公式化解決genetic problems (Type 5 examples part 2)

e.g. 7 2002 HKAL IB (13b)

Eye colour and wing morphology of a certain species of fly are controlled by two different genes. A student made a cross using flies he collected from the field. The cross and its F1 progeny are shown below:

From part (a), red eye colour is the dominant trait while the yellow eye colour is the recessive trait.
In this species of fly, X and Y are the sex chromosomes. XY confers male phenotype and XX confers female phenotype. From the given data, what evidence supports the hypothesis that ‘the gene for eye colour is NOT on the X chromosome’?

If eye colour gene is on the X-chromosome, the red-eyed male parent must have one dominant red eye allele on his only X-chromosomes for the manifestation of red eye colour.
This X-chromosome with the dominant allele must have been inherited to all F1 females.
As the dominant allele will express its phenotype, all F1 females should be red-eyed.
However, some F1 females have yellow eyes. Hence, the eye colour gene must not be located on the X-chromosome.

If eye colour gene is on the X-chromosome, the yellow-eyed females in the F1 progeny must have 2 recessive yellow eye alleles on her X-chromosomes for the manifestation of yellow eye colour.
One of the X-chromosomes with the recessive allele of the daughter must have been inherited from the male parent.
As the male parent only possesses one X-chromosome, he should have manifested yellow eye if the eye colour gene is sex-linked.
However, the male parent has red eye. Hence, the eye colour gene must not be located on the X-chromosome.

e.g. 8 2011 HKAL II (1d)

In a family of four, both the father and the mother are normal but both their son and daughter suffer from a genetic disease. Without using genetic diagrams,

(i) Deduce whether the genetic defect is a dominant trait or a recessive trait.

Both father and mother are normal. They must have at least one normal allele.
The son is diseased. He must have at least one defective allele.
One of the diseased alleles must have been inherited from either one of the parents.
Hence, either one of the parents is heterozygous.
Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the heterozygous parents are normal, normal allele must be dominant while the defective allele must be recessive.

(ii) Deduce whether the gene for the genetic defect is located on the X-chromosome.

Genotype of female is XX. If the disease is sex-linked, the daughter must have 2 recessive alleles on her X-chromosomes for the manifestation of the disease.
One of the X-chromosomes with the recessive allele of the daughter must have been inherited from her father.
Genotype of male is XY. As the father only possesses one X-chromosome, he should have manifested the disease if it is sex-linked.
However, the father is normal. Hence, the defective allele must not be located on the X-chromosome / is located on the autosome.

e.g. 9 2004 October / November 9700 / 04 GCE AL IV (4: modified)

The diagram below shows a pedigree for the inheritance of sickle cell anaemia.

(a) Deduce whether sickle cell anaemia is a dominant trait or a recessive trait.

Both A and B are normal. They must have at least one normal allele.
C / F is diseased. He / she must have at least one defective allele.
One of the diseased alleles must have been inherited from either one of the parents.
Hence, either one of the parents is heterozygous.
Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the heterozygous parents are normal, normal allele must be dominant while the defective allele must be recessive.

(b) Deduce whether the gene for the genetic defect is located on the X-chromosome.

Genotype of female is XX. If the disease is sex-linked, F must have 2 recessive alleles on her X-chromosomes for the manifestation of the disease.
One of the X-chromosomes with the recessive allele of the daughter must have been inherited from her father B.
Genotype of male is XY. As B only possesses one X-chromosome, he should have manifested the disease if it is sex-linked.
However, B is normal. Hence, the defective allele must not be located on the X-chromosome / is located on the autosome.

e.g. 10 2008 May / June 9700 / 41 GCE AL IV (6: modified)

Colour blindness is a condition characterised by the inability of the brain to perceive certain colours accurately. The most common form is termed red-green colour blindness (RGC). The diagram below shows a pedigree for the inheritance of RGC.

(a) It is given that that 1 and 8 came from families with no history of RGC. Deduce the dominance or recessiveness of the allele of RGC.

Individual 1 and 2 (7 and 8) are normal. They must have at least one normal allele.
Individual 6 (10) is affected. He must have at least one defective allele.
Since 1 (8) has no family history of this diseases, he is unlikely to have the defective allele. So the defective allele must have been inherited from 2 (7).
Hence, 1 (7) is heterozygous.
Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since heterozygous 1 (7) is normal, normal allele must be dominant while the defective allele is recessive.

(b) Deduce if RGC is X-linked or autosomal.

If RGC is autosomal, individual 6 (10) must have 2 recessive alleles on his autosomes for manifestation of defect.
One of the autosomes must come from 1 and the other from 2 (from 7 and the other from 8).
However, 1 (8) has no family history of the disease, so he is homozygous and unlikely to have the defective allele. It is impossible for him to transmit a defective allele to individual 6 (10).
2 (7) is female and has a genotype of XX.
If the defective allele is located on X-chromosome, only one X-chromosome with the defective allele from 2 (7) has to be present in individual 6 (10) for the manifestation of defect. Hence, the defective allele is likely to be X-linked.

e.g. 11 Edexcel IGCSE 2014 June 1BR (5: modified)

Huntington’s disease causes damage to the nervous system. It is an inherited condition caused by a dominant allele. The diagram below shows a pedigree for the inheritance of Huntington’s disease.

Based on the pedigree above, deduce with reasons whether or not the inheritance of Huntington’s disease is sex-linked.

Genotype of male is XY. If the trait is sex-linked, individual 9 (14) must have one dominant allele on his only X-chromosomes for the manifestation of the trait.
This X-chromosome with the dominant allele must have been inherited to individual 15 (inherited from individual 10).
Genotype of female is XX. As the dominant allele will express its phenotype, 15 (10) should have manifested the trait if it is sex-linked.
However, 15 (10) is normal. Hence, the allele for the trait must not be located on the X-chromosome.

Genotype of female is XX. If the trait is sex-linked, individual 2 (5) must have at least one X-chromosomes with the dominant allele for the manifestation of the trait.
This X-chromosome with the dominant allele must have been inherited to individual 6 and 7 (inherited from individual 1).
Genotype of male is XY. As the dominant allele will express its phenotype, 6 and 7 (1) should have manifested the trait if it is sex-linked.
However, 6 and 7 are normal (1 is normal). Hence, the allele for the trait must not be located on the X-chromosome.

e.g. 12 Edexcel IGCSE 2016 June 2B (3: modified)

Familial hypercholesterolemia (FH) is an inherited condition caused by a dominant allele. People with FH have high blood cholesterol level. The diagram below shows a pedigree for the inheritance of FH.

Based on the pedigree above, deduce with reasons whether or not the inheritance of Huntington’s disease is sex-linked.

Genotype of male is XY. If the trait is sex-linked, individual 5 (individual 14) must have one dominant allele on his only X-chromosomes for the manifestation of the trait.
This X-chromosome with the dominant allele must have been inherited from his mother individual 2 (individual 7)
Genotype of female is XX. As the dominant allele will express its phenotype, 2 (7) should have manifested the trait if it is sex-linked.
However, 2 (7) is normal. Hence, the allele for the trait must not be located on the X-chromosome.

Genotype of male is XY. If the trait is sex-linked, individual 10 must have one dominant allele on his only X-chromosomes for the manifestation of the trait.
This X-chromosome with the dominant allele must have been inherited to his daughters individual 17 and 18.
Genotype of female is XX. As the dominant allele will express its phenotype, 17 and 18 should have manifested the trait if it is sex-linked.
However, 17 and 18 are normal. Hence, the allele for the trait must not be located on the X-chromosome.