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2015年7月6日

Type 3: 利用一個交配cross做deduction




再講Mendel 做嘅monohybrid experiment。佢一開始將一棵高homozygous plant 同另一棵矮homozygous plant cross, 出到黎全部都係heterozygous 高 既plant。係呢種題型會出現類似情況,可以用一句口訣總括:父母異型生雜種。即係兩個有吾同phenotype 既父母cross出嚟既offspring全部都有同樣既phenotype,咁全部offspring都係heterozygous


Type 2 type 3 既相同之處都係最要既步驟係prove某個人係heterozygous,只要做到呢步就知邊個trait dominant 而邊個係recessive


Formula of case 1:
  • Father (mother) with the trait must have at least one allele for the trait.
  • Mother (father) with normal trait must have at least one normal allele.
  • When they are crossed, all offspring are normal.
  • Since the offspring must have received an allele for the trait and a normal allele, they are heterozygous.
  • Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since all offspring are normal, normal allele must be dominant while the allele for the trait must be recessive.

Formula of case 2:
  • Father (mother) with the trait must have at least one allele for the trait.
  • Mother (father) with normal trait must have at least one normal allele.
  • When they are crossed, all offspring have the trait.
  • Since the offspring must have received an allele for the trait and a normal allele, they are heterozygous.
  • Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since all offspring have the trait, the allele for the trait must be dominant while the normal allele must be recessive.

e.g. 1 2012 HKDSE IB (8bi) 


The two traits of the kernels are controlled by genes located on different homologous chromosomes.


  • Pure-bred purple and smooth kernels must have at least one allele for purple and smooth kernels respectively.
  • Pure-bred yellow wrinkled kernels must have at least one allele for yellow and wrinkled kernels respectively. 
  • When they are crossed, all offspring produce purple and smooth kernels.
  • Since the F1 offspring must have received an allele for purple colour, yellow colour, smooth appearance and wrinkled appearance, they are heterozygous for grain colour and appearance.
  • Under heterozygous conditions, only the dominant alleles will be expressed while the recessive alleles will be masked. Since all offspring can produce purple and smooth kernels, alleles for purple colour and smooth appearance must be dominant while the alleles for yellow colour and wrinkled appearance must be recessive.

e.g. 2 1996 HKCE I (3ai)

  • Consider cross B.
  • Parent plants with pear-shaped fruits must have at least one allele for pear-shaped fruits.
  • Parent plants with spherical fruits must have at least one allele for spherical fruits.
  • When they are crossed, all offspring produce spherical fruits.
  • Since the offspring must have received an allele for spherical fruit and an allele for pear-shaped fruit, they are heterozygous.
  • Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since all offspring can produce spherical fruit, allele for spherical fruit must be dominant while the allele for pear-shaped fruit must be recessive.

e.g. 3 2011 HKCE I (6a)


In a breeding programme, a leopard and a black panther which are both homozygous for their coat patterns, were allowed to interbreed. The diagram below shows the programme and the results:


  • Leopards must have at least one allele for spotted coat.
  • Black panthers must have at least one allele for black coat.
  • When they are crossed, all offspring are leopards with spotted coat.
  • Since the offspring must have received an allele for spotted coat and an allele for black coat, they are heterozygous.
  • Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since all offspring can have spotted coat, allele for spotted coat must be dominant while the allele for black coat must be recessive.

e.g. 4 1999 HKASL (5a)


A purebred red-eyed and long-winged insect species was crossed with a purebred white-eyed and short-winged insect of the same species. Their F1 offspring were all red-eyed and long-winged. Each of the characters is governed by a pair of alleles.

What are the dominant characters? Explain your answer WITHOUT the use of genetic diagrams.
  • Pure-bred red-eyed and long-winged insects must have at least one allele for red eye and long wing respectively.
  • Pure-bred white-eyed and short-winged must have at least one allele for white eye and short wing respectively. 
  • When they are crossed, all offspring are red-eyed and long-winged.
  • Since the F1 offspring must have received an allele for red eye, white eye, long wing and short wing, they are heterozygous for eye colour and wing morphology. 
  • Under heterozygous conditions, only the dominant alleles will be expressed while the recessive alleles will be masked. Since all offspring are red-eyed and long-winged, alleles for red eye and long wing must be dominant while the alleles for white eye and short wing must be recessive.


Type 4:         推斷一個人既genotype


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