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In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer?
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In a population of 100 individuals in hardy Weinberg's equilibrium,49 ...
Ryt ans is c acc. to h.w. equilibrium p+q=1here p2=49 so p=7 now put in equation p+q=1 so. we get -6
so p*q=heterozygous individual
so 6*7=42 Ans .
so p*q=heterozygous individual
so 6*7=42 Ans .
Community Answer
In a population of 100 individuals in hardy Weinberg's equilibrium,49 ...
**Homozygous dominant individuals in the population:**
In Hardy-Weinberg equilibrium, the frequency of homozygous dominant individuals can be represented by the equation p^2 = 49/100, where p represents the frequency of the dominant allele in the population.
Solving for p, we take the square root of both sides of the equation:
p = √(49/100) = 7/10
**Heterozygous individuals in the population:**
In Hardy-Weinberg equilibrium, the frequency of heterozygous individuals can be represented by the equation 2pq, where p represents the frequency of the dominant allele and q represents the frequency of the recessive allele.
Since we know the frequency of the dominant allele is 7/10, we can substitute this value into the equation:
2(7/10)q = 2(7q/10) = 14q/10 = 7q/5
We also know that p + q = 1, so we can substitute 7/10 for p in the equation:
7/10 + q = 1
q = 1 - 7/10
q = 3/10
Now we can substitute this value of q back into the equation for heterozygous individuals:
7q/5 = 7(3/10)/5 = 21/50
Thus, the number of heterozygous individuals in the population is 21/50.
**Answer:**
The correct answer is not provided in the given options.
In Hardy-Weinberg equilibrium, the frequency of homozygous dominant individuals can be represented by the equation p^2 = 49/100, where p represents the frequency of the dominant allele in the population.
Solving for p, we take the square root of both sides of the equation:
p = √(49/100) = 7/10
**Heterozygous individuals in the population:**
In Hardy-Weinberg equilibrium, the frequency of heterozygous individuals can be represented by the equation 2pq, where p represents the frequency of the dominant allele and q represents the frequency of the recessive allele.
Since we know the frequency of the dominant allele is 7/10, we can substitute this value into the equation:
2(7/10)q = 2(7q/10) = 14q/10 = 7q/5
We also know that p + q = 1, so we can substitute 7/10 for p in the equation:
7/10 + q = 1
q = 1 - 7/10
q = 3/10
Now we can substitute this value of q back into the equation for heterozygous individuals:
7q/5 = 7(3/10)/5 = 21/50
Thus, the number of heterozygous individuals in the population is 21/50.
**Answer:**
The correct answer is not provided in the given options.
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In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer?
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In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer?.
In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a population of 100 individuals in hardy Weinberg's equilibrium,49 individuals are homozygous dominant. The number of heterozygous individuals in that population would be? a) 21 b) 9 c) 42 d) 51 What's the answer?.
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