Match the syndromes listed in Column I with the cause/symptoms listed in Column IIColumn IColumn IIP. Prader-Willi syndrome(i) a collection of signs and symptoms due to prolonged exposure to corticosteroids like cortisolQ. Down syndrome(ii) a syndrome of inadequate reabsorption in the proximal renal tubule of the kidneyR. Cushing syndrome(iii) a genetic disorder usually caused by the deletion of a part of chromosome 15S. Turner syndrome(iv) a genetic disorder caused by the presence of all or part of a third copy of chromosome 21T. Fanconi syndrome(v) a genetic condition in which a female has partially or completely missing an X chromosome

The match is P-(iii); Q-(iv); R-(i); S-(v); T-(ii).Prader-Willi syndrome is usually caused by loss of the paternal copy of a region of chromosome 15 (15q11-q13).Down syndrome is caused by a third copy of all or part of chromosome 21.Cushing syndrome results from prolonged exposure to high levels of corticosteroids such as cortisol.Turner syndrome is a female with one X chromosome partly or wholly missing (45,X).Fanconi syndrome is a failure of reabsorption in the proximal renal tubule.The other options swap the chromosomal disorders or attach the kidney or steroid descriptions to genetic syndromes.

Human Genetics Questions and Answers

Q: Red-Green color blindness is an X-linked recessive disorder. In a population which is in the Hardy-Weinberg equilibrium, the incidence of occurrence of this in males is 1:1000. What will be the expected incidence of affected homozygous females?

The expected incidence is 1 in 1000000.For an X-linked recessive trait, a male has a single X chromosome, so he is affected whenever that X carries the allele. The frequency of affected males therefore equals the allele frequency: \(q = 1/1000\).A female is affected only if she is homozygous, receiving the allele on both X chromosomes. Under Hardy-Weinberg equilibrium that has frequency\[ q^{2} = \left(\frac{1}{1000}\right)^{2} = \frac{1}{1000000} \]This is why X-linked recessive conditions are far more common in men than in women. 1 in 1002000 would come from treating 1:1000 as 1 affected to 1000 unaffected (q = 1/1001), which the key does not use; 1 in 2000000 halves the correct value, and 1 in 1001000 matches no standard calculation.

Q: Haemophilia disease is more commonly seen in males because

Haemophilia is commoner in males because it is an X-linked disease.The genes for clotting factors VIII and IX lie on the X chromosome, and haemophilia is X-linked recessive. A male has only one X chromosome, so a single defective copy causes the disease. A female has two X chromosomes and is usually only a carrier unless both are affected, which is rare.An autosomal disease would affect both sexes equally. A Y-linked trait would pass only from father to son, which is not haemophilia's pattern. It is a blood clotting disorder, not a disease of the reproductive system.

Q: Which of the following represents triplet repeat disorder ?

The triplet repeat disorder is Huntington's disease.Huntington's disease is caused by expansion of a CAG repeat in the HTT gene: more than about 36 repeats give an expanded polyglutamine tract in the huntingtin protein, which damages neurons. Like other repeat disorders it shows anticipation, with longer repeats and earlier onset in later generations.Cystic fibrosis is usually caused by a three-base deletion (ΔF508) in CFTR, not a repeat expansion. Xeroderma pigmentosum is a defect of nucleotide excision repair, and phenylketonuria results from point mutations in the phenylalanine hydroxylase gene.

Q: In human, hypertrichosis is inherited as Y-linked trait. If a man marries a normal woman, which of the following types of offspring may they have ?

The answer is that all sons will have hypertrichosis but none of the daughters.A Y-linked (holandric) gene is carried only on the Y chromosome, which passes from father to every son and to no daughter. An affected father therefore passes the trait to all his sons, and his daughters, who receive his X chromosome, are unaffected. The mother, having no Y chromosome, contributes nothing to the trait.Daughters cannot be affected, which rules out the first option. 'Half the sons' would fit an X-linked gene from a carrier mother, not a Y-linked gene from the father, and 'none of the children' ignores that every son inherits his father's Y.

Q: Hemophilia and colour blindness are :

Both are X-linked recessive inheritance.The genes for clotting factor VIII (haemophilia A) and for the red and green opsins (red-green colour blindness) lie on the X chromosome, and the disease alleles are recessive. A male, with one X, shows the trait whenever he inherits the allele; a female is usually an unaffected carrier. So these conditions are far more common in men and pass from carrier mothers to sons.They are not autosomal, since their genes are on the X, and they are not Y-linked; the Y carries very few genes and Y-linked traits would pass only from father to son. Neither is dominant.

Q: Double lines in pedigree analysis shows :

A double line shows marriages between related individuals.In standard pedigree symbols a single horizontal line joins a mating pair, and a double line marks a consanguineous mating, between relatives such as cousins. Such matings matter because they raise the chance that a child inherits two copies of the same rare recessive allele.Sex unspecified is shown by a diamond, unaffected individuals by unshaded symbols, and a mating between unrelated people by an ordinary single line.

Q: One of the common symptoms of a metabolic disorder is that the patients have lighter hair and skin colour than their siblings. What is the name of this disorder ?

The disorder is Phenylketonuria.In phenylketonuria the enzyme phenylalanine hydroxylase is deficient, so phenylalanine is not converted to tyrosine. Tyrosine is the starting material for melanin, and phenylalanine also inhibits melanin synthesis, so untreated patients often have fairer skin and hair than their siblings, along with intellectual disability.Alkaptonuria causes urine that darkens on standing and dark pigment in cartilage. Tay-Sachs disease is a ganglioside storage disorder affecting the nervous system, and Cori's disease is a glycogen storage disease caused by deficiency of the debranching enzyme.

Q: Klinefelter’s Syndrome is identified by one of the following karyotypes ?

Klinefelter's syndrome has the karyotype 47, XXY.Males with Klinefelter's syndrome carry an extra X chromosome, usually from nondisjunction in a parent's meiosis, giving 47 chromosomes. Typical features are small testes, infertility and tall stature.45, XO is Turner's syndrome, in females with a single X. 45, XY and 45, XX would each mean a missing autosome, which is not Klinefelter's syndrome.

Q: If a woman having blood group A with |A|O genotype married a man having a blood group with |B|B genotype, the expected proportion of heterozygotes among their children is :

The expected proportion of heterozygotes is 100%.The woman's genotype is IAIO (printed |A|O) and the man's is IBIB (printed |B|B). She passes on IA or IO with equal chance; he can pass on only IB:\[ \begin{aligned} I^A \times I^B &\rightarrow I^A I^B\ (\text{AB}) \\ I^O \times I^B &\rightarrow I^B I^O\ (\text{B}) \end{aligned} \]Each child carries two different alleles, so every child is heterozygous: half with blood group AB, half with blood group B. 50% is the trap of counting only the AB children, and 25% or 75% would need both parents to be heterozygous.

Red-Green color blindness is an X-linked recessive disorder. In a population which is in the Hardy-Weinberg equilibrium, the incidence of occurrence of this in males is 1:1000. What will be the expected incidence of affected homozygous females?

A.1 in 1002000
B.1 in 2000000
C.1 in 1001000
D.1 in 1000000

Exam Year(s)

GATE Life Sciences - 2012,

Answer: D

The expected incidence is 1 in 1000000.

For an X-linked recessive trait, a male has a single X chromosome, so he is affected whenever that X carries the allele. The frequency of affected males therefore equals the allele frequency: \(q = 1/1000\).

A female is affected only if she is homozygous, receiving the allele on both X chromosomes. Under Hardy-Weinberg equilibrium that has frequency

\[ q^{2} = \left(\frac{1}{1000}\right)^{2} = \frac{1}{1000000} \]

This is why X-linked recessive conditions are far more common in men than in women. 1 in 1002000 would come from treating 1:1000 as 1 affected to 1000 unaffected (q = 1/1001), which the key does not use; 1 in 2000000 halves the correct value, and 1 in 1001000 matches no standard calculation.

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Match the syndromes listed in Column I with the cause/symptoms listed in Column II

Column I	Column II
P. Prader-Willi syndrome	(i) a collection of signs and symptoms due to prolonged exposure to corticosteroids like cortisol
Q. Down syndrome	(ii) a syndrome of inadequate reabsorption in the proximal renal tubule of the kidney
R. Cushing syndrome	(iii) a genetic disorder usually caused by the deletion of a part of chromosome 15
S. Turner syndrome	(iv) a genetic disorder caused by the presence of all or part of a third copy of chromosome 21
T. Fanconi syndrome	(v) a genetic condition in which a female has partially or completely missing an X chromosome

A.P - (ii); Q - (v); R - (iv); S - (iii); T - (i)
B.P - (iv); Q - (iii); R - (i); S - (ii); T - (v)
C.P - (iii); Q - (iv); R - (i); S - (v); T - (ii)
D.P - (v); Q - (iv); R - (ii); S - (i); T - (iii)

Exam Year(s)

GATE Life Sciences - 2023,

Answer: C

The match is P-(iii); Q-(iv); R-(i); S-(v); T-(ii).

Prader-Willi syndrome is usually caused by loss of the paternal copy of a region of chromosome 15 (15q11-q13).
Down syndrome is caused by a third copy of all or part of chromosome 21.
Cushing syndrome results from prolonged exposure to high levels of corticosteroids such as cortisol.
Turner syndrome is a female with one X chromosome partly or wholly missing (45,X).
Fanconi syndrome is a failure of reabsorption in the proximal renal tubule.

The other options swap the chromosomal disorders or attach the kidney or steroid descriptions to genetic syndromes.

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Haemophilia disease is more commonly seen in males because

A.It is X-linked disease
B.It is an autosomal disease
C.It is Y-linked disease
D.It is a disease of male reproductive system/organ

Exam Year(s)

MPPSC Forest Service - 2022,

Answer: A

Haemophilia is commoner in males because it is an X-linked disease.

The genes for clotting factors VIII and IX lie on the X chromosome, and haemophilia is X-linked recessive. A male has only one X chromosome, so a single defective copy causes the disease. A female has two X chromosomes and is usually only a carrier unless both are affected, which is rare.

An autosomal disease would affect both sexes equally. A Y-linked trait would pass only from father to son, which is not haemophilia's pattern. It is a blood clotting disorder, not a disease of the reproductive system.

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Which of the following represents triplet repeat disorder ?

A.Huntington's disease
B.Cystic fibrosis
C.Xeroderma pigmentosum
D.Phenylketonuria

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The triplet repeat disorder is Huntington's disease.

Huntington's disease is caused by expansion of a CAG repeat in the HTT gene: more than about 36 repeats give an expanded polyglutamine tract in the huntingtin protein, which damages neurons. Like other repeat disorders it shows anticipation, with longer repeats and earlier onset in later generations.

Cystic fibrosis is usually caused by a three-base deletion (ΔF508) in CFTR, not a repeat expansion. Xeroderma pigmentosum is a defect of nucleotide excision repair, and phenylketonuria results from point mutations in the phenylalanine hydroxylase gene.

Discuss in Forum

In human, hypertrichosis is inherited as Y-linked trait. If a man marries a normal woman, which of the following types of offspring may they have ?

A.All sons and daughters will have hypertrichosis
B.All sons will have hypertrichosis but none of the daughters.
C.Only half of their sons will have hypertrichosis but none of the daughters
D.None of the children will have hypertrichosis.

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The answer is that all sons will have hypertrichosis but none of the daughters.

A Y-linked (holandric) gene is carried only on the Y chromosome, which passes from father to every son and to no daughter. An affected father therefore passes the trait to all his sons, and his daughters, who receive his X chromosome, are unaffected. The mother, having no Y chromosome, contributes nothing to the trait.

Daughters cannot be affected, which rules out the first option. 'Half the sons' would fit an X-linked gene from a carrier mother, not a Y-linked gene from the father, and 'none of the children' ignores that every son inherits his father's Y.

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Hemophilia and colour blindness are :

A.Autosomal dominant inheritance patterns
B.Y-linked recessive inheritance
C.X-linked recessive inheritance
D.Y- and X-linked dominant inheritance respectively

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

Both are X-linked recessive inheritance.

The genes for clotting factor VIII (haemophilia A) and for the red and green opsins (red-green colour blindness) lie on the X chromosome, and the disease alleles are recessive. A male, with one X, shows the trait whenever he inherits the allele; a female is usually an unaffected carrier. So these conditions are far more common in men and pass from carrier mothers to sons.

They are not autosomal, since their genes are on the X, and they are not Y-linked; the Y carries very few genes and Y-linked traits would pass only from father to son. Neither is dominant.

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Double lines in pedigree analysis shows :

A.Sex unspecific
B.Unaffected offspring
C.Marriages between related individuals
D.Marriages between unrelated individuals

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

A double line shows marriages between related individuals.

In standard pedigree symbols a single horizontal line joins a mating pair, and a double line marks a consanguineous mating, between relatives such as cousins. Such matings matter because they raise the chance that a child inherits two copies of the same rare recessive allele.

Sex unspecified is shown by a diamond, unaffected individuals by unshaded symbols, and a mating between unrelated people by an ordinary single line.

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One of the common symptoms of a metabolic disorder is that the patients have lighter hair and skin colour than their siblings. What is the name of this disorder ?

A.Alkaptonuria
B.Tay-Sachs disease
C.Phenylketonuria
D.Cori’s disease

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The disorder is Phenylketonuria.

In phenylketonuria the enzyme phenylalanine hydroxylase is deficient, so phenylalanine is not converted to tyrosine. Tyrosine is the starting material for melanin, and phenylalanine also inhibits melanin synthesis, so untreated patients often have fairer skin and hair than their siblings, along with intellectual disability.

Alkaptonuria causes urine that darkens on standing and dark pigment in cartilage. Tay-Sachs disease is a ganglioside storage disorder affecting the nervous system, and Cori's disease is a glycogen storage disease caused by deficiency of the debranching enzyme.

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Klinefelter’s Syndrome is identified by one of the following karyotypes ?

A.45, XO
B.47, XXY
C.45, XY
D.45, XX

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

Klinefelter's syndrome has the karyotype 47, XXY.

Males with Klinefelter's syndrome carry an extra X chromosome, usually from nondisjunction in a parent's meiosis, giving 47 chromosomes. Typical features are small testes, infertility and tall stature.

45, XO is Turner's syndrome, in females with a single X. 45, XY and 45, XX would each mean a missing autosome, which is not Klinefelter's syndrome.

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10.

If a woman having blood group A with |A|O genotype married a man having a blood group with |B|B genotype, the expected proportion of heterozygotes among their children is :

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

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The expected proportion of heterozygotes is 100%.

The woman's genotype is I^AI^O (printed |A|O) and the man's is I^BI^B (printed |B|B). She passes on I^A or I^O with equal chance; he can pass on only I^B:

\[ \begin{aligned} I^A \times I^B &\rightarrow I^A I^B\ (\text{AB}) \\ I^O \times I^B &\rightarrow I^B I^O\ (\text{B}) \end{aligned} \]

Each child carries two different alleles, so every child is heterozygous: half with blood group AB, half with blood group B. 50% is the trap of counting only the AB children, and 25% or 75% would need both parents to be heterozygous.

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