Phenylketonuria (PKU) is a human metabolic disease caused by the recessive allele k. If 2 heterozygous carriers of the allele marry and plan a family of five children:
a) what is the chance that all their children will be unaffected?
b) what is the chance that four children will be unaffected and one affected with phenylketonuria?
c) what is the chance that the first child will be an unaffected girl?
hardy weinberg problem
To determine the chances in each scenario, we need to understand the inheritance pattern of phenylketonuria (PKU). PKU is caused by a recessive allele, which means that an individual needs to inherit two copies of the allele (one from each parent) in order to have the disease. With this information, let's calculate the chances for each scenario:
a) To determine the chance that all their children will be unaffected, we need to calculate the probability of each child being unaffected. Since both parents are heterozygous carriers (meaning they have one copy of the disease allele "k" and one copy of the normal allele "K"), they have a 50% chance of passing on either allele to their children.
The probability for each child to be unaffected is:
- For the first child, the chance of being unaffected is 75% (3 out of 4 possibilities: KK, Kk, Kk, kk).
- For the second child, the chance is also 75%.
- For the third child, the chance remains 75%.
- For the fourth child, the chance is still 75%.
- For the fifth child, the chance remains 75%.
To calculate the overall probability of all the children being unaffected, we multiply the probabilities for each child together:
0.75 * 0.75 * 0.75 * 0.75 * 0.75 = 0.2373 (approximately)
Therefore, there is approximately a 23.73% chance that all five children will be unaffected.
b) To determine the chance that four children will be unaffected and one will be affected with PKU, we follow a similar approach. In this scenario, we want to calculate the probability of four children being unaffected and one being affected.
The probability for each child to be unaffected is still 75% (as mentioned in scenario a), and the probability for one child to be affected is 25% (one out of four possibilities: kk). To calculate the overall probability, we multiply the probabilities together:
0.75 * 0.75 * 0.75 * 0.75 * 0.25 = 0.0703 (approximately)
So, there is approximately a 7.03% chance that four children will be unaffected and one child will be affected with PKU.
c) To determine the chance that the first child will be an unaffected girl, we need to consider the probability of each event occurring. The probability of a child being unaffected is 75% (explained in scenario a), and the probability of a child being a girl is 50% (assuming equal chances for boys and girls).
To calculate the overall probability, we multiply these probabilities together:
0.75 * 0.50 = 0.375 (approximately)
Therefore, there is approximately a 37.5% chance that the first child will be an unaffected girl.