
Parents are the carrier for albinism. What will be the genotype of the first three children?
(a) Some normal, heterozygous, and albino
(b) All normal
(c) All heterozygous albino
(d) All albino
Answer
557.7k+ views
Hint: Albinism is an autosomal recessive disorder. An autosomal recessive disorder means two copies of an abnormal gene must be present in order for the disease or trait to develop. A normal individual will have genotype AA and a carrier of albinism will have genotype Aa.
Complete answer:
Since albinism is an autosomal recessive disorder, it means that only the homozygous recessive genotype will show the disorder and the heterozygous genotype will be normal but will be the carrier as the recessive is masked due to the presence of the normal dominant gene.
Now, in the question, it is given that parents are a carrier which means they are heterozygous normal.
So, the parent generation is Aa (carrier father) x Aa (carrier mother)
F1 generation: 1 normal child (AA) : 2 normal carrier (Aa) : 1 albino
This means that the first three children will be normal, heterozygous, and albino in the ratio of 1:2:1.
Additional information:
Albinism is a congenital disorder in humans that is characterized by a partial or complete absence of pigment in eyes, skin, and hair. It is due to the absence or defect of tyrosinase, a copper-containing enzyme that is involved in the production of melanin. This lack of pigmentation in the skin makes the individual more prone to sunburns and skin cancers.
So, the correct answer is ‘Some normal, heterozygous, and albino’.
Note: If both the parents are carriers, then the genotype of the first three children will be some normal, heterozygous, and albino. If only one of the parents is a carrier, then the genotype will be some normal and some heterozygous, none of them will be an albino.
Complete answer:
Since albinism is an autosomal recessive disorder, it means that only the homozygous recessive genotype will show the disorder and the heterozygous genotype will be normal but will be the carrier as the recessive is masked due to the presence of the normal dominant gene.
Now, in the question, it is given that parents are a carrier which means they are heterozygous normal.
So, the parent generation is Aa (carrier father) x Aa (carrier mother)
| Aa | A | a |
| A | AA | Aa |
| a | Aa | aa |
F1 generation: 1 normal child (AA) : 2 normal carrier (Aa) : 1 albino
This means that the first three children will be normal, heterozygous, and albino in the ratio of 1:2:1.
Additional information:
Albinism is a congenital disorder in humans that is characterized by a partial or complete absence of pigment in eyes, skin, and hair. It is due to the absence or defect of tyrosinase, a copper-containing enzyme that is involved in the production of melanin. This lack of pigmentation in the skin makes the individual more prone to sunburns and skin cancers.
So, the correct answer is ‘Some normal, heterozygous, and albino’.
Note: If both the parents are carriers, then the genotype of the first three children will be some normal, heterozygous, and albino. If only one of the parents is a carrier, then the genotype will be some normal and some heterozygous, none of them will be an albino.
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