Question:
In a population of 1000 moths, 450 moths were brown-colored with homozygous genotype and 350 albino moths that were homozygous for the recessive allele. Assuming random mating within the population, the frequency of heterozygotes in the next generation would be:
In a population of 1000 moths, 450 moths were brown-colored with homozygous genotype and 350 albino moths that were homozygous for the recessive allele. Assuming random mating within the population, the frequency of heterozygotes in the next generation would be:
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Even if a population is not initially in Hardy-Weinberg equilibrium, just one generation of random mating is sufficient to bring the offspring generation into equilibrium. Always calculate allele frequencies from the parental gene pool first.
Updated On: Jun 16, 2026
- 0.495
- 0.200
- 0.800
- 0.505
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The question asks for the frequency of heterozygotes in the next generation of a population of moths with given genotype numbers, assuming random mating.
Step 2: Key Formula or Approach:
Under the Hardy-Weinberg principle, allele frequencies in a population can be calculated from genotype numbers.
If $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele: \[ p = \frac{2 \times N_{AA} + N_{Aa}}{2N} \] \[ q = \frac{2 \times N_{aa} + N_{Aa}}{2N} \]
After one generation of random mating, the genotype frequencies will reach Hardy-Weinberg equilibrium, where the heterozygote frequency is given by: \[ f(Aa) = 2pq \]
Step 3: Detailed Explanation:
• We are given a population of $N = 1000$ moths.
• The number of homozygous dominant (brown-colored) moths is $N_{AA} = 450$.
• The number of homozygous recessive (albino) moths is $N_{aa} = 350$.
• The remaining moths are heterozygous, so: \[ N_{Aa} = 1000 - 450 - 350 = 200 \]
• Let us calculate the allele frequencies: \[ p = f(A) = \frac{2(450) + 200}{2000} = \frac{900 + 200}{2000} = \frac{1100}{2000} = 0.55 \] \[ q = f(a) = \frac{2(350) + 200}{2000} = \frac{700 + 200}{2000} = \frac{900}{2000} = 0.45 \]
• Now, assuming random mating occurs, the population will establish Hardy-Weinberg equilibrium in the next generation.
• The expected frequency of heterozygous individuals ($Aa$) in the next generation is: \[ f(Aa) = 2pq = 2 \times 0.55 \times 0.45 = 0.495 \]
Step 4: Final Answer:
Therefore, the frequency of heterozygotes in the next generation will be 0.495.
The question asks for the frequency of heterozygotes in the next generation of a population of moths with given genotype numbers, assuming random mating.
Step 2: Key Formula or Approach:
Under the Hardy-Weinberg principle, allele frequencies in a population can be calculated from genotype numbers.
If $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele: \[ p = \frac{2 \times N_{AA} + N_{Aa}}{2N} \] \[ q = \frac{2 \times N_{aa} + N_{Aa}}{2N} \]
After one generation of random mating, the genotype frequencies will reach Hardy-Weinberg equilibrium, where the heterozygote frequency is given by: \[ f(Aa) = 2pq \]
Step 3: Detailed Explanation:
• We are given a population of $N = 1000$ moths.
• The number of homozygous dominant (brown-colored) moths is $N_{AA} = 450$.
• The number of homozygous recessive (albino) moths is $N_{aa} = 350$.
• The remaining moths are heterozygous, so: \[ N_{Aa} = 1000 - 450 - 350 = 200 \]
• Let us calculate the allele frequencies: \[ p = f(A) = \frac{2(450) + 200}{2000} = \frac{900 + 200}{2000} = \frac{1100}{2000} = 0.55 \] \[ q = f(a) = \frac{2(350) + 200}{2000} = \frac{700 + 200}{2000} = \frac{900}{2000} = 0.45 \]
• Now, assuming random mating occurs, the population will establish Hardy-Weinberg equilibrium in the next generation.
• The expected frequency of heterozygous individuals ($Aa$) in the next generation is: \[ f(Aa) = 2pq = 2 \times 0.55 \times 0.45 = 0.495 \]
Step 4: Final Answer:
Therefore, the frequency of heterozygotes in the next generation will be 0.495.
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