Question

Consider rare disorders in a population caused by an autosomal recessive mutation. From the frequencies of the disorder in the population given, calculate the percentage of heterozygous carriers: (a) 0.0064 (b) 0.000081 (c) 0.09 (d) 0.01 (e) 0.10

Short answer

Question: Calculate the percentage of heterozygous carriers for the following autosomal recessive disorder frequencies: (a) 0.0064, (b) 0.000081, (c) 0.09, (d) 0.01, and (e) 0.10. Answer: The percentage of heterozygous carriers for each disorder frequency is: (a) 14.72%, (b) 1.78%, (c) 42.00%, (d) 18.00%, and (e) 43.21%.

Step-by-step solution

Understand autosomal recessive mutations and Hardy-Weinberg equilibrium

Autosomal recessive disorders are caused by mutations in both copies of a gene. In such cases, only those individuals with two copies of the mutant allele (homozygous recessive) will exhibit the disorder. Heterozygous carriers have one normal gene copy and one mutated gene copy, so they do not show the disorder's symptoms but can pass the mutated gene to their offspring. In the Hardy-Weinberg equilibrium, p represents the frequency of the normal (dominant) allele (A) and q represents the frequency of the mutant (recessive) allele (a). According to the equation: p^2 + 2pq + q^2 = 1 Step 2: Calculate the frequency of the recessive allele

Calculate the frequency of the recessive allele (q)

The frequency of individuals with the disorder (homozygous recessive) is given by q^2. Using this information, we can determine the frequency of the mutant (recessive) allele (q) by taking the square root of the disorder frequency: q = sqrt(q^2) Step 3: Calculate the frequency of the dominant allele

Calculate the frequency of the dominant allele (p)

Since p and q must add up to 1 in the Hardy-Weinberg equilibrium: p = 1 - q Step 4: Determine the percentage of heterozygous carriers

Determine the percentage of heterozygous carriers (2pq)

The frequency of heterozygous carriers (Aa) can be calculated using the formula: 2pq = 2 * (p) * (q) To find the percentage, multiply the result by 100. Now, let's apply these steps to each of the given disorder frequencies: (a) q^2 = 0.0064: 1. q = sqrt(0.0064) = 0.08 2. p = 1 - 0.08 = 0.92 3. 2pq = 2 * (0.92) * (0.08) = 0.1472 Percentage of carriers = 14.72% (b) q^2 = 0.000081: 1. q = sqrt(0.000081) = 0.009 2. p = 1 - 0.009 = 0.991 3. 2pq = 2 * (0.991) * (0.009) = 0.017838 Percentage of carriers = 1.78% (c) q^2 = 0.09: 1. q = sqrt(0.09) = 0.3 2. p = 1 - 0.3 = 0.7 3. 2pq = 2 * (0.7) * (0.3) = 0.42 Percentage of carriers = 42.00% (d) q^2 = 0.01: 1. q = sqrt(0.01) = 0.1 2. p = 1 - 0.1 = 0.9 3. 2pq = 2 * (0.9) * (0.1) = 0.18 Percentage of carriers = 18.00% (e) q^2 = 0.10: 1. q = sqrt(0.10) = 0.3162 2. p = 1 - 0.3162 = 0.6838 3. 2pq = 2 * (0.6838) * (0.3162) = 0.4321 Percentage of carriers = 43.21% So, the percentage of heterozygous carriers for each disorder frequency is: (a) 14.72% (b) 1.78% (c) 42.00% (d) 18.00% (e) 43.21%

Key concepts

Hardy-Weinberg Equilibrium

The Hardy-Weinberg equilibrium is a fundamental principle in genetics, describing a model where allele frequencies in a large population remain constant from one generation to the next, assumming there is no mutation, gene flow, genetic drift, or selection pressure. This concept helps to predict the genetic variation within a population when these conditions are met.
This equilibrium relies on the equation: \[ p^2 + 2pq + q^2 = 1 \]In this equation, \( p \) represents the frequency of the dominant allele, while \( q \) represents the frequency of the recessive allele. The term \( p^2 \) corresponds to the proportion of the population that is homozygous dominant, \( 2pq \) represents the proportion of heterozygous carriers, and \( q^2 \) represents the proportion that is homozygous recessive. When the population is in Hardy-Weinberg equilibrium, these proportions can be used to calculate the allele frequencies and predict the genetic makeup of future generations.

Heterozygous Carriers

Heterozygous carriers play a key role in the transmission of autosomal recessive disorders. Individuals who are heterozygous possess one normal allele and one mutated allele. They typically do not exhibit any symptoms of the disorder because the normal allele can compensate for the mutated one. However, these carriers can pass the mutated allele to their offspring, who may be affected if they inherit two mutated alleles.
The frequency of these heterozygous carriers in the population, denoted as \( 2pq \), is an important quantity to know because it assesses the risk of the disorder being transmitted to future generations. The proportion of heterozygous carriers is often higher than that of affected individuals since being a carrier does not usually impact an individual's health or reproductive success.

Mutation Frequency Calculation

Mutation frequency calculation is a crucial aspect of genetic studies, providing insight into how often a specific genetic variant occurs within a population. In the context of autosomal recessive disorders, we use the frequency of individuals who have the disorder to calculate the frequency of the recessive allele \( q \), and subsequently the frequency of carriers.
By taking the square root of the frequency of affected individuals (who are homozygous recessive), we find \( q \), the frequency of the recessive allele. The formula is as follows: \[ q = \sqrt{q^2} \]Then, we calculate the frequency of the dominant allele \( p \) by subtracting \( q \) from 1 (since p and q must total 1). Finally, to determine the percentage of heterozygous carriers in the population, we use the equation \( 2pq \) and multiply the result by 100. This calculation helps predict the distribution of genetic traits within a population, and is critical for understanding the potential for rare disorders to be passed on to subsequent generations.