Question

Albinism is an autosomal recessive condition characterized by the absence of melanin pigment from the skin, eye, and hair. Two carriers of albinism marry and plan to have FIVE children. Assume a 1:1 sex ratio. What is the probability that at least 3 children will be normal?

A. 0.264
B. 0.399
C. 0.663
D. 0.750

Answer 1

The probability that at least 3 out of 5 children will be normal when both parents are carriers of the autosomal recessive condition of albinism is calculated using the binomial distribution and the principles of autosomal recessive inheritance.

Step-by-step explanation:

The question pertains to calculating the probability that at least 3 children will be normal (not have albinism) when two carriers of the autosomal recessive condition have five children. Each child has a 75% chance of being normal (3 out of 4), as two carrier parents have a 25% chance of producing a child with albinism (rr genotype) according to autosomal recessive inheritance. To find the probability of at least 3 children being normal, we need to add the probabilities of having exactly 3, exactly 4, and exactly 5 normal children.

• 
  
• P(3 normal) = 10 x (0.75^3) x (0.25^2)
• 
  
• P(4 normal) = 5 x (0.75^4) x (0.25)
• 
  
• P(5 normal) = (0.75^5)

The coefficients (10, 5, 1) are binomial coefficients from the binomial theorem, representing the different ways the normal and affected children can occur. Calculating these probabilities and adding them gives us the probability of at least 3 normal children.