1. a. It is estimated that 1 out of 20 people in Europe are carriers of cystic f
ID: 48094 • Letter: 1
Question
1.
a. It is estimated that 1 out of 20 people in Europe are carriers of cystic fibrosis. What is the probability that both a husband and a wife in Europe are carriers?
b. If both parents are carriers, what is the probability that their first child will also be a carrier?
c. If both parents are carriers, what is the probability that their first child will also be afflicted?
d. Given the frequency of this allele in the European population, calculate the probability that a child is born with cystic fibrosis in this population.
Explanation / Answer
a. The ratio is 1:20. This gives a ratio of 5:100. So for every 100 persons, there are 5 carriers. So the probability that both a husband a wife are carriers would be:
the probability of wife being a carrier + probability of husband being a carrier
= 5% + 5%
= 10%.
b.If both the parents are carriers, they both will be heterozygous(Cc) for the gene. The combinations they can produce in their children are:
1 CC
2 Cc
1 cc
So the chance that the baby is heterozygous and a carrier is 50%.
c. IF both the parents are carriers, the combinations they can produce in their children are:
1 CC
2 Cc
1 cc
Hence, the chance that the baby would be afflicted is 25%.
d. The frequency of the homozygous recessive allele is 1 out of 20 which is 0.05. This is q2. The probabiltiy that a child is born with CF would be 0.05 x 0.25( probability of being born from two heterozygous parents)
= 0.012.
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