What is gene and mutation?

CFTR (cystic fibrosis transmembrane conductance regulator), chromosome 7. The most common mutation is F508, a three-base-pair deletion that removes phenylalanine at position 508 of the CFTR protein.

Autosomal recessive. Two unaffected carrier parents (Cc × Cc) have a 25 percent chance of an affected child. Carrier frequency in Australians of Northern European ancestry is approximately 1 in 25.

What is pathophysiology?

CFTR is a chloride channel in the apical membrane of epithelial cells. Loss of function reduces chloride and water secretion onto epithelial surfaces, producing thick viscous mucus. The mucus obstructs:

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Inquiry Question 2: Do non-infectious diseases cause more deaths than infectious diseases?

Investigate the causes and effects of named genetic diseases on humans, including cystic fibrosis, sickle cell anaemia and Huntington's disease, and analyse pedigrees showing their inheritance

HSC Biology Module 8 answer on named genetic disorders - cystic fibrosis (autosomal recessive, CFTR), sickle cell anaemia (autosomal recessive, HBB) and Huntington's disease (autosomal dominant, HTT) - with their molecular causes, effects, and pedigree and Punnett-square analysis.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-28

Reviewed by: AI editorial process; not yet individually human-reviewed

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NESA wants you to describe the cause, inheritance pattern and effects of named genetic disorders, and to read pedigrees identifying them. Cystic fibrosis, sickle cell anaemia and Huntington's disease are the three most commonly examined.

The answer

Cystic fibrosis (autosomal recessive)

Gene and mutation: CFTR (cystic fibrosis transmembrane conductance regulator), chromosome 7. The most common mutation is $\Delta F508$ , a three-base-pair deletion that removes phenylalanine at position 508 of the CFTR protein.
Inheritance: Autosomal recessive. Two unaffected carrier parents ( $Cc \times Cc$ ) have a 25 percent chance of an affected child. Carrier frequency in Australians of Northern European ancestry is approximately 1 in 25.
Pathophysiology: CFTR is a chloride channel in the apical membrane of epithelial cells. Loss of function reduces chloride and water secretion onto epithelial surfaces, producing thick viscous mucus. The mucus obstructs:

Lungs. Mucus traps bacteria (Pseudomonas aeruginosa, Staphylococcus aureus), causing recurrent infection, inflammation and progressive lung damage.
Pancreas. Blocked ducts prevent digestive enzyme delivery to the intestine, causing malabsorption and failure to thrive.
Sweat glands. Excessive salt loss (the basis of the sweat chloride test).
Reproductive tract. Congenital bilateral absence of the vas deferens in males causes infertility.

Treatment. Airway clearance physiotherapy, inhaled antibiotics, pancreatic enzyme replacement, high-calorie diet, and CFTR modulator drugs (e.g. ivacaftor for G551D, elexacaftor/tezacaftor/ivacaftor for $\Delta F508$ ). Lung transplant for end-stage disease. Gene therapy and gene editing (CRISPR) are in trial.

Sickle cell anaemia (autosomal recessive)

Gene and mutation: HBB (beta-globin) on chromosome 11. A single point mutation (GAG to GTG) changes glutamate to valine at position 6 of the beta-globin chain (the HbS allele).
Inheritance: Autosomal recessive. Homozygotes ( $ss$ ) have sickle cell disease; heterozygotes ( $Ss$ ) have sickle cell trait, which is largely asymptomatic but offers partial protection against malaria. This heterozygote advantage explains why the disorder persists at high frequency in malaria-endemic regions: in parts of West, Central and East Africa the carrier (HbAS) frequency reaches about 10 to 25 percent, while the HbS allele frequency itself is roughly half that (around 5 to 15 percent).
Pathophysiology: Mutant haemoglobin (HbS) polymerises under low oxygen tension, deforming red blood cells into rigid sickle shapes. Sickle cells:

Block capillaries, causing painful vaso-occlusive crises and tissue infarction.
Are destroyed prematurely, causing haemolytic anaemia.
Predispose to bacterial infection (functional asplenia) and stroke.

Treatment. Hydration, pain relief during crises, hydroxyurea (boosts fetal haemoglobin), blood transfusions, prophylactic antibiotics. Allogeneic bone marrow transplant is curative. CRISPR-based gene therapy (Casgevy, approved 2023) edits the BCL11A gene to reactivate fetal haemoglobin and effectively cures the disease.

Huntington's disease (autosomal dominant)

Gene and mutation: HTT (huntingtin), chromosome 4. The mutation is an expanded CAG trinucleotide repeat in exon 1. Fewer than 27 repeats is normal; 36 or more causes disease. The expanded repeat encodes a long polyglutamine tract that makes the huntingtin protein toxic to neurons.
Inheritance: Autosomal dominant. One affected heterozygote parent ( $Hh$ ) has a 50 percent chance per child of passing the affected allele. The repeat can expand further when transmitted, particularly through the father (anticipation): each generation may have earlier onset.
Pathophysiology: Mutant huntingtin causes selective neurodegeneration in the basal ganglia (caudate, putamen) and cortex. Symptoms include:

Motor. Chorea (involuntary jerky movements), dystonia, later rigidity.
Cognitive. Executive dysfunction, dementia.
Psychiatric. Depression, irritability, psychosis.

Symptoms typically begin between ages 30 and 50, after most affected individuals have already had children, which is why the allele persists in populations. Death typically 15 to 20 years after onset, often from pneumonia.

Treatment. No disease-modifying therapy. Symptomatic management with tetrabenazine for chorea, antipsychotics, antidepressants. Antisense oligonucleotide trials (tominersen) target huntingtin mRNA.

Pedigree analysis for these conditions

Autosomal recessive pedigree (cystic fibrosis, sickle cell).

Trait often skips generations (unaffected carriers).
Both sexes affected equally.
Two unaffected parents can have affected children.
Consanguinity raises risk.
Affected child cross $Cc \times Cc$ : 25 percent affected, 50 percent carriers, 25 percent unaffected non-carriers.

Autosomal dominant pedigree (Huntington's).

Trait appears in every generation.
Roughly 50 percent of offspring of an affected parent are affected.
Both sexes affected equally.
Affected father can transmit to son (rules out X-linked).
New mutations are uncommon; almost all cases have an affected parent.

Punnett squares (autosomal recessive carriers).

	C	c
C	CC	Cc
c	Cc	cc

Ratio 1 CC : 2 Cc : 1 cc (25 percent affected, 50 percent carrier, 25 percent homozygous unaffected).

Worked exam answers

Worked example 1 - the sickle cell mutation (3 marks)

Question. Describe the mutation that causes sickle cell anaemia and explain how it leads to the disease.

Model answer. A single base substitution in the HBB (beta-globin) gene changes codon 6 from GAG to GTG, replacing glutamate with valine in the beta-globin chain. This produces abnormal haemoglobin (HbS). Because valine is hydrophobic, HbS molecules polymerise under low oxygen tension, deforming red blood cells into rigid sickle shapes that block capillaries (vaso-occlusive crises) and are destroyed prematurely (haemolytic anaemia).

Marker's note: 1 mark for the base/codon change (GAG to GTG), 1 mark for the amino-acid change (Glu to Val at position 6), 1 mark for linking HbS polymerisation to the sickled cell and a clinical effect. Stating only "a mutation in haemoglobin" without the specific substitution caps at 1 mark.

Worked example 2 - cystic fibrosis carrier cross (4 marks)

Question. A couple of Northern European ancestry have one child with cystic fibrosis and one unaffected child. Neither parent has CF. Determine the parents' genotypes and the probability that the next child is affected, and state two reproductive options for the couple.

Model answer. CF is autosomal recessive, so the affected child is $cc$ and both unaffected parents must be heterozygous carriers ( $Cc$ ). The cross $Cc \times Cc$ gives $1\ CC : 2\ Cc : 1\ cc$ , so each pregnancy has a 25% ( $\tfrac{1}{4}$ ) chance of an affected ( $cc$ ) child, independent of previous children. Reproductive options include preimplantation genetic testing (PGT-M during IVF) to select unaffected embryos, prenatal testing (chorionic villus sampling), or natural conception with neonatal screening - with genetic counselling throughout.

Marker's note: 1 mark for both parent genotypes ( $Cc$ ), 1 mark for the correct cross/ratio, 1 mark for the 25% per-pregnancy risk (it does not change with prior children), 1 mark for two valid reproductive options. (The unaffected sibling has a two-thirds chance of being a carrier, but only conditional on already being unaffected - do not apply that to a new pregnancy.)

Worked example 3 - reading a Huntington's pedigree (4 marks)

Question. A pedigree shows a disorder present in every generation; about half the children of each affected parent are affected, an affected father has an affected daughter, and both sexes are affected. Deduce the inheritance pattern and justify your answer.

Model answer. The pattern is autosomal dominant (consistent with Huntington's disease). It appears in every generation with no skipping, ruling out a recessive trait that would pass silently through carriers. Roughly 50% of an affected parent's children are affected, matching a heterozygous ( $Hh$ ) parent crossed with an unaffected ( $hh$ ) partner. Father-to-daughter transmission and an equal sex ratio rule out X-linked inheritance, confirming the allele is autosomal.

Marker's note: 1 mark for naming autosomal dominant, 1 mark for the "every generation / ~50%" justification, 1 mark for excluding recessive, 1 mark for excluding X-linked using father-to-child transmission of both sexes. A bare label with no pedigree evidence caps at 1 to 2 marks.

Examples in context

Example 1. Cystic fibrosis carrier screening in NSW pregnancy planning. Around 1 in 25 Australians of European ancestry is a heterozygous carrier (Aa) of a CFTR mutation, mostly F508del. RANZCOG guidelines since 2018 recommend reproductive carrier screening for all couples planning pregnancy. If both partners are heterozygous, their probability of an affected (aa) child is 1 in 4 per pregnancy. The NSW pre-pregnancy carrier screening program now includes CFTR alongside SMA and Fragile X. Detected carrier couples are offered options including PGT-M (preimplantation genetic testing during IVF) to select unaffected embryos, prenatal testing, or accepting the recurrence risk. The pedigree pattern - two unaffected carrier parents with an affected child - is the autosomal recessive signature.

Example 2. Huntington's disease predictive testing at Royal Melbourne Hospital. Huntington's disease is autosomal dominant, so an affected parent has a 50 percent chance of passing the expanded HTT allele to each child. The Royal Melbourne Hospital Predictive Testing Program offers presymptomatic genetic testing to adult children of affected parents, including extensive pre-test counselling (typically three sessions over months) to ensure informed consent. The test counts CAG repeats: under 27 normal, 27 to 35 intermediate, 36 to 39 reduced penetrance, 40 or more fully penetrant. Roughly 15 percent of eligible at-risk individuals choose to test. Those who test positive face complex life decisions including reproductive choices, career and insurance, all underpinned by the predictable autosomal dominant pedigree pattern (every generation affected).

Exam technique

Lead with the inheritance pattern, gene and mutation type. For any named disorder, state the pattern (autosomal recessive/dominant), the gene and chromosome, and the kind of mutation (CFTR deletion, HBB point substitution, HTT repeat expansion). Precise terms lift a Band 4 answer to Band 6.

For a "read the pedigree" item, test each pattern against the evidence. Does the trait skip generations (recessive) or appear in every one (dominant)? Are both sexes affected (autosomal)? Does an affected father pass it to a son or daughter (rules out X-linked)? State the clue, then the conclusion - do not just assert the pattern.

Show Punnett-square working and keep per-pregnancy risk separate from conditional risk. $Cc \times Cc$ gives 25% affected each pregnancy. The "two-thirds carrier" figure applies only to a sibling already known to be unaffected - markers penalise mixing the two.

Match depth to the command word. "Describe" wants the cause and effects in order; "compare" wants the same features for each disorder; "evaluate" (e.g. genetic testing) wants a justified judgement weighing benefits, limitations and ethics.

Note

Your genes are an instruction manual, with two copies of every page - one from each parent. Cystic fibrosis and sickle cell anaemia are like a typo that only causes trouble if both copies of the page are wrong; if just one is wrong you are a healthy "carrier" who can pass it on. Sickle cell is a single-letter typo (one base changes) that swaps one building block in a blood protein, so red blood cells turn from soft discs into stiff crescents that jam in blood vessels. Huntington's disease is different: a single faulty page is enough to cause it, so there are no silent carriers - if a parent has it, each child has a one-in-two chance. Because Huntington's usually starts after age 30, people often have children before they know, which is why this fatal fault still gets passed down. Reading a family tree is just spotting these patterns: a fault that hides and skips generations is recessive; one that shows up in every generation is dominant.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksState the inheritance pattern, gene and chromosome for (a) cystic fibrosis, (b) sickle cell anaemia, and (c) Huntington's disease.

Show worked solution →

1 mark - cystic fibrosis: Autosomal recessive; gene CFTR on chromosome 7.
1 mark - sickle cell anaemia: Autosomal recessive; gene HBB (beta-globin) on chromosome 11.
1 mark - Huntington's disease: Autosomal dominant; gene HTT (huntingtin) on chromosome 4.

Each mark needs the inheritance pattern correctly paired with its gene; naming the disorder without the pattern, or pairing the wrong chromosome, loses that mark.

foundation2 marksDescribe the specific mutation that causes sickle cell anaemia and the change it produces in the protein.

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1 mark - the DNA/codon change. A single base substitution in the HBB gene changes the codon GAG to GTG (a point mutation), so the sixth codon of the beta-globin chain is altered.

1 mark - the protein change. This swaps glutamate (Glu) for valine (Val) at position 6 of the beta-globin chain, producing the abnormal haemoglobin HbS.

Full marks require both the base/codon change AND the resulting amino-acid substitution. An answer that says only "a mutation in haemoglobin" without the Glu to Val swap caps at 1 mark.

foundation2 marksExplain why cystic fibrosis is described as a multi-system disease rather than only a lung disease.

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1 mark - the underlying cause. CFTR is a chloride channel in the apical membrane of epithelial cells throughout the body, so loss of function produces thick mucus wherever those epithelia line a surface or duct.

1 mark - named systems affected. Therefore it affects more than the lungs: it also obstructs the pancreatic ducts (malabsorption), affects the sweat glands (salt loss) and the reproductive tract (male infertility). Naming at least two non-lung systems earns this mark.

core4 marksTwo parents of Northern European ancestry, neither of whom has cystic fibrosis, have a child diagnosed with the disease. Using a Punnett square, determine the genotypes of the parents and calculate the probability that their next child is an unaffected carrier.

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1 mark - deduce parent genotypes. CF is autosomal recessive, so an affected child is cc. Each parent must carry a c allele but is unaffected, so both parents are heterozygous carriers, Cc.

1 mark - construct the Punnett square. $Cc \times Cc$ gives:

	C	c
C	CC	Cc
c	Cc	cc

1 mark - genotypic ratio. Offspring are $1\ CC : 2\ Cc : 1\ cc$ , i.e. 25% CC, 50% Cc, 25% cc.

1 mark - the required probability. The chance the next child is an unaffected carrier (Cc) is $\tfrac{2}{4} = \tfrac{1}{2}$ = 50%.

The trap is answering "two-thirds": that is the carrier probability only when you are told the child is unaffected. For a next pregnancy with no such condition, the carrier probability is 50%.

core5 marksCompare the inheritance and molecular cause of sickle cell anaemia and Huntington's disease, and explain how each pattern would appear in a pedigree.

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Award up to 5 marks for genuine comparison (both similarities and differences) across inheritance, molecular cause and pedigree appearance.

Inheritance (1-2 marks): Sickle cell anaemia is autosomal recessive - only homozygotes (ss) are affected and heterozygotes (Ss) are unaffected carriers. Huntington's disease is autosomal dominant - a single copy of the expanded allele (Hh) causes disease, so there are no unaffected carriers.
Molecular cause (1-2 marks): Sickle cell is a point mutation (GAG to GTG) in HBB, swapping Glu for Val at position 6 of beta-globin. Huntington's is an expanded CAG trinucleotide repeat in HTT (36 or more repeats) encoding a toxic polyglutamine tract - a different class of mutation (repeat expansion, not substitution).
Pedigree appearance (1-2 marks): Recessive sickle cell tends to skip generations: two unaffected carrier parents can have an affected child, and consanguinity raises risk. Dominant Huntington's appears in every generation, with roughly 50% of an affected parent's children affected and no skipping. Both affect both sexes equally (autosomal) - the shared feature.

A response that describes only one disorder, or omits the pedigree contrast, cannot reach full marks.

core3 marksTwo heterozygous carriers of the CFTR mutation (Cc) plan to have three children. Calculate (a) the probability that all three are unaffected and (b) the probability that at least one is affected. Show your working.

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1 mark - per-child probabilities: From $Cc \times Cc$ , each child is affected (cc) with probability $\tfrac{1}{4}$ and unaffected with probability $\tfrac{3}{4}$ . Pregnancies are independent events.
1 mark - part (a): $P(\text{all three unaffected}) = \left(\tfrac{3}{4}\right)^3 = \tfrac{27}{64} \approx 0.42$ .
1 mark - part (b): $P(\text{at least one affected}) = 1 - \tfrac{27}{64} = \tfrac{37}{64} \approx 0.58$ .

Using the complement ( $1 - P(\text{none affected})$ ) for part (b) is the expected method; adding the separate cases must also give $\tfrac{37}{64}$ .

exam6 marksThe pedigree below tracks a disorder. Generation I: an affected father (I-1) and an unaffected mother (I-2). Generation II: of their four children, two are affected and two are unaffected; the disorder appears in every generation down to generation III. Both sexes are affected. Deduce the most likely mode of inheritance, justify your reasoning by ruling out the alternatives, and state the genotype of the affected father.

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Target a sequenced response that names the pattern, then rules out each alternative with evidence from the pedigree.

Identify the pattern (1-2 marks): The disorder appears in every generation, roughly half the children of an affected parent are affected, and both sexes are affected - this is the signature of autosomal dominant inheritance (consistent with Huntington's disease).
Rule out recessive (1-2 marks): If the allele were recessive, it would tend to skip generations (carriers unaffected); appearing in every generation without consanguinity makes recessive inheritance very unlikely.
Rule out X-linked (1-2 marks): Because an affected father (I-1) has affected daughters and the trait passes father to child of both sexes, it is not X-linked recessive (an affected father would pass his X only to daughters, who would be carriers, not affected) and the equal sex ratio argues against X-linkage generally - so autosomal is supported.
State the genotype (1 mark): The affected father is a heterozygote, Hh (most affected individuals are heterozygous; the expanded allele is dominant), which also explains the roughly 50% transmission to his children.

Band 6 answers do not just label the pattern - they cite the specific pedigree features (every generation, ~50%, father-to-daughter transmission) that exclude the alternatives.

exam7 marksGenetic testing is available for both cystic fibrosis (carrier screening) and Huntington's disease (predictive testing). Evaluate the use of genetic testing for these two disorders, considering the different inheritance patterns and the consequences of a result for the individual.

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"Evaluate" requires a judgement weighing benefits against limitations and ethical costs for each disorder - not just a description.

Cystic fibrosis - carrier screening (2 marks): CF is autosomal recessive, so testing identifies unaffected heterozygous carriers (Cc). Reproductive carrier screening lets couples who are both carriers (25% risk per pregnancy) make informed reproductive choices - PGT-M during IVF, prenatal testing, or accepting the risk with neonatal screening. The result concerns a future child's risk, not the tested adult's own health, so the psychological stakes are comparatively lower.
Huntington's disease - predictive testing (2 marks): HD is autosomal dominant with high penetrance, so a positive predictive test (36 or more CAG repeats) tells an asymptomatic adult they will almost certainly develop an incurable, fatal neurodegenerative disease. This is information about the tested person's own future, with major implications for psychological wellbeing, relationships, employment, insurance and reproductive decisions - hence the requirement for extensive pre- and post-test genetic counselling.
Benefits and limitations weighed (1-2 marks): Benefits: informed reproductive and life planning, early management, reduced uncertainty for some. Limitations and costs: testing cannot currently cure or prevent HD; a positive result may cause distress or discrimination; CF carrier status alone has no health consequence for the carrier; results affect blood relatives who may not wish to know.
Judgement (1-2 marks): A supported conclusion, e.g.: genetic testing is highly valuable for both, but its purpose and ethical weight differ - CF screening chiefly informs reproductive choice with low personal-health stakes, whereas HD predictive testing delivers certain personal prognosis for an untreatable disease and is therefore justified only with rigorous counselling and informed consent. An answer listing pros and cons without an explicit, justified judgement caps below full marks.