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 \(\frac{1}{4}\).

Step 1: Find the genotypes of the non-albino child and the carrier spouse

Since the parents are normal and they have one albino child, they must be carriers (A,a). So, they have the following genotype combinations: \((A,a) \times (A,a)\). The non-albino child receives at least one A gene. Therefore, the genotype for the non-albino child must be (A,a), as the child is normal. The carrier spouse occurs when the individual has at least one A gene and one a gene (A,a).

Step 2: Calculate the possible genetic combinations for the offspring

The genetic combinations for the offspring from the non-albino child and carrier spouse are: \((A,a) \times (A,a)\). The Punnett square for this combination will give us four possible outcomes: (A,A), (A,a), (a,A), (a,a).

Step 3: Calculate the probability of the first offspring being an albino

Albino offspring are the ones with the (a,a) genotype. In the Punnett square, we have one out of four outcomes as (a,a). Thus, the probability of the first offspring being an albino is: P(First albino) = \(\frac{1}{4}\)

Step 4: Calculate the conditional probability for the second offspring

First, let's determine the unconditional probability of having a non-albino offspring. From the Punnett square, three out of four outcomes are non-albino: (A,A), (A,a), and (a,A). P(First non-albino) = \(\frac{3}{4}\) Now, given that the firstborn is not an albino, the genetic combination for subsequent offspring remains the same as in Step 2. Thus, the conditional probability of the second offspring being an albino is equal to the unconditional probability of any offspring being an albino. P(Second albino | First non-albino) = P(First albino) = \(\frac{1}{4}\) #Conclusion#: (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 \(\frac{1}{4}\).

Key Concepts

Punnett Square

The Punnett square is a visual tool in genetics that helps predict the genotype combinations of offspring from two parents. It's a grid that shows how different alleles from each parent can combine during fertilization. For albinism, a recessive trait represented by the allele 'a', a Punnett square helps to clarify how two carrier parents, with genotype (A,a), can produce offspring with the various combinations of alleles.

Each parent contributes one allele to their child, and the Punnett square outlines all potential combinations of these alleles. When dealing with simple Mendelian inheritance like albinism, it's a reliable way to visualize the probability of an offspring inheriting a particular trait.

Genotype Combinations

Genotype combinations refer to the set of alleles an individual inherits from their parents, one from the mother and one from the father. For albinism, the key genotypes are AA, Aa, and aa, where 'A' symbolizes the normal pigmentation allele and 'a' the albinism allele. A person with 'AA' or 'Aa' genotypes will not exhibit albinism, while 'aa' results in the albino phenotype.

Through the Punnett square, we can determine that the offspring of two 'Aa' carriers have a 25% chance of being 'aa' (albino), a 50% chance of being 'Aa' (carrier), and a 25% chance of being 'AA' (normal without carrier status).

Inheritance Patterns

Inheritance patterns describe how traits are transmitted from parents to their children through their genes. Albinism is an example of a recessive inheritance pattern, where two copies of the recessive allele 'a' are needed for the trait to be expressed.

In the case of our exercise, the nonalbino child from carrier parents must have inherited at least one dominant allele, which we represent as 'A'. Understanding these patterns is crucial when predicting the likelihood of an offspring inheriting certain traits, like albinism.

Conditional Probability

Conditional probability in genetics accounts for the likelihood of an event, given that another event has already occurred. In our exercise, the calculation of the probability that the second child is an albino, subjected to the first child not being an albino, is an application of conditional probability.

However, since the genotype possibilities for each child are independent of each other, the conditional probability equals the unconditional probability. This means that the occurrence of the first child's phenotype does not influence the genetic outcomes of the second child.

Albinism Gene Inheritance

Albinism gene inheritance refers to how the albinism trait, caused by a recessive allele 'a', is passed down from one generation to the next. An individual has to inherit two copies of the recessive allele (one from each parent) to exhibit the characteristic lack of pigmentation.

The probability of an albino child from two carriers is 25%, as the recessive allele must be inherited from both the mother and the father. As such, when analyzing genetic problems involving albinism, one must account for the recessive nature of the gene and the carrier status of the parents to accurately assess the probabilities of inheritance.