability that she is a carrier is 1/9, about 0.111. With one unaffected son it would be 1/3, and with two it would be 1/5. Each unaffected son lowers the probability but never makes it zero.
Example 5 (Core): recurrence risk for a recessive trait
An unaffected woman has a sibling with a recessive trait. Her partner is unaffected and unrelated, and one person in 50 in the population is a carrier. What is the probability that their first child is affected?
- Her parents are both carriers, so she is NN, NS or SS with probabilities 1/4, 1/2, 1/4. She is unaffected, so she is NS with probability 2/3.
- Her partner is a carrier with probability 1/50.
- If both are carriers, a child is affected with probability 1/4.
- Risk = 2/3 × 1/50 × 1/4 = 1/300.
The risk is about 0.0033, or 1 in 300. This is 75 times lower than the 1/4 risk for two known carriers, because the partner is unlikely to carry the allele at all.
Example 6 (Core): comparing two hypotheses with a likelihood ratio
A family has five children and two are affected. Hypothesis H1: autosomal dominant with one heterozygous affected parent (each child affected with probability 1/2). Hypothesis H2: autosomal recessive with two carrier parents (each child affected with probability 1/4). Compare the likelihoods.
- The binomial coefficient C(5, 2) = 10 is the same under both, so it cancels in the ratio.
- H1: (1/2)5 = 0.03125.
- H2: (1/4)2 × (3/4)3 = 0.0625 × 0.421875 = 0.02637.
- Ratio H1 : H2 = 0.03125 / 0.02637 = 1.185.
The ratio is 1.19, close to 1, so five children do not separate these hypotheses. More family members, especially a skipped generation or an affected person with unaffected parents, would settle it.
Example 7 (Extension): first cousins
Two unaffected first cousins want a child. Their shared grandparents had a child with a recessive trait, so both grandparents are known carriers. Each cousin is the child of a different unaffected child of those grandparents. Find the chance that the cousins' child is affected.
- The grandparents are NS × NS. Each of their unaffected children is a carrier with probability 2/3.
- Each cousin receives the carrier allele from that parent with probability 1/2, so each cousin is a carrier with probability 2/3 × 1/2 = 1/3.
- Both are carriers with probability 1/3 × 1/3 = 1/9.
- If both are carriers, the child is affected with probability 1/4. Risk = 1/9 × 1/4 = 1/36.
The risk is 1/36, about 0.028, or 2.8%. Compare with Example 5: unrelated partners had a risk of 1/300. Relatedness raises the chance of sharing a rare allele.
Example 8 (Extension): a population carrier frequency
A recessive trait affects 1 person in 2500. What fraction of the population are unaffected carriers?
- Incidence q2 = 1/2500 = 0.0004, so q = 0.02.
- Carrier frequency = 2q(1 − q) = 2 × 0.02 × 0.98 = 0.0392.
About 3.9% of the population are carriers, roughly 1 in 25.5. There are about 98 carriers for every affected person.
E. Graphs and diagrams
Reading a pedigree like a graph
Foundation
A pedigree is a diagram, not a graph with axes, but it should be read systematically in the same way. Start with the generations (rows), then look for the pattern of affected individuals. Ask four questions in order: is the trait present in every generation? Is it more common in one sex? Do affected children have unaffected parents? Do affected fathers have affected sons?
Graphs in the simulator
Core
- Likelihood by mode. Bars show the base-10 logarithm of the likelihood under each of the four modes. A bar at the floor is a mode with likelihood exactly 0. Because the scale is logarithmic, a difference of 1 means a ratio of 10.
- Affected by generation. The fraction affected in each generation. A dominant trait tends to show a high fraction in every generation; a recessive trait often has a low fraction until carriers marry.
- Observed against expected. For every recorded pedigree, the expected number of affected children (from the hidden parental genotypes) is plotted against the observed number. Mendel's rules predict a line through the origin with gradient 1.
- Accuracy over runs. The share of mystery pedigrees you identified correctly, plotted as a running proportion.
Line of best fit through the origin
Extension
When a relationship must pass through (0, 0), as here (zero expected affected children means zero observed on average), the least-squares gradient is m = Σxy / Σx2. The standard error of m is s / √(Σx2), where s2 = Σ(y − mx)2 / (n − 1). If m is within about two standard errors of 1, the data are consistent with the Mendelian rules.
F. Real-world limits
What the simulation leaves out
Core
The simulation uses clean, idealised rules so that you can learn the reasoning. Real families are messier.
- Reduced penetrance. Some people with a dominant genotype never show the trait. Their phenotype looks unaffected, so an apparent "skipped generation" in a dominant trait does not always rule it out. The likelihood tool has a penetrance setting to explore this.
- Variable expressivity. The same allele can have mild effects in one person and strong effects in another, so affected status can be hard to assign.
- New mutations. A child may be the first affected person in a family. This is why "affected child of two unaffected parents" does not completely rule out a dominant mode in real life, even though the simulator treats it as impossible.
- Genetic heterogeneity. Different genes can produce the same phenotype, and a single family may mix causes.
- Other modes. Y-linked, mitochondrial (maternal), polygenic and multifactorial inheritance do not follow the four patterns here. Mitochondrial inheritance passes from mothers to all children.
- Small pedigrees. A small family often fits more than one mode. Real counsellors extend the chart, confirm diagnoses and use laboratory tests.
- Incomplete information. Real people die young, are adopted, or are not known to be unaffected. A person's status may change with age if a trait appears late.
- Consanguinity and population structure. The founders in the simulator are unrelated people drawn at random. In real populations, relatives marry more often than chance would predict.
G. Planning and carrying out the investigation
Research question
A good question names what you will compare and what you expect. For example: "Can the pattern of affected individuals in a three-generation pedigree identify the mode of inheritance, and how often does a small pedigree leave more than one mode possible?" This is testable, uses the simulator's hidden answer as a check, and lets you measure your own accuracy.
Variables
| Type | Variable | How it is handled |
|---|---|---|
| Independent | Mode of inheritance used to build the family | Set to one of four modes, or Mystery so that you do not know it in advance. |
| Dependent | Number of modes still possible, and whether your hypothesis was correct | Read from the likelihood check and the recorded table. |
| Control | Number of generations, children per couple, allele frequency | Keep the same settings while you compare modes, and change one at a time afterwards. |
| Control | Method of reasoning | Use the same four questions on every pedigree (see section E). |
| Random | The seed | Each seed gives one family. Record it, because the same seed reproduces the same family. |
Method for the virtual lab
- Choose Mystery. Press New pedigree. Look at the chart and write down which modes you think are possible, with a reason.
- Select individuals and read their relatives. Ask the four questions in section E.
- Choose a hypothesis and press Test hypothesis. Broken rules are outlined in red. Test all four modes.
- Record the pedigree. The simulator reveals the true mode and the hidden carriers, which you should compare with your reasoning.
- Repeat for at least ten pedigrees, balancing the four modes by choosing them directly for some runs.
- Open the Analyse section and run the likelihood check, the recurrence-risk calculation and the Mendelian fit on your recorded data.
Method for a real investigation of a family
A real family history is collected by a trained professional. It starts with the affected person, then asks about parents, siblings, children, grandparents, aunts, uncles and cousins on both sides, noting the age, sex, health and, where relevant, cause of death. Whether each person is truly affected should be confirmed from medical records, not memory alone. Consent is needed from the family member being asked, and relatives are not contacted without permission.
Sample size
Core
One pedigree is one data point about a mode, and chance affects small families a great deal. To judge your accuracy, record many pedigrees. If you identified the mode correctly in 6 of 10 mystery pedigrees, the standard error of that proportion is about √(0.6 × 0.4 / 10) = 0.15, so the true accuracy could be 0.3 to 0.9. Twenty or more pedigrees narrow that range.
H. Data analysis and uncertainty
Where uncertainty comes from
The simulator has no measurement error, because phenotypes are recorded exactly. The uncertainty is of three kinds.
- Chance in inheritance. Each child's genotype is random, so counts of affected children vary even when the model is exactly right. For n children with probability p, the number affected has mean np and standard deviation √(np(1 − p)).
- Ambiguity in the pedigree. Several modes may fit. The likelihood ratio quantifies how strongly the data favour one over another.
- Uncertain assumptions. The population allele frequency q and the penetrance are assumed. The recurrence-risk tool shows the range of the answer for q of 2%, 5% and 10%.
Binomial uncertainty and the standard error
Core
If you observe a proportion p̂ in N independent trials, its standard error is √(p(1 − p) / N). In the prediction calculator, simulating 2000 children with p = 0.25 gives a standard error of √(0.25 × 0.75 / 2000) = 0.0097. An observed proportion between 0.231 and 0.269 is within two standard errors and is therefore not surprising.
The Mendelian fit
The Analyse section plots observed against expected affected children for each recorded pedigree and fits a line through the origin. The gradient m = Σxy / Σx2 should be near 1, with the standard error computed from the residuals. It is also useful to compute a z-score for the total: z = (O − E) / √V, where O is the observed total, E the expected total and V the sum of p(1 − p) over all children. Values of |z| less than 2 are within chance variation.
Likelihood and support
Extension
Given equal prior belief in the four modes, the support for a mode is its likelihood divided by the sum of the four likelihoods. This is a posterior probability under the stated assumptions. A mode with likelihood 0 has support 0. When one mode has support above 0.95 you can be fairly confident; when two modes share the support, you need more evidence. A likelihood ratio between 1 and 3 is usually called weak, 3 to 20 moderate and above 20 strong, though these labels are only conventions.
Recording and presenting data
- Record the seed, true mode, your hypothesis and the number of modes still possible, so that others can repeat your work.
- Show probabilities as fractions, decimals or percentages consistently, to a sensible number of significant figures (usually 2 to 3).
- State the assumptions behind every risk: mode of inheritance, allele frequency, penetrance and the partner's status.
- Download the CSV file to analyse your results further in a spreadsheet.
I. Apparatus and measurement methods
The virtual apparatus
Foundation
- Pedigree generator. Builds a random family from a founder couple. Each child receives alleles by chance according to the chosen mode. People who marry in are drawn from the population allele frequency.
- Canvas chart. Draws squares and circles, couples and sibships with Roman-numeral generations. You can select an individual by clicking or by the arrow keys.
- Hypothesis test. Marks individuals who break the inheritance rules of the mode you chose.
- Likelihood tool. Calculates the probability of the pedigree for each mode by summing over all hidden genotypes.
- Risk calculator. Gives the recurrence risk for a chosen individual, with a Bayesian table shown step by step.
How real family histories are gathered
Core
The first step in a real study is a structured conversation and review of records. The three-generation chart is the usual standard, because it shows inheritance patterns without being overwhelming. The most useful information comes from confirming diagnoses, as people may be unsure what condition a relative had or may use different names for it.
How real laboratories identify carriers
Extension
When a gene is known, a sample of blood or saliva can be tested for the specific allele. A positive result in a relative converts a probability into near certainty. This is called carrier testing, and results can then replace the Bayesian estimates in a pedigree, which usually changes the risk a great deal. Other kinds of testing include looking at chromosome structure under a microscope, and sequencing a gene. The details depend on the condition and are outside this laboratory, which uses a fictional trait with no real test.
Genetic counselling (introductory)
Core
A genetic counsellor helps individuals and families understand a condition, how it is inherited, the chances of it occurring again and what options exist. Typical stages are: gather the family history; confirm the diagnosis; assess the mode of inheritance; calculate risks and explain what they mean; describe testing and choices; and support the family emotionally. Professional guidelines describe counselling as non-directive: the counsellor gives balanced information and respects the person's decisions, rather than telling people what to do.
Explaining a risk
A risk of 1 in 300 is easier to understand than 0.33%, and a count such as "about 3 in 1000 families like this one" is more concrete still. Counsellors also say what the risk is a risk of, over what time, and how it compares with the background risk that every pregnancy carries. A good explanation names the uncertainty: the risk is an estimate under assumptions, and new information can change it.
J. Safety and limits of the simulation
Safety
This is a computer simulation with no physical hazards. The main safety issues are about people: family history can be personal and sometimes sad, and genetic information about one person can affect the whole family.
Ethics and privacy of genetic information
Core
Genetic information is unlike most personal data, because a result about one person says something about their relatives, who did not choose to take part. Think about the following ideas whenever you work with pedigrees.
- Informed consent. A person decides whether to share family information or to take a test, after understanding what the result might mean. Consent can be withdrawn.
- Confidentiality. Results are shared only with the people the person agrees to, and are stored securely. Sharing a pedigree can reveal who is a carrier.
- The right not to know. Some people choose not to learn their carrier status or the risk of a late-onset condition. Counselling respects that choice.
- Duty to relatives. A result may matter to a relative's health. Counsellors usually encourage a person to tell relatives themselves, and professional rules limit when anyone else may do it.
- Discrimination. Information about genes could be used to treat people unfairly, for instance in employment or insurance. Many countries have laws that limit this, and the details differ, so people should check the rules where they live.
- Autonomy and non-directive support. Decisions about testing, having children and treatment belong to the individual and family.
- Fairness and stigma. A trait should never be used to judge a person's worth, and a carrier is not "damaged". Everyone carries some alleles that could cause problems in particular combinations.
Limits of the simulation
- The trait is fictional and the families are random. Real conditions have details, variation and treatments that a model cannot capture.
- The generator builds families that fit one clean mode. It does not include new mutations, mosaicism, incomplete penetrance in the generated data, adoption or non-paternity.
- Founders are unrelated and the generator has no consanguinity. Real risks for related partners are higher (see example 7).
- The likelihood tool needs a pedigree without loops and complete information on phenotypes. Real charts can have loops and missing data.
- The numbers are for learning. They are not a medical or legal basis for any decision.
K. Common mistakes
| Mistake | Why it is wrong | Do this instead |
|---|---|---|
| "Dominant means common." | Dominance describes how an allele shows in heterozygotes. A dominant allele can be very rare. | Use "dominant" only for the phenotype of heterozygotes. |
| "Recessive traits always skip generations." | They often do, but if a carrier partners with a carrier or affected person, the trait may appear in consecutive generations. | Look for affected children of unaffected parents as the clue, not just skipping. |
| "Passing every rule proves the mode." | Rules exclude modes. A small pedigree can be consistent with two or three modes. | Compare likelihoods, and extend the family if needed. |
| "An affected father with an affected son rules out every X-linked mode." | A father gives his son a Y chromosome, so the son's X comes from his mother. For a recessive X-linked trait this is possible if the mother is a carrier or is affected. It rules out an X-linked dominant mode only when the mother is unaffected. | Check the mother's status before concluding. |
| "Two carriers of a recessive allele have a 1/2 chance of an affected child, because each passes the allele with probability 1/2." | Both parents must pass the allele, so the probabilities multiply: 1/2 × 1/2 = 1/4. For an X-linked trait, split the answer by sex: a son of a carrier mother and an unaffected father is affected with probability 1/2 and a daughter with probability 0. | Multiply for events that must both happen, and treat sons and daughters separately for X-linked traits. |
| "After three affected children, the next is unlikely." | Children are independent. The risk stays the same for every child. | Use the same probability for each child. |
| Forgetting to condition on the unaffected phenotype. | An unaffected sibling of an affected child is NN or NS, not NN, NS or SS. The carrier probability is 2/3, not 1/2. | Remove impossible genotypes and renormalise. |
| Thinking that unaffected sons prove a woman is not a carrier. | Each unaffected son of a carrier mother has probability 1/2 under the carrier hypothesis, so he lowers the carrier probability but never eliminates it. | Use Bayes' theorem with the correct conditional probability. |
| Adding probabilities that should be multiplied. | For two independent events that both must happen, multiply. For alternatives, add. | Draw a probability tree. |
| Mixing up the symbols. | A filled symbol is affected, not carrier. A dot or half-shading is a carrier. | Check the key before you read the chart. |
L. Exam technique
Command words
| Command word | What it asks for |
|---|---|
| State, name | A short fact or label, with no explanation. |
| Identify | Pick the mode, individual or genotype that fits, usually with a reason. |
| Deduce | Reach a conclusion from the evidence shown in the pedigree, naming the evidence. |
| Explain | Give a reason using the rules of inheritance. |
| Calculate | Work out a number and show the steps. |
| Discuss | Give several points, including advantages, limits or ethical issues. |
A reliable method for pedigree questions
- Read the key and note which symbols are used for carriers, if any.
- Label individuals by generation and number so you can refer to them (II-3, III-5).
- Test each mode in turn. Quote the individuals that exclude a mode: for instance "II-4 and II-5 are unaffected but have an affected child III-3, so the trait cannot be dominant".
- When you deduce a genotype, write it down on the chart in pencil, using N and S or the letters the question gives.
- Mark obligate carriers (for example unaffected parents of an affected child in a recessive trait).
- For a probability, state the assumptions, set out the table or the tree, and finish with a sentence.
Calculations
- Write the formula, substitute numbers, and give the answer with units or as a probability between 0 and 1.
- Give fractions as well as decimals when they are simple (1/9 and 0.111).
- Do not round early. Keep at least three significant figures until the last step.
- A probability must lie between 0 and 1. Check this on every answer.
Essay and discussion questions
Extension
For discussion on ethics, structure your answer by stakeholder (the individual, relatives, society), give a benefit and a concern for each, and finish with a reasoned view. Use specific terms such as consent, confidentiality, autonomy and discrimination, and say how a counsellor would put them into practice.
M. Curriculum map
The table lists topic names only. It shows where pedigree analysis commonly appears in the four course families this laboratory supports. Courses change, so check your own current specification or syllabus, and your teacher's guidance, for the exact content, the notation your exams use and the equations you are given.
| Course family | Typical topic names where this laboratory applies |
|---|---|
| UK and Cambridge (GCSE, IGCSE, AS and A Level) | Inheritance; monogenic inheritance; genetic diagrams; sex linkage; pedigree diagrams; probability in genetics; Hardy-Weinberg principle (A Level); genetic screening and counselling; ethical issues in genetics. |
| WAEC and WASSCE (West Africa) | Genetics and heredity; Mendelian inheritance; sex determination and sex-linked traits; monohybrid crosses; family pedigrees; human genetic variation and disorders. |
| International Baccalaureate Diploma (Biology) | Inheritance; Mendelian genetics; sex linkage; pedigree charts; probability in genetic crosses; genetic screening; ethics in genetics (Higher Level extends the content). |
| US and North America (high school, AP Biology, introductory university) | Heredity; Mendelian genetics; non-Mendelian inheritance and sex linkage; human pedigrees; probability rules; population genetics and Hardy-Weinberg; genetic counselling and ethics. |
Levels used in these notes
- Foundation: symbols, simple crosses, dominant against recessive.
- Core: exclusion rules for four modes, probabilities, the binomial formula, Bayes' theorem with one piece of evidence, counselling principles.
- Extension: likelihoods, Hardy-Weinberg, consanguinity, combined risk calculations and ethics in depth.
N. Applications
Genetic counselling
Pedigree analysis is the central tool of genetic counselling. A counsellor draws the family tree, decides which mode of inheritance fits, identifies who could be a carrier and calculates risks. The calculation that you can do in the Analyse section, prior probability updated by a family observation, is the same kind of reasoning that counsellors apply, with real data, tests and care.
Medical research and gene discovery
Large pedigrees in which a trait is common have been used to locate genes. When many affected relatives share a stretch of DNA that unaffected relatives do not, the gene is likely to lie in that stretch. This is called linkage analysis. The first step is always to establish the mode of inheritance from the pedigree.
Newborn and carrier screening
When a recessive condition is common in a population, health services sometimes offer carrier screening to couples. The numbers in notes C and D, such as a carrier frequency of 2q(1 − q), explain why such a programme can be worthwhile in some populations and not in others.
Plant and animal breeding
Breeders use pedigrees to track traits and to avoid mating relatives who both carry a harmful recessive allele. Kennel clubs, horse registries and crop breeders keep detailed pedigrees for this reason.
Family history and public health
Even without gene tests, a family health history helps doctors to tailor screening. People with a close relative with certain conditions may be offered checks earlier. This is a use of pedigree reasoning in everyday medicine.
Genealogy and privacy
Family trees are also used in genealogy, and consumer genetic databases can link relatives who never meet. This raises the privacy issues of note J: a person's data can reveal facts about relatives who did not agree to share them.
Forensic and legal uses
Relationships in a pedigree can be checked with DNA to settle questions of parentage or identity. This must be done with proper legal and ethical safeguards, and the people tested must be treated respectfully.
O. Extension
Exact likelihood by peeling
Extension
The likelihood tool works by peeling. For each person it computes, for every possible genotype, the probability of their own phenotype and of everything below them in the chart. It then works upward, multiplying over children and summing over the genotypes of married-in partners. For a pedigree with no loops, this gives the exact likelihood without listing every combination of genotypes. The calculation is checked in the self-test against a brute-force enumeration of every genotype in small families, and the results agree to rounding error.
Conditioning on more evidence
The Bayesian table separates three parts. The prior uses the relatives above and beside a person. The conditional part uses the person's own phenotype and their descendants. The posterior multiplies the two and renormalises. This order shows why an unaffected child helps a little, and several unaffected children help more. Each unaffected child multiplies the conditional probability for a carrier mother of an X-linked trait by 1/2 for sons.
Consanguinity
Related parents are more likely to carry the same rare allele, because they share ancestors. The inbreeding coefficient F is the probability that the two alleles at a locus in a child are identical by descent. For first cousins F = 1/16. The risk of an affected child with a rare recessive allele of frequency q is approximately q2 + Fq(1 − q) for related parents, compared with q2 for unrelated parents. For a rare allele the second term dominates, which is why the risk rises sharply.
Reduced penetrance and age of onset
If a genotype shows the trait with probability less than 1, observing an unaffected person does not rule out the genotype. Likelihoods can include this with a penetrance parameter, as the simulator does. For late-onset traits the penetrance depends on age, so an unaffected young relative carries less information than an unaffected older relative.
Other modes of inheritance
- Y-linked: passes from father to all sons.
- Mitochondrial: passes from mothers to all of their children, and fathers do not transmit it.
- Polygenic and multifactorial: many genes and environment contribute, so no clean Mendelian ratios appear.
- Genomic imprinting: the effect of an allele depends on which parent it came from.
Questions to investigate
- How does the number of children per couple change the number of modes that remain possible?
- For which mode is the first pedigree most often ambiguous? Use your recorded table to find out.
- How does the risk in the Bayesian calculator change as q rises from 2% to 10%, and why?
- Why is a pedigree with four generations often more informative than one with three?
P. Key points
- A pedigree uses squares (male), circles (female) and filled shapes (affected). The proband is marked with an arrow.
- Autosomal dominant: every generation, affected people usually have an affected parent, males and females equal.
- Autosomal recessive: unaffected carrier parents can have affected children, trait can skip generations, males and females equal.
- X-linked recessive: more affected males, no father-to-son transmission, an affected daughter needs an affected father.
- X-linked dominant: all daughters of an affected father are affected, none of his sons.
- Exclusion rules can rule a mode out. Passing them does not prove a mode.
- For one child: carrier × carrier (autosomal recessive) gives 1/4 affected. Heterozygote × unaffected (autosomal dominant) gives 1/2.
- For several independent children: P(r of n) = C(n, r) pr (1 − p)n − r.
- Bayes' theorem updates a prior with a conditional probability: posterior = joint ÷ total.
- An unaffected sibling of an affected child in a recessive trait is a carrier with probability 2/3.
- Recurrence risk combines the individual's carrier probability, the partner's carrier probability and the cross probability.
- Hardy-Weinberg: carrier frequency 2q(1 − q) with q = √incidence.
- Likelihood is the probability of the observed pedigree under a mode. A likelihood of 0 means the mode is impossible.
- Genetic counselling is non-directive. Consent, confidentiality, the right not to know and fairness guide the use of genetic information.
- The silver-streak trait is fictional. The simulator is for learning, not for medical decisions.
Q. Glossary
- Allele
- One version of a gene.
- Autosome
- A chromosome that is not a sex chromosome.
- Bayes' theorem
- A rule for updating a probability in the light of new evidence.
- Binomial probability
- The probability of r successes in n independent trials with the same chance p each.
- Carrier
- A person with one copy of a recessive allele who does not show the trait.
- Conditional probability
- The probability of an event given that another event has occurred.
- Consanguinity
- A relationship in which two people share recent ancestors.
- Dominant
- An allele whose effect shows when one copy is present.
- Exclusion rule
- A rule that makes a mode of inheritance impossible if a pedigree breaks it.
- Expressivity
- The degree to which a genotype affects a person's phenotype.
- Founder
- A person in a pedigree with no parents shown.
- Genetic counselling
- Help for people to understand inherited conditions and make informed decisions.
- Genotype
- The alleles a person carries at a gene.
- Hardy-Weinberg principle
- The relationship between allele frequency and genotype frequency in a randomly mating population.
- Hemizygous
- Having only one copy of a gene, as in males for X-linked genes.
- Heterozygote
- A person with two different alleles at a gene.
- Homozygote
- A person with two identical alleles at a gene.
- Incidence
- How often a trait occurs in a population.
- Informed consent
- Agreement given freely after understanding the purpose, benefits and risks.
- Likelihood
- The probability of the observed data if a model is true.
- Likelihood ratio
- The ratio of the likelihoods of two models for the same data.
- Mode of inheritance
- The pattern in which a trait is passed from parents to children.
- Non-directive counselling
- Counselling that gives information and support without telling people what to decide.
- Obligate carrier
- A person who must carry an allele, deduced from the pedigree.
- Pedigree
- A family tree that records a trait with standard symbols.
- Penetrance
- The probability that a person with a genotype shows the trait.
- Phenotype
- The observable trait of a person.
- Posterior probability
- The probability after taking new evidence into account.
- Prior probability
- The probability before taking new evidence into account.
- Proband
- The affected person through whom a family is first identified.
- Recessive
- An allele whose effect shows only when two copies are present.
- Recurrence risk
- The probability that a condition occurs again in a family.
- Sex linkage
- Inheritance of a gene on the X or Y chromosome.
- Sibship
- A set of brothers and sisters with the same parents.
- Standard error
- A measure of how much a sample estimate would vary from sample to sample.
- X-linked
- Located on the X chromosome.
R. References and further reading
Only titles and publishers are given, together with stable home pages. Search each site for the title to reach the exact page.
- OpenStax, Biology 2e, chapters on Mendel's experiments and on modern understandings of inheritance. Free and open textbook. https://openstax.org/
- OpenStax, Concepts of Biology, chapter on heredity. Free textbook with an introduction to pedigrees. https://openstax.org/
- National Human Genome Research Institute (NHGRI), educational resources on genetics and genomics, including the Talking Glossary of Genomic and Genetic Terms. https://www.genome.gov/
- National Society of Genetic Counselors, resources on genetic counselling as a profession. https://www.nsgc.org/
- World Health Organization, publications on genomics, genetics and ethics in public health. https://www.who.int/
- Hartl, D. L. and Clark, A. G., Principles of Population Genetics, Sinauer Associates (extension reading on Hardy-Weinberg and inbreeding).
- Ott, J., Analysis of Human Genetic Linkage, Johns Hopkins University Press (extension reading on likelihood methods for pedigrees).
The silver-streak trait used in the simulator is invented. It does not describe any real condition.
8Practise: extra problems
All traits are fictional. N is the normal allele and S the trait allele. Attempt each problem before opening the answer.
Problem 1 (Foundation)
Two unaffected carriers (NS and NS) of an autosomal recessive trait have a child. Find the probability that the child is (a) affected, (b) an unaffected carrier, (c) unaffected and not a carrier.
Show the answer
The child is SS, NS or NN with probabilities 1/4, 1/2 and 1/4. (a) 1/4 = 0.25. (b) 1/2 = 0.5. (c) 1/4 = 0.25.
Problem 2 (Core)
A carrier woman and an unaffected man have children and the trait is X-linked recessive. Find the probability that a son is affected, that a daughter is a carrier, and that two sons are both unaffected.
Show the answer
A son receives the mother's S with probability 1/2, so he is affected with probability 0.5. A daughter receives the father's normal X and one of the mother's, so she is a carrier with probability 0.5. Two unaffected sons: (1/2) × (1/2) = 0.25.
Problem 3 (Core)
Two carriers of an autosomal recessive trait plan five children. What is the probability that at least one is affected?
Show the answer
P(none affected) = (3/4)5 = 0.2373. P(at least one affected) = 1 − 0.2373 = 0.7627, about 76%.
Problem 4 (Core)
An unaffected man has a sister with an autosomal recessive trait. His unaffected, unrelated partner is a carrier with probability 1/25. What is the probability that their first child is affected?
Show the answer
His parents are both carriers, so as an unaffected person he is a carrier with probability 2/3. Risk = 2/3 × 1/25 × 1/4 = 1/150 = 0.0067, or about 0.67%.
Problem 5 (Core)
A recessive trait affects 1 person in 40 000. Find the allele frequency q and the proportion of the population who are unaffected carriers.
Show the answer
q2 = 1/40 000, so q = 0.005. Carriers: 2q(1 − q) = 2 × 0.005 × 0.995 = 0.00995, which is about 1 in 100.
Problem 6 (Extension)
A family has four children, of whom one is affected. Compare H1 (each child affected with probability 1/2) with H2 (each child affected with probability 1/4) using a likelihood ratio. What do you conclude?
Show the answer
H1: (1/2)4 = 0.0625. H2: (1/4) × (3/4)3 = 0.1055. The binomial coefficient 4 is common to both and cancels. The ratio H1 : H2 = 0.593, so H2 is favoured by about 1.7 to 1. That is weak evidence, so four children are not enough to decide.