Phenylketonuria, a Metabolic disease in humans, is caused by a recessive allele, k. If two heterozygotes marry and plan a family of six children: a)what is the probability that all will be unaffected? b)what is the probability that three will be unaffected and three will be affected? c)what is the probability that at least two will be unaffected? d)what is the probability that the first child will be an unaffected girl?
I don't even know how to approach this problem. Much help is needed. Thanks
I don't get b or c at alll. What are the u's and a's? What are the exclamation points?
To solve this problem, we need to use the principles of Mendelian genetics and probability.
Given information:
- Phenylketonuria (PKU) is caused by a recessive allele, k.
- Assuming the two parents are heterozygotes (carriers) for the PKU allele.
a) Probability that all six children will be unaffected:
To determine this probability, we need to calculate the chance of each child being unaffected. Since the parents are heterozygotes, the probability of having an unaffected child is 75% (because 25% of the possible genotypes will result in PKU).
Therefore, the probability that any individual child is unaffected is 0.75.
To calculate the probability of all six children being unaffected, we multiply the probabilities of each outcome together.
P(unaffected child) = 0.75
P(all six children unaffected) = (0.75)^6 ≈ 0.178 (approximately 17.8%)
b) Probability that three children will be unaffected and three will be affected:
To determine this probability, we need to calculate the probability for each specific combination of three unaffected and three affected children.
We can use the binomial distribution formula to calculate this:
P(X = k) = (nCk) * (p^k) * (q^(n-k))
where:
n = total number of children (6)
k = number of unaffected children (3)
p = probability of success (0.75 for unaffected child)
q = probability of failure (0.25 for affected child)
P(3 unaffected and 3 affected) = (6C3) * (0.75^3) * (0.25^3)
= (6! / (3! * 3!)) * (0.75^3) * (0.25^3)
= 20 * 0.421875 * 0.015625
≈ 0.131 (approximately 13.1%)
c) Probability that at least two children will be unaffected:
To solve this, we can calculate the probability of having exactly 2 unaffected children, exactly 3 unaffected children, and so on, up to having all six children unaffected. Then we sum up these probabilities.
P(at least two unaffected children) = P(2 unaffected) + P(3 unaffected) + ... + P(all 6 unaffected)
= P(2 unaffected) + P(3 unaffected) + P(4 unaffected) + P(5 unaffected) + P(all 6 unaffected)
= (6C2) * (0.75^2) * (0.25^4) + (6C3) * (0.75^3) * (0.25^3) + (6C4) * (0.75^4) * (0.25^2) + (6C5) * (0.75^5) * (0.25^1) + (6C6) * (0.75^6) * (0.25^0)
≈ 0.812 (approximately 81.2%)
d) Probability that the first child will be an unaffected girl:
Since the probability of an unaffected child is 0.75, and gender is independent of the inherited alleles, the probability of the first child being an unaffected girl is also 0.75.
Therefore, the probability that the first child will be an unaffected girl is 0.75 or 75%.
To solve these problems, we will use principles of probability and Punnett squares. Phenylketonuria (PKU) is caused by a recessive allele, k. Heterozygotes carry one normal allele (K) and one disease-causing allele (k). Let's address each question step by step:
a) What is the probability that all six children will be unaffected?
To determine the probability that each child will be unaffected, we need to find the probability that each child will inherit at least one normal allele. Since both parents (heterozygotes) carry one normal allele each, the Punnett square can help us visualize the possible outcomes:
| K | k |
-----------------
K | KK | Kk |
-----------------
k | Kk | kk |
From the Punnett square, we see that there are four possible outcomes: KK, Kk, Kk, and kk.
Out of these four outcomes, three result in unaffected children: KK, Kk, and Kk.
The probability that each child will be unaffected is therefore 3/4, assuming that the parents are both heterozygotes.
b) What is the probability that three children will be unaffected, and three will be affected?
To find this probability, we need to consider the combinations that result in three unaffected children and three affected children. We can use combinatorics to help us calculate this.
Let's calculate the probability for three unaffected children and three affected children:
P(unaffected child) = 3/4
P(affected child) = 1/4
The number of ways we can arrange three unaffected children and three affected children in a family of six children is given by the combination formula: C(6,3) = 6! / (3! * (6-3)!) = 20.
The probability of having three unaffected and three affected children is given by:
P(3 unaffected, 3 affected) = (3/4)^3 * (1/4)^3 * C(6,3) = 27/256 * 20 = 135/256.
Therefore, the probability that three children will be unaffected, and three will be affected is 135/256.
c) What is the probability that at least two children will be unaffected?
To calculate this probability, we need to consider three scenarios: exactly two, exactly three, and exactly four unaffected children.
P(2 unaffected children) = (3/4)^2 * (1/4)^4 * C(6,2)
P(3 unaffected children) = (3/4)^3 * (1/4)^3 * C(6,3)
P(4 unaffected children) = (3/4)^4 * (1/4)^2 * C(6,4)
We can calculate these probabilities and sum them to obtain the probability that at least two children will be unaffected.
d) What is the probability that the first child will be an unaffected girl?
To find the probability that the first child will be an unaffected girl, we need to consider all the possible outcomes for the first child.
From the Punnett square:
| K | k |
-----------------
K (unaffected) | KK | Kk |
-----------------
k (affected) | Kk | kk |
As we want an unaffected child, we need to consider two possibilities: KK and Kk (unaffected girls).
The probability of having an unaffected girl as the first child is:
P(unaffected girl) = P(KK) + P(Kk) = (1/4) + (1/2) = 3/4.
Therefore, the probability that the first child will be an unaffected girl is 3/4.
By using these principles, you can solve the different scenarios and calculate the probabilities in question.
Prob each is affected= 1/2*1/2= 1/4
prob all unaffected= (3/4)^6
prob three affected= (1/4)^3 * (3/4)^3 * ways.
now ways = 6!/3!3!
at least two will be unaffected
combination ways
uuaaaa 6!/(2!4!)= 15
uuuaaa 6!/(3!3!)=40
uuuuaa 6!/(2!4!) =15
uuuuua 6!/5!1! =6
uuuuuu 6!/6! =1
check those, I did them in my head
Pr(at least two)= sum prob*combinations
= 15((3/4)^2(1/4)^4)+40((3/4)^3(1/4)^3)+ ....
pr first child ug?
Pr= (3/4)*1/2