Type 5: 判斷一個trait係咪sex-linked

3.7.15

呢種可以話係genetics problems 中最高階既題型，aim high 既同學可以留意，呢隻既題型係出自 a level 既 biology。之後果啲human biology 同外國試題係我改過黎。

係講呢種題型之前想先講下一啲基本概念。sex-linked trait 可以有三種吾同情況： Y-linked, X-linked recessive 同 X-linked dominant。

再講下sex-linkage 推理用到既概念。

1.        XX係女性而XY係男性
2.        老母會將佢x chromosome 遺傳俾仔或女。老豆會將佢y chromosome 遺傳俾個仔，將佢x chromosome 遺傳俾個女
3.        一個dominant 既allele 係男定女都可以將phenotype express 出黎。一個recessive 既allele 係男可以將phenotype express 出黎，但係女人度只有係homozygous 既情況下先可以express 度個phenotype。


再講一講一陣prove 用到既手法：prove by contradiction 反證法。反證法其實係數學一種常見既prove手法，讀core既同學可能比較陌生，但係a level pure maths 就經常用到。個原理係咁：我要prove A事件正確。利用呢個手法，我會假設A係錯誤既，再睇下利用呢個假設會有啲咩嘢發生，做到一個地步同事實衝突、矛盾，我地就會推翻前設，即推翻A係錯誤呢個假設，由此證明A係正確。

有咗呢啲概念同手法就可以開始處理呢種題型。呢種題型圍繞5.1 Y-linked, 5.2 X-linked recessive 同 5.3 X-linked dominant 三種 trait，每種都會叫你prove 係定吾係sex-link。通常prove 親都吾係sex-linked. 而呢種題型通常會係之前加1 part type 2 既題型，叫你prove 個trait係dominant 定recessive


解決呢類要記住先前講過既兩句口訣。
1. 相生相剋
2. 異性相吸


今次可以淨係用第一句，或者用曬兩句都得。淨係用第一句就係搵出有吾同phenotype既母女或父子。兩句用曬即係搵一pair 有吾同phenotype 既父女或母子。

5.1.1 prove 個trait 吾係Y-linked




第一步：       先揾出推翻個trait係Y-linked既證據
- 個女有嘢 - 老豆有嘢但個仔無事 - 老豆無嘢但個仔有事

第二步：       用反證法，假設個trait係Y-linked


Formula of case 1 and 2:

  • Genotype of male is XY. If the trait is Y-linked, the father (son) must have an allele on his Y-chromosomes for the manifestation of the trait.
  • This Y-chromosome with the allele must have been inherited to his son (from his father).
  • However, the father (son) is normal. Hence, the allele must not be located on the Y-chromosome. 


Formula of case 3:

  • The genotype of female is XX. If the trait is Y-linked, all sons will be affected and all daughters should be unaffected as daughters have no Y-chromosome.
  • However, there is an affected daughter. Hence, the allele must not be located on the Y-chromosome.


5.1.2 prove 個trait 係Y-linked


第一步：       先揾出證明個trait係Y-linked既證據（同上面係相反既）
- 無女有個trait - 老豆同個仔都有個 trait

第二步：       解釋呢啲現象


Formula:
  • If the trait is Y-linked, the father (son) must have an allele on his Y-chromosome for the manifestation of the trait. This Y-chromosome with the allele must have been inherited to his son (from his father).
  • Genotype of female is XX. As the daughter has no Y-chromosome, she should not manifest the trait. 
  • Genotype of male is XY. As the father (son) only possesses one Y-chromosome, he should manifest the trait. Hence, the allele is located on the Y-chromosome.

5.2.1 prove 個trait 吾係 X-linked recessive


第一步：       先揾出推翻個trait係X-linked recessive 既證據
-  個女有嘢但個老豆無事 - 老母有嘢但個仔無事
第二步：       用反證法，假設個trait係X-linked recessive

第三步：       由個有trait既人開始做deduction 然後到無事既人


Formula:
  • Genotype of female is XX. If the trait is X-linked recessive, the daughter (mother) must have 2 recessive alleles on her X-chromosomes for the manifestation of the trait.
  • One of the X-chromosomes with the recessive allele of the daughter (mother) must have been inherited from her father (to her son).
  • Genotype of male is XY. As the father (son) only possesses one X-chromosome, he should have manifested the trait if it is X-linked recessive.
  • However, the father (son) is normal. Hence, the allele must not be located on the X-chromosome. 

5.2.2 prove 個trait 係X-linked recessive 


Case 1 & 2

第一步：       先揾出證明個trait係X-linked dominant 既證據（同上面係相反既）
-  個女同老豆有嘢 - 老母同個仔有嘢

第二步：       解釋呢啲現象


Formula:
  • Genotype of female is XX. If the trait is X-linked recessive, the daughter (mother) must have two X-chromosomes with the recessive alleles for the manifestation of the trait.
  • One of the X-chromosomes with the recessive allele of the daughter (mother) must have been inherited from her father (to her son).
  • Genotype of male is XY. As the father (son) only possesses one X-chromosome, he should manifest the trait. Hence, the allele is located on the X-chromosome.


Case 3 係special case，要見過先識處理
第一步：       搵出暗示老豆無allele for trait 既證據, 例如佢無果個trait既家族遺傳歷史
第二步：       用反證法，假設個trait係autosomal

第三步：       由個有trait既仔開始做deduction 然後到無事既父母

Formula:
  • If the trait is autosomal, the son must have 2 recessive alleles on his autosomes for manifestation of trait. 
  • One of the autosomes must come from the father and the other from the mother.  
  • However, the father has no family history of the trait, so he is homozygous and unlikely to have the allele. It is impossible for him to transmit the allele of the trait to the son.
  • The mother is female and has a genotype of XX.
  • If the allele is located on X-chromosome, only one X-chromosome with the allele from the mother has to be present in the son for the manifestation of disease. Hence, the defective allele is likely to be X-linked.

5.3.1 prove 個trait 吾係 X-linked dominant


 

第一步：       先揾出推翻個trait係X-linked dominant 既證據
-  個老豆有嘢但個女無事
- 個仔有嘢但個老母無事
- 個老母有嘢但個仔同女無嘢（或者淨係個女無嘢）
- 個女有嘢但老豆無嘢
第二步：       用反證法，假設個trait係X-linked dominant

第三步：       由個有trait既人開始做deduction 然後到無事既人


Formula of case 1:
  • Genotype of male is XY. If the trait is X-linked dominant, the father (son) 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 daughter (from his mother).
  • Genotype of female is XX. As the dominant allele will express its phenotype, his daughter (his mother) should have manifested the trait if it is X-linked dominant.
  • However, the daughter (the mother) is normal. Hence, the allele must not be located on the X-chromosome. 

Formula of case 2:
  • Genotype of female is XX. If the trait is X-linked dominant, the mother (the daughter) 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 her daughter and son (from her father).
  • Genotype of male is XY. As the dominant allele will express its phenotype, his daughter and son (the father) should have manifested the trait if it is X-linked dominant.
  • However, the son and daughter are normal (the father is normal). Hence, the allele must not be located on the X-chromosome. 

5.3.2 prove 個trait 係X-linked dominant



第一步：       先揾出證明個trait係X-linked dominant 既證據（同上面係相反既）
 -  個女同老豆有嘢 - 老母同個仔同女有嘢
第二步：       解釋呢啲現象


Formula of case 1:
  • Genotype of male is XY. If the trait is X-linked dominant, the father must have one X-chromosome with the dominant allele for the manifestation of the trait.
  • This X-chromosome with the dominant allele of the father must have been inherited to his daughter.
  • Genotype of female is XX. As the dominant allele will express its phenotype, the daughter should manifest the trait. Hence, the allele must be located on the X-chromosome.
Formula of case 2:
  • Genotype of female is XX. If the trait is X-linked dominant, the mother 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 her daughter and son.
  • Genotype of male is XY. As the dominant allele will express its phenotype, his daughter and son should manifest the trait if it is sex-linked. Hence, the allele must be located on the X-chromosome.