249. In a Random mating population of 28,000
individuals percentage of dominant homozygous
individuals is 49% find out the percentage of
heterozygous individual -
(1) 21% (2) 42% (3) 32% (4) 9%
With explanation. Urgent
Answers
Answer:
According to the Hardy Weinberg law, the allele and genotype frequencies in a population remain constant under absence of factors responsible for evolution. It states that the sum of all genotype frequencies can be represented as the binomial expansion of the square of the sum of p and q. This sum is equal to one. (p + q)2 = p2 + 2pq + q2 = 1.
Here, "p" is the frequency of dominant allele and q is frequency of recessive allele. The "2pq" in equation shows frequency of heterozygotes in the population.
In the question, the frequency of dominant homozygous individuals p
2
= 49% or 0.49.
Thus, frequency of dominant allele (p)= "0.7".
Since, p+q=1, so q=1-p.
The frequency of recessive allele (q)= 1-p=1-0.7=0.3.
Thus, the frequency of heterozygotes = "2pq= 2 X 0.7 X 0.3 = 0.42 or 42%". So, option B is correct.
Answer:
According to the Hardy Weinberg law, the allele and genotype frequencies in a population remain constant under absence of factors responsible for evolution. It states that the sum of all genotype frequencies can be represented as the binomial expansion of the square of the sum of p and q. This sum is equal to one. (p + q)2 = p2 + 2pq + q2 = 1.
Here, "p" is the frequency of dominant allele and q is frequency of recessive allele. The "2pq" in equation shows frequency of heterozygotes in the population.
In the question, the frequency of dominant homozygous individuals p
2
= 49% or 0.49.
Thus, frequency of dominant allele (p)= "0.7".
Since, p+q=1, so q=1-p.
The frequency of recessive allele (q)= 1-p=1-0.7=0.3.
Thus, the frequency of heterozygotes = "2pq= 2 X 0.7 X 0.3 = 0.42 or 42%". So, option B is correct.