WFU

   5** 操 Past Paper 正確方法       BIO 5** 是如何考出來      5**讀書態度及技巧分享       數學 MC 極速解題技巧
線上:

2015年7月5日

Type 4.1: 推斷一個人既genotype


type 2 一樣呢種題型可再細分做兩類:4.1 佢無講係autosomal sex-linked (通常當佢係 autosomal)4.2 佢講明係sex-linked


先講4.1。呢種係cert 間吾中出下,但係cert level human biology 係好常見。解決呢類要記住之前講過嘅一句口訣:相生相剋。即係搵出一pair 有吾同phenotype 既父母子女,未必要吾同性,但要吾同型。再由homozygous recessive individual 做推論(叫呢個人做A),再去到同A有吾同phenotype individual (叫呢個人做B) B heterozygous



Formula of case 1:
  • A has the trait. He / she must have 2 recessive alleles for the trait, i.e. he / she must be homozygous recessive for the manifestation of the trait. 
  • One of the recessive alleles must have been inherited to / from B.
  • B is normal. He / she must have at least one dominant normal allele.
  • Hence, B is heterozygous.

Formula of case 2:
  • A is normal in phenotype. He / she must have 2 recessive normal alleles, i.e. he / she must be homozygous recessive for the manifestation of the normal phenotype. 
  • One of the recessive alleles must have been inherited to / from B.
  • B is normal. He / she must have at least one dominant normal allele.
  • Hence, B is heterozygous.

e.g.1  1997 HKCEE I (1cii)


Mediterranean anaemia is a human blood disease. Whether a person has this disease or not is determined by a pair of alleles. The allele for the normal character is dominant, while the allele for the disease is recessive. The pedigree below shows the inheritance of this disease in a family:


Deduce, with reasons, the genotype of Peter.
  • John has the disease. He must have 2 recessive diseased alleles, i.e. he must be homozygous recessive for the manifestation of the disease. 
  • One of the recessive alleles must have been inherited from Peter.
  • Peter is normal. He must have at least one dominant normal allele.
  • Hence, Peter is heterozygous.

e.g.2  2007 HKCEE I (5b)


The inheritance of the shape of the little finger is controlled by a pair of alleles. The following pedigree shows the inheritance of this trait in a family:


Provided that the allele for the bent little fingers is dominant, deduce the possible genotype(s) of individual 4. 

  • Individual 1 has straight little fingers. He must have 2 recessive alleles, i.e. she must be homozygous recessive for the manifestation of the straight little fingers. 
  • One of the recessive alleles must have been inherited to individual 4.
  • Individual 4 has bent little fingers. He must have at least one dominant allele for bent little fingers.
  • Hence, individual 4 is heterozygous.

e.g.3  2009 HKCEE I (3ai)


It is known that G6PD deficiency is controlled by a pair of alleles and allele causing G6PD deficiency is recessive. Deduce the possible genotype(s) of individual 7.
  • Individual 8 is G6PD deficient. She must have 2 recessive diseased alleles, i.e. she must be homozygous recessive for the manifestation of the disease. 
  • One of the recessive alleles must have been inherited from individual 7.
  • Individual 7 is normal. She must have at least one dominant normal allele.
  • Hence, individual 7 is heterozygous.
 

e.g.4  1995 HKCEE Human Biology I (1ci)


The diagram below shows a pedigree for the inheritance of ABO blood groups which is determined by three alleles, IA, IB and i. Both IA and IB are dominant and i is recessive.


Deduce, with reasons, the genotype(s) of Andy without using a genetic diagram.

  • Fred has blood group O. He must have 2 recessive alleles ii / homozygous recessive for blood group O.
  • One of the i alleles must have been inherited from Andy and the other from Betty.
  • David has blood group A. He must have at least 1 dominant allele IA.
  • Betty has blood group B. She must have at least 1 dominant allele IB. Therefore, Betty has the genotype of IB i. Hence, the IA allele of David must have been inherited from Andy but not Betty.
  • Hence, Andy has the genotype of  IAi / is heterozygous for blood group A.

e.g.5  2000 HKCEE Human Biology I (3ai & ii)




Mr. and Mrs. Ko are the biological parents of Joe. They also claim to be the biological parents of a pair of twins, Ken and Ada. The table below lists the blood groups of these five persons. The inheritance of blood groups is determined by three alleles, IA, IB and i. Both IA and IB are dominant, and i is recessive.


(i) Deduce, with reasons, the genotype of Mr. Ko.

  • Joe has blood group O. He must have 2 recessive alleles ii / homozygous recessive for blood group O.
  • One of the i alleles must have been inherited from Mr. Ko.
  • Mr. Ko has blood group B. He must have at least 1 dominant allele IB.
  • Therefore, Mr. Ko has the genotype of IB i / heterozygous for blood group B.


(ii) Based on the given information, determine whether Mr. and Mrs. Ko are the biological parents of the twins. Explain your answer without using a genetic diagram.

  • Ken has blood group AB. He has a genotype of IA IB. Mrs. Ko has blood group O. She must have 2 recessive alleles ii / homozygous recessive for blood group O. Mr. Ko has a genotype of IB i.
  • If they are the biological parents of the twins, IA allele of Ken should have been inherited from either Mr. Ko or Mrs. Ko. However, both of them lack IA. Hence, they are not biological parents of the twin. 

e.g.6  2005 HKCEE Human Biology I (1bi) 


Heredity is one of the causes of astigmatism. The inheritance of this eye defect is controlled by a pair of alleles, which are not sex-linked. The allele for this eye defect (A) is dominant, while that for normal vision (a) is recessive. The pedigree below shows the inheritance of astigmatism in a family:

State the genotype(s) of individual 1. Explain how you arrive at your answer. 
  • Individual 3 has normal vision. He must have 2 recessive alleles, i.e. he must be homozygous recessive for the manifestation of the normal vision. 
  • One of the recessive alleles must have been inherited from individual 1.
  • Individual 1 has astigmatism. She must have at least one dominant allele for astigmatism.
  • Hence, individual 1 is heterozygous.

e.g.7  2012 HKALE II (1ci)


β thalassaemia is caused by autosomal gene mutation. Persons who are homozygous for the mutated allele are known as β thalassaemia major patients. Persons who are heterozygous appear normal.



Deduce which of the above individuals are heterozygous.

  • Individual 4 is homozygous recessive for the disease. He must have passed the mutated allele to each of the offspring, i.e. individual 8, 9 and 10.
  • Individual 11 is homozygous for the disease. Individual 7 and 8 must have one mutated allele passed to individual 11.
  • Individual 7, 8, 9 and 10 is normal in phenotype. They must have at least one dominant normal allele. Hence, they are heterozygous.


Type 4.2:         推斷一個人既genotype (sex-linked) 



沒有留言:

張貼留言

歡迎留言

Cut-off 補習