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In hardyweinberg population of 1500 individuals dominant and rescessive ratio are 21:4 calculate no of heterozygous individuals in the population?a)720 b)480 c)1260 d)240?
Most Upvoted Answer
In hardyweinberg population of 1500 individuals dominant and rescessiv...
720 is the answer..
q^2=4/25...
q=2/5...
p+q=1..
p=3/5...
No. of heterozygotes in the population=2pq×given population..
=2×2/5×3/5×1500..
=720.
Community Answer
In hardyweinberg population of 1500 individuals dominant and rescessiv...
Solution:

In Hardy-Weinberg equilibrium, the frequency of alleles and genotypes in a population remains constant from generation to generation.

The dominant genotype is represented by AA, and the recessive genotype is represented by aa.

Let's assume that the frequency of the dominant allele A is p and the frequency of the recessive allele a is q.

p + q = 1 (because there are only two alleles in a population)

According to the given information:

The ratio of dominant to recessive individuals is 21:4

This means that:

The frequency of the dominant genotype AA is (21/25)^2 = 0.7056

The frequency of the recessive genotype aa is (4/25)^2 = 0.0256

The frequency of the heterozygous genotype Aa is 1 - (0.7056 + 0.0256) = 0.2688

Therefore, the number of heterozygous individuals in the population can be calculated as:

Heterozygous frequency = 0.2688

Total population = 1500

Number of heterozygous individuals = Heterozygous frequency x Total population

Number of heterozygous individuals = 0.2688 x 1500 = 403.2

Rounding off to the nearest whole number, the number of heterozygous individuals is 403.

Therefore, the answer is not given in the options provided.
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In hardyweinberg population of 1500 individuals dominant and rescessive ratio are 21:4 calculate no of heterozygous individuals in the population?a)720 b)480 c)1260 d)240?
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