Answer:
Hardy-Weinberg Equilibrium, also referred to as the Hardy-Weinberg principle, is used to compare
allele frequencies in a given population over a period of time. A population of alleles must meet five
rules in order to be considered “in equilibrium”:
1) No gene mutations may occur and therefore allele changes do not occur.
2) There must be no migration of individuals either into or out of the population.
3) Random mating must occur, meaning individuals mate by chance.
4) No genetic drift, a chance change in allele frequency, may occur.
5) No natural selection, a change in allele frequency due to environment, may occur.
Hardy-Weinberg Equilibrium never occurs in nature because there is always at least one rule being
violated. Hardy-Weinberg Equilibrium is an ideal state that provides a baseline against which
scientists measure gene evolution in a given population. The Hardy-Weinberg equations can be used
for any population; the population does not need to be in equilibrium.
There are two equations necessary to solve a Hardy-Weinberg Equilibrium question:
P+Q=1
[tex]P^{2}+2PQ+Q^{2}=1[/tex]
P is the frequency of the dominant allele.
Q is the frequency of the recessive allele.
P² is the frequency of individuals with the homozygous dominant genotype.
2PQ is the frequency of individuals with the heterozygous genotype.
Q² is the frequency of individuals with the homozygous recessive genotype.
Example 1a:
A population of cats can be either black or white; the black allele (B) has
complete dominance over the white allele (b). Given a population of 1,000 cats, 840 black and
160 white, determine the allele frequency, the frequency of individuals per genotype, and
number of individuals per genotype.
To solve this problem, solve for all the preceding variables P,Q,P²,Q² and 2PQ
Step 1: Find the frequency of white cats, the homozygous recessive genotype, as they have only
one genotype, bb. Black cats can have either the genotype Bb or the genotype BB, and therefore,
the frequency cannot be directly determined.
Frequency of individuals=individuals/ total population
=160/1000=0.16
Frequency of white cats = 0.16; therefore, Q² = 0.16
Step 2: Find Q by taking the square root of Q².
√(Q²) = √(0.16)
Q = 0.4
Step 3: Use the first Hardy-Weinberg equation (P + Q= 1) to solve for P.
P + Q = 1
P= 1 − Q = 1 − (0.4)
P= 0.6
Now that the allele frequencies in the population are known, solve for the remaining
frequency of individuals by using P² + 2PQ + Q² = 1.
Step 4: Square P to find P².
P = 0.6
P² = (0.6)²
P² = 0.36
Step 5: Multiply 2 × P × Q to get 2PQ
2PQ = 2(0.6)(0.4)
2PQ = 0.48
Therefore:
The frequency of the dominant alleles: P = 0.6
The frequency of the recessive alleles: Q = 0.4
The frequency of individuals with the dominant genotype: P² = 0.36
The frequency of individuals with the heterozygous genotype: 2PQ = 0.48
The frequency of individuals with the recessive genotype: Q² = 0.16
Remember: Frequencies can be checked by substituting the values above back into the
Hardy-Weinberg equations.
0.6 + 0.4 = 1
0.36 + 0.48 + 0.16 = 1
Step 6: Multiply the frequency of individuals (P², 2PQ and Q²) by the total population to get
the number of individuals with that given genotype.
P² × Total population= 0.36 × 1,000 = 360 black cats, BB genotype.
2PQ × Total population = 0.48 × 1,000 = 480 black cats, Bb genotype.
P² × Total population = 0.16 × 1,000 = 160 white cats, bb genotype.
Comparing Generations
To know if a population is in Hardy-Weinberg Equilibrium scientists have to observe at least two
generations. If the allele frequencies are the same for both generations then the population is in
Hardy-Weinberg Equilibrium.
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Verified answer
Answer:
Hardy-Weinberg Equilibrium, also referred to as the Hardy-Weinberg principle, is used to compare
allele frequencies in a given population over a period of time. A population of alleles must meet five
rules in order to be considered “in equilibrium”:
1) No gene mutations may occur and therefore allele changes do not occur.
2) There must be no migration of individuals either into or out of the population.
3) Random mating must occur, meaning individuals mate by chance.
4) No genetic drift, a chance change in allele frequency, may occur.
5) No natural selection, a change in allele frequency due to environment, may occur.
Hardy-Weinberg Equilibrium never occurs in nature because there is always at least one rule being
violated. Hardy-Weinberg Equilibrium is an ideal state that provides a baseline against which
scientists measure gene evolution in a given population. The Hardy-Weinberg equations can be used
for any population; the population does not need to be in equilibrium.
There are two equations necessary to solve a Hardy-Weinberg Equilibrium question:
P+Q=1
[tex]P^{2}+2PQ+Q^{2}=1[/tex]
P is the frequency of the dominant allele.
Q is the frequency of the recessive allele.
P² is the frequency of individuals with the homozygous dominant genotype.
2PQ is the frequency of individuals with the heterozygous genotype.
Q² is the frequency of individuals with the homozygous recessive genotype.
Example 1a:
A population of cats can be either black or white; the black allele (B) has
complete dominance over the white allele (b). Given a population of 1,000 cats, 840 black and
160 white, determine the allele frequency, the frequency of individuals per genotype, and
number of individuals per genotype.
To solve this problem, solve for all the preceding variables P,Q,P²,Q² and 2PQ
Step 1: Find the frequency of white cats, the homozygous recessive genotype, as they have only
one genotype, bb. Black cats can have either the genotype Bb or the genotype BB, and therefore,
the frequency cannot be directly determined.
Frequency of individuals=individuals/ total population
=160/1000=0.16
Frequency of white cats = 0.16; therefore, Q² = 0.16
Step 2: Find Q by taking the square root of Q².
√(Q²) = √(0.16)
Q = 0.4
Step 3: Use the first Hardy-Weinberg equation (P + Q= 1) to solve for P.
P + Q = 1
P= 1 − Q = 1 − (0.4)
P= 0.6
Now that the allele frequencies in the population are known, solve for the remaining
frequency of individuals by using P² + 2PQ + Q² = 1.
Step 4: Square P to find P².
P = 0.6
P² = (0.6)²
P² = 0.36
Step 5: Multiply 2 × P × Q to get 2PQ
2PQ = 2(0.6)(0.4)
2PQ = 0.48
Therefore:
The frequency of the dominant alleles: P = 0.6
The frequency of the recessive alleles: Q = 0.4
The frequency of individuals with the dominant genotype: P² = 0.36
The frequency of individuals with the heterozygous genotype: 2PQ = 0.48
The frequency of individuals with the recessive genotype: Q² = 0.16
Remember: Frequencies can be checked by substituting the values above back into the
Hardy-Weinberg equations.
0.6 + 0.4 = 1
0.36 + 0.48 + 0.16 = 1
Step 6: Multiply the frequency of individuals (P², 2PQ and Q²) by the total population to get
the number of individuals with that given genotype.
P² × Total population= 0.36 × 1,000 = 360 black cats, BB genotype.
2PQ × Total population = 0.48 × 1,000 = 480 black cats, Bb genotype.
P² × Total population = 0.16 × 1,000 = 160 white cats, bb genotype.
Comparing Generations
To know if a population is in Hardy-Weinberg Equilibrium scientists have to observe at least two
generations. If the allele frequencies are the same for both generations then the population is in
Hardy-Weinberg Equilibrium.
[tex]\large\text{\underline{Answer:-}}[/tex]
Hardy-Weinberg Equilibrium, also referred to as the Hardy-Weinberg principle, is used to compare
allele frequencies in a given population over a period of time. A population of alleles must meet five
rules in order to be considered “in equilibrium”:
1) No gene mutations may occur and therefore allele changes do not occur.
2) There must be no migration of individuals either into or out of the population.
3) Random mating must occur, meaning individuals mate by chance.
4) No genetic drift, a chance change in allele frequency, may occur.
5) No natural selection, a change in allele frequency due to environment, may occur.