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AN74.1-4 | Patterns of Inheritance — Gate Quiz

Graded 10 questions · 20 min · 3 attempts

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Q1 AN74.1 1 pt

Both parents are phenotypically normal but their first child has phenylketonuria (an autosomal recessive disorder). What is the probability that their next child will be phenotypically normal?

A 25%
B 50%
C 75%
D 100%

Correct. Both parents are carriers (Aa). Punnett square: AA (25%) + Aa (50%) + aa (25%). Phenotypically normal = AA + Aa = 75%.

For an autosomal recessive disorder with both parents as carriers: 25% affected (aa), 50% carrier (Aa, normal), 25% homozygous normal (AA). Phenotypically normal = 75%.

Recall the carrier × carrier cross (Aa × Aa). Four outcomes: 1 AA : 2 Aa : 1 aa. 'Affected' = aa = 25%. 'Normal phenotype' = all others = 75%.

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Q2 AN74.1 1 pt

A man with Marfan syndrome (autosomal dominant) marries a woman without the condition. What proportion of their children are expected to have Marfan syndrome?

A 0%
B 25%
C 50%
D 100%

Correct. Autosomal dominant: affected father is heterozygous (Aa). Cross: Aa × aa → 50% Aa (affected) + 50% aa (unaffected).

In autosomal dominant disorders, an affected heterozygous parent transmits the condition to 50% of offspring regardless of the child's sex.

Marfan syndrome is autosomal dominant. The affected father is almost always heterozygous (Aa). Each child has a 50% chance of inheriting the abnormal allele.

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Q3 AN74.2 1 pt

A carrier woman (X^H X^h) for haemophilia A marries a normal man (X^H Y). What is the probability that their son will have haemophilia?

A 0%
B 25%
C 50%
D 100%

Correct. Sons get Y from father and either X^H or X^h from mother. Probability of X^h Y (haemophilia) = 50% of sons = 25% of all children.

For X-linked recessive: carrier mother × normal father → 50% of sons affected, 50% of daughters carriers. There is NO father-to-son transmission.

Sons receive their Y chromosome from their father and one X chromosome from their mother. The mother has equal probability of passing X^H or X^h. So 50% of sons will be affected.

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Q4 AN74.2 1 pt

A father with an X-linked dominant condition (e.g., Vitamin D-resistant rickets) marries a normal woman. Which statement about their offspring is CORRECT?

A All sons will be affected
B All daughters will be affected
C 50% of sons and 50% of daughters will be affected
D No offspring will be affected

Correct. An affected father (X^D Y) passes X^D to ALL daughters (who all get X from father) and Y to ALL sons. Sons are unaffected; all daughters are affected.

X-linked dominant key rule: Affected father → ALL daughters affected, NO sons affected. This is the hallmark that distinguishes X-linked dominant from autosomal dominant (which would affect 50% of both sexes).

In X-linked dominant inheritance, an affected father passes his X^D to ALL daughters and Y to ALL sons. Sons cannot inherit X-linked condition from father.

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Q5 AN74.3 1 pt

A couple is first cousins from a South Indian community where the thalassaemia carrier rate is 8%. Both are carriers of beta-thalassaemia. They are expecting their first child. What is the probability of the child having thalassaemia major?

A 4%
B 8%
C 25%
D 50%

Correct. Both parents are known carriers (Aa × Aa). Probability of affected child (aa) = 25%, regardless of the population carrier rate.

Once both parents are CONFIRMED carriers, the population carrier rate is irrelevant. Carrier × Carrier = 25% affected, 50% carrier, 25% normal — every pregnancy independently.

The population carrier rate tells you the risk of a RANDOM individual being a carrier. But here, BOTH parents are KNOWN carriers. The Punnett square for Aa × Aa gives 25% affected (aa).

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Q6 AN74.1 1 pt

A woman with Leber's hereditary optic neuropathy (mitochondrial inheritance) has 4 children. What is expected regarding her children?

A 50% of children affected regardless of sex
B All children may be affected; sex ratio 1:1
C Only daughters transmit to next generation; all children may be affected
D Only sons are affected

Correct. Mitochondrial inheritance: all children of an affected mother may be affected (mitochondria transmitted via ovum only). However, only DAUGHTERS transmit to the next generation — sons do not pass on mitochondrial DNA.

Mitochondrial inheritance: exclusively maternal (through egg); ALL children of affected mother may be affected; only daughters transmit to subsequent generations; no male-to-offspring transmission.

Mitochondrial DNA is transmitted exclusively through the egg (maternal inheritance). All children of an affected mother may be affected, but only daughters can pass it to THEIR children.

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Q7 AN74.2 1 pt

In a pedigree, a disorder affects individuals in every generation. Both males and females are equally affected. An affected father has an affected son. What is the most likely inheritance pattern?

A X-linked recessive
B X-linked dominant
C Autosomal dominant
D Autosomal recessive

Correct. Autosomal dominant: every generation affected (vertical transmission), both sexes affected equally, and father-to-son transmission is possible (ruling out X-linked).

Key pedigree analysis rules: (1) Father-to-son transmission → autosomal (not X-linked). (2) Every generation affected → dominant. (3) Both sexes equally → autosomal. Together: Autosomal Dominant.

Father-to-son transmission rules out X-linked inheritance (fathers give Y to sons). Every generation affected with both sexes points to autosomal dominant.

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Q8 AN74.3 1 pt

A couple has three sons, all of whom have Duchenne muscular dystrophy (X-linked recessive). The mother is a confirmed carrier. What is the probability that their NEXT child (sex unknown) will be an unaffected carrier?

A 12.5%
B 25%
C 50%
D 75%

Correct. Cross: X^D X^d (carrier mother) × X^D Y (normal father). Four outcomes: X^D X^D (normal girl, 25%), X^D X^d (carrier girl, 25%), X^D Y (normal boy, 25%), X^d Y (affected boy, 25%). Unaffected carrier = carrier girl = 25%.

X-linked recessive carrier mother × normal father: 4 equal outcomes. Unaffected carriers are exclusively carrier daughters = 25% of all children. This is essential for counselling X-linked families.

List all 4 possible offspring: normal girl (25%), carrier girl (25%), normal boy (25%), affected boy (25%). 'Unaffected carrier' = only carrier girls = 25%.

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Q9 AN74.1 1 pt

Which of the following features in a pedigree most strongly suggests autosomal recessive inheritance?

A Affected individuals in every generation
B Only males affected, transmitted through carrier females
C Consanguinity and multiple affected siblings with unaffected parents
D All daughters of an affected father are affected

Correct. Autosomal recessive hallmarks: affected siblings with phenotypically normal parents (horizontal transmission), and consanguinity increases the likelihood that two carriers of a rare recessive allele marry.

Autosomal recessive pedigree features: (1) Horizontal pattern — siblings affected, parents unaffected. (2) Consanguinity in pedigree. (3) Both sexes equally affected. (4) 25% recurrence risk.

Autosomal recessive shows horizontal transmission (sibs affected, parents normal), often with consanguinity, and affects both sexes equally.

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Q10 AN74.3 1 pt

A child with achondroplasia (autosomal dominant, FGFR3 mutation) is born to phenotypically normal parents with no family history. What is the most likely explanation?

A Both parents are carriers of a recessive allele
B The condition skipped a generation — this is autosomal recessive
C De novo (new) mutation in the FGFR3 gene
D X-linked inheritance from an unaffected carrier mother

Correct. Approximately 80% of achondroplasia cases arise from de novo mutations (new mutations in FGFR3), most commonly on the paternal chromosome. Both parents are phenotypically and genotypically normal.

De novo mutations are responsible for 80% of achondroplasia cases (paternal origin, associated with advanced paternal age). When an AD condition appears with no family history, always consider de novo mutation before questioning diagnosis.

When an autosomal dominant condition appears with no family history, the most likely explanation is a de novo (new) mutation — particularly for achondroplasia where 80% of cases are new mutations.

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