Question

First child of a normal couple is albinic, then what would be the probability of their second child to be albinic again? (Albinism is an autosomal recessive disorder)

Answer 1

Hint: We know that certain diseases are inherited by children from their parents. These diseases are called genetic diseases. Genetic diseases are classified into two groups. They are Mendelian disorders.

Complete answer: We know that certain diseases are caused because of mutations that occur in a single gene. These disorders are called Mendelian disorders. Chromosomal disorders are caused by changes in the number and arrangements in the chromosomes. Keeping in mind that albinism is an autosomal recessive disorder, let’s look into the current scenario.
We know that the homozygous recessive condition is needed for a person to be albinic. As the first child is albinic, we should understand that both the parents are carriers of this trait.
Let’s represent the parents as \[A'A\], as both of them carry the trait. Keep in mind that $A'$ represents the mutated condition and $A$ represents the normal condition. Crossing them, we get
$A'A \times A'A$.
The children would be $A'A',A'A,A'A$ and $AA$. Here, only one among the four children would have the disease and two of them are carriers. So, the probability of their second child to be albinic again is $25\% $. The probability of the child to be a carrier of the disease is $50\% $ and the child to have a completely normal genotype is $25\% $.
Hence, the correct answer is $25\% $.

Note: When mutations occur in sex chromosomes, the disease is said to be sex-linked. Haemophilia and colour blindness are examples of sex-linked disorders. In haemophilia, the patient will have continuous bleeding from minor injuries. In colour blind individuals, they fail to distinguish red and green colour.