Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 21: Problem 9
If 4 percent of a population in equilibrium expresses a recessive trait, what is the probability that the offspring of two individuals who do not express the trait will express it?
Short Answer
Expert verified
Answer: The probability that the offspring of two non-trait-expressing individuals will express the recessive trait is approximately 2.78%.
Step by step solution
01
1. Understand Hardy-Weinberg equilibrium principle and formula
The Hardy-Weinberg equilibrium principle states that in a large, randomly mating population with no factors such as mutation, migration, or natural selection affecting it, the frequencies of alleles and genotypes will remain constant from generation to generation.
The formula for Hardy-Weinberg equilibrium is:
p^2 + 2pq + q^2 = 1
Where p is the frequency of the dominant allele (A), q is the frequency of the recessive allele (a), p^2 represents the frequency of the homozygous dominant genotype (AA), 2pq represents the frequency of the heterozygous genotype (Aa), and q^2 represents the frequency of the homozygous recessive genotype (aa).
02
2. Determine the frequency of the recessive genotype (q^2)
We are given that 4% of the population, or 0.04, expresses the recessive trait. This means that 0.04 is the frequency of the homozygous recessive genotype (aa) in the population. Therefore, q^2 = 0.04.
03
3. Calculate the frequency of the recessive allele (q)
To find the frequency of the recessive allele (a), we take the square root of q^2:
q = sqrt(q^2) = sqrt(0.04) = 0.2
04
4. Calculate the frequency of the dominant allele (p)
Since p and q must sum to 1, we can find the frequency of the dominant allele (A) by subtracting the frequency of the recessive allele (q) from 1:
p = 1 - q = 1 - 0.2 = 0.8
05
5. Calculate the probability that the offspring of two non-trait-expressing individuals will express the recessive trait
If two individuals do not express the recessive trait, they must have either the AA or Aa genotype. To find the probability that their offspring will express the recessive trait, we will perform a Punnett square analysis.
The probabilities of the parental genotypes are:
P(AA) = p^2 = (0.8)^2 = 0.64
P(Aa) = 2pq = 2(0.8)(0.2) = 0.32
P(aa) = q^2 = 0.04
Since the parents do not express the recessive trait, they must have either AA or Aa genotypes. Their combined frequency is 0.64 + 0.32 = 0.96.
Now we will calculate the probability of their offspring being aa (expressing the trait) using the Punnett square:
Aa x Aa --> 1/4 AA, 1/2 Aa, 1/4 aa
As aa is the only genotype that expresses the trait, the offspring's probability is 1/4.
However, we must account for the probability that both parents are carriers (Aa) based on the combined frequency of AA and Aa parents (0.96):
Probability(Aa parent) = 0.32 / 0.96 = 1/3
Thus, the probability that both parents are carriers (Aa) and that their offspring will express the recessive trait (aa) is:
(1/3) * (1/3) * (1/4) = 1/36 鈮� 0.0278
Therefore, the probability that the offspring of two non-trait-expressing individuals will express the recessive trait is approximately 2.78%.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Recessive Trait Probability
In understanding recessive trait probability, it's crucial to delve into the genetic makeup of individuals within a population. A recessive trait is only expressed when an individual inherits two copies of the recessive allele, one from each parent. This means that for a trait to appear in the offspring, the genotype must be homozygous recessive (aa).
When considering the probability of offspring expressing a recessive trait, we examine the combinations of parental genotypes that could lead to this outcome. In the context of Hardy-Weinberg equilibrium, understanding these combinations provides insight into how often the recessive trait might appear.
- If both parents are heterozygous (Aa), there is a 25% chance that their offspring will inherit the recessive trait and be aa. - If one parent is homozygous dominant (AA) and the other is heterozygous (Aa), there's no chance for aa offspring. - When both parents are AA, the recessive trait will not be expressed.
Throughout these probabilities, it's important to consider the overall frequency of these genotypes in the population, as they impact the likelihood of two non-expressive parents having expressive offspring.
When considering the probability of offspring expressing a recessive trait, we examine the combinations of parental genotypes that could lead to this outcome. In the context of Hardy-Weinberg equilibrium, understanding these combinations provides insight into how often the recessive trait might appear.
- If both parents are heterozygous (Aa), there is a 25% chance that their offspring will inherit the recessive trait and be aa. - If one parent is homozygous dominant (AA) and the other is heterozygous (Aa), there's no chance for aa offspring. - When both parents are AA, the recessive trait will not be expressed.
Throughout these probabilities, it's important to consider the overall frequency of these genotypes in the population, as they impact the likelihood of two non-expressive parents having expressive offspring.
Allele Frequency Calculation
Calculating allele frequencies is a foundational aspect of genetic analysis in a population. Using the Hardy-Weinberg equilibrium model, these frequencies help in determining the genetic diversity present and predicting how traits may pass on through generations. The formula uses three key components:
- **p**: the frequency of the dominant allele (A) - **q**: the frequency of the recessive allele (a) - Both frequencies sum to 1, represented as \(p + q = 1\)
Given that the frequency of the recessive genotype (aa) is \(q^2\), we can take the square root of this value to determine **q**. For example, if \(q^2 = 0.04\), then \(q = \sqrt{0.04} = 0.2\).
Once \(q\) is known, you can easily find **p** using \(p = 1 - q\). Following the example where \(q = 0.2\), then \(p = 1 - 0.2 = 0.8\). These allele frequencies are crucial as they underpin further calculations about the population's genotypic ratios and the likelihood of offspring expressing various traits.
- **p**: the frequency of the dominant allele (A) - **q**: the frequency of the recessive allele (a) - Both frequencies sum to 1, represented as \(p + q = 1\)
Given that the frequency of the recessive genotype (aa) is \(q^2\), we can take the square root of this value to determine **q**. For example, if \(q^2 = 0.04\), then \(q = \sqrt{0.04} = 0.2\).
Once \(q\) is known, you can easily find **p** using \(p = 1 - q\). Following the example where \(q = 0.2\), then \(p = 1 - 0.2 = 0.8\). These allele frequencies are crucial as they underpin further calculations about the population's genotypic ratios and the likelihood of offspring expressing various traits.
Genotype Punnett Square Analysis
A Punnett square is a simple tool for visualizing genetic crosses and predicting the genotypes of offspring. It uses the parental alleles to show potential genetic combinations and the likelihood of different genotypes. In our context, it aids in understanding how two non-recessive trait carriers can produce recessive trait offspring.
To perform this analysis, consider the potential parental genotypes: - **AA**: homozygous dominant, does not express the trait. - **Aa**: heterozygous, carries the recessive allele but does not express it.
For instance, when both parents are Aa, the Punnett square illustrates the chances of deriving each genotype: - 25% AA - 50% Aa - 25% aa
The presence of aa offspring results in the recessive trait being expressed. However, this 25% probability assumes both parents are heterozygous. Calculating this probability accurately involves understanding current genotypic frequencies in the population and applying these calculations contextually, ensuring the broader socio-genetic dynamics are factored into the Hardy-Weinberg model.
To perform this analysis, consider the potential parental genotypes: - **AA**: homozygous dominant, does not express the trait. - **Aa**: heterozygous, carries the recessive allele but does not express it.
For instance, when both parents are Aa, the Punnett square illustrates the chances of deriving each genotype: - 25% AA - 50% Aa - 25% aa
The presence of aa offspring results in the recessive trait being expressed. However, this 25% probability assumes both parents are heterozygous. Calculating this probability accurately involves understanding current genotypic frequencies in the population and applying these calculations contextually, ensuring the broader socio-genetic dynamics are factored into the Hardy-Weinberg model.
