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BIOXII07:EVOLUTION

346565 In a population of 800 rabbits showing Hardy– Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

1 0.48

2 0.84

3 0.36

4 0.4

KCET - 2023

BIOXII07:EVOLUTION

346566 The Hardy-Weinberg law of equilibrium was based on the following

1 Random mating, natural selection, gene flow

2 Random mating, genetic drift, gene flow

3 Non-random mating, mutation, gene flow

4 Random mating, no mutation, no gene flow

BIOXII07:EVOLUTION

346567 A gene locus has two alleles $A$ , a. If the frequency of dominant allele $A$ is 0.4 , then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

1

0.36 (AA); 0.48 (Aa); 0.16 (aa)

2

0.16 (AA); 0.24 (Aa); 0.36 (aa)

3

0.16 (AA); 0.48 (Aa); 0.36 (aa)

4

0.16 (AA); 0.36 (Aa); 0.48 (aa)

Explanation:

Frequency of dominant allele (say p) $= 0.4$
Frequency of recessive allele (say q) $= 1 - 0.4 = 0.6$
Frequency of homozygous dominant individuals (AA)
$= p^{2} = (0.4)^{2} = 0.16$
Frequency of heterozygous individuals (Aa)
$= 2 pq = 2 (0.4) (0.6) = 0.48$
Frequency of homozygous recessive individuals (aa)
$= q^{2} = (0.6)^{2} = 0.36$

NEET - 2019

BIOXII07:EVOLUTION

346568 ${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ represents an equation used in

1 Population genetics

2 Mendelian genetics

3 Biometrics

4 Molecular genetics

Explanation:

${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ is a mathematical representation of 'Hardy-Weinberg principle' used in population genetics. It states that the allelic and genotypic frequencies of a population are stable and is constant over generations.

NCERT Exemplar

BIOXII07:EVOLUTION

346569 Hardy - Weinberg principle is expresed by following equation:

1

p^{2} - q^{2}

2

p^{2} + q^{2}

3

p^{2} + 2 p q + q^{2}

4

p^{2} - 2 pq - q^{2}

Explanation:

In the equation, $p^{2}$ represents the frequency of the homozygous genotype AA, $q^{2}$ represents the frequency of the homozygous genotype aa, and $2 pq$ represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so $p + q = 1$ .

BIOXII07:EVOLUTION

346565 In a population of 800 rabbits showing Hardy– Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

1 0.48

2 0.84

3 0.36

4 0.4

KCET - 2023

BIOXII07:EVOLUTION

346566 The Hardy-Weinberg law of equilibrium was based on the following

1 Random mating, natural selection, gene flow

2 Random mating, genetic drift, gene flow

3 Non-random mating, mutation, gene flow

4 Random mating, no mutation, no gene flow

BIOXII07:EVOLUTION

346567 A gene locus has two alleles $A$ , a. If the frequency of dominant allele $A$ is 0.4 , then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

1

0.36 (AA); 0.48 (Aa); 0.16 (aa)

2

0.16 (AA); 0.24 (Aa); 0.36 (aa)

3

0.16 (AA); 0.48 (Aa); 0.36 (aa)

4

0.16 (AA); 0.36 (Aa); 0.48 (aa)

Explanation:

Frequency of dominant allele (say p) $= 0.4$
Frequency of recessive allele (say q) $= 1 - 0.4 = 0.6$
Frequency of homozygous dominant individuals (AA)
$= p^{2} = (0.4)^{2} = 0.16$
Frequency of heterozygous individuals (Aa)
$= 2 pq = 2 (0.4) (0.6) = 0.48$
Frequency of homozygous recessive individuals (aa)
$= q^{2} = (0.6)^{2} = 0.36$

NEET - 2019

BIOXII07:EVOLUTION

346568 ${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ represents an equation used in

1 Population genetics

2 Mendelian genetics

3 Biometrics

4 Molecular genetics

Explanation:

${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ is a mathematical representation of 'Hardy-Weinberg principle' used in population genetics. It states that the allelic and genotypic frequencies of a population are stable and is constant over generations.

NCERT Exemplar

BIOXII07:EVOLUTION

346569 Hardy - Weinberg principle is expresed by following equation:

1

p^{2} - q^{2}

2

p^{2} + q^{2}

3

p^{2} + 2 p q + q^{2}

4

p^{2} - 2 pq - q^{2}

Explanation:

In the equation, $p^{2}$ represents the frequency of the homozygous genotype AA, $q^{2}$ represents the frequency of the homozygous genotype aa, and $2 pq$ represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so $p + q = 1$ .

BIOXII07:EVOLUTION

346565 In a population of 800 rabbits showing Hardy– Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

1 0.48

2 0.84

3 0.36

4 0.4

KCET - 2023

BIOXII07:EVOLUTION

346566 The Hardy-Weinberg law of equilibrium was based on the following

1 Random mating, natural selection, gene flow

2 Random mating, genetic drift, gene flow

3 Non-random mating, mutation, gene flow

4 Random mating, no mutation, no gene flow

BIOXII07:EVOLUTION

346567 A gene locus has two alleles $A$ , a. If the frequency of dominant allele $A$ is 0.4 , then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

1

0.36 (AA); 0.48 (Aa); 0.16 (aa)

2

0.16 (AA); 0.24 (Aa); 0.36 (aa)

3

0.16 (AA); 0.48 (Aa); 0.36 (aa)

4

0.16 (AA); 0.36 (Aa); 0.48 (aa)

Explanation:

Frequency of dominant allele (say p) $= 0.4$
Frequency of recessive allele (say q) $= 1 - 0.4 = 0.6$
Frequency of homozygous dominant individuals (AA)
$= p^{2} = (0.4)^{2} = 0.16$
Frequency of heterozygous individuals (Aa)
$= 2 pq = 2 (0.4) (0.6) = 0.48$
Frequency of homozygous recessive individuals (aa)
$= q^{2} = (0.6)^{2} = 0.36$

NEET - 2019

BIOXII07:EVOLUTION

346568 ${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ represents an equation used in

1 Population genetics

2 Mendelian genetics

3 Biometrics

4 Molecular genetics

Explanation:

${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ is a mathematical representation of 'Hardy-Weinberg principle' used in population genetics. It states that the allelic and genotypic frequencies of a population are stable and is constant over generations.

NCERT Exemplar

BIOXII07:EVOLUTION

346569 Hardy - Weinberg principle is expresed by following equation:

1

p^{2} - q^{2}

2

p^{2} + q^{2}

3

p^{2} + 2 p q + q^{2}

4

p^{2} - 2 pq - q^{2}

Explanation:

In the equation, $p^{2}$ represents the frequency of the homozygous genotype AA, $q^{2}$ represents the frequency of the homozygous genotype aa, and $2 pq$ represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so $p + q = 1$ .

BIOXII07:EVOLUTION

346565 In a population of 800 rabbits showing Hardy– Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

1 0.48

2 0.84

3 0.36

4 0.4

KCET - 2023

BIOXII07:EVOLUTION

346566 The Hardy-Weinberg law of equilibrium was based on the following

1 Random mating, natural selection, gene flow

2 Random mating, genetic drift, gene flow

3 Non-random mating, mutation, gene flow

4 Random mating, no mutation, no gene flow

BIOXII07:EVOLUTION

346567 A gene locus has two alleles $A$ , a. If the frequency of dominant allele $A$ is 0.4 , then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

1

0.36 (AA); 0.48 (Aa); 0.16 (aa)

2

0.16 (AA); 0.24 (Aa); 0.36 (aa)

3

0.16 (AA); 0.48 (Aa); 0.36 (aa)

4

0.16 (AA); 0.36 (Aa); 0.48 (aa)

Explanation:

Frequency of dominant allele (say p) $= 0.4$
Frequency of recessive allele (say q) $= 1 - 0.4 = 0.6$
Frequency of homozygous dominant individuals (AA)
$= p^{2} = (0.4)^{2} = 0.16$
Frequency of heterozygous individuals (Aa)
$= 2 pq = 2 (0.4) (0.6) = 0.48$
Frequency of homozygous recessive individuals (aa)
$= q^{2} = (0.6)^{2} = 0.36$

NEET - 2019

BIOXII07:EVOLUTION

346568 ${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ represents an equation used in

1 Population genetics

2 Mendelian genetics

3 Biometrics

4 Molecular genetics

Explanation:

${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ is a mathematical representation of 'Hardy-Weinberg principle' used in population genetics. It states that the allelic and genotypic frequencies of a population are stable and is constant over generations.

NCERT Exemplar

BIOXII07:EVOLUTION

346569 Hardy - Weinberg principle is expresed by following equation:

1

p^{2} - q^{2}

2

p^{2} + q^{2}

3

p^{2} + 2 p q + q^{2}

4

p^{2} - 2 pq - q^{2}

Explanation:

In the equation, $p^{2}$ represents the frequency of the homozygous genotype AA, $q^{2}$ represents the frequency of the homozygous genotype aa, and $2 pq$ represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so $p + q = 1$ .

BIOXII07:EVOLUTION

346565 In a population of 800 rabbits showing Hardy– Weinberg equilibrium, the frequency of recessive individuals was 0.16. What is the frequency of heterozygous individuals?

1 0.48

2 0.84

3 0.36

4 0.4

KCET - 2023

BIOXII07:EVOLUTION

346566 The Hardy-Weinberg law of equilibrium was based on the following

1 Random mating, natural selection, gene flow

2 Random mating, genetic drift, gene flow

3 Non-random mating, mutation, gene flow

4 Random mating, no mutation, no gene flow

BIOXII07:EVOLUTION

346567 A gene locus has two alleles $A$ , a. If the frequency of dominant allele $A$ is 0.4 , then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the population?

1

0.36 (AA); 0.48 (Aa); 0.16 (aa)

2

0.16 (AA); 0.24 (Aa); 0.36 (aa)

3

0.16 (AA); 0.48 (Aa); 0.36 (aa)

4

0.16 (AA); 0.36 (Aa); 0.48 (aa)

Explanation:

Frequency of dominant allele (say p) $= 0.4$
Frequency of recessive allele (say q) $= 1 - 0.4 = 0.6$
Frequency of homozygous dominant individuals (AA)
$= p^{2} = (0.4)^{2} = 0.16$
Frequency of heterozygous individuals (Aa)
$= 2 pq = 2 (0.4) (0.6) = 0.48$
Frequency of homozygous recessive individuals (aa)
$= q^{2} = (0.6)^{2} = 0.36$

NEET - 2019

BIOXII07:EVOLUTION

346568 ${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ represents an equation used in

1 Population genetics

2 Mendelian genetics

3 Biometrics

4 Molecular genetics

Explanation:

${(p + q)}^{2} = p^{2} + 2 p q + q^{2} = 1$ is a mathematical representation of 'Hardy-Weinberg principle' used in population genetics. It states that the allelic and genotypic frequencies of a population are stable and is constant over generations.

NCERT Exemplar

BIOXII07:EVOLUTION

346569 Hardy - Weinberg principle is expresed by following equation:

1

p^{2} - q^{2}

2

p^{2} + q^{2}

3

p^{2} + 2 p q + q^{2}

4

p^{2} - 2 pq - q^{2}

Explanation:

In the equation, $p^{2}$ represents the frequency of the homozygous genotype AA, $q^{2}$ represents the frequency of the homozygous genotype aa, and $2 pq$ represents the frequency of the heterozygous genotype Aa. In addition, the sum of the allele frequencies for all the alleles at the locus must be 1, so $p + q = 1$ .