Type 2: 推斷一個trait 係dominant 定recessive sex-linked
呢種題型可再細分做兩類:2.1 佢無講係autosomal 定sex-linked (通常當佢係 autosomal);2.2 佢講明係sex-linked。
先講2.1 。呢種題型係cert level 非常典型嘅題種,cert 出得最多就係呢隻。解決呢類要記住兩句口訣。
1. 求同存異
2.父母生仔女,父母大過仔女
2.父母生仔女,父母大過仔女
第一句可以幫你揾出用邊個人做推論,即係搵出兩個有相同phenotype既父母,再搵出同父母有吾同phenotype 既子女呢個就係求同存異既意思。
第二句係可以幫你知道邊個trait 係dominant 同邊個係recessive。我係咁記,父母大過仔女,所以父母有既trait 係dominant 而仔女既trait 係recessive。呢句仲可以話俾你知應由父母再到子女做deduction
Formula of case 1:
- Both parents are normal so they must have at least one normal allele.
- The son / daughter is affected so he / she must have at least one allele for the trait.
- His / her allele of the trait must have been inherited from either one of the parents.
- Hence, either one of the 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 the trait must be recessive.
Formula of case 2:
- Both parents have the trait so they must have at least one allele for the trait.
- The son / daughter is normal so he / she must have at least one normal allele.
- His / her normal allele must have been inherited from either one of the parents.
- Hence, either one of the 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 have the trait, the allele for the trait is dominant while normal allele must be recessive.
e.g. 1 1991 HKCE I (1ci)
- Both individual 3 and 4 are normal so they must have at least one normal allele.
- Individual 10 has albino so she must have at least one albino allele.
- Her albino allele must have been inherited from either individual 3 or 4.
- Hence, either 3 or 4 must be 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 are normal, normal allele is dominant while albino allele must be recessive.
用第一句口訣,就會知道應該用右邊個pedigree而吾係左邊,而且應該用3,4 同10做deduction。用第二句口訣,就會知道父母有既trait 係dominant ,所以normal allele 係 dominant 。仔女有既trait 係recessive ,所以albino allele 係 recessive。
下面個啲題目都係好類似,可以嘗試應用下條formula去解決。
e.g. 2 1993 HKCE I (1bi)
- Both individual 1 and 2 are six-toed so they must have at least one allele for six toes.
- Individual 3 / 6 is normal so she / he must have at least one normal allele.
- His / her normal allele must have been inherited from either individual 1 or 2.
- Hence, either 1 or 2 must be heterozygous.
- In heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the both individual 1 and 2 are six-toed, allele for six toes is dominant while normal allele must be recessive.
e.g. 3 1995 HKCE I (1bi)
- Both individual 3 and 4 have blood group A so they must have at least one antigen A allele.
- Individual 11 / 12 has blood group O so she / he must have at least one allele for blood group O.
- Her / his allele for blood group O must have been inherited from either individual 3 or 4.
- Hence, either 3 or 4 must be 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 blood group A, allele for antigen A is dominant while allele for blood group O must be recessive.
e.g. 4 2002 HKCE I (3ci)
- Both individual 1 and 2 have short fingers so they must have at least one allele for short fingers.
- Individual 3 / 6 has normal fingers so she / he must have at least one allele for normal fingers.
- Her / his allele for normal fingers must have been inherited from either individual 1 or 2.
- Hence, either 1 or 2 must be heterozygous.
- In heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since the both individual 1 and 2 have short fingers, allele for short fingers is dominant while allele for normal fingers must be recessive.
e.g. 5 2003 HKCE I (2ai)
- Both parents produce purple flowers so they must have at least one allele for purple flowers.
- The offspring produce white flowers so it must have at least one allele for white flowers.
- The allele for white flowers must have been inherited from either one of the parents.
- Hence, either one of the 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 produce purple flowers, allele for purple flowers is dominant while allele for white flowers must be recessive.
e.g. 6 2002 HKAL IB (13a)
Eye colour and wing morphology of a certain species of fly are controlled by two different genes. A student made a cross using flies he collected from the field. The cross and its F1 progeny are shown below:
Deduce the recessive trait for eye colour and wing morphology. Explain your deduction.
- Both parents have red eyes and normal-wings so they must have at least one allele for red eye and normal wing respectively.
- Some progenies / offspring have yellow eyes and vestigial wings so they must have at least one allele for yellow eye and vestigial wings respectively.
- The alleles for yellow eyes and vestigial wings must have been inherited from either one of the parents.
- Hence, either one of the parents must be heterozygous for eye colour and wing morphology.
- Under heterozygous condition, only the dominant alleles will be expressed while the recessive alleles will be masked. Since both parents have red eyes and normal wings, yellow eye and vestigial wing are recessive traits.