Hardy - Weinberg Principle Questions in English

(B) According to the Hardy-Weinberg principle,for a population in genetic equilibrium,the genotype frequencies are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Here,$p$ is the frequency of the dominant allele $(A)$ and $q$ is the frequency of the recessive allele $(a)$.
Given: $p = 0.6$ and $q = 0.4$.
The frequency of heterozygotes $(Aa)$ is represented by $2pq$.
Substituting the values: $2 \times 0.6 \times 0.4 = 2 \times 0.24 = 0.48$.
Therefore,the frequency of heterozygotes in the population is $0.48$.

(C) The Hardy-Weinberg principle states that in a large,randomly mating population,where no mutations,gene flow,or natural selection occur,the allele and genotype frequencies remain constant from generation to generation. This is known as genetic equilibrium. This principle was independently proposed by $G$.$H$. Hardy and Wilhelm Weinberg in $1908$.

(B) According to the Hardy-Weinberg principle,the sum of the frequencies of all alleles in a population is always equal to $1$.
For two alleles $A_1$ and $A_2$,the equation is: $p + q = 1$,where $p$ is the frequency of $A_1$ and $q$ is the frequency of $A_2$.
Given that the frequency of $A_1$ $(p)$ is $0$,we substitute this into the equation:
$0 + q = 1$
Therefore,$q = 1$.
Thus,the frequency of the allele $A_2$ is $1$.

(B) According to the Hardy-Weinberg principle,the genetic equilibrium is represented by the equation $(p + q)^2 = p^2 + 2pq + q^2 = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.
Given that the frequency of the recessive allele $(q)$ is $80\%$,we have $q = 0.8$.
Consequently,the frequency of the dominant allele $(p)$ is $p = 1 - q = 1 - 0.8 = 0.2$.
Carrier individuals are those who are heterozygous for the recessive trait,represented by the term $2pq$ in the Hardy-Weinberg equation.
Substituting the values: $2pq = 2 \times 0.2 \times 0.8 = 0.32$.
Converting this to a percentage: $0.32 \times 100 = 32\%$.
Therefore,the frequency of carrier individuals in the population is $32\%$.

(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.
Given that the frequency of the dominant allele $p = 0.7$,we can calculate the frequency of the recessive allele $q$ as follows:
$q = 1 - p = 1 - 0.7 = 0.3$.
The frequency of the recessive phenotype in a population is represented by the Hardy-Weinberg equation term $q^2$ (homozygous recessive genotype).
Therefore,the frequency of the recessive phenotype = $q^2 = (0.3)^2 = 0.09$.

(C) According to the Hardy-Weinberg principle,the genetic equilibrium in a population is represented by the equation: $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Here,$p$ represents the frequency of the dominant allele '$A$',and $q$ represents the frequency of the recessive allele '$a$'.
Given: $p = 0.6$ and $q = 0.4$.
The frequency of heterozygotes $(Aa)$ is represented by the term $2pq$.
Substituting the values: $2pq = 2 \times 0.6 \times 0.4 = 0.48$.
Therefore,the frequency of heterozygotes in the population is $0.48$.

(C) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary influences. This is expressed by the equation $p^2 + 2pq + q^2 = 1$,where $p$ and $q$ represent the frequencies of two alleles in a population.

(D) According to the Hardy-Weinberg principle,$p + q = 1$ and $p^2 + 2pq + q^2 = 1$.
Here,$q$ represents the frequency of the recessive allele,and $p$ represents the frequency of the dominant allele.
Given: $q = 0.20$.
Therefore,$p = 1 - 0.20 = 0.80$.
The individuals showing the dominant trait are those with genotypes $AA$ $(p^2)$ and $Aa$ $(2pq)$.
The frequency of the dominant phenotype is $p^2 + 2pq = 1 - q^2$.
Calculating $q^2$: $(0.20)^2 = 0.04$.
Frequency of dominant phenotype = $1 - 0.04 = 0.96$.
To express this as a percentage: $0.96 \times 100 = 96\%$.

(B) According to the Hardy-Weinberg principle,the genetic equilibrium is represented by the equation $(p + q)^2 = p^2 + 2pq + q^2 = 1$,where $p^2$ is the frequency of dominant homozygous individuals,$2pq$ is the frequency of heterozygous individuals,and $q^2$ is the frequency of recessive homozygous individuals.
Given that the percentage of dominant homozygous individuals $(p^2)$ is $49\%$,we have $p^2 = 0.49$.
Taking the square root,we get $p = \sqrt{0.49} = 0.7$.
Since $p + q = 1$,we can find $q$ as $q = 1 - p = 1 - 0.7 = 0.3$.
The frequency of heterozygous individuals is represented by $2pq$.
Substituting the values,$2pq = 2 \times 0.7 \times 0.3 = 0.42$.
Converting this to a percentage,$0.42 \times 100 = 42\%$.
Therefore,the percentage of heterozygous individuals in the population is $42\%$.

(C) In a population,the gene frequencies are expected to remain constant according to the Hardy-Weinberg equilibrium principle. However,this equilibrium is disturbed by factors such as mutation,migration (gene flow),and natural selection. When these factors act on a small population,the change in gene frequency is often attributed to 'sampling error' or 'genetic drift'. Genetic drift refers to random fluctuations in allele frequencies in a population due to chance,which is more pronounced in smaller populations.

(D) The $Hardy-Weinberg$ principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces.
Factors that disturb this genetic equilibrium and lead to evolution include:
$1$. Gene migration or gene flow: Movement of alleles into or out of a population.
$2$. Genetic drift: Random changes in allele frequencies due to chance.
$3$. Mutation: Sudden heritable changes in the $DNA$ sequence.
$4$. Genetic recombination: Shuffling of genes during sexual reproduction.
$5$. Natural selection: Differential survival and reproduction of individuals.
Since all these factors contribute to genetic variation and disturb the equilibrium,the correct answer is $D$.

(C) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces like mutation,selection,or genetic drift. This means that even recessive alleles,which may not be expressed in the phenotype of heterozygotes,are maintained within the gene pool and are not lost over time. This concept is a fundamental pillar of population genetics.

(B) The correct answer is $2pq$.
In a stable population,for a gene with two alleles,'$A$' (dominant) and '$a$' (recessive),if the frequency of '$A$' is $p$ and the frequency of '$a$' is $q$,then the frequencies of the three possible genotypes ($AA$,$Aa$,and $aa$) are expressed by the Hardy-Weinberg equation:
$p^2 + 2pq + q^2 = 1$
Where:
$p^2$ = Frequency of $AA$ (homozygous dominant) individuals.
$q^2$ = Frequency of $aa$ (homozygous recessive) individuals.
$2pq$ = Frequency of $Aa$ (heterozygous) individuals.

(C) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation under specific conditions,which include random mating,a large population size,no migration (gene flow),no mutations,and no natural selection.
If individuals mate selectively (non-random mating),the allele frequencies will change over generations,and the population will not be in Hardy-Weinberg equilibrium.
Therefore,the correct option is $C$.

(C) The total number of individuals is $1000$.
Each individual has $2$ alleles, so the total number of alleles in the population is $1000 \times 2 = 2000$.
The genotype $AA$ has $360$ individuals, contributing $360 \times 2 = 720$ alleles of $A$.
The genotype $Aa$ has $480$ individuals, contributing $480$ alleles of $A$.
The genotype $aa$ has $160$ individuals, contributing $0$ alleles of $A$.
Total number of $A$ alleles = $720 + 480 + 0 = 1200$.
Frequency of allele $A$ $(p)$ = $\frac{\text{Total number of } A \text{ alleles}}{\text{Total number of alleles}} = \frac{1200}{2000} = 0.6$.

(B) The $Hardy-Weinberg$ principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces such as mutation, gene flow, genetic drift, recombination, and natural selection.
$Random$ mating is one of the key assumptions of this principle.
If mating is not random (non-random mating), it leads to changes in genotype frequencies, thereby disturbing the genetic equilibrium.
Therefore, the correct option is $B$.

(B) According to the $Hardy-Weinberg$ principle,the allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary influences.
These influences include mutation,natural selection,genetic drift,gene flow,and non-random mating.
Therefore,$Random$ $mating$ is one of the essential conditions for maintaining genetic equilibrium,as it ensures that alleles are combined by chance without bias.

(D) According to the $Hardy-Weinberg$ principle, allele frequencies in a population remain constant in the absence of evolutionary forces.
$Natural selection$ is the only evolutionary force that causes a change in allele frequencies in a specific, adaptive direction.
$Migration$ (gene flow) and $Mutation$ can change allele frequencies but are not inherently directional in terms of adaptation.
$Genetic drift$ causes random, non-directional changes in allele frequencies, especially in small populations.

(D) According to the $Hardy-Weinberg$ principle,the allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces.
These forces include gene flow (migration),genetic drift,mutation,recombination,and natural selection.
$Non-random$ mating is a factor that disrupts the genetic equilibrium because it changes the genotype frequencies,even if allele frequencies might remain the same in some cases.
Therefore,$non-random$ mating is a cause of disturbance in genetic equilibrium.

(B) The Hardy-Weinberg principle states that the allele frequencies in a population are stable and remain constant from generation to generation. The gene pool remains constant. This is called genetic equilibrium.
In a diploid organism,the sum of allele frequencies $p$ and $q$ is $1$ (i.e.,$p + q = 1$).
When we expand the binomial $(p + q)^2$,we get $p^2 + 2pq + q^2 = 1$.
Here,$p^2$ represents the frequency of homozygous dominant individuals (e.g.,$AA$).
$2pq$ represents the frequency of heterozygous individuals (e.g.,$Aa$).
$q^2$ represents the frequency of homozygous recessive individuals (e.g.,$aa$).

(A) In statistics,when the mean,median,and mode of a data set are equal,the distribution is perfectly symmetrical.
This symmetrical distribution is known as a $Normal$ $distribution$ (or Gaussian distribution),which is represented by a bell-shaped curve.
In a skewed distribution,the mean,median,and mode are typically different from each other.
Therefore,the equality of mean and median indicates a normal distribution.

(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,where $p$ is the frequency of allele $A$ and $q$ is the frequency of allele $a$.
Given $p = 0.4$,then $q = 1 - 0.4 = 0.6$.
The genotype frequencies are given by the expansion of $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Here,$p^2$ represents the frequency of homozygous dominant individuals $(AA)$,
$2pq$ represents the frequency of heterozygous individuals $(Aa)$,
and $q^2$ represents the frequency of homozygous recessive individuals $(aa)$.
Calculating the values:
$p^2 = (0.4)^2 = 0.16 (AA)$
$2pq = 2 \times 0.4 \times 0.6 = 0.48 (Aa)$
$q^2 = (0.6)^2 = 0.36 (aa)$
Therefore,the frequencies are $0.16(AA), 0.48(Aa),$ and $0.36(aa)$.

(B) For a gene with two alleles,$A$ and $a$,if the frequency of $A$ is $p$ and the frequency of $a$ is $q$,then the frequencies of the three possible genotypes ($AA, Aa,$ and $aa$) are expressed by the Hardy-Weinberg equation: $p^2 + 2pq + q^2 = 1$.
Here,$p$ is the frequency of allele $A$ and $q$ is the frequency of allele $a$.
Given that $p = 0.7$,we know that $p + q = 1$,so $q = 1 - 0.7 = 0.3$.
The frequency of the heterozygous genotype $Aa$ is represented by $2pq$.
Substituting the values: $2pq = 2 \times 0.7 \times 0.3 = 0.42$.

(A) The Hardy-Weinberg principle states that allele frequencies in a population remain constant in the absence of evolutionary forces.
The assumptions for Hardy-Weinberg equilibrium include:
$1.$ Large population size.
$2.$ Random mating.
$3.$ No mutation.
$4.$ No gene flow (migration).
$5.$ No natural selection.
Comparing these with the given list:
- $iv.$ Occurrence of natural selection: This violates the principle because natural selection causes changes in allele frequencies.
- $v.$ Small size of population: This violates the principle because small populations are subject to genetic drift,which changes allele frequencies.
Therefore,$iv$ and $v$ are the conditions that do not meet the criteria for Hardy-Weinberg equilibrium.

(N/A) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation. The gene pool (total genes and their alleles in a population) remains constant,which is called genetic equilibrium.
In a diploid organism,let the frequency of allele $A$ be $p$ and the frequency of allele $a$ be $q$.
The sum of all allele frequencies is $1$,so $p + q = 1$.
The frequency of individuals with genotype $AA$ is $p^2$,the frequency of individuals with genotype $aa$ is $q^2$,and the frequency of individuals with genotype $Aa$ is $2pq$.
Thus,the algebraic equation is $p^2 + 2pq + q^2 = 1$,which is the expansion of $(p + q)^2 = 1$.
When the measured allele frequencies differ from the expected values,it indicates evolutionary change. Disturbance in genetic equilibrium or Hardy-Weinberg equilibrium is interpreted as the result of evolutionary change in a population.

(N/A) Five factors are known to affect the Hardy-Weinberg equilibrium. These are gene migration or gene flow, genetic drift, mutation, genetic recombination, and natural selection.
When migration of a section of population to another place and population occurs, gene frequencies change in the original as well as in the new population. New genes/alleles are added to the new population and these are lost from the old population. There would be gene flow if this gene migration happens multiple times.
If the same changes occur by chance, it is called genetic drift. Sometimes the change in allele frequency is so different in the new sample of population that they become a different species. The original drifted population becomes founders and the effect is called the $founder effect$.

(A) According to the Hardy-Weinberg principle,the frequency of an allele in a population can be calculated from the genotype frequencies.
$1$. Frequency of $B$ allele $(p)$: This is calculated as the frequency of the homozygous dominant genotype $(BB)$ plus half the frequency of the heterozygous genotype $(Bb)$.
Frequency of $B = \text{Frequency of } BB + \frac{1}{2} \times \text{Frequency of } Bb$
Frequency of $B = 22\% + \frac{1}{2} \times 62\% = 22\% + 31\% = 53\% = 0.53$.
$2$. Frequency of $b$ allele $(q)$: This is calculated as the frequency of the homozygous recessive genotype $(bb)$ plus half the frequency of the heterozygous genotype $(Bb)$.
Frequency of $b = \text{Frequency of } bb + \frac{1}{2} \times \text{Frequency of } Bb$
Frequency of $b = 16\% + \frac{1}{2} \times 62\% = 16\% + 31\% = 47\% = 0.47$.
Thus,the frequency of $B$ is $0.53$ and $b$ is $0.47$.

(A) The five factors that affect the Hardy-Weinberg equilibrium are:
$1$. Gene flow
$2$. Genetic drift
$3$. Genetic recombination
$4$. Mutation
$5$. Natural selection
Therefore,the other two factors are natural selection and mutation.
Mutation is a sudden heritable change in an organism,which is generally due to a change in the base sequence of the nucleic acid of the organism's genome.
Natural selection is a phenomenon by which organisms possessing heritable variations that enable their better survival reproduce and possess a greater number of progeny than their counterparts.

(N/A) Mendelian population is a group of interbreeding individuals of the same species that share a common gene pool. The three most characteristic criteria are:
$1$. The population must be sufficiently large to allow for the free flow of genetic material among individuals through sexual reproduction.
$2$. Mating within the population must occur randomly (panmixia).
$3$. The population must be free from evolutionary forces such as mutation,migration (gene flow),and natural selection,which ensures that the allele frequencies remain constant over generations (Hardy-Weinberg equilibrium).

(N/A) The $Hardy-Weinberg$ principle states that the sum of allelic frequencies in a population remains constant.
Five factors are known to affect $Hardy-Weinberg$ equilibrium:
$(i)$ Gene migration or gene flow: When migration of a section of populations to another place occurs,gene frequencies change in the original as well as in the new population. New genes/alleles are added to the new population and these are lost from the old population.
$(ii)$ Genetic drift: It refers to the elimination of the genes of certain traits when a section of population migrates or dies due to natural calamity.
$(iii)$ Mutation: The sudden heritable change in a gene that alters the genetic makeup of an individual.
$(iv)$ Genetic recombination: It occurs during gamete formation when chromosomes pass from parent to offspring.
$(v)$ Natural selection: It is a phenomenon by which individuals of the population having traits that enable them to grow and reproduce at a higher rate are favoured.

(N/A) The gene pool is defined as the sum total of all the genes and their alleles present in a population at any given time. It represents the complete genetic diversity available to a population of organisms.

(N/A) In $1908$,$G.H. Hardy$ and $W. Weinberg$ established a mathematical relationship to study gene frequencies,known as the Hardy-Weinberg principle.
This principle states that allele frequencies in a population are stable and remain constant from generation to generation,meaning the gene pool remains constant. This state is called genetic equilibrium.
Mathematically,the sum total of all allelic frequencies is $1$. For a gene with two alleles $A$ and $a$ having frequencies $p$ and $q$ respectively,the frequency of individuals in a population is represented by the binomial expansion $(p + q)^2 = p^2 + 2pq + q^2 = 1$.
Here,$p^2$ represents the frequency of homozygous dominant individuals $(AA)$,$q^2$ represents the frequency of homozygous recessive individuals $(aa)$,and $2pq$ represents the frequency of heterozygous individuals $(Aa)$.
When the measured frequency differs from the expected values,the difference indicates the extent of evolutionary change. Any disturbance in this equilibrium,such as a change in allele frequency,is interpreted as the result of evolution.
Five factors are known to affect the Hardy-Weinberg equilibrium: gene migration (gene flow),genetic drift,mutation,genetic recombination,and natural selection.

(C) The $Hardy-Weinberg$ principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary influences.
The equation is expressed as $p^2 + 2pq + q^2 = 1$,where $p$ and $q$ represent the frequencies of the dominant and recessive alleles,respectively.
Therefore,this principle is primarily used to calculate and determine the allele frequencies within a population.

(D) The total sum of all genes and their alleles present in a population at any given time is known as the $Gene \ pool$.
This concept is fundamental to the $Hardy-Weinberg$ principle,which states that allele frequencies in a population remain constant from generation to generation in the absence of evolutionary influences.

(B) The $Hardy-Weinberg$ principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces such as mutation,natural selection,genetic drift,gene flow,and non-random mating. This is known as genetic equilibrium. Therefore,the sum of all allele frequencies in a population is always $1$.

(D) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces.
This is represented by the algebraic equation $(p + q)^{2} = p^{2} + 2pq + q^{2} = 1$,where:
$p$ represents the frequency of the dominant allele.
$q$ represents the frequency of the recessive allele.
$p^{2}$ represents the frequency of homozygous dominant individuals.
$2pq$ represents the frequency of heterozygous individuals.
$q^{2}$ represents the frequency of homozygous recessive individuals.
Therefore,the correct expression is $p^{2} + 2pq + q^{2} = 1$.

(C) According to the $Hardy-Weinberg$ principle,the allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces.
This state is known as genetic equilibrium.
If the measured allele frequency differs from the expected value calculated using the $Hardy-Weinberg$ equation $(p^2 + 2pq + q^2 = 1)$,it indicates that the population is undergoing evolutionary change.
Genetic recombination is one of the primary factors that contribute to variation and changes in allele frequencies over time,leading to a deviation from the expected equilibrium.

(C) The Hardy-Weinberg equilibrium states that allele frequencies in a population are stable and remain constant from generation to generation. The gene pool remains constant. This is called genetic equilibrium. Sum total of all the allelic frequencies is $1$.
Five factors are known to affect the Hardy-Weinberg equilibrium:
$1$. Gene migration or gene flow
$2$. Genetic drift
$3$. Mutation
$4$. Genetic recombination
$5$. Natural selection

(B) The Hardy-Weinberg equation is given by $p^{2} + 2pq + q^{2} = 1$.
In this equation:
$1$. $p^{2}$ represents the frequency of homozygous dominant individuals (e.g.,$AA$).
$2$. $2pq$ represents the frequency of heterozygous individuals (e.g.,$Aa$).
$3$. $q^{2}$ represents the frequency of homozygous recessive individuals (e.g.,$aa$).
Therefore,the term representing the frequency of heterozygous individuals is $2pq$.

(C) The Hardy-Weinberg equation is given by $p^{2} + 2pq + q^{2} = 1$.
In this equation:
$1$. $p^{2}$ represents the frequency of the homozygous dominant genotype (e.g.,$AA$).
$2$. $2pq$ represents the frequency of the heterozygous genotype (e.g.,$Aa$).
$3$. $q^{2}$ represents the frequency of the homozygous recessive genotype (e.g.,$aa$).
Therefore,the term that represents the homozygous recessive genotype is $q^{2}$.

(D) The $Hardy-Weinberg$ principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of disturbing factors.
Factors that disturb this equilibrium include:
$1$. Gene flow (migration of individuals into or out of a population).
$2$. Genetic drift (random changes in allele frequencies).
$3$. Mutation (introduction of new alleles).
$4$. Genetic recombination.
$5$. Natural selection.
Since gene flow,mutation,and natural selection are all factors that disrupt the equilibrium,the correct answer is $All$ $of$ $the$ $above$.

(D) The $Hardy-Weinberg$ principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary forces.
Evolutionary forces that disrupt this genetic equilibrium include:
$1$. $Mutation$: Sudden heritable changes in $DNA$ sequences.
$2$. $Genetic \text{ } drift$: Random fluctuations in allele frequencies due to chance events, especially in small populations.
$3$. $Recombination$: The formation of new combinations of alleles during meiosis.
$4$. $Gene \text{ } flow$ (migration) and $Natural \text{ } selection$ are also key factors.
Since mutation, genetic drift, and recombination all contribute to changing allele frequencies, the correct answer is $All \text{ } of \text{ } the \text{ } above$.

(C) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of disturbing factors.
These disturbing factors include:
$1$. Gene migration or gene flow.
$2$. Genetic drift.
$3$. Mutation.
$4$. Genetic recombination.
$5$. Natural selection.
Since natural selection is one of the key factors that disrupts the equilibrium,statement $C$ is correct.
Statement $A$ is incorrect because there are five factors,not four.
Statement $B$ is incorrect because genetic drift significantly affects the equilibrium.
Statement $D$ is incorrect because migration is called gene flow,not mutation.

(A) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype is represented by $q^2$.
Given that $25\%$ of the population exhibits the recessive trait,we have $q^2 = 25\% = 0.25$.
To find the frequency of the recessive allele $(q)$,we take the square root of $q^2$:
$q = \sqrt{0.25} = 0.5$.
Therefore,the frequency of the recessive allele in the gene pool is $0.5$.

(B) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype is represented by $q^2$.
Given that the recessive trait frequency is $36 \%$,we have $q^2 = 0.36$.
Taking the square root of both sides,we get $q = \sqrt{0.36} = 0.6$.
Here,$q$ represents the frequency of the recessive allele.
The sum of the frequencies of the dominant allele $(p)$ and the recessive allele $(q)$ is $1$,i.e.,$p + q = 1$.
Therefore,the frequency of the dominant allele is $p = 1 - q = 1 - 0.6 = 0.4$.
Thus,the correct option is $B$.

(A) According to the Hardy-Weinberg principle,the frequency of the recessive phenotype $(q^2)$ is given by the ratio of individuals expressing the recessive trait to the total population.
Total population = $200$.
Number of individuals with recessive phenotype = $98$.
Frequency of recessive phenotype $(q^2)$ = $98 / 200 = 0.49$.
Therefore,the frequency of the recessive allele $(q)$ = $\sqrt{0.49} = 0.7$.
Since $p + q = 1$,the frequency of the dominant allele $(p)$ = $1 - 0.7 = 0.3$.
The frequency of heterozygotes is given by $2pq$.
$2pq = 2 \times 0.3 \times 0.7 = 0.42$.
To express this as a percentage,multiply by $100$: $0.42 \times 100 = 42\%$.

(C) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,where $p$ is the frequency of the dominant allele and $q$ is the frequency of the recessive allele.
Individuals with at least one recessive allele include both homozygous recessive $(q^2)$ and heterozygous $(2pq)$ genotypes.
The frequency of individuals with at least one recessive allele is given by $q^2 + 2pq = 0.51$.
Since $p^2 + 2pq + q^2 = 1$,we can write $p^2 = 1 - (q^2 + 2pq)$.
Substituting the given value: $p^2 = 1 - 0.51 = 0.49$.
The frequency of the dominant phenotype is represented by $p^2$ (homozygous dominant) plus $2pq$ (heterozygous),which is $1 - q^2$. However,the question asks for the frequency of the dominant phenotype,which is $p^2 + 2pq = 1 - q^2$. Wait,the dominant phenotype is $p^2 + 2pq$. Since $p^2 + 2pq + q^2 = 1$,the dominant phenotype frequency is $1 - q^2$.
Actually,the frequency of the recessive phenotype is $q^2$. Given $q^2 + 2pq = 0.51$,we know $p^2 = 1 - (q^2 + 2pq) = 1 - 0.51 = 0.49$.
Since $p^2 = 0.49$,$p = 0.7$. Then $q = 1 - 0.7 = 0.3$.
The dominant phenotype frequency is $p^2 + 2pq = (0.7)^2 + 2(0.7)(0.3) = 0.49 + 0.42 = 0.91$.

(A) According to the Hardy-Weinberg principle,the sum of allele frequencies is $p + q = 1$,and the genotype frequencies are $p^2 + 2pq + q^2 = 1$.
Here,$p^2$ represents the frequency of homozygous dominant individuals,and $2pq$ represents the frequency of heterozygous individuals.
The sum of these two,$p^2 + 2pq$,represents the frequency of individuals possessing at least one dominant allele.
Given that $p^2 + 2pq = 51\% = 0.51$.
Since the total frequency of all genotypes is $p^2 + 2pq + q^2 = 1$,the frequency of recessive individuals $(q^2)$ is calculated as:
$q^2 = 1 - (p^2 + 2pq)$
$q^2 = 1 - 0.51 = 0.49$.
Therefore,the frequency of recessive individuals in the population is $0.49$.

(D) The Hardy-Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation in the absence of evolutionary influences.
Factors that disturb this equilibrium include:
$1$. Natural selection: Differential survival and reproduction of individuals.
$2$. Mutation: Introduction of new alleles into the gene pool.
$3$. Genetic drift: Random changes in allele frequencies due to chance events.
$4$. Gene flow (migration): Movement of alleles into or out of the population.
$5$. Non-random mating: Individuals choosing mates based on specific phenotypes.
Since natural selection,mutation,and genetic drift are all factors that disrupt the equilibrium,the correct answer is $D$.