Within a certain island population (that is in Hardy - Weinberg equilibrium) the frequency of the $L$ allele of the $L M$ blood group is 0.6 and the frequency of the $M$ allele is 0.4 . $L$ and $M$ alleles are co - dominant. There are 1000 people on the island. How many are heterozygous in this population?

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∻Answer∻ ⚹ 480 ⚹ ∻Solution∻ ‖ a ⋮ Calculate the frequency of the heterozygous genotype (Lm) using the Hardy-Weinberg equation: $p^2 + 2pq + q^2 = 1$, where $p$ is the frequency of the L allele, $q$ is the frequency of the M allele, and $2pq$ represents the frequency of the heterozygous genotype. ‖ ‖ b ⋮ Given $p = 0.6$ and $q = 0.4$, calculate $2pq$: $2pq = 2 imes 0.6 imes 0.4$. ‖ ‖ c ⋮ Perform the calculation: $2pq = 2 imes 0.6 imes 0.4 = 0.48$. ‖ ‖ d ⋮ Determine the number of heterozygous individuals by multiplying the frequency of the heterozygous genotype by the total population size: $0.48 imes 1000$. ‖ ‖ e ⋮ Perform the calculation: $0.48 imes 1000 = 480$. ‖ ∻Key Concept∻ ⚹ Hardy-Weinberg equilibrium and heterozygote frequency calculation ⚹ ∻Explanation∻ ⚹ The Hardy-Weinberg equation is used to calculate the expected frequency of genotypes in a population that is not evolving. The term $2pq$ represents the frequency of heterozygous individuals. By multiplying this frequency by the total population size, we can determine the number of heterozygous individuals. ⚹◊What is the formula to calculate the frequency of heterozygous individuals in a population using the Hardy-Weinberg equilibrium principle?⍭ Generate me a similar question◊

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