AP Biology Flashcards: Hardy Weinberg Equilibrium

What this deck covers

This deck focuses on Hardy Weinberg Equilibrium, giving you a quick way to review the definitions, rules, and examples that matter most for AP Biology.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What does the variable $p$ represent in the Hardy-Weinberg principle?

Answer: Frequency of the dominant allele. $p$ is the proportion of dominant alleles in the gene pool.

Flashcard 2: What does the Hardy-Weinberg law predict about allele frequencies?

Answer: They remain constant over generations. Frequencies stay stable when all equilibrium conditions are met.

Flashcard 3: What does it mean if a population is not in Hardy-Weinberg equilibrium?

Answer: Evolution is occurring. Deviations indicate one or more evolutionary forces are acting.

Flashcard 4: What is the significance of the Hardy-Weinberg principle in biology?

Answer: It provides a model for genetic variation in populations. Serves as a null hypothesis to detect evolutionary change.

Flashcard 5: What is the sum of genotype frequencies in Hardy-Weinberg equilibrium?

Answer:

1. All possible genotypes must account for 100% of the population.

Flashcard 6: What is the significance of the Hardy-Weinberg principle in biology?

Answer: It provides a model for genetic variation in populations. Serves as a null hypothesis to detect evolutionary change.

Flashcard 7: What does it mean if a population is not in Hardy-Weinberg equilibrium?

Answer: Evolution is occurring. Deviations indicate one or more evolutionary forces are acting.

Flashcard 8: What is the Hardy-Weinberg equation for genotype frequencies?

Answer: $p^2 + 2pq + q^2 = 1$. Represents all possible genotype combinations from $(p+q)^2$ expansion.

Flashcard 9: List another condition necessary for Hardy-Weinberg equilibrium.

Answer: No mutation. New alleles cannot be created or existing ones changed.

Flashcard 10: What is the Hardy-Weinberg equation for genotype frequencies?

Answer: $p^2 + 2pq + q^2 = 1$. Represents all possible genotype combinations from $(p+q)^2$ expansion.

Flashcard 11: What is the sum of allele frequencies in Hardy-Weinberg equilibrium?

Answer:

1. All possible alleles must account for 100% of the gene pool.

Flashcard 12: Calculate $p^2$ if $p = 0.7$.

Answer: $p^2 = 0.49$. Square the allele frequency: $(0.7)^2 = 0.49$.

Flashcard 13: If $p = 0.8$, what is $q$ and what does it represent?

Answer: $q = 0.2$, frequency of the recessive allele. Since $p + q = 1$, $q = 0.2$ and represents recessive allele frequency.

Flashcard 14: What happens to genotype frequencies under Hardy-Weinberg equilibrium?

Answer: They remain constant over generations. Genotype proportions remain unchanged across generations.

Flashcard 15: What is another condition for Hardy-Weinberg equilibrium?

Answer: No gene flow (migration). No individuals can enter or leave the population.

Flashcard 16: If $p = 0.6$, what is $q$ in a Hardy-Weinberg population?

Answer: $q = 0.4$. Since $p + q = 1$, then $q = 1 - 0.6 = 0.4$.

Flashcard 17: List another condition necessary for Hardy-Weinberg equilibrium.

Answer: No mutation. New alleles cannot be created or existing ones changed.

Flashcard 18: If $q = 0.4$, what is the frequency of the homozygous recessive genotype?

Answer: $q^2 = 0.16$. Square the allele frequency: $(0.4)^2 = 0.16$.

Flashcard 19: What does the Hardy-Weinberg law predict about allele frequencies?

Answer: They remain constant over generations. Frequencies stay stable when all equilibrium conditions are met.

Flashcard 20: What is the sum of allele frequencies in Hardy-Weinberg equilibrium?

Answer:

1. All possible alleles must account for 100% of the gene pool.

Flashcard 21: What is the frequency of the recessive phenotype if $q^2 = 0.04$?

Answer: 0.04. The recessive phenotype frequency equals $q^2$.

Flashcard 22: Identify the term for $q^2$ in the Hardy-Weinberg equation.

Answer: Frequency of homozygous recessive genotype. $q^2$ represents individuals with two copies of the recessive allele.

Flashcard 23: What is the main assumption of the Hardy-Weinberg equilibrium model?

Answer: No evolution occurs in the population. Equilibrium assumes no evolutionary forces change allele frequencies.

Flashcard 24: Determine $q$ if the frequency of $aa$ is $0.25$.

Answer: $q = 0.5$. Since $q^2 = 0.25$, then $q = \sqrt{0.25} = 0.5$.

Flashcard 25: Identify the term for $p^2$ in the Hardy-Weinberg equation.

Answer: Frequency of homozygous dominant genotype. $p^2$ represents individuals with two copies of the dominant allele.

Flashcard 26: In Hardy-Weinberg, what is the expected frequency of the $Aa$ genotype?

Answer: $2pq$. The frequency of heterozygous individuals.

Flashcard 27: If $q = 0.4$, what is the frequency of the homozygous recessive genotype?

Answer: $q^2 = 0.16$. Square the allele frequency: $(0.4)^2 = 0.16$.

Flashcard 28: Determine the frequency of $AA$ if $p = 0.6$.

Answer: $p^2 = 0.36$. Square the allele frequency: $(0.6)^2 = 0.36$.

Flashcard 29: What happens to genotype frequencies under Hardy-Weinberg equilibrium?

Answer: They remain constant over generations. Genotype proportions remain unchanged across generations.

Flashcard 30: Provide another condition for Hardy-Weinberg equilibrium.

Answer: No natural selection. All genotypes must have equal survival and reproductive success.

Flashcard 31: Find $q^2$ if the recessive allele frequency is $0.2$.

Answer: $q^2 = 0.04$. Square the allele frequency: $(0.2)^2 = 0.04$.

Flashcard 32: Determine the frequency of $AA$ if $p = 0.6$.

Answer: $p^2 = 0.36$. Square the allele frequency: $(0.6)^2 = 0.36$.

Flashcard 33: Find the frequency of the dominant phenotype if $p = 0.8$ and $q = 0.2$.

Answer: 0.96. Dominant phenotype = $p^2 + 2pq = 0.64 + 0.32 = 0.96$.

Flashcard 34: What is the sum of genotype frequencies in Hardy-Weinberg equilibrium?

Answer:

1. All possible genotypes must account for 100% of the population.

Flashcard 35: Find $2pq$ if $p = 0.8$ and $q = 0.2$.

Answer: $2pq = 0.32$. Multiply: $2 \times 0.8 \times 0.2 = 0.32$.

Flashcard 36: If $p = 0.7$, what is the frequency of the recessive allele?

Answer: $q = 0.3$. Since $p + q = 1$, then $q = 1 - 0.7 = 0.3$.

Flashcard 37: Find $q^2$ if the recessive allele frequency is $0.2$.

Answer: $q^2 = 0.04$. Square the allele frequency: $(0.2)^2 = 0.04$.

Flashcard 38: If $q = 0.3$, what is the frequency of the homozygous dominant genotype?

Answer: $p^2 = 0.49$. If $q = 0.3$, then $p = 0.7$, so $p^2 = (0.7)^2 = 0.49$.

Flashcard 39: What does the variable $q$ represent in the Hardy-Weinberg principle?

Answer: Frequency of the recessive allele. $q$ is the proportion of recessive alleles in the gene pool.

Flashcard 40: List one condition necessary for Hardy-Weinberg equilibrium.

Answer: Random mating. Mating must be independent of genotype for equilibrium.

Flashcard 41: State the Hardy-Weinberg equation for allele frequencies.

Answer: $p + q = 1$. The two allele frequencies must sum to 1 in a two-allele system.

Flashcard 42: What is the frequency of the recessive phenotype if $q^2 = 0.04$?

Answer: 0.04. The recessive phenotype frequency equals $q^2$.

Flashcard 43: Which condition of Hardy-Weinberg equilibrium involves mate selection?

Answer: Random mating. Assumes individuals choose mates regardless of their genotype.

Flashcard 44: Calculate $q^2$ if $q = 0.3$.

Answer: $q^2 = 0.09$. Square the allele frequency: $(0.3)^2 = 0.09$.

Flashcard 45: What does the variable $q$ represent in the Hardy-Weinberg principle?

Answer: Frequency of the recessive allele. $q$ is the proportion of recessive alleles in the gene pool.

Flashcard 46: Calculate the frequency of heterozygotes if $p = 0.5$ and $q = 0.5$.

Answer: $2pq = 0.5$. Calculate: $2 \times 0.5 \times 0.5 = 0.5$.

Flashcard 47: What is the main assumption of the Hardy-Weinberg equilibrium model?

Answer: No evolution occurs in the population. Equilibrium assumes no evolutionary forces change allele frequencies.

Flashcard 48: If $p = 0.7$, what is the frequency of the recessive allele?

Answer: $q = 0.3$. Since $p + q = 1$, then $q = 1 - 0.7 = 0.3$.

Flashcard 49: If $q = 0.3$, what is the frequency of the homozygous dominant genotype?

Answer: $p^2 = 0.49$. If $q = 0.3$, then $p = 0.7$, so $p^2 = (0.7)^2 = 0.49$.

Flashcard 50: Identify the term for $2pq$ in the Hardy-Weinberg equation.

Answer: Frequency of heterozygous genotype. $2pq$ represents individuals with one copy of each allele type.

Flashcard 51: State the Hardy-Weinberg equation for allele frequencies.

Answer: $p + q = 1$. The two allele frequencies must sum to 1 in a two-allele system.

Flashcard 52: If $p = 0.6$, what is $q$ in a Hardy-Weinberg population?

Answer: $q = 0.4$. Since $p + q = 1$, then $q = 1 - 0.6 = 0.4$.

Flashcard 53: State one reason why a population might deviate from Hardy-Weinberg equilibrium.

Answer: Mutation. Changes in DNA can alter allele frequencies over time.

Flashcard 54: What does the variable $p$ represent in the Hardy-Weinberg principle?

Answer: Frequency of the dominant allele. $p$ is the proportion of dominant alleles in the gene pool.

Flashcard 55: Calculate the frequency of heterozygotes if $p = 0.5$ and $q = 0.5$.

Answer: $2pq = 0.5$. Calculate: $2 \times 0.5 \times 0.5 = 0.5$.

Flashcard 56: Identify the term for $2pq$ in the Hardy-Weinberg equation.

Answer: Frequency of heterozygous genotype. $2pq$ represents individuals with one copy of each allele type.

Flashcard 57: Identify the term for $q^2$ in the Hardy-Weinberg equation.

Answer: Frequency of homozygous recessive genotype. $q^2$ represents individuals with two copies of the recessive allele.

Flashcard 58: What is another condition for Hardy-Weinberg equilibrium?

Answer: No gene flow (migration). No individuals can enter or leave the population.

Flashcard 59: Provide another condition for Hardy-Weinberg equilibrium.

Answer: No natural selection. All genotypes must have equal survival and reproductive success.

Flashcard 60: Identify the term for $p^2$ in the Hardy-Weinberg equation.

Answer: Frequency of homozygous dominant genotype. $p^2$ represents individuals with two copies of the dominant allele.

Flashcard 61: In Hardy-Weinberg, what is the expected frequency of the $aa$ genotype?

Answer: $q^2$. The frequency of homozygous recessive individuals.

Flashcard 62: List one condition necessary for Hardy-Weinberg equilibrium.

Answer: Random mating. Mating must be independent of genotype for equilibrium.

Flashcard 63: In Hardy-Weinberg, what is the expected frequency of the $AA$ genotype?

Answer: $p^2$. The frequency of homozygous dominant individuals.

Flashcard 64: What is the result if a population is in Hardy-Weinberg equilibrium?

Answer: Allele frequencies remain constant. No evolutionary forces act to change the genetic composition.

Flashcard 65: Determine $q$ if the frequency of $aa$ is $0.25$.

Answer: $q = 0.5$. Since $q^2 = 0.25$, then $q = \sqrt{0.25} = 0.5$.

Flashcard 66: Find the frequency of the dominant phenotype if $p = 0.8$ and $q = 0.2$.

Answer: 0.96. Dominant phenotype = $p^2 + 2pq = 0.64 + 0.32 = 0.96$.

Flashcard 67: State another condition necessary for Hardy-Weinberg equilibrium.

Answer: Large population size. Large size prevents random sampling effects (genetic drift).

Flashcard 68: What is the result if a population is in Hardy-Weinberg equilibrium?

Answer: Allele frequencies remain constant. No evolutionary forces act to change the genetic composition.

Flashcard 69: Which condition of Hardy-Weinberg equilibrium involves mate selection?

Answer: Random mating. Assumes individuals choose mates regardless of their genotype.

Flashcard 70: State one reason why a population might deviate from Hardy-Weinberg equilibrium.

Answer: Mutation. Changes in DNA can alter allele frequencies over time.

Flashcard 71: State another condition necessary for Hardy-Weinberg equilibrium.

Answer: Large population size. Large size prevents random sampling effects (genetic drift).

Flashcard 72: Calculate $q^2$ if $q = 0.3$.

Answer: $q^2 = 0.09$. Square the allele frequency: $(0.3)^2 = 0.09$.

Flashcard 73: If $p = 0.8$, what is $q$ and what does it represent?

Answer: $q = 0.2$, frequency of the recessive allele. Since $p + q = 1$, $q = 0.2$ and represents recessive allele frequency.

Flashcard 74: In Hardy-Weinberg, what is the expected frequency of the $AA$ genotype?

Answer: $p^2$. The frequency of homozygous dominant individuals.