Type 2.2: 推斷一個trait 係dominant 定recessive (sex-linked)

再講2.2。解決呢類要記住兩句口訣。

1. 相生相剋
2. 異性相吸

兩句夾埋即係搵一pair 有吾同phenotype 既父女或母子。搵到之後就由男性做推斷再去到女性。

Formula of case 1:

The father (the son) has the trait and being a male he has the genotype of XY. His X-chromosome must bear the allele for the trait on his only X-chromosome.
This X-chromosome with the allele for the trait must have been inherited to the daughter (from his mother).
Genotype of female is XX. The daughter (the mother) is normal in phenotype. Her other X-chromosome must bear the normal allele for manifestation of the normal phenotype.
Hence, the daughter (the mother) is heterozygous.
Under heterozygous condition, only the dominant allele will be expressed while the recessive allele will be masked. Hence, the allele for trait is recessive.

Formula of case 2:

The father (the son) is normal and being a male he has the genotype of XY. His X-chromosome must bear the normal allele on his only X-chromosome.
This X-chromosome with the normal allele must have been inherited to the daughter (from his mother).
Genotype of female is XX. The daughter (the mother) has the trait. Her other X-chromosome must bear the allele for the trait for manifestation of the trait.
Hence, the daughter (the mother) is heterozygous.
Under heterozygous condition, only the dominant allele will be expressed while the recessive allele will be masked. Hence, the allele for trait is dominant.

e.g. 1 2001 HKCE Human Biology I (3biii)

The pedigree below shows the inheritance of a disease caused by a sex-linked gene.

Without using a genetic diagram, deduce whether the allele for this disease is dominant or recessive.

Individual 3 has the disease and being a male he has the genotype of XY. His X-chromosome must bear the defective allele on his only X-chromosome.
This X-chromosome with the defective allele must have been inherited from individual 1.
Genotype of female is XX. Individual 1 is normal in phenotype. Her other X-chromosome must bear the normal allele for manifestation of the normal phenotype.
Hence, 1 is heterozygous.
Under heterozygous condition, only the dominant allele will be expressed while the recessive allele will be masked. Hence, the defective gene is recessive.

e.g. 2 2010 May / June 9700 / 04 GCE AL IV (6b: modified)

The pedigree below shows the inheritance of a rare form of rickets caused X-linked allele.

Without using a genetic diagram, deduce whether the allele for this disease is dominant or recessive.

7 is normal and being a male he has the genotype of XY. His X-chromosome must bear the normal allele on his only X-chromosome.
This X-chromosome with the normal allele must have been inherited to 9.
Genotype of female is XX. 9 has rickets. Her other X-chromosome must bear the allele for the trait for manifestation of rickets.
Hence, 9 is heterozygous.
Under heterozygous condition, only the dominant allele will be expressed while the recessive allele will be masked. Hence, the allele for rickets is dominant.

呢條可以考慮 6 同 11.

e.g. 3 1993 HKAL II (1b)

The occurrence of colour blindness and haemophilia in a family is shown by the pedigree below. The alleles for the two traits are found on the X chromosome.

Deduce the dominance or recesssiveness of the allele which leads to

(1) Colour blindness

Individual A has colour blindness and being a male he has the genotype of XY. His X-chromosome must bear the defective allele on his only X-chromosome.
This X-chromosome with the defective allele must have been inherited to individual D.
Genotype of female is XX. Individual D is normal in phenotype / not colour blind. Her other X-chromosome must bear the normal allele for manifestation of the normal phenotype.
Hence, D is heterozygous.
Under heterozygous condition, only the dominant allele will be expressed while the recessive allele will be masked. Hence, allele for colour blindness is recessive.

(2) Haemophilia

Individual C has haemophilia and being a male he has the genotype of XY. His X-chromosome must bear the defective allele on his only X-chromosome.
This X-chromosome with the defective allele must have been inherited to individual F.
Genotype of female is XX. Individual F is normal in phenotype. Her other X-chromosome must bear the normal allele for manifestation of the normal phenotype.
Hence, F is heterozygous.
Under heterozygous condition, only the dominant allele will be expressed while the recessive allele will be masked. Hence, allele for haemophilia is recessive.

Type 3: 利用一個交配cross做deduction