BiologyCBSE Class 12 · Principles of Inheritance and Variation

Pedigree Analysis and Mendelian Disorders

Updated 15 October 2026 · Fact-checked

Pedigree analysis traces a trait through several generations of a family to tell whether it is dominant or recessive, autosomal or sex-linked. Mendelian disorders are caused by a mutation in a single gene: haemophilia and colour blindness are X-linked recessive; sickle-cell anaemia, phenylketonuria and thalassaemia are autosomal recessive and appear when both parents are carriers.

Two unaffected parents with an affected child show the trait is recessive, so both parents must be carriers.

Illustration drawn for QuestionBuddy.

Understand Pedigree Analysis and Mendelian Disorders

Why pedigrees. Controlled crosses are impossible in humans, so we study the family record of a trait over generations: pedigree analysis. It is used in genetic counselling to predict the chance of a disorder.

Symbols (learn to read them; diagrams in the paper use these):

square = male; circle = female; diamond = sex unspecified;
filled (shaded) symbol = affected; half-filled or dotted = carrier (heterozygote), where shown;
horizontal line between a male and a female = mating; double line = consanguineous mating (between relatives);
vertical line down to a horizontal sibship line = offspring, in order of birth from left to right;
generations numbered I, II, III; individuals 1, 2, 3 within each.

Reading a pedigree: four tests.
1. Two unaffected parents with an affected child → the trait is recessive (both parents are carriers).
2. Two affected parents with an unaffected child → the trait is dominant.
3. Mostly males affected, passed from an unaffected mother to sons, never father to son → X-linked recessive.
4. Males and females affected about equally, father-to-son transmission possible → autosomal.

Mendelian disorders are mainly caused by alteration or mutation in a single gene and are inherited by Mendel's rules. Common ones: haemophilia, cystic fibrosis, sickle-cell anaemia, colour blindness, phenylketonuria, thalassaemia. They may be dominant or recessive, autosomal or sex-linked.

1. Colour blindness (X-linked recessive). A defect in the red or green cone of the eye, so red and green cannot be told apart. The genes are on the X chromosome. It occurs in about 8 per cent of males but only about 0.4 per cent of females, because a male has one X (one copy decides) while a female needs two copies. A carrier woman ( $X^{c} X$ ) has normal vision but each son has a 50 per cent chance of being colour blind. A daughter is colour blind only if she receives $X^{c}$ from both parents: her father must be colour blind and her mother at least a carrier.

2. Haemophilia (X-linked recessive). A protein in the cascade of blood-clotting proteins is affected, so a simple cut can bleed non-stop. It passes from an unaffected carrier female to some of her sons. A haemophilic female is extremely rare: her mother must be at least a carrier and her father haemophilic. Queen Victoria was a carrier, and her pedigree shows many haemophilic descendants.

3. Sickle-cell anaemia (autosomal recessive). Controlled by a pair of alleles $H b^{A}$ and $H b^{S}$ . Only $H b^{S} H b^{S}$ people have the disease; $H b^{A} H b^{S}$ people look unaffected but are carriers (sickle-cell trait) with a 50 per cent chance of passing on $H b^{S}$ . Cause: glutamic acid (Glu) is replaced by valine (Val) at the sixth position of the β-globin chain, because of a single base substitution in the sixth codon, GAG → GUG. The mutant haemoglobin polymerises under low oxygen tension, changing red blood cells from biconcave discs into elongated sickle shapes.

4. Phenylketonuria (PKU, autosomal recessive). An inborn error of metabolism: the person lacks the enzyme that converts phenylalanine to tyrosine. Phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives, which build up in the brain causing mental retardation, and are excreted in urine. PKU also reduces hair and skin pigmentation: one gene affecting several traits is pleiotropy.

5. Thalassaemia (autosomal recessive). Mutation or deletion reduces the rate of synthesis of one of the globin chains (α or β), forming abnormal haemoglobin and causing anaemia. α-thalassaemia is controlled by two closely linked genes HBA1 and HBA2 on chromosome 16; the more of the four copies affected, the less α-globin made. β-thalassaemia is controlled by the HBB gene on chromosome 11. Thalassaemia is a quantitative problem (too few globin molecules); sickle-cell anaemia is qualitative (incorrectly functioning globin).

X-linked cross to remember (carrier mother × normal father):

	X	Y
$X^{h}$	$X^{h} X$ carrier daughter	$X^{h} Y$ haemophilic son
X	XX normal daughter	XY normal son

Sons: 1/2 affected; daughters: 1/2 carriers, none affected.

Draw and label: a pedigree for an autosomal recessive trait. Checklist: generation I, an unaffected couple (square and circle joined by a line); generation II, three children: one shaded square or circle (affected) and two unshaded. Labels: I-1, I-2, II-1 to II-3, mating line, sibship line, affected, unaffected (carrier parents can be half-shaded).

Pedigree Analysis and Mendelian Disorders: key facts

Two carriers (autosomal recessive): $A a \times A a \to \frac{1}{4} AA : \frac{1}{2} A a : \frac{1}{4} aa$; Chance of an affected child 1/4; 2/3 of unaffected children are carriers.
Carrier mother × normal father (X-linked): $X^{h} X \times X Y \to sons \frac{1}{2} affected; daughters \frac{1}{2} carriers$; Same for colour blindness.
Affected father × normal (non-carrier) mother (X-linked): $X X \times X^{h} Y \to all daughters carriers; all sons normal$; No father-to-son transmission: sons get his Y.
Sickle-cell mutation: $G A G \to G U G : Gl u \to V a l at position 6 of β -globin$; Single base substitution (point mutation).
PKU block: $phenylalanine ↛ tyrosine \Rightarrow phenylpyruvic acid accumulates$; Brain damage (mental retardation); excreted in urine.

How to answer Pedigree Analysis and Mendelian Disorders questions

Pedigree and disorder questions are probability problems in disguise. Decide the mode of inheritance, assign genotypes, then cross.

1Decide dominant or recessive: look for unaffected parents with an affected child (recessive) or affected parents with an unaffected child (dominant).
2Decide autosomal or X-linked: father-to-son transmission or an affected daughter of an unaffected father rules out X-linked recessive; mostly affected males through carrier mothers suggests X-linked.
3Write genotypes, starting with the affected (recessive: aa or $X^{h} Y$ / $X^{h} X^{h}$ ), then their parents and children who must carry the allele.
4For X-linked crosses write the Y explicitly; a son's X always comes from his mother.
5Draw a Punnett square and give probabilities separately for sons and daughters if the question asks by sex.
6For disorder descriptions, give: gene or chromosome, molecular defect, consequence, and the inheritance pattern.

Pedigree Analysis and Mendelian Disorders in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Genetic counselling for thalassaemia

Suppose two cousins, both healthy, plan to marry. Blood tests show that each is a carrier of β-thalassaemia, and they ask a genetic counsellor about their future children.

What is the chance that a child will have thalassaemia, will be a carrier, or will be completely free of the allele? If their first child is healthy, what is the chance that this child is a carrier?

Try it first, then show the solutionSolution

Both are carriers: Tt × Tt (t = thalassaemia allele, autosomal recessive).
Children: 1/4 TT (free), 1/2 Tt (carrier), 1/4 tt (thalassaemic).
A healthy child is TT or Tt; among healthy children carriers are (1/2) ÷ (3/4) = 2/3.

Answer: Thalassaemic 1/4, carrier 1/2, free 1/4 for each child; a healthy child is a carrier with probability 2/3.

Memory hook: Two carriers: one in four affected, two in three healthy ones carry it.

Tricks to remember Pedigree Analysis and Mendelian Disorders

Short memory aids. Say each one aloud once, then use it on a question the same day.

Criss-cross for X
X-linked traits zig-zag: grandfather → mother (carrier) → grandson.
Use it when: Recognising X-linked pedigrees.
Thala = too little, Sickle = wrong shape
Thalassaemia: quantity of globin low; sickle-cell: quality wrong.
Use it when: Differentiating the two blood disorders.
GAG to GUG at six
Glu becomes Val at position six because GAG becomes GUG.
Use it when: Sickle-cell molecular basis.

Pedigree Analysis and Mendelian Disorders: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Thalassaemia carriers · 1 mark · MCQBoard level

If both parents are carriers of thalassaemia, what is the chance that a child will be affected?

Try it first, then show the solutionSolution

Thalassaemia is autosomal recessive; carriers are Tt.
Tt × Tt gives tt with probability 1/4.

Answer: 25%.

Example 2: Colour-blind family · 3 marks · Short answerHarder board level

In a family, the father, a daughter and a son are colour blind, and the mother has normal vision. Did both children inherit the condition from their father? Work out the cross.

Try it first, then show the solutionSolution

Father $X^{c} Y$ . The daughter is $X^{c} X^{c}$ : she needs one $X^{c}$ from each parent, so the mother must be a carrier, $X^{c} X$ (1 mark).
Cross $X^{c} X$ × $X^{c} Y$ : daughters $X^{c} X^{c}$ (colour blind) and $X^{c} X$ (carrier); sons $X^{c} Y$ (colour blind) and XY (normal) (1 mark).
The daughter got one $X^{c}$ from the father and one from the mother. The son got his $X^{c}$ from the mother only; from the father he received Y. So only the daughter inherited it (partly) from the father (1 mark).

Answer: Mother is a carrier; the daughter inherited from both parents, the son only from the mother.

Example 3: Colour-blind mother · 1 mark · MCQHarder board level

A colour-blind woman marries a man with normal vision. Which of their children are expected to be colour blind?
(a) all sons and no daughters (b) half the sons and half the daughters (c) all daughters and no sons (d) none of the children

Try it first, then show the solutionSolution

Mother $X^{c} X^{c}$ gives $X^{c}$ to every child; father XY gives X to daughters and Y to sons.
Sons $X^{c} Y$ : all colour blind. Daughters $X^{c} X$ : carriers with normal vision.

Answer: (a) all sons and no daughters (the daughters are carriers).

Example 4: Sickle-cell trait in a family · 5 marks · Long answerHarder board level

A woman with sickle-cell trait ( $H b^{A} H b^{S}$ ) marries a man with normal haemoglobin ( $H b^{A} H b^{A}$ ). (a) What change in the β-globin chain does her $H b^{S}$ allele cause, and what happens to red blood cells carrying only the mutant haemoglobin at low oxygen? (b) Why is she called a carrier? (c) Work out the cross and give the chances of a child with the disease and with the trait. (d) Why is sickle-cell anaemia called a qualitative defect, unlike thalassaemia?

Try it first, then show the solutionSolution

(a) A single base substitution in the sixth codon of the β-globin gene, GAG → GUG, replaces glutamic acid by valine at the sixth position; such haemoglobin polymerises under low oxygen tension and RBCs change from biconcave discs to elongated sickle shapes (1 mark).
(b) She has one normal and one mutant allele: she is apparently unaffected but can pass $H b^{S}$ to her children (50 per cent of her gametes) (1 mark).
(c) Gametes: woman $H b^{A}$ , $H b^{S}$ ; man $H b^{A}$ only. Children: 1/2 $H b^{A} H b^{A}$ (normal) : 1/2 $H b^{A} H b^{S}$ (trait/carrier) (1 mark).
Chance of a child with the disease ( $H b^{S} H b^{S}$ ) = 0; chance of a carrier child = 1/2 (1 mark).
(d) Sickle-cell: globin is made but works incorrectly (qualitative); thalassaemia: too few globin molecules are made (quantitative) (1 mark).

Answer: Glu → Val at β-6 (GAG → GUG), sickling at low O₂; carrier = one $H b^{S}$ allele; children 1/2 normal, 1/2 carriers, 0 diseased; qualitative vs quantitative.

Example 5: Reading a pedigree · 3 marks · Short answerHOTS / competency

Suppose a pedigree shows: generation I, an affected man married to an unaffected woman; generation II, two unaffected sons and two daughters, both unaffected; generation III, one daughter of generation II marries an unaffected man and has two affected sons and one unaffected son and an unaffected daughter. (a) Dominant or recessive? (b) Autosomal or X-linked? Give reasons. (c) Give the genotype of the generation II daughter.

Try it first, then show the solutionSolution

(a) Recessive: the trait skips generation II and reappears in III from unaffected parents (1 mark).
(b) Likely X-linked recessive: an affected grandfather passes it through his unaffected daughter to her sons (criss-cross), only males are affected, and there is no father-to-son transmission (1 mark).
(c) She received her father's $X^{a}$ , so she is a carrier, $X^{a} X$ (1 mark).

Answer: (a) Recessive (b) X-linked (criss-cross, affected males only) (c) $X^{a} X$ , carrier.

Example 6: PKU and pleiotropy · 3 marks · Short answerHarder board level

How does a single gene defect cause phenylketonuria, and why is PKU an example of pleiotropy?

Try it first, then show the solutionSolution

A mutation (autosomal recessive) means the enzyme that converts phenylalanine to tyrosine is not made (1 mark).
Phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives, which build up in the brain (mental retardation) and are excreted in urine (1 mark).
The same single gene defect also reduces hair and skin pigmentation, so one gene affects several phenotypic traits: pleiotropy, because the enzyme's product feeds several pathways (1 mark).

Answer: Missing enzyme → phenylalanine and phenylpyruvic acid build up → mental retardation and reduced pigmentation; one gene, many effects.

Where students lose marks in Pedigree Analysis and Mendelian Disorders

The slips examiners see most often, and what to write instead.

Mistake: Father-to-son transmission of an X-linked gene
Students forget that sons receive Y from the father.
Wrong working
The colour-blind son got the gene from his colour-blind father.
Right working
A son's X comes from his mother; the father gives Y.
Do this: Write Y explicitly in every X-linked cross.
Usually costs: 1-2 marks.
Mistake: Calling thalassaemia qualitative
The two blood disorders are mixed up.
Wrong working
Thalassaemia: defective haemoglobin structure due to Glu → Val.
Right working
Thalassaemia: reduced synthesis of α or β globin (quantitative); Glu → Val is sickle-cell.
Do this: Thala = too little.
Usually costs: 1 mark.
Mistake: Wrong codon change
Students write GUG → GAG.
Wrong working
GUG is replaced by GAG.
Right working
GAG (Glu) → GUG (Val) at the sixth codon.
Do this: Normal G-A-G becomes G-U-G: A to U in the middle.
Usually costs: 1 mark.
Mistake: Not giving a reason when identifying a pedigree
Students write only 'recessive'.
Wrong working
The trait is recessive and autosomal.
Right working
Recessive, because unaffected parents I-1 and I-2 have an affected child; autosomal, because an affected daughter has an unaffected father.
Do this: Quote the specific individuals as evidence.
Usually costs: 1 mark per part.

Exam tips

Use the notation $X^{h}$ , $X^{c}$ , $H b^{A}$ , $H b^{S}$ exactly and keep it consistent.
In pedigree answers, name the individuals (e.g. II-2) that prove your conclusion.
For X-linked problems, give separate ratios for sons and daughters.
Learn the numbers: 8% and 0.4% (colour blindness), position 6 and GAG → GUG (sickle-cell), chromosome 16 and 11 (thalassaemia genes).
Name Queen Victoria as the well-known carrier when asked for an example of haemophilia in a pedigree.

Practice questions on Pedigree Analysis and Mendelian Disorders

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): A man always inherits haemophilia from his mother.
Reason (R): The gene for haemophilia lies on the X chromosome.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Where does a son's X come from?
Answer
(a). A son's only X comes from his mother.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In thalassaemia, an abnormal myoglobin chain is formed.
Reason (R): α-thalassaemia is controlled by the genes HBA1 and HBA2 on chromosome 16.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which protein is affected?
Answer
(d). Thalassaemia reduces synthesis of globin chains of haemoglobin, not myoglobin; R is true.
Q3 · 2 marks · Very short answerBoard level
Why is colour blindness far more common in men (about 8%) than in women (about 0.4%)?
Hint
Number of X chromosomes.
Answer
The gene is recessive and on the X chromosome. Men have one X, so a single recessive allele makes them colour blind. Women have two X chromosomes and are affected only if both carry the allele, which is much rarer; with one copy they are carriers with normal vision.
Q4 · 3 marks · Short answerHOTS / competency
Suppose a haemophilic man marries a woman who is a carrier. (a) Give the possible genotypes of their children. (b) What fraction of their daughters will be haemophilic? (c) Why is such a daughter extremely rare in the population?
Hint
$X^{h} Y$ × $X^{h} X$ .
Answer
(a) Daughters: $X^{h} X^{h}$ (haemophilic) or $X^{h} X$ (carrier); sons: $X^{h} Y$ (haemophilic) or XY (normal). (b) 1/2 of daughters. (c) It needs a haemophilic father and an at least carrier mother; haemophilic men are rare and many may not survive to later life, so such matings are rare.
Q5 · 3 marks · Short answerHarder board level
Differentiate between sickle-cell anaemia and thalassaemia on three points.
Hint
Nature of defect, genes, result.
Answer
Sickle-cell: qualitative defect (globin made but wrong); single point mutation in HBB (Glu → Val at β-6); RBCs become sickle-shaped under low oxygen. Thalassaemia: quantitative defect (too little α or β globin); mutation or deletion in HBA1/HBA2 (chromosome 16) or HBB (chromosome 11); abnormal haemoglobin and anaemia. Both are autosomal recessive.
Q6 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a pedigree shows two unaffected parents (I-1, I-2) with four children: a son II-1 affected, a daughter II-2 affected, and II-3, II-4 unaffected. II-2 marries an unaffected man with no family history and has two unaffected children.
(a) Is the trait dominant or recessive? (1)
(b) Is it autosomal or X-linked? Give one reason. (1)
(c) What are the genotypes of I-1, I-2 and the children of II-2? What is the chance that II-3 is a carrier? (2)
Hint
Affected daughter with an unaffected father.
Answer
(a) Recessive: unaffected parents have affected children. (b) Autosomal: an affected daughter (II-2) has an unaffected father; with X-linked recessive inheritance her father would have to be affected. (c) I-1 and I-2 are Aa; II-2 is aa and her husband AA, so their children are all Aa (carriers). II-3 is unaffected, so the chance of being a carrier is 2/3.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which square symbol means an affected male?Show answer
A filled (shaded) square.
2.What does a double line between partners indicate?Show answer
Consanguineous mating.
3.Chance of a colour-blind son from a carrier mother and normal father?Show answer
50% of sons.
4.Which amino acid replaces glutamic acid in sickle-cell haemoglobin?Show answer
Valine.
5.Which enzyme function is missing in PKU?Show answer
Conversion of phenylalanine to tyrosine.
6.On which chromosome is HBB, the β-thalassaemia gene?Show answer
Chromosome 11.

Pedigree Analysis and Mendelian Disorders: frequently asked questions

What is pedigree analysis?

The study of a trait in several generations of a family, drawn with standard symbols, to find whether it is dominant or recessive and autosomal or sex-linked.

Why are haemophilia and colour blindness more common in males?

Their genes are recessive and on the X chromosome. Males have one X, so one recessive allele is enough; females need two.

What causes sickle-cell anaemia?

A point mutation (GAG → GUG) that replaces glutamic acid with valine at the sixth position of the β-globin chain, so haemoglobin polymerises at low oxygen and RBCs become sickle-shaped.

How is thalassaemia different from sickle-cell anaemia?

Thalassaemia is a quantitative defect (too little α or β globin is made); sickle-cell anaemia is a qualitative defect (a faulty globin is made).

What is phenylketonuria?

An autosomal recessive inborn error of metabolism in which phenylalanine cannot be converted to tyrosine, so phenylpyruvic acid accumulates and causes mental retardation.