公式化解決genetic problems (Type 5 examples part 1)

e.g. 1 2010 HKCE I (9ai & iii)

Disease A is a kind a genetic disorder. Its occurrence is controlled by a pair of alleles. The following pedigree shows the inheritance of disease A in a family.

(i) Based on the pedigree above, a student deduced that the allele for disease A is dominant. Explain how he arrive at his answer.

Individual 3 and 4 have disease A. They must have at least one allele for disease A.
Individual 8 is normal. She must have at least one normal allele.
The normal allele must have been inherited from either 3 or 4.
Hence, either 3 or 4 is heterozygous.
In heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the both individual 3 and 4 have disease A, allele for disease A is dominant while normal allele must be recessive.

(iii) Using the information in the pedigree for individuals 2, 5 and 6, explain why it is not possible for the allele of disease A to be located on the X chromosome or the Y chromosome.

Individual 2, having disease A, must have at least one dominant allele for disease A.
If the allele for disease A is X-linked, individual 5 must receive defective allele in X-chromosome and have manifested the disease. This X-chromosome should have been inherited from 2.
If the allele for disease A is Y-linked, individual 6 must receive defective allele in Y-chromosome and have manifested the disease.
However, both 5 and 6 have not manifested the disease and have a normal phenotype.
Hence, it is impossible for allele of disease A to be located on X or Y chromosome.

e.g. 2 1994 HKCE Human Biology I (2d: modified)

The diagram below shows a pedigree for the inheritance of short-sightedness.

(i) Deduce whether the allele for short-sightedness is dominant or recessive.

Both individual A and B (F and G) are normal so they must have at least one normal allele.
Individual E (K) is short-sighted so he must have at least one allele for short-sightedness.
His allele for short-sightedness must have been inherited from either individual A or B (F or G).
Hence, either A or B (F or G) must be heterozygous.
In heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the both individual A and B (F or G) are normal, normal allele is dominant while allele for short-sightedness must be recessive.

(ii) Deduce whether the allele for short sight is X-linked or autosomal.

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

Genotype of female is XX. If the trait is sex-linked, C must have 2 recessive alleles on her X-chromosomes for the manifestation of short-sightedness.
One of her X-chromosomes with the recessive allele must have been inherited to her son G / J.
Genotype of male is XY. As G / J only possesses one X-chromosome, he should have manifested short-sightedness if it is sex-linked.
However, G / J is normal. Hence, the allele for short-sightedness must not be located on the X-chromosome / is located on the autosome.

e.g. 3 1996 HKCE Human Biology I (1ai & ii: modified)

The diagram below shows a pedigree for the inheritance of a human genetic disease.

(i) It is given that that 3 came from family with no history of this hereditary disease. Deduce the dominance or recessiveness of the allele of the disease.

Individual 3 and 4 are normal. They must have at least one normal allele.
Individual 8 is affected. He must have at least one defective allele.
Since 3 have no family history of this diseases, he is unlikely to have the defective allele. So the defective allele must have been inherited from 4.
Hence, 4 is heterozygous.
Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since heterozygous 4 is normal, normal allele must be dominant while the defective allele is recessive.

(ii) Deduce if the disease is X-linked or autosomal.

If the disease is autosomal, individual 8 must have 2 recessive alleles on his autosomes for manifestation of disease.
One of the autosomes must come from 3 and the other from 4.
However, 3 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 8.
4 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 4 has to be present in individual 8 for the manifestation of disease. Hence, the defective allele is likely to be X-linked.

e.g. 4 2003 HKCE Human Biology I (1bii (1) & (2): modified)

The occurrence of cystic fibrosis (CF) is controlled by a pair of allele. In a family, a girl is born with CF but both of its parents are normal.

(1) Deduce whether the allele for CF is dominant or recessive.

Both parents are normal so they must have at least one normal allele.
The girl has CF so he must have at least one allele for CF.
Her allele for CF must have been inherited from either one of her parents.
Hence, either one of her parents must be heterozygous.
In heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the both parents are normal, normal allele is dominant while allele for CF must be recessive.

(2) Deduce whether the allele for CF is X-linked or autosomal.

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

e.g. 5 1999 HKAL II (3c)

The following pedigree shows the occurrence of a hereditary disease D among the members of a family:

Assuming that disease D is controlled by a single gene, deduce why this disease is NOT sex-linked.

XX is female. XY is male.
A mother only transmits X sex chromosome to offspring. A father segregates either X or Y sex chromosome to an offspring.
A dominant allele will always express its phenotype in both a male and a female individual. A recessive allele will also express its phenotype in a male individual but it requires homozygosity for expressing the controlled phenotype in a female individual.

If the disease-causing allele is Y-linked,

It is expected that all male progeny of an affected father will be affected individuals. However, for the pedigree, male individual 6 in F1 was not affected although his father (individual 2) was affected. Thus, the disease-causing allele is not Y-linked.

It is expected that all affected individuals must be male. However, from the pedigree, female individuals 4 and 5 in F1 and female individual 11 and 12 in F2 were also affected. Thus, the disease-causing allele is not Y-linked.

If the disease-causing allele is X-linked and recessive,

It is expected that all affected females inherit the disease-causing alleles from both her father and mother. However, from the pedigree, affected female individuals 11 and 12 in F2 were fathered by unaffected individual 3 and affected mother 4. Thus, the diseases-causing allele is not recessive X-linked.

It is expected that an affected female produces affected male progeny. However, from the pedigree, male individuals 9 and 10 in F2 were unaffected, and they were sons of the affected mother (individual 4). Thus, the disease-causing allele is not recessive X-linked.

If the disease-causing allele is X-linked and dominant,

It is expected that an affected son must inherit his disease-causing allele from his mother who is an affected individual too. However, from the pedigree, affected male (individual 7 in F1) was mothered by unaffected individual. Thus, the disease-causing allele is not dominant X-linked.

It is expected that the female progeny of an affected father will be affected. However, from the pedigree, female individuals 15 and 16 in F2 were unaffected although their father (individual 7) was affected. Thus, the disease-causing allele is not dominant X-linked.

Thus, from the above dedication, the disease-causing allele is neither X nor Y-linked.

e.g. 6 2001 HKASL (3: modified)

The pedigree below shows the inheritance of a rare hereditary disease in a family. Whether a person has this disease is controlled by a pair of allele.

(a) It is given that that 2 and 12 came from families with no history of this hereditary disease. Deduce the dominance or recessiveness of the allele of the disease.

Individual 11 and 12 are normal. They must have at least one normal allele.
Individual 14 and 17 are affected. They must have at least one defective allele.
Since 12 have no family history of this diseases, he is unlikely to have the defective allele. So the defective allele must have been inherited from 11.
Hence, 11 is heterozygous.
Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since heterozygous 11 is normal, normal allele must be dominant while the defective allele is recessive.

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

If the disease is autosomal, individual 14 / 17 must have 2 recessive alleles on his autosomes for manifestation of disease.
One of the autosomes must come from 2 and the other from 12.
However, 12 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 14.
11 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 11 has to be present in individual 14 and 17 for the manifestation of disease. Hence, the defective allele is likely to be X-linked.