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In an isolated population, the frequencies of the \(I^{A}, I^{B}\) and \(i\) alleles of the \(A-B-O\) blood type gene are, respectively, \(0.15,0.25,\) and \(0.60 .\) If the genotypes of the \(\mathrm{A}-\mathrm{B}-\mathrm{O}\) blood type gene are in Hardy-Weinberg proportions, what fraction of the people who have type \(A\) blood in this population are expected to be homozygous for the \(F^{A}\) allele?

Short Answer

Expert verified
Approximately 11.1% of type A individuals are homozygous for \( I^A I^A \).

Step by step solution

01

Understanding Hardy-Weinberg Equilibrium

Under Hardy-Weinberg equilibrium, the genotype frequencies can be calculated using allele frequencies. The general formula for the genotype frequency is \( p^2 + 2pq + q^2 = 1 \) when there are two alleles with frequencies \( p \) and \( q \). With three alleles, we calculate each genotype frequency separately.
02

Calculate Frequencies Associated with A Allele

The allele frequencies for blood type A genotypes are: homozygous \( I^A I^A \), and heterozygous \( I^A i \). For the \( I^A I^A \) genotype, frequency is \((0.15)^2 = 0.0225\). For \( I^A i \), frequency is \(2 \times 0.15 \times 0.60 = 0.18\). Another possible blood type A genotype is \(I^A I^B\), but since it results in blood type AB, it doesn't contribute to type A directly.
03

Calculate Total Type A Blood Type Frequency

The total frequency of individuals with Type A blood is the sum of the frequencies of \( I^A I^A \) and \( I^A i \). Thus, total Type A frequency is \(0.0225 + 0.18 = 0.2025\).
04

Calculate the Fraction of Homozygous Type A Individuals

The desired fraction of individuals with Type A blood that are homozygous is the frequency of \( I^A I^A \) divided by the total frequency of Type A. So, the fraction is \(\frac{0.0225}{0.2025}=0.111\).

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Key Concepts

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

Allele Frequency
Allele frequency is an essential concept in population genetics, as it refers to how common an allele is within a population. In this case, we are looking at the alleles related to blood type genetics:
  • \( I^A \) with a frequency of 0.15,
  • \( I^B \) with a frequency of 0.25,
  • \( i \) with a frequency of 0.60.
These frequencies sum up to 1, ensuring that all alleles for the gene in the population are accounted for. This allows us to calculate genotype frequencies under Hardy-Weinberg equilibrium by applying the allele frequencies within a formula. The Hardy-Weinberg principle is a way to predict how genetic variation is maintained over time without influence from external factors like selection or mutation. It provides an idealized model to help understand real genetic variation.
Genotype Frequency
Genotype frequency determines how often a particular genetic combo or genotype appears in a population. For blood type, these frequencies can be worked out using the Hardy-Weinberg equation, incorporating the known allele frequencies. For example:
  • The homozygous genotype frequency for \( I^A I^A \) is calculated as \((0.15)^2 = 0.0225\), meaning only 2.25% of the population is expected to have this genotype.
  • The heterozygous genotype \( I^A i \) appears with frequency \(2 imes 0.15 imes 0.60 = 0.18\) or 18% under Hardy-Weinberg equilibrium.
  • The genotype \( I^A I^B \), although it results in blood type AB, is essential for understanding diversity but not crucial for calculating Type A blood.
These frequencies add up to inform the expected distribution of blood types in the population. Hardy-Weinberg equilibrium helps estimate these frequencies without evolutionary forces acting on the population.
Blood Type Genetics
Blood type genetics is governed by the A-B-O blood group system, which is determined by three alleles: \( I^A, I^B, \) and \( i \).
  • Type A blood can result from either the homozygous genotype \( I^A I^A \) or the heterozygous \( I^A i \).
  • Understanding these combinations helps predict the prevalence of each blood type in a given set under Hardy-Weinberg equilibrium.
The total frequency of Type A blood in the population combines both \( I^A I^A \) and \( I^A i \) genotypes, totaling \( 0.0225 + 0.18 = 0.2025 \). To find the proportion of the population with Type A blood that is homozygous \( I^A I^A \), you calculate \( \frac{0.0225}{0.2025} = 0.111 \). Thus, around 11.1% of all individuals with Type A blood are expected to have the homozygous genotype, illustrating the interaction between allele and genotype frequencies within a population.

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Most popular questions from this chapter

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