To answer this question, we need to use the Hardy-Weinberg equilibrium principle, which states that the frequency of alleles in a population will remain constant from generation to generation in the absence of other evolutionary influences.

The formula for Hardy-Weinberg equilibrium is p^2 + 2pq + q^2 = 1, where p is the frequency of one allele, q is the frequency of the other allele, p^2 is the frequency of homozygous dominant individuals, 2pq is the frequency of heterozygous individuals, and q^2 is the frequency of homozygous recessive individuals.

In this case, we know that the disease is autosomal recessive and very rare, affecting only 1 in 40,000 individuals. This means that q^2 = 1/40,000 = 0.000025. To find q, we take the square root of 0.000025, which is approximately 0.005.

The frequency of heterozygotes (2pq) in the population can be found by doubling the product of p and q. However, because the disease is so rare, p is approximately 1 (since p + q = 1). Therefore, 2pq is approximately 2 * 1 * 0.005 = 0.01, or about 1%.

So, the expected frequency of heterozygotes in this population is about 1%.

Question

To answer this question, we need to use the Hardy-Weinberg equilibrium principle, which states that the frequency of alleles in a population will remain constant from generation to generation in the absence of other evolutionary influences.

The formula for Hardy-Weinberg equilibrium is p^2 + 2pq + q^2 = 1, where p is the frequency of one allele, q is the frequency of the other allele, p^2 is the frequency of homozygous dominant individuals, 2pq is the frequency of heterozygous individuals, and q^2 is the frequency of homozygous recessive individuals.

In this case, we know that the disease is autosomal recessive and very rare, affecting only 1 in 40,000 individuals. This means that q^2 = 1/40,000 = 0.000025. To find q, we take the square root of 0.000025, which is approximately 0.005.

The frequency of heterozygotes (2pq) in the population can be found by doubling the product of p and q. However, because the disease is so rare, p is approximately 1 (since p + q = 1). Therefore, 2pq is approximately 2 * 1 * 0.005 = 0.01, or about 1%.

So, the expected frequency of heterozygotes in this population is about 1%.

Knowee AI · Accepted Answer

To answer this question, we need to use the Hardy-Weinberg equilibrium principle, which states that the frequency of alleles in a population will remain constant from generation to generation in the absence of other evolutionary influences.

The formula for Hardy-Weinberg equilibrium is p^2 + 2pq + q^2 = 1, where p is the frequency of one allele, q is the frequency of the other allele, p^2 is the frequency of homozygous dominant individuals, 2pq is the frequency of heterozygous individuals, and q^2 is the frequency of homozygous recessive individuals.

In this case, we know that the disease is autosomal recessive and very rare, affecting only 1 in 40,000 individuals. This means that q^2 = 1/40,000 = 0.000025. To find q, we take the square root of 0.000025, which is approximately 0.005.

The frequency of heterozygotes (2pq) in the population can be found by doubling the product of p and q. However, because the disease is so rare, p is approximately 1 (since p + q = 1). Therefore, 2pq is approximately 2 * 1 * 0.005 = 0.01, or about 1%.

So, the expected frequency of heterozygotes in this population is about 1%.

Consider a very rare autosomal recessive disease, which affects only 1 in 40,000 individuals in a population. What is the expected frequency of heterozygotes in this population?Multiple choice question.about 1%about 2%about 5%about 0.005%

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