Problem 61 - A First Course in Probability

Question

Genes relating to albinism are denoted by \(A\) and \(a\). Only those people who receive the \(a\) gene from both parents will be albino. Persons having the gene pair \(A, a\) are normal in appearance and, because they can pass on the trait to their offspring, are called carriers. Suppose that a normal couple has two children, exactly one of whom is an albino. Suppose that the nonalbino child mates with a person who is known to be a carrier for albinism. (a) What is the probability that their first offspring is an albino? (b) What is the conditional probability that their second offspring is an albino given that their firstborn is not?

Step-by-Step Solution

Answer

(a) The probability that their first offspring is an albino is \(\frac{1}{4}\). (b) The conditional probability that their second offspring is an albino given that their firstborn is not is \(\frac{1}{3}\).

Step 1: Determine the genotype of each parent

The couple in question has two children - one albino and one non-albino. Carriers for albinism are those who have an \(A\) and an \(a\) gene. Let the normal couple be \(P1\). As they have an albino child, at least one parent should be a carrier. If only one parent was a carrier, the probability of having an albino child would be 0 which means both \(P1\) parents must be carriers, having the \(Aa\) genotype. The nonalbino child (child 1) mates with a person known to be a carrier for albinism (person 2). So, we now have the genotypes of both parents: Child 1 is \(Aa\) (since they are normal), and person 2 is \(Aa\) (known to be a carrier).

Step 2: Calculate the probability of the first offspring being an albino

To find the probability of the first offspring being an albino, we can use a Punnett square with the genotypes of child 1 and person 2. \(A\) \(a\) \(A\) \(AA\) \(Aa\) \(a\) \(Aa\) \(aa\) From the Punnett square, we can see that there are four possible outcomes for their offspring, and one of them is albino (\(aa\)). Therefore, the probability of their first offspring being an albino is: \[\frac{1}{4}\]

Step 3: Calculate the conditional probability of the second offspring being an albino given that the firstborn is not

We are now asked to find the probability of their second offspring being an albino given that their firstborn is not an albino. Since the genotypes of both parents \((Aa)\) did not change between the firstborn and second offspring, the probability ratio for their offspring remains the same: \(AA : Aa : Aa : aa = 1 : 2 : 1\) However, we are told that the firstborn is not an albino. Therefore, the firstborn must have one of the following genotypes: \(AA, Aa, Aa\). Now, the probability ratio for the second offspring is based on the genotypes of the second offspring excluding the firstborn: \(Aa : Aa : aa = 2 : 1\) Thus, the conditional probability of the second offspring being an albino is: \[\frac{1}{3}\] In summary: (a) The probability that their first offspring is an albino is \(1/4\). (b) The conditional probability that their second offspring is an albino given that their firstborn is not is \(1/3\).

Key Concepts

Punnett Square Analysis

Genetics can often seem like a complex subject, but breaking down genetic probabilities can be much more manageable with tools like the Punnett Square. Developed by Reginald Punnett, the Punnett Square is a graphical representation that predicts the genotype of offspring from parental DNA. In determining traits like albinism, the Punnett Square visually outlines all possible combinations of parents' genes.

In our example, parents with genotypes Aa (carriers of albinism) have a particular probability of their offspring being albino, denoted by aa. The square, which looks like a simple grid, aligns one parent's possible gametes along the top and the other's along the side. The boxes within the grid showcase the potential genetic mixes. For albinism, a child must receive the a gene from both parents to exhibit the trait. The Punnett Square analysis, therefore, reveals that there is (with a genotype of aa) a 1 in 4 chance for the couple in question to have an albino child.

Genetics and Probability

Understanding the probability in genetics requires the recognition that genetic inheritance follows specific statistical rules. These rules align with the fundamental principles of inheritance first outlined by Gregor Mendel. The probability of an offspring inheriting a particular trait is based upon the Mendelian inheritance patterns.

For traits that are governed by simple dominant and recessive genetics, such as the gene for albinism in our exercise, we calculate probabilities using combinations of alleles. Alleles are the variations of a gene—in this case, A for normal pigmentation and a for albinism. If a trait is recessive, both alleles must be recessive (aa) for the trait to manifest. This probabilistic approach allows us to predict the likelihood of a trait appearing in offspring, such as a 1 in 4 chance for albino offspring from two carrier parents.

Conditional Probability in Genetics

Conditional probability comes into play in genetics when calculating the likelihood of an event (like producing an albino child) under a certain condition (like the first child not being albino). Genetics problems often require an understanding of this concept because the outcome of one event can affect the probability of another.

In the given exercise, we use conditional probability to determine the chance of the second child being albino given the first is not. Since three out of the four genetic combinations result in a non-albino child, we exclude the albino combination when considering the second child. This recalculates the odds to only considering the remaining possibilities where the first child is not albino, resulting in a 1 in 3 chance for the subsequent child to be albino. Conditional probability refines our predictions by incorporating new information, thus narrowing down possibilities and changing the odds.