HARDY-WEINBERG EQUILIBRIUM
CALCULATIONS
1 MATHEMATICAL PROBLEMS ON HARDY-WEINBERG
EQUILIBRIUM
1.1 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.2 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.3 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.4 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.5 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.6 PROBLEM 6: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.7 PROBLEM 7: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.8 PROBLEM 8: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.9 PROBLEM 9: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.10 PROBLEM 10: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.11 PROBLEM 11: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.12 PROBLEM 12: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.13 PROBLEM 13: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.14 PROBLEM 14: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.15 PROBLEM 15: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.16 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.17 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.18 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.19 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.20 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.21 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.22 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.23 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.24 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.25 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.26 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.27 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.28 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.29 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.30 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.31 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.32 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.33 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.34 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.35 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.36 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.37 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.38 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.39 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.40 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.41 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.42 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.43 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.44 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.45 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.46 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.47 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.48 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.49 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.50 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.51 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.52 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.53 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.54 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.55 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.56 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.57 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.58 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.59 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.60 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.61 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.62 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.63 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.64 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.65 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.66 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.67 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.68 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.69 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.70 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.71 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.72 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.73 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.74 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.75 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.76 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.77 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.78 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.79 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.80 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.81 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.82 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.83 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.84 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.85 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.86 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.87 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.88 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.89 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.90 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.91 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.92 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.93 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.94 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.95 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.96 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.97 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.98 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.99 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.100 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.101 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.102 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.103 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.104 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.105 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.106 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.107 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.108 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.109 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.110 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.111 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.112 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.113 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.114 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.115 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.116 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.117 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.118 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.119 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.120 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.121 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.122 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.123 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.124 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.125 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.126 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.127 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.128 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.129 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.130 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.131 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.132 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.133 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.134 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.135 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.136 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.137 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.138 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.139 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.140 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.141 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.142 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.143 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.144 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.145 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.146 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.147 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.148 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.149 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.150 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.151 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.152 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.153 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.154 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.155 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.156 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.157 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.158 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.159 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.160 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.161 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.162 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.163 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.164 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.165 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.166 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.167 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.168 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.169 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.170 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.171 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.172 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.173 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.174 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.175 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.176 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.177 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.178 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.179 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.180 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.181 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.182 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.183 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.184 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.185 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.186 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.187 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.188 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.189 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.190 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.191 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.192 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.193 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.194 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.195 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.196 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.197 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.198 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.199 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.200 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.201 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.202 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.203 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.204 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.205 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.206 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.207 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.208 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.209 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.210 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.211 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.212 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.213 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.214 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.215 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.216 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.217 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.218 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.219 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.220 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.221 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.222 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.223 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.224 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.225 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.226 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.227 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.228 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.229 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.230 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.231 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.232 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.233 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.234 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.235 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.236 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.237 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.238 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.239 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.240 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.241 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.242 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.243 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.244 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.245 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.246 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.247 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.248 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.249 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.250 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.251 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.252 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.253 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.254 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.255 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.256 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.257 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.258 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.259 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.260 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.261 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.262 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.263 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.264 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.265 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.266 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.267 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.268 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.269 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.270 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.271 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.272 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.273 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.274 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.275 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.276 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.277 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.278 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.279 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.280 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.281 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.282 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.283 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.284 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.285 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.286 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.287 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.288 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.289 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.290 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
1.291 PROBLEM 1: BASIC ALLELE FREQUENCY CALCULATION
In a large population of flowers, 16% of the plants have white flowers (recessive trait) and 84%
have purple flowers (dominant trait). Assuming Hardy-Weinberg equilibrium, calculate the
frequencies of the dominant (P) and recessive (p) alleles.
Solution:
Given:
• Frequency of white flowers (recessive phenotype) = 16% = 0.16
• Frequency of purple flowers (dominant phenotype) = 84% = 0.84
Step 1: In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q²) equals
the frequency of the homozygous recessive genotype. q² = 0.16
Step 2: Calculate q (frequency of recessive allele) q = √0.16 = 0.4
Step 3: Calculate p (frequency of dominant allele) p + q = 1 p = 1 - q = 1 - 0.4 = 0.6
Therefore, the frequency of the dominant allele (P) is 0.6, and the frequency of the recessive
allele (p) is 0.4.
1.292 PROBLEM 2: GENOTYPE FREQUENCY CALCULATION
In a population of 10,000 individuals, 4,900 have genotype AA, 4,200 have genotype Aa, and
900 have genotype aa. Calculate the allele frequencies and determine if the population is in
Hardy-Weinberg equilibrium.
Solution:
Step 1: Calculate genotype frequencies Total individuals = 4,900 + 4,200 + 900 = 10,000 f(AA)
= 4,900 / 10,000 = 0.49 f(Aa) = 4,200 / 10,000 = 0.42 f(aa) = 900 / 10,000 = 0.09
Step 2: Calculate allele frequencies p (frequency of A) = f(AA) + ½f(Aa) = 0.49 + (0.5 × 0.42)
= 0.70 q (frequency of a) = 1 - p = 1 - 0.70 = 0.30
Step 3: Check Hardy-Weinberg equilibrium If in H-W equilibrium: f(AA) should equal p² = 0.70²
= 0.49 f(Aa) should equal 2pq = 2(0.70)(0.30) = 0.42 f(aa) should equal q² = 0.30² = 0.09
These calculated frequencies match the observed frequencies, so the population is in Hardy-
Weinberg equilibrium.
1.293 PROBLEM 3: TESTING FOR HARDY-WEINBERG EQUILIBRIUM
A population of 1000 individuals was genotyped for a biallelic locus. The results showed 360 AA,
480 Aa, and 160 aa individuals. Use a chi-square test to determine if this population is in Hardy-
Weinberg equilibrium. Use a significance level of 0.05.
Solution:
Step 1: Calculate allele frequencies p (frequency of A) = (2 × 360 + 480) / (2 × 1000) = 0.60 q
(frequency of a) = 1 - p = 0.40
Step 2: Calculate expected genotype frequencies under H-W equilibrium Expected AA = p² ×
1000 = 0.60² × 1000 = 360 Expected Aa = 2pq × 1000 = 2 × 0.60 × 0.40 × 1000 = 480
Expected aa = q² × 1000 = 0.40² × 1000 = 160
Step 3: Calculate chi-square statistic χ² = Σ (Observed - Expected)² / Expected χ² = (360 -
360)² / 360 + (480 - 480)² / 480 + (160 - 160)² / 160 = 0
Step 4: Determine degrees of freedom df = number of genotypes - number of alleles = 3 - 2 =
1
Step 5: Compare to critical value At α = 0.05 and df = 1, the critical value is 3.84 Since 0 <
3.84, we fail to reject the null hypothesis.
Therefore, there is no significant evidence to suggest that the population is not in Hardy-
Weinberg equilibrium.
1.294 PROBLEM 4: PREDICTING GENOTYPE FREQUENCIES AFTER ONE
GENERATION
In a population of plants, the frequency of a dominant allele A is 0.7, and the frequency of the
recessive allele a is 0.3. Assuming random mating and no other evolutionary forces, what will
be the frequency of heterozygotes (Aa) in the next generation?
Solution:
Given: p (frequency of A) = 0.7 q (frequency of a) = 0.3
In Hardy-Weinberg equilibrium, the frequency of heterozygotes is given by 2pq.
Frequency of Aa = 2pq = 2(0.7)(0.3) = 0.42
Therefore, in the next generation, 42% of the population will be heterozygotes (Aa).
1.295 PROBLEM 5: EFFECT OF INBREEDING ON HARDY-WEINBERG
EQUILIBRIUM
Consider a large population with allele frequencies p = 0.6 for allele A and q = 0.4 for allele a.
If the inbreeding coefficient (F) in this population is 0.1, calculate the frequencies of genotypes
AA, Aa, and aa.
Solution:
Given: p = 0.6 q = 0.4 F = 0.1
The genotype frequencies under inbreeding are given by: f(AA) = p² + Fpq f(Aa) = 2pq(1-F)
f(aa) = q² + Fpq
Step 1: Calculate f(AA) f(AA) = 0.6² + 0.1(0.6)(0.4) = 0.36 + 0.024 = 0.384
Step 2: Calculate f(Aa) f(Aa) = 2(0.6)(0.4)(1-0.1) = 0.48 × 0.9 = 0.432
Step 3: Calculate f(aa) f(aa) = 0.4² + 0.1(0.6)(0.4) = 0.16 + 0.024 = 0.184
Step 4: Verify that frequencies sum to 1 0.384 + 0.432 + 0.184 = 1
Therefore, under inbreeding: Frequency of AA = 0.384 or 38.4% Frequency of Aa = 0.432 or
43.2% Frequency of aa = 0.184 or 18.4%
