Question

1. A boat carrying 2,000 passengers sets sail for an island with a population of 23,000. Both of these populations consist of an equal number of males and females.

21 male passengers on the ship have an X-linked recessive condition. 15 of the male island inhabitants have the same condition, and no females on the ship or the island show the phenotype.

1. What is the frequency of the disease-causing allele on the ship?

2. How many females on the ship would you expect to be carriers for this condition?

3. The ship docks, and a woman from the island falls pregnant to an unaffected man from the ship. What are the chances their first son will be affected with the condition?

4. An affected man from the island and a woman from the ship have a girl. What is the daughter’s chance of being a carrier?

Answer 1

a).

The frequency of X-linked recessive allele = frequency of the X-linked recessive trait in males (as males have only one X-chromosome)

Frequency of recessive trait in males = 21/2000 = 0.0105

Frequency of the disease – causing allele = 0.0105

b).

Frequency of the disease – causing allele = 0.0105

Frequency of the normal allele = 1- 0.0105 = 0. 9895

Frequency of carrier females = 2 * 0.9895 * 0.0105 = 0.0208

No. of carriers = 0.0208 * 2000 = 42

c).

Frequency of recessive trait in males = 15/23000 = 0.00065

Frequency of the disease – causing allele = 0.00065

Frequency of normal allele = 1 – 0.00065 = 0.99935

Frequency of carrier females = 2 * 0.99935 * 0.00065 = 0.0013

If the woman is normal (homozygous), there would be no chance of affected son.

If woman is carrier…..

XAY (man) x (woman) XAXa (woman) ---Parents

	XA (0.99935)	Xa (0.00065)
XA (0.9895)	XAXA (normal daughter-0.99)	XAXa (carrier daughter = 0.0006)
Y	XAY = normal son (0.99935)	XaY (affected son = 0.00065)

The chances their first son will be affected with the condition = 0.00065

d).

If the woman is normal (homozygous), there would be no chance of carrier female.

If woman is carrier…..

XAY (man) x (woman) XAXa (woman) ---Parents

	XA (0.9895)	Xa (0.0105)
XA (0.99935)	XAXA (normal daughter-0.99)	XAXa (carrier daughter = 0.01)
Y	XAY = normal son (0.9835)	XaY (affected son = 0.01)

The chances their daughter with carrier genotype = 0.01