Question

Albinism is caused by a recessive autosomal allele. A man and a woman, both normally pigmented, have an albino first born child together. The mother is now pregnant for a second time and her doctor tells her she is having fraternal twins.

(a) Use a genetic diagram to represent this cross and the possible offspring genotypes.
(b) Calculate the probability that the twins will have normal pigmentation.
(c) Calculate the probability that the fourth born will be a girl and albino.
(d) If the couple has a total of five children together, calculate the probability that two wil be albino.

Answer 1

(a) Genetic Diagram: NN (parents) x NN (parents) → Nn, Nn, nn, Nn

(b) Probability that the twins will have normal pigmentation: 9/16

(c) Probability that the fourth born will be a girl and albino: 1/8

(d) Probability that two out of five children will be albino: 135/512

(a) Genetic Diagram:

Parents: NN (normally pigmented) x NN (normally pigmented)

Possible Offspring Genotypes:

Nn (normally pigmented)

Nn (normally pigmented)

nn (albino)

Nn (normally pigmented)

(b) Probability that the twins will have normal pigmentation:

Since the mother and father are both normally pigmented (NN), the probability for each child being normally pigmented is 3/4. For fraternal twins, the probability is (3/4) * (3/4) = 9/16.

(c) Probability that the fourth born will be a girl and albino:

The probability of being a girl is 1/2, and the probability of being albino is 1/4. Therefore, the combined probability is (1/2) * (1/4) = 1/8.

(d) Probability that two out of five children will be albino:

Using the binomial probability formula:

P(X = k) = C(n, k) * p^k * q^(n-k)

where n is the number of trials, k is the number of successes, p is the probability of success, and q is the probability of failure.

For two albino children out of five (k = 2):

P(X = 2) = C(5, 2) * (1/4)^2 * (3/4)^3 =