Principles of Inheritance and Variation · TOPIC LESSON
Follow chromosome transmission, test pedigree models and keep conditional inheritance probabilities distinct.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
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No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
A pedigree records biological relationships and the distribution of a defined trait across generations. It can support an inheritance model and constrain possible genotypes, but a small family often fits more than one model. The aim is to test compatibility, not to recognise a pattern from one memorable symbol.
This lesson uses fictional single-locus examples. Unless a case says otherwise, assume complete penetrance, reliable phenotype information, ordinary Mendelian transmission, no new mutation and no environmental phenocopy. A phenocopy is a similar phenotype produced by another cause. Real families can violate these simplifying assumptions.
Affected means that an individual displays the trait being tracked; it does not describe their whole health, abilities or worth. Unaffected means that the tracked phenotype is absent, not necessarily that the relevant allele is absent. For a recessive trait, an unaffected heterozygote can transmit the allele.
Pedigree reasoning is educational and cannot establish a personal diagnosis, parentage claim or reproductive recommendation. Actual clinical interpretation may require a detailed history, molecular evidence and qualified genetic counselling. The calculations here concern specified models rather than anyone’s family.
02 / UNDERSTAND & APPLY
Conventional introductory pedigrees use a square for a male, a circle for a female and a diamond when sex is unspecified. Filled symbols show the defined phenotype. A horizontal connection joins parents, a vertical descent line leads to a sibship line, and separate vertical branches connect siblings.
Generations are commonly labelled with Roman numerals, while individuals within a generation receive Arabic numbers. These identifiers make a statement such as II-3 unambiguous. An arrow can identify the proband, the person through whom a family comes to attention, rather than indicating gene flow.
Carrier notation varies: a dot or partial shading may be used in some diagrams, but only when the legend defines it. Do not assume that every unfilled symbol is known to be a noncarrier. Likewise, a slash usually indicates death and says nothing by itself about the cause.
The conventional symbols simplify biological sex for these inheritance examples and do not describe gender identity or every sex-chromosome variation. Our crosses explicitly use ordinary XX and XY transmission at non-pseudoautosomal X or Y loci where stated. The diagram’s legend and model govern the calculation.
03 / UNDERSTAND & APPLY
For an autosomal recessive model with alleles A and a, affected individuals are aa, while unaffected individuals can be AA or Aa. If two unaffected parents have an affected child under the stated no-new-mutation model, each parent must have contributed a, so both are Aa.
The cross Aa × Aa gives probabilities one-quarter AA, one-half Aa and one-quarter aa. These are probabilities for each independent conception, not a requirement that every group of four children contain one of each category. Both sexes have the same genotype probabilities when sex and the autosomal locus segregate independently.
An unaffected child of these parents has a two-thirds probability of being a carrier, because the aa outcome has been excluded. Among the three remaining equally weighted genotype routes, two are Aa and one is AA. Using one-half here incorrectly retains the affected outcomes in the denominator.
Recessive traits can skip visible generations, but skipping is not mandatory. An aa × aa cross gives only aa offspring in this simple model. A pedigree with affected people in several successive generations can therefore still be recessive; vertical appearance alone is not proof of dominance.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
04 / UNDERSTAND & APPLY
For a fully penetrant autosomal dominant model, a single D allele produces the phenotype, while dd is unaffected. A heterozygous Dd parent and dd parent produce affected Dd and unaffected dd offspring with equal probability. The genotype of the affected parent must be stated or inferred before assigning a probability.
Two affected heterozygotes can have an unaffected child because Dd × Dd includes dd with probability one-quarter. Conversely, two unaffected dd parents cannot produce D under the simple no-new-mutation model. This is a compatibility statement about that model, not a universal rule for all real dominant disorders.
Father-to-son transmission is compatible with autosomal inheritance because a father can pass an autosomal allele to a son. Both sexes can transmit the allele. A small family need not display equal numbers of affected males and females even when their probabilities are equal.
An affected child with apparently unaffected parents can occur in real dominant conditions through a new variant, reduced penetrance, mosaicism or age-dependent expression. Therefore the introductory vertical pattern is a useful clue only after the assumptions are made explicit.
05 / UNDERSTAND & APPLY
For a locus on the non-pseudoautosomal region of the X chromosome, an ordinary XY father passes his X to daughters and his Y to sons. An ordinary XX mother passes one of her X chromosomes to each child. Thus a son’s X-linked allele comes from his mother in this model.
An XY individual has one copy of the usual X-linked locus and is hemizygous, not homozygous recessive. Writing XᵃY records that the variant lies on X; Y is not a dominant alternative allele at the same locus. The absence of a second X copy changes the expression logic.
No father-to-son transmission of that X-linked allele occurs through the father’s X. However, an affected father and affected son can still appear in one family if the son inherits the relevant allele from his mother. A visible father–son pair is not automatically proof that the father transmitted the allele.
Genes in pseudoautosomal regions behave differently because corresponding regions of X and Y can pair and recombine. Sex-limited expression is also different from sex linkage: an autosomal gene can affect a trait expressed mainly in one sex. Chromosome location and phenotype restriction are separate questions.
06 / UNDERSTAND & APPLY
Consider an X-linked recessive model with Xᴬ representing the reference allele and Xᵃ the recessive trait allele. A heterozygous mother XᴬXᵃ and a reference father XᴬY can produce XᴬXᴬ daughters, XᴬXᵃ daughters, XᴬY sons and XᵃY sons in equal proportions under equal gamete and sex probabilities.
Among sons, one-half are expected to display the trait; among daughters, one-half are heterozygous and none are affected in the simplest complete-recessivity model. Among all children, the probability of an affected son is one-quarter. The difference between one-half and one-quarter comes from conditioning, not contradictory genetics.
An affected father XᵃY and a reference homozygous mother XᴬXᴬ produce heterozygous daughters and reference sons. The mother’s genotype is essential: replacing her with a heterozygote changes the outcomes. It is unsafe to state that every affected father has unaffected sons without specifying the maternal contribution.
An affected XᵃXᵃ mother with a reference father gives affected sons and heterozygous daughters in this model. Females can display X-linked recessive conditions; two variant copies are one route, while X-inactivation and chromosome differences can complicate expression in real cases. Carrier must not be assumed to mean no symptoms in every real X-linked condition.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
07 / UNDERSTAND & APPLY
For a simple fully penetrant X-linked dominant trait, an affected father XᴰY and unaffected mother XᵈXᵈ give the trait allele to every daughter and to no son through the paternal X. A heterozygous mother XᴰXᵈ and unaffected father give a one-half chance of transmitting Xᴰ to either a daughter or a son.
For a locus restricted to the non-pseudoautosomal Y, an ordinary XY father transmits that Y to sons and not daughters. A fully penetrant trait associated with such a locus can show father-to-son continuity. The model presumes that the individuals reproduce and does not imply that every Y-linked condition permits unaffected fertility.
Mitochondrial inheritance follows a different genome. In the usual human pattern, mitochondria and mitochondrial DNA are inherited through the egg. A father does not ordinarily transmit his mitochondrial variant to children, while a mother can transmit mitochondrial variants to both sons and daughters.
Not every child necessarily receives the same variant fraction or displays the same phenotype. Heteroplasmy means that more than one mitochondrial DNA variant can coexist, and segregation and threshold effects can alter expression. A maternal pattern is a clue to investigate, not permission to replace all mitochondrial probabilities with a universal all-affected rule.
08 / UNDERSTAND & APPLY
Under the simple assumptions, an affected daughter with an unaffected ordinary XY father is incompatible with the standard fully recessive X-linked model if her affected state requires two Xᵃ copies. Her father would need to contribute Xᵃ. Autosomal recessive inheritance can explain such a daughter when both unaffected parents are carriers.
A fully penetrant X-linked dominant model predicts that an affected father and reference mother have affected daughters and unaffected sons. A reference daughter in that cross challenges the model. Yet a small family containing only affected daughters does not uniquely prove X linkage, because an autosomal model can produce the same small sample by chance.
An all-male set of affected relatives is compatible with several explanations: X linkage, Y linkage, sex-limited expression, biased ascertainment or chance. Ask which parent transmits, whether daughters can transmit, and whether any obligate genotype follows from the complete pedigree.
Outsiders are not automatically noncarriers. Some classroom problems explicitly assume a rare allele and unrelated marrying-in individuals are noncarriers; others do not. Use that assumption only when stated or clearly part of the specified model, and label conclusions that depend on it.
09 / UNDERSTAND & APPLY
If a known heterozygous mother and reference father have the X-linked recessive cross described earlier, the probability an already-known son is affected is one-half. The probability the next child is both a son and affected is one-half times one-half, or one-quarter, when the sex probabilities are assumed equal.
For independent conceptions with a fixed one-quarter trait probability, the chance of no affected child among three is (3/4)³ = 27/64. The chance of at least one is the complement, 1 − 27/64 = 37/64. This avoids separately adding the one-, two- and three-affected cases.
Exactly one affected child among three has probability 3 × (1/4) × (3/4)² = 27/64. The factor three counts which child is affected. A specified order, such as affected first and unaffected next twice, has only one of those arrangements and probability 9/64.
These calculations assume independence conditional on the parental genotypes and model. If a parent’s genotype is uncertain, observations of earlier children can update that uncertainty, so unconditional outcomes need not be independent. Identical twins also should not be treated as two independent conceptions.
10 / UNDERSTAND & APPLY
Suppose a fictional woman has a prior one-half probability of being an X-linked recessive carrier. If she is a carrier, a known son has a one-half probability of being unaffected; if she is a reference homozygote, that probability is one. With a reference father and the simple model, observing one unaffected son changes the relative weights.
The carrier-and-observation weight is (1/2)(1/2) = 1/4. The noncarrier-and-observation weight is (1/2)(1) = 1/2. After normalisation, the carrier probability becomes (1/4)/(3/4) = 1/3. A second independent unaffected son would reduce it to 1/5 under the same model.
This is Bayesian conditioning on evidence, not a change in the probability that a known carrier transmits one of her X chromosomes. If carrier status were established, previous unaffected sons would not change the one-half transmission probability to another son.
Clinical carrier probabilities can depend on molecular findings, family structure, penetrance and other information. The fictional prior and simplified likelihoods here are supplied to teach conditional probability; they must not be substituted for an assessment of a real family.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
11 / UNDERSTAND & APPLY
Penetrance is the fraction of individuals with a relevant genotype who display the associated phenotype under a defined context. Reduced penetrance can make a trait appear to skip a generation even when a dominant allele was transmitted. Age and how carefully a phenotype is assessed can affect the apparent pattern.
Variable expressivity concerns differences in how a phenotype appears among individuals who express it. It is not the same as whether expression occurs at all. Different severity in relatives need not imply different inheritance modes.
X-inactivation can produce cellular mosaics in individuals with two X chromosomes. Skewed patterns can affect expression of X-linked variants. This is one reason real X-linked conditions may not fit an absolute unaffected-carrier versus affected-person split.
New variants, germline mosaicism, heterogeneous causes and incomplete records add further limits. The right response to a mismatch is to re-examine the model and evidence, not to accuse a family member or declare a relationship impossible from a classroom pedigree.
12 / UNDERSTAND & APPLY
All families and priors are fictional. Use only the assumptions stated in each question.
ORIGINAL WORKED EXAMPLE
Two unaffected parents have an affected daughter under a fully penetrant autosomal recessive model with no new mutation. Infer parental genotypes.
Answer Both parents are Aa.
ORIGINAL WORKED EXAMPLE
An unaffected child is born to Aa × Aa parents. What is the conditional carrier probability?
Answer 2/3.
ORIGINAL WORKED EXAMPLE
A known carrier XᴬXᵃ mother and XᴬY father have a simple X-linked recessive trait. Compare risk among sons with the event affected son among all children.
Answer 1/2 among sons; 1/4 for an affected son among all children.
ORIGINAL WORKED EXAMPLE
An affected XᵃY father and XᴬXᴬ mother have children. State daughter and son genotypes.
Answer All daughters XᴬXᵃ; all sons XᴬY in this model.
ORIGINAL WORKED EXAMPLE
Can two affected parents in a fully penetrant autosomal dominant model have an unaffected child?
Answer Yes, if both affected parents are heterozygous; the model probability is 1/4.
ORIGINAL WORKED EXAMPLE
With a fixed independent 1/4 probability per conception, find the chance of at least one affected child among three.
Answer 37/64, approximately 57.81%.
ORIGINAL WORKED EXAMPLE
An affected father and unaffected mother have two affected daughters. Does this alone prove X-linked dominant inheritance?
Answer No. Compatibility is not unique identification.
ORIGINAL WORKED EXAMPLE
A fictional woman starts with carrier probability 1/2 and has one known unaffected son. Use the simple X-linked recessive model and reference father to update her carrier probability.
Answer 1/3; a known carrier’s transmission probability itself has not changed.
13 / UNDERSTAND & APPLY
Read the event carefully: all children, known sons, known daughters or unaffected individuals can require different denominators.
ORIGINAL PRACTICE · 1
No.
It indicates the tracked phenotype is absent unless the legend adds genotype information.
ORIGINAL PRACTICE · 2
1/2.
This uses all outcomes, unlike the 2/3 conditional probability among unaffected children.
ORIGINAL PRACTICE · 3
Yes.
Appropriate crosses, including aa × aa, can produce affected offspring in each generation.
ORIGINAL PRACTICE · 4
No.
There is one copy of the ordinary X-linked locus, not two recessive copies.
ORIGINAL PRACTICE · 5
The mother.
The father supplies Y to the son.
ORIGINAL PRACTICE · 6
No.
The son’s relevant X allele can come from his mother.
ORIGINAL PRACTICE · 7
1/2.
Daughters receive paternal Xᴬ and either maternal X with equal probability.
ORIGINAL PRACTICE · 8
All sons.
Every son receives maternal Xᵃ and paternal Y.
ORIGINAL PRACTICE · 9
All daughters and no sons.
This follows the paternal chromosome route.
ORIGINAL PRACTICE · 10
Yes.
Sex-limited expression does not establish sex-chromosome location.
ORIGINAL PRACTICE · 11
No.
Heteroplasmy, segregation and threshold effects can modify expression.
ORIGINAL PRACTICE · 12
27/64.
3 × (1/4) × (3/4)² counts the three possible positions.
ORIGINAL PRACTICE · 13
9/64.
(1/4) × (3/4)² contains only the specified order.
ORIGINAL PRACTICE · 14
No.
Independent conceptions retain the same conditional probability.
ORIGINAL PRACTICE · 15
Whether expression occurs versus how it appears among those expressing it.
Reduced penetrance can hide transmission; variable expressivity changes manifestation.
ORIGINAL PRACTICE · 16
Not automatically.
Unknown genotype should remain uncertain unless evidence or an explicit assumption resolves it.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
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