Chapter 25: Problem 20
A total of 6129 North A mer ican Caucasians were blood typed for the MN locus, which is determined by two codominant alleles, \(L^{\text {M }}\) and \(L^{\text {N. }}\). The following data were obtained: $$ \begin{array}{cc} \text { Blood type } & \text { Number } \\ \mathrm{M} & 1787 \\ \mathrm{MN} & 3039 \\ \mathrm{N} & 1303 \end{array} $$ Carry out a chi-square test to determine whether this population is in Hardy- Weinberg equilibrium at the MN locus.
Short Answer
Step by step solution
Determine Observed Genotype Frequencies
Calculate Allele Frequencies
Calculate Expected Genotype Frequencies
Calculate Expected Genotype Numbers
Perform Chi-Square Calculation
Draw Conclusions
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Allele Frequencies
To calculate the allele frequencies in our population:
- Frequency of \( L^{M} \) (denoted as \( p \)) is found by considering not just the homozygous MM individuals but also the heterozygous MN individuals. The formula used is \( p = \frac{2 \times 1787 + 3039}{2 \times 6129} \). Here, we multiply the number of MM individuals by 2 because each has two copies of \( L^{M} \), and we add the MN count, considering MN has one copy of each allele.
- Similarly, frequency of \( L^{N} \) (denoted as \( q \)) accounts for both NN and MN individuals using the formula \( q = \frac{2 \times 1303 + 3039}{2 \times 6129} \).
Genotype Frequencies
- \( p^2 \) represents the expected frequency of homozygous MM individuals (genotype MM)
- \( 2pq \) represents the expected frequency of heterozygous MN individuals (genotype MN)
- \( q^2 \) represents the expected frequency of homozygous NN individuals (genotype NN)
If there is a significant difference between the observed and expected genotype frequencies, it may suggest factors such as selection, mutation, or genetic drift are influencing the population.
Chi-Square Test
To perform a chi-square test:
- First, calculate the expected number of individuals for each genotype by multiplying the expected genotype frequencies by the total number of individuals.
- Use the formula for chi-square: \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) is the observed frequency of each genotype and \( E_i \) is the expected frequency. This formula measures the discrepancies between observed and expected values across the genotypes.
- The calculated \( \chi^2 \) value is then compared to the critical value from a chi-square distribution table, considering 1 degree of freedom.
Codominant Alleles
With codominance:
- An individual with the genotype MM would exhibit only the M antigen, displaying the M phenotype.
- Likewise, an NN individual would have the N phenotype.
- However, an individual with the MN genotype expresses phenotypic traits of both alleles, showing both M and N antigens equally.