The frequency of individuals affected with an autosomally recessive inherited disorder is 1 in 4. Assuming that the population affected by this disease is in Hardy-Weinberg equilibrium, what is the proportion of carrier individuals in this population?
视频信息
答案文本
视频字幕
We have a Hardy-Weinberg equilibrium problem. The frequency of individuals affected with an autosomal recessive disorder is 1 in 4. This means that q-squared, representing the frequency of homozygous recessive individuals, equals 0.25. We need to find the proportion of carrier individuals, which are the heterozygotes.
Now let's calculate the allele frequencies step by step. Since q-squared equals 0.25, we can find q by taking the square root, which gives us 0.5. Since p plus q must equal 1 in Hardy-Weinberg equilibrium, p also equals 0.5. So both alleles have equal frequencies of 0.5 in this population.
Using the Hardy-Weinberg formula, we can calculate all genotype frequencies. The formula is p plus q squared, which expands to p-squared plus 2pq plus q-squared. Substituting our values of 0.5 for both p and q, we get: p-squared equals 0.25 for AA individuals, 2pq equals 0.5 for Aa carriers, and q-squared equals 0.25 for aa affected individuals.
Let's visualize this in a population of 100 individuals. We would have 25 AA normal individuals shown in blue, 50 Aa carriers shown in green, and 25 aa affected individuals shown in red. The carriers, highlighted here, make up exactly 50% of the population. Therefore, the proportion of carrier individuals is 0.5 or 50%.
In conclusion, the proportion of carrier individuals in this population is 0.5, which equals 50% or one-half. This result demonstrates an important principle of Hardy-Weinberg equilibrium: when 25% of a population is affected by an autosomal recessive disorder, exactly 50% are carriers who do not show symptoms but can pass the recessive allele to their offspring.