QBManager

Evolution

185095 In a population of 800 rabbits showing HardyWeinberg equilibrium, the frequency of recessive individuals was 0.16 . What is the frequency of heterozygous individuals?

1 0.36

2 0.4

3 0.48

4 0.84

Explanation:

According to the Hardy-Weinberg principle,
$p^{2} + 2 p q + q^{2} = 1$
Where $p$ and $q$ represent the individual allele frequencies.
$p^{2} =$ frequency of homozygous condition represented $p$ .
$q^{2} =$ frequency of homozygous condition represented $q$ .
$2 pq =$ frequency of hetrozygous condition.
Given,
$p^{2} = 0.16$
Hence, $p = 0.4$
Using
$p + q = 1$
$0.4 + q = 1$
$q = 1 - 0.4$
$q = 0.6$
So, allelic frequency of hetrozygous individual is $2 pq = 2 \times 0.4 \times 0.6 = 0.48$

Karnataka CET-20.05.2023

Evolution

185097 Match the phenomenon listed under column I with those listed under column II. Select the correct answer from the options given.
| | Column I | | Column II |
| :---: | :---: | :---: | :---: |
| A | Warburg effect | p | Change in gene frequency by chance |
| B | Pasteur effect | q | Postponing severance in the leaves by applying cytokinin |
| C | Emerson effect | $r$ | Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration |
| D | Wright effect | s | Inhibitory effect of $O_{2}$ on photosynthesis |
| | | t | Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light |

1

A = t, B = s, C = p, D = q

2

A = s, B = r, C = t, D = p

3

A = s, B = t, C = q, D = r

4

A = t, B = r, C = p, D = s

Explanation:

| Column-I | Column-II |
| ----------- | --------- |
| A. Warburg effect | (s) Inhibitory effect of $O_{2}$ on photosynthesis. |
| B. Pasteur effect | (r) Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration. |
| C. Emerson effect | (t) Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light. |
| D. Wright effect | (p) Change in gene frequency by chance. |

Karnataka CET-2008

Evolution

185103 Which of the following defines Hardy Weinberg's law?

1

p^{2} + 2 p q + q^{2} = 1

2

p^{2} + 3 p q + q^{2} = 1

3

P^{2} + 2 pq + q^{2} = 0

4

q^{2} + p^{2} + 2 p q = 0

Explanation:

$p^{2} + 2 pq + q^{2} = 1$
According to Hardy Weinberg's law, It is possible to calculate all alleles and genotype frequencies using the expressions allele frequency $p + q = 1$
and genotype frequency $p^{2} + 2 pq + q^{2} = 1$ .
Where, $p^{2} =$ Frequency of homozygous dominant genotype (AA).
$pq =$ Frequency of heterozygous genotype (Aa)
$q^{2} =$ Frequency of homozygous recessive genotype (aa)

DUMET-2008

Evolution

185106 Frequency of an autosomal lethal gene is 0.4 . The frequency of carrier in a population or 200 individual is:

1 72

2 96

3 104

4 36

Explanation:

96 :- The frequency of the recessive allele is $q = 0.4$ (given)
hence, $p = 1 - 0.4 = 0.6$ (as P = $1 - q)$
According to Hardy Weinberg principle
$p^{2} + 2 pq + q^{2} = 1$
$p^{2} + (0.6)^{2} = 0.36$
$q^{2} = (0.4)^{2} = 0.16$
$2 pq = 0.36 + 0.16 = 0.48$
Since, $2 pq$ is the frequency of heterozygous carrier
$= 0.48 \times 200$
$= 96$ individuals.

VMMC-2006

Evolution

185109 A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele ' $A$ ' is 0.7 , genotype frequency of ' $Aa$ ' is:

1 0.21

2 0.42

3 0.36

4 0.7

Explanation:

The Hardy - Weinberg equation is a mathematical equation that describes the expected genotype frequencies in a population under some condition like no mutation, no migration, large population size and no natural selection is essential. The
Hardy - Weinberg equation for a gene with two alleles is $p^{2} + 2 p q + q^{2} = 1$
Where $p$ is the frequency of the dominant allele A, $q$ is the frequency of the recessive allele a, according to formula
$p + q = 1$
$= 0.7 + q = 1$
$q = 1 - 0.7$
$= 0.3$
$2 pq$ , is the frequency of the heterozygous genotypes So, according to formula. $2 pq = 2 \times 0.7 \times 0.3$
$2 pq = 0.42$

AP EAMCET-2003

Evolution

185095 In a population of 800 rabbits showing HardyWeinberg equilibrium, the frequency of recessive individuals was 0.16 . What is the frequency of heterozygous individuals?

1 0.36

2 0.4

3 0.48

4 0.84

Explanation:

According to the Hardy-Weinberg principle,
$p^{2} + 2 p q + q^{2} = 1$
Where $p$ and $q$ represent the individual allele frequencies.
$p^{2} =$ frequency of homozygous condition represented $p$ .
$q^{2} =$ frequency of homozygous condition represented $q$ .
$2 pq =$ frequency of hetrozygous condition.
Given,
$p^{2} = 0.16$
Hence, $p = 0.4$
Using
$p + q = 1$
$0.4 + q = 1$
$q = 1 - 0.4$
$q = 0.6$
So, allelic frequency of hetrozygous individual is $2 pq = 2 \times 0.4 \times 0.6 = 0.48$

Karnataka CET-20.05.2023

Evolution

185097 Match the phenomenon listed under column I with those listed under column II. Select the correct answer from the options given.
| | Column I | | Column II |
| :---: | :---: | :---: | :---: |
| A | Warburg effect | p | Change in gene frequency by chance |
| B | Pasteur effect | q | Postponing severance in the leaves by applying cytokinin |
| C | Emerson effect | $r$ | Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration |
| D | Wright effect | s | Inhibitory effect of $O_{2}$ on photosynthesis |
| | | t | Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light |

1

A = t, B = s, C = p, D = q

2

A = s, B = r, C = t, D = p

3

A = s, B = t, C = q, D = r

4

A = t, B = r, C = p, D = s

Explanation:

| Column-I | Column-II |
| ----------- | --------- |
| A. Warburg effect | (s) Inhibitory effect of $O_{2}$ on photosynthesis. |
| B. Pasteur effect | (r) Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration. |
| C. Emerson effect | (t) Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light. |
| D. Wright effect | (p) Change in gene frequency by chance. |

Karnataka CET-2008

Evolution

185103 Which of the following defines Hardy Weinberg's law?

1

p^{2} + 2 p q + q^{2} = 1

2

p^{2} + 3 p q + q^{2} = 1

3

P^{2} + 2 pq + q^{2} = 0

4

q^{2} + p^{2} + 2 p q = 0

Explanation:

$p^{2} + 2 pq + q^{2} = 1$
According to Hardy Weinberg's law, It is possible to calculate all alleles and genotype frequencies using the expressions allele frequency $p + q = 1$
and genotype frequency $p^{2} + 2 pq + q^{2} = 1$ .
Where, $p^{2} =$ Frequency of homozygous dominant genotype (AA).
$pq =$ Frequency of heterozygous genotype (Aa)
$q^{2} =$ Frequency of homozygous recessive genotype (aa)

DUMET-2008

Evolution

185106 Frequency of an autosomal lethal gene is 0.4 . The frequency of carrier in a population or 200 individual is:

1 72

2 96

3 104

4 36

Explanation:

96 :- The frequency of the recessive allele is $q = 0.4$ (given)
hence, $p = 1 - 0.4 = 0.6$ (as P = $1 - q)$
According to Hardy Weinberg principle
$p^{2} + 2 pq + q^{2} = 1$
$p^{2} + (0.6)^{2} = 0.36$
$q^{2} = (0.4)^{2} = 0.16$
$2 pq = 0.36 + 0.16 = 0.48$
Since, $2 pq$ is the frequency of heterozygous carrier
$= 0.48 \times 200$
$= 96$ individuals.

VMMC-2006

Evolution

185109 A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele ' $A$ ' is 0.7 , genotype frequency of ' $Aa$ ' is:

1 0.21

2 0.42

3 0.36

4 0.7

Explanation:

The Hardy - Weinberg equation is a mathematical equation that describes the expected genotype frequencies in a population under some condition like no mutation, no migration, large population size and no natural selection is essential. The
Hardy - Weinberg equation for a gene with two alleles is $p^{2} + 2 p q + q^{2} = 1$
Where $p$ is the frequency of the dominant allele A, $q$ is the frequency of the recessive allele a, according to formula
$p + q = 1$
$= 0.7 + q = 1$
$q = 1 - 0.7$
$= 0.3$
$2 pq$ , is the frequency of the heterozygous genotypes So, according to formula. $2 pq = 2 \times 0.7 \times 0.3$
$2 pq = 0.42$

AP EAMCET-2003

Evolution

185095 In a population of 800 rabbits showing HardyWeinberg equilibrium, the frequency of recessive individuals was 0.16 . What is the frequency of heterozygous individuals?

1 0.36

2 0.4

3 0.48

4 0.84

Explanation:

According to the Hardy-Weinberg principle,
$p^{2} + 2 p q + q^{2} = 1$
Where $p$ and $q$ represent the individual allele frequencies.
$p^{2} =$ frequency of homozygous condition represented $p$ .
$q^{2} =$ frequency of homozygous condition represented $q$ .
$2 pq =$ frequency of hetrozygous condition.
Given,
$p^{2} = 0.16$
Hence, $p = 0.4$
Using
$p + q = 1$
$0.4 + q = 1$
$q = 1 - 0.4$
$q = 0.6$
So, allelic frequency of hetrozygous individual is $2 pq = 2 \times 0.4 \times 0.6 = 0.48$

Karnataka CET-20.05.2023

Evolution

185097 Match the phenomenon listed under column I with those listed under column II. Select the correct answer from the options given.
| | Column I | | Column II |
| :---: | :---: | :---: | :---: |
| A | Warburg effect | p | Change in gene frequency by chance |
| B | Pasteur effect | q | Postponing severance in the leaves by applying cytokinin |
| C | Emerson effect | $r$ | Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration |
| D | Wright effect | s | Inhibitory effect of $O_{2}$ on photosynthesis |
| | | t | Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light |

1

A = t, B = s, C = p, D = q

2

A = s, B = r, C = t, D = p

3

A = s, B = t, C = q, D = r

4

A = t, B = r, C = p, D = s

Explanation:

| Column-I | Column-II |
| ----------- | --------- |
| A. Warburg effect | (s) Inhibitory effect of $O_{2}$ on photosynthesis. |
| B. Pasteur effect | (r) Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration. |
| C. Emerson effect | (t) Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light. |
| D. Wright effect | (p) Change in gene frequency by chance. |

Karnataka CET-2008

Evolution

185103 Which of the following defines Hardy Weinberg's law?

1

p^{2} + 2 p q + q^{2} = 1

2

p^{2} + 3 p q + q^{2} = 1

3

P^{2} + 2 pq + q^{2} = 0

4

q^{2} + p^{2} + 2 p q = 0

Explanation:

$p^{2} + 2 pq + q^{2} = 1$
According to Hardy Weinberg's law, It is possible to calculate all alleles and genotype frequencies using the expressions allele frequency $p + q = 1$
and genotype frequency $p^{2} + 2 pq + q^{2} = 1$ .
Where, $p^{2} =$ Frequency of homozygous dominant genotype (AA).
$pq =$ Frequency of heterozygous genotype (Aa)
$q^{2} =$ Frequency of homozygous recessive genotype (aa)

DUMET-2008

Evolution

185106 Frequency of an autosomal lethal gene is 0.4 . The frequency of carrier in a population or 200 individual is:

1 72

2 96

3 104

4 36

Explanation:

96 :- The frequency of the recessive allele is $q = 0.4$ (given)
hence, $p = 1 - 0.4 = 0.6$ (as P = $1 - q)$
According to Hardy Weinberg principle
$p^{2} + 2 pq + q^{2} = 1$
$p^{2} + (0.6)^{2} = 0.36$
$q^{2} = (0.4)^{2} = 0.16$
$2 pq = 0.36 + 0.16 = 0.48$
Since, $2 pq$ is the frequency of heterozygous carrier
$= 0.48 \times 200$
$= 96$ individuals.

VMMC-2006

Evolution

185109 A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele ' $A$ ' is 0.7 , genotype frequency of ' $Aa$ ' is:

1 0.21

2 0.42

3 0.36

4 0.7

Explanation:

The Hardy - Weinberg equation is a mathematical equation that describes the expected genotype frequencies in a population under some condition like no mutation, no migration, large population size and no natural selection is essential. The
Hardy - Weinberg equation for a gene with two alleles is $p^{2} + 2 p q + q^{2} = 1$
Where $p$ is the frequency of the dominant allele A, $q$ is the frequency of the recessive allele a, according to formula
$p + q = 1$
$= 0.7 + q = 1$
$q = 1 - 0.7$
$= 0.3$
$2 pq$ , is the frequency of the heterozygous genotypes So, according to formula. $2 pq = 2 \times 0.7 \times 0.3$
$2 pq = 0.42$

AP EAMCET-2003

Evolution

185095 In a population of 800 rabbits showing HardyWeinberg equilibrium, the frequency of recessive individuals was 0.16 . What is the frequency of heterozygous individuals?

1 0.36

2 0.4

3 0.48

4 0.84

Explanation:

According to the Hardy-Weinberg principle,
$p^{2} + 2 p q + q^{2} = 1$
Where $p$ and $q$ represent the individual allele frequencies.
$p^{2} =$ frequency of homozygous condition represented $p$ .
$q^{2} =$ frequency of homozygous condition represented $q$ .
$2 pq =$ frequency of hetrozygous condition.
Given,
$p^{2} = 0.16$
Hence, $p = 0.4$
Using
$p + q = 1$
$0.4 + q = 1$
$q = 1 - 0.4$
$q = 0.6$
So, allelic frequency of hetrozygous individual is $2 pq = 2 \times 0.4 \times 0.6 = 0.48$

Karnataka CET-20.05.2023

Evolution

185097 Match the phenomenon listed under column I with those listed under column II. Select the correct answer from the options given.
| | Column I | | Column II |
| :---: | :---: | :---: | :---: |
| A | Warburg effect | p | Change in gene frequency by chance |
| B | Pasteur effect | q | Postponing severance in the leaves by applying cytokinin |
| C | Emerson effect | $r$ | Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration |
| D | Wright effect | s | Inhibitory effect of $O_{2}$ on photosynthesis |
| | | t | Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light |

1

A = t, B = s, C = p, D = q

2

A = s, B = r, C = t, D = p

3

A = s, B = t, C = q, D = r

4

A = t, B = r, C = p, D = s

Explanation:

| Column-I | Column-II |
| ----------- | --------- |
| A. Warburg effect | (s) Inhibitory effect of $O_{2}$ on photosynthesis. |
| B. Pasteur effect | (r) Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration. |
| C. Emerson effect | (t) Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light. |
| D. Wright effect | (p) Change in gene frequency by chance. |

Karnataka CET-2008

Evolution

185103 Which of the following defines Hardy Weinberg's law?

1

p^{2} + 2 p q + q^{2} = 1

2

p^{2} + 3 p q + q^{2} = 1

3

P^{2} + 2 pq + q^{2} = 0

4

q^{2} + p^{2} + 2 p q = 0

Explanation:

$p^{2} + 2 pq + q^{2} = 1$
According to Hardy Weinberg's law, It is possible to calculate all alleles and genotype frequencies using the expressions allele frequency $p + q = 1$
and genotype frequency $p^{2} + 2 pq + q^{2} = 1$ .
Where, $p^{2} =$ Frequency of homozygous dominant genotype (AA).
$pq =$ Frequency of heterozygous genotype (Aa)
$q^{2} =$ Frequency of homozygous recessive genotype (aa)

DUMET-2008

Evolution

185106 Frequency of an autosomal lethal gene is 0.4 . The frequency of carrier in a population or 200 individual is:

1 72

2 96

3 104

4 36

Explanation:

96 :- The frequency of the recessive allele is $q = 0.4$ (given)
hence, $p = 1 - 0.4 = 0.6$ (as P = $1 - q)$
According to Hardy Weinberg principle
$p^{2} + 2 pq + q^{2} = 1$
$p^{2} + (0.6)^{2} = 0.36$
$q^{2} = (0.4)^{2} = 0.16$
$2 pq = 0.36 + 0.16 = 0.48$
Since, $2 pq$ is the frequency of heterozygous carrier
$= 0.48 \times 200$
$= 96$ individuals.

VMMC-2006

Evolution

185109 A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele ' $A$ ' is 0.7 , genotype frequency of ' $Aa$ ' is:

1 0.21

2 0.42

3 0.36

4 0.7

Explanation:

The Hardy - Weinberg equation is a mathematical equation that describes the expected genotype frequencies in a population under some condition like no mutation, no migration, large population size and no natural selection is essential. The
Hardy - Weinberg equation for a gene with two alleles is $p^{2} + 2 p q + q^{2} = 1$
Where $p$ is the frequency of the dominant allele A, $q$ is the frequency of the recessive allele a, according to formula
$p + q = 1$
$= 0.7 + q = 1$
$q = 1 - 0.7$
$= 0.3$
$2 pq$ , is the frequency of the heterozygous genotypes So, according to formula. $2 pq = 2 \times 0.7 \times 0.3$
$2 pq = 0.42$

AP EAMCET-2003

Evolution

185095 In a population of 800 rabbits showing HardyWeinberg equilibrium, the frequency of recessive individuals was 0.16 . What is the frequency of heterozygous individuals?

1 0.36

2 0.4

3 0.48

4 0.84

Explanation:

According to the Hardy-Weinberg principle,
$p^{2} + 2 p q + q^{2} = 1$
Where $p$ and $q$ represent the individual allele frequencies.
$p^{2} =$ frequency of homozygous condition represented $p$ .
$q^{2} =$ frequency of homozygous condition represented $q$ .
$2 pq =$ frequency of hetrozygous condition.
Given,
$p^{2} = 0.16$
Hence, $p = 0.4$
Using
$p + q = 1$
$0.4 + q = 1$
$q = 1 - 0.4$
$q = 0.6$
So, allelic frequency of hetrozygous individual is $2 pq = 2 \times 0.4 \times 0.6 = 0.48$

Karnataka CET-20.05.2023

Evolution

185097 Match the phenomenon listed under column I with those listed under column II. Select the correct answer from the options given.
| | Column I | | Column II |
| :---: | :---: | :---: | :---: |
| A | Warburg effect | p | Change in gene frequency by chance |
| B | Pasteur effect | q | Postponing severance in the leaves by applying cytokinin |
| C | Emerson effect | $r$ | Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration |
| D | Wright effect | s | Inhibitory effect of $O_{2}$ on photosynthesis |
| | | t | Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light |

1

A = t, B = s, C = p, D = q

2

A = s, B = r, C = t, D = p

3

A = s, B = t, C = q, D = r

4

A = t, B = r, C = p, D = s

Explanation:

| Column-I | Column-II |
| ----------- | --------- |
| A. Warburg effect | (s) Inhibitory effect of $O_{2}$ on photosynthesis. |
| B. Pasteur effect | (r) Decline in the consumption of respiratory substrate due to a change from anaerobic to aerobic respiration. |
| C. Emerson effect | (t) Enhancement of photosynthesis by subjecting chlorophyll to the effect of two different wavelengths of light. |
| D. Wright effect | (p) Change in gene frequency by chance. |

Karnataka CET-2008

Evolution

185103 Which of the following defines Hardy Weinberg's law?

1

p^{2} + 2 p q + q^{2} = 1

2

p^{2} + 3 p q + q^{2} = 1

3

P^{2} + 2 pq + q^{2} = 0

4

q^{2} + p^{2} + 2 p q = 0

Explanation:

$p^{2} + 2 pq + q^{2} = 1$
According to Hardy Weinberg's law, It is possible to calculate all alleles and genotype frequencies using the expressions allele frequency $p + q = 1$
and genotype frequency $p^{2} + 2 pq + q^{2} = 1$ .
Where, $p^{2} =$ Frequency of homozygous dominant genotype (AA).
$pq =$ Frequency of heterozygous genotype (Aa)
$q^{2} =$ Frequency of homozygous recessive genotype (aa)

DUMET-2008

Evolution

185106 Frequency of an autosomal lethal gene is 0.4 . The frequency of carrier in a population or 200 individual is:

1 72

2 96

3 104

4 36

Explanation:

96 :- The frequency of the recessive allele is $q = 0.4$ (given)
hence, $p = 1 - 0.4 = 0.6$ (as P = $1 - q)$
According to Hardy Weinberg principle
$p^{2} + 2 pq + q^{2} = 1$
$p^{2} + (0.6)^{2} = 0.36$
$q^{2} = (0.4)^{2} = 0.16$
$2 pq = 0.36 + 0.16 = 0.48$
Since, $2 pq$ is the frequency of heterozygous carrier
$= 0.48 \times 200$
$= 96$ individuals.

VMMC-2006

Evolution

185109 A population is in Hardy-Weinberg equilibrium for a gene with only two alleles. If the gene frequency of an allele ' $A$ ' is 0.7 , genotype frequency of ' $Aa$ ' is:

1 0.21

2 0.42

3 0.36

4 0.7

Explanation:

The Hardy - Weinberg equation is a mathematical equation that describes the expected genotype frequencies in a population under some condition like no mutation, no migration, large population size and no natural selection is essential. The
Hardy - Weinberg equation for a gene with two alleles is $p^{2} + 2 p q + q^{2} = 1$
Where $p$ is the frequency of the dominant allele A, $q$ is the frequency of the recessive allele a, according to formula
$p + q = 1$
$= 0.7 + q = 1$
$q = 1 - 0.7$
$= 0.3$
$2 pq$ , is the frequency of the heterozygous genotypes So, according to formula. $2 pq = 2 \times 0.7 \times 0.3$
$2 pq = 0.42$

AP EAMCET-2003