Two parents who are each known to be carriers of an autosomal recessive allele have four children. None of the children has the recessive condition. What is the probability that three or more of the children are the carriers of the recessive allele?
To calculate the probability, we first need to determine the possible genotypes of the children based on the information given.
Since both parents are carriers of an autosomal recessive allele, we can represent their genotypes as follows:
Parent 1: Aa (carrier)
Parent 2: Aa (carrier)
Now, let's consider the possible genotypes of the children assuming independent assortment:
Child 1: 25% chance of being AA (not a carrier)
50% chance of being Aa (carrier)
25% chance of being aa (affected)
Child 2: Same probabilities as Child 1
Child 3: Same probabilities as Child 1
Child 4: Same probabilities as Child 1
To calculate the probability that three or more of the children are carriers, we need to consider the different combinations of children who are carriers:
1. Three children are carriers and one child is unaffected:
- There are four possibilities for the unaffected child (AA genotype)
- There are four possibilities for the carrier children (Aa genotype)
- So, the probability for this combination is 4 * 4 = 16
2. Four children are carriers:
- There is only one possibility for all four children being carriers (Aa genotype)
- So, the probability for this combination is 1
Now, let's calculate the total number of possible combinations:
- Each child has 3 possible genotypes (AA, Aa, aa)
- Since there are four children, the total number of combinations is 3 * 3 * 3 * 3 = 81
Now, we can calculate the probability:
Probability (three or more children carriers) = (number of desired combinations) / (total number of combinations)
= (16 + 1) / 81
= 17 / 81
≈ 0.21
Therefore, the probability that three or more of the children are carriers of the recessive allele is approximately 0.21, or 21%.
To calculate the probability that three or more of the children are carriers of the recessive allele, we can use the binomial probability formula. The binomial probability formula calculates the probability of obtaining a certain number of successes in a fixed number of independent Bernoulli trials (where each trial has only two possible outcomes, success or failure).
In this case, the Bernoulli trials refer to each child either being a carrier (success) or not being a carrier (failure) of the recessive allele. The probability of each child being a carrier is 1/2, since each parent is a known carrier of the allele.
To calculate the probability, we need to calculate the probability of each possible outcome (3 carriers, 4 carriers). Then, we sum these probabilities to get the total probability of three or more children being carriers.
Let's calculate the probabilities for each outcome:
The probability of exactly three children being carriers is calculated as follows:
P(3 carriers) = C(4,3) * (1/2)^3 * (1/2)^(4-3)
= 4 * (1/2)^3 * (1/2)^1
= 4 * 1/8 * 1/2
= 4/16
= 1/4
Similarly, the probability of exactly four children being carriers is:
P(4 carriers) = C(4,4) * (1/2)^4 * (1/2)^(4-4)
= 1 * (1/2)^4 * (1/2)^0
= 1 * 1/16 * 1
= 1/16
To get the total probability of three or more children being carriers, we sum these probabilities:
P(3 or more carriers) = P(3 carriers) + P(4 carriers)
= 1/4 + 1/16
= 4/16 + 1/16
= 5/16
Therefore, the probability that three or more of the children are carriers of the recessive allele is 5/16.