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Chapter 3: Problem 116

Suppose a birth defect has a recessive form of inheritance. In a study population, the recessive gene (a) initially has a prevalence of \(25 \% .\) A subject has the birth defect if both maternal and paternal genes are of type a. Suppose that in 25\% of marriages both people are from population \(A,\) in \(65 \%\) both are from population \(B\), and in \(10 \%\) there is one partner from population \(A\) and one from population B. What is the probability of a birth defect in the next generation?

Short Answer

Expert verified
The probability of a birth defect in the next generation is 0.0625.

Step by step solution

01

Define Recessive Inheritance

In a recessive inheritance pattern, an individual expresses the recessive trait only if both alleles are the recessive form (aa). This means a subject will only have the birth defect if they inherit allele 'a' from both parents.
02

Determine Recessive Allele Frequency

The recessive gene a has a prevalence of 25%, which means that the frequency, q, of the allele a is 0.25. Therefore, the frequency of the dominant allele A, denoted by p, is `p = 1 - q = 0.75`.
03

Calculate Genotype Frequencies

Using Hardy-Weinberg equilibrium, the frequencies of the genotypes AA, Aa, and aa are calculated as: \( P(AA) = p^2 = 0.75^2 = 0.5625 \), \( P(Aa) = 2pq = 2 \times 0.75 \times 0.25 = 0.375 \), and \( P(aa) = q^2 = 0.25^2 = 0.0625 \). Only individuals with aa have the birth defect.
04

Probability of Birth Defect from Population A

If both parents are from population A, then the probability that their child has the defect (aa) is just the probability of aa: 0.0625.
05

Probability of Birth Defect from Population B

Similarly, if both parents come from population B, the probability of a child with aa remains 0.0625 since the given prevalence and genetic distribution seem to be the same across the study population.
06

Probability of Mixed Population Marriages

In mixed marriages (one partner from A and one from B), assuming allele frequencies do not depend on the subpopulation (and children inherit alleles from each parent randomly), the probability of a child being aa remains the same, which is 0.0625.
07

Calculate Overall Probability of Birth Defect

Using the weighted average formula: \( P(\text{defect}) = 0.25 \times 0.0625 + 0.65 \times 0.0625 + 0.10 \times 0.0625 = 0.0625 \). The probability remains constant as all groups have the same allele distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recessive Inheritance
Recessive inheritance is a genetic concept that refers to how certain traits or conditions are passed down from parents to their offspring through genes. With recessive inheritance, a condition only manifests when an individual inherits two copies of the recessive allele, one from each parent. For instance, if the allele causing a certain birth defect is represented as 'a,' then the individual must have the genotype 'aa' to exhibit the trait.

In simpler terms:
  • Both parents must contribute the recessive allele ('a') for the child to show the trait.
  • If an individual has only one recessive allele ('a') and one dominant allele ('A'), the dominant trait appears instead.
This type of inheritance explains why some genetic conditions can "skip" generations or only appear under certain parental pairings. In the given exercise, discovering how prevalent the "aa" genotype is helps determine how widespread a condition is in a population.
Hardy-Weinberg Equilibrium
The Hardy-Weinberg equilibrium is a principle that explains how allele and genotype frequencies remain constant from generation to generation in an ideal population, assuming no other evolutionary influences, such as natural selection or mutations, are acting on the population. This state is crucial in understanding genetic inheritance.

Learning about Hardy-Weinberg can help us determine the consistency of allele distribution within a population over time. The equations associated with this principle鈥�
  • \( p^2 + 2pq + q^2 = 1 \)
  • \( p + q = 1 \)
鈥� allow us to calculate the expected frequencies of different genotypes, such as homozygous dominant (AA), heterozygous (Aa), and homozygous recessive (aa). In the exercise, the equilibrium applies to estimate the occurrence of the birth defect, by maintaining these frequencies when analyzing the prevalence of the recessive trait in a population.
Genotype Frequencies
Genotype frequencies describe how common different genetic makeups (genotypes) are in a population. Using Hardy-Weinberg equilibrium, we established three basic genotype frequencies:
  • Homozygous Dominant (AA): Calculated as \( p^2 \)
  • Heterozygous (Aa): Calculated as \( 2pq \)
  • Homozygous Recessive (aa): Calculated as \( q^2 \)
These frequencies give us a vital insight into how many individuals might express a trait linked to certain genotypes.

For instances like the exercise, where understanding the likelihood of a recessive birth defect is key, we zero in on the 'aa' genotype. This helps predict how these traits manifest in large populations, as well as in specific family trees. Knowing which genotype corresponds to which trait or condition allows for better genetic risk assessments and understanding of genetic disorders.

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Most popular questions from this chapter

Consider a family with a mother, father, and two children. Let \(A_{1}=\left\\{\text { mother has influenzal, } A_{2}=\\{ \text { father has influenzal, }\right.\) \(A_{3}=\left\\{\text { first child has influenzal, } A_{4}=\\{ \text { second child has influ- }\right.\) enzal, \(B=\) lat least one child has influenzal, \(C=\) lat least one parent has influenzal, and \(D=\\{\) at least one person in the family has influenzal.What does \(A_{1} \cap A_{2}\) mean?

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Consider a family with a mother, father, and two children. Let \(A_{1}=\left\\{\text { mother has influenzal, } A_{2}=\\{ \text { father has influenzal, }\right.\) \(A_{3}=\left\\{\text { first child has influenzal, } A_{4}=\\{ \text { second child has influ- }\right.\) enzal, \(B=\) lat least one child has influenzal, \(C=\) lat least one parent has influenzal, and \(D=\\{\) at least one person in the family has influenzal.What does \(A_{3} \cap B\) mean?

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Consider a family with a mother, father, and two children. Let \(A_{1}=\left\\{\text { mother has influenzal, } A_{2}=\\{ \text { father has influenzal, }\right.\) \(A_{3}=\left\\{\text { first child has influenzal, } A_{4}=\\{ \text { second child has influ- }\right.\) enzal, \(B=\) lat least one child has influenzal, \(C=\) lat least one parent has influenzal, and \(D=\\{\) at least one person in the family has influenzal. Represent \(\bar{D}\) in terms of \(B\) and \(C .\)
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