Question

Albinism in humans is controlled by a recessive gene 'a'. If both parents were known to be carriers (Aa) for albinism, what are the chances of 1 normal and 3 albinos in a family of four?
A. 12/256
B. 81/256
C. 108/256
D. 54 /256
E. 1/256

Answer 1

Hint:-When both parents are heterozygous in each pregnancy there are 25% of the offspring will show recessive traits and 75% of the offspring will show normal trait.The probability is calculated for the four offsprings.

Complete step-by-step solution:- Albinism is a condition in which there is lack of pigmentation in eyes, skin and hair. This is an inherited condition .The combination of recessive alleles are passed from both the parents. In this condition parent 1 will have Aa and the parent 2 will also have Aa genotype. If the genotype are crossed the offsprings produced will be AA,Aa,aa,Aa.So in this condition there are chances of 25% of offspring with albino and 75% of patients with normal condition.From this condition

- Probability of birth of all normal offspring is calculated as ¾*3/4*3/4*3/4
- Probability of birth of albino offspring is. ¼*1/4*1/4*1/4
- Probability of 2 normal and 2 albino ¾*3/4*1/4*1/4
- Probability of 1 normal and 3 albino. ¾*1/4*1/4/*1/4

So from the question the chances of the offspring is given ¾*1/4*1/4*1/4 power of 3 is 12/256


Correct option : A. 12/256

Note:- In most of the cases, all types of OCA and some types of OA. Albinism is a condition mostly passed as an autosomal recessive inheritance pattern. This means a child has to inherit two copies of the faulty gene each from one parent to show the condition dominant.