MCAT Biological and Biochemical Foundations of Living Systems Flashcards: 1c Population Genetics Hardy Weinberg

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This deck focuses on 1c Population Genetics Hardy Weinberg, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Biological and Biochemical Foundations of Living Systems.

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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: Identify the expected genotype frequencies under Hardy–Weinberg in terms of $p$ and $q$.

Answer: $AA: p^2$, $Aa: 2pq$, $aa: q^2$. These frequencies represent the expected proportions of homozygous dominant, heterozygous, and homozygous recessive genotypes under equilibrium.

Flashcard 2: If $p = 0.70$ in Hardy–Weinberg, what is the expected heterozygote frequency?

Answer: $2pq = 0.42$. Heterozygote frequency is $2pq$, with $q = 1 - p$ under Hardy-Weinberg conditions.

Flashcard 3: If $p = 0.80$ under Hardy–Weinberg, what are the expected frequencies of $AA$ and $aa$?

Answer: $AA: p^2 = 0.64$; $aa: q^2 = 0.04$. Homozygous frequencies are squares of their respective allele frequencies in equilibrium.

Flashcard 4: Identify the formula for allele frequency $p$ from genotype counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$.

Answer: $p = \frac{2N_{AA}+N_{Aa}}{2(N_{AA}+N_{Aa}+N_{aa})}$. This formula counts $A$ alleles from homozygotes and heterozygotes, divided by total alleles in the population.

Flashcard 5: Identify the formula for allele frequency $q$ from genotype counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$.

Answer: $q = \frac{2N_{aa}+N_{Aa}}{2(N_{AA}+N_{Aa}+N_{aa})}$. This formula tallies $a$ alleles from homozygotes and heterozygotes, normalized by total alleles.

Flashcard 6: Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of random mating?

Answer: Nonrandom mating (assortative mating or inbreeding). Nonrandom mating disrupts random union of gametes, altering genotype frequencies from Hardy-Weinberg expectations.

Flashcard 7: Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no migration?

Answer: Gene flow. Gene flow introduces alleles from other populations, changing allele frequencies and violating isolation.

Flashcard 8: Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no mutation?

Answer: Mutation. Mutation alters allele sequences, directly changing allele frequencies in the population.

Flashcard 9: Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no selection?

Answer: Natural selection. Natural selection favors certain genotypes, shifting allele frequencies away from equilibrium.

Flashcard 10: If observed genotype frequencies differ significantly from $p^2:2pq:q^2$, what is the correct conclusion?

Answer: The population is not in Hardy–Weinberg equilibrium at that locus. Deviation from expected ratios indicates violation of one or more Hardy-Weinberg assumptions at the locus.

Flashcard 11: If a recessive allele is lethal in homozygotes, which genotype is selected against most strongly?

Answer: $aa$. Homozygous recessives express the lethal phenotype, experiencing complete selection against viability.

Flashcard 12: In a bottleneck event, which population genetics force is primarily responsible for allele frequency changes?

Answer: Genetic drift. Bottlenecks reduce population size, amplifying random sampling effects of genetic drift on allele frequencies.

Flashcard 13: Which genotype frequency equals the heterozygote frequency under Hardy–Weinberg?

Answer: $2pq$. Heterozygote frequency is calculated as twice the product of allele frequencies, accounting for both allele combinations.

Flashcard 14: Which evolutionary mechanism most directly violates the Hardy–Weinberg assumption of a very large population?

Answer: Genetic drift (including founder and bottleneck effects). Genetic drift causes random allele frequency fluctuations in finite populations, especially small ones.

Flashcard 15: Under inbreeding (nonrandom mating), which Hardy–Weinberg genotype class typically decreases relative to expectation?

Answer: Heterozygotes decrease (homozygotes increase). Inbreeding increases homozygosity by promoting mating between relatives, reducing heterozygote proportions.

Flashcard 16: What do $p$ and $q$ represent in Hardy–Weinberg for a two-allele locus?

Answer: $p$ and $q$ are allele frequencies (e.g., $A$ and $a$) in the population. In Hardy-Weinberg, $p$ and $q$ denote the proportions of dominant and recessive alleles, respectively, summing to 1.

Flashcard 17: State the Hardy–Weinberg allele frequency equation for two alleles.

Answer: $p + q = 1$. This equation ensures the sum of allele frequencies for two alleles at a locus equals unity in Hardy-Weinberg equilibrium.

Flashcard 18: State the Hardy–Weinberg genotype frequency equation for two alleles.

Answer: $p^2 + 2pq + q^2 = 1$. This equation describes the expected distribution of genotypes in a population under Hardy-Weinberg equilibrium for a locus with two alleles.

Flashcard 19: What are the five assumptions required for Hardy–Weinberg equilibrium to hold?

Answer: Large population, random mating, no mutation, no migration, no selection. These assumptions prevent evolutionary forces from altering allele frequencies, maintaining equilibrium across generations.

Flashcard 20: If a population is in Hardy–Weinberg equilibrium, how do allele frequencies change over generations?

Answer: They remain constant across generations. Under Hardy-Weinberg equilibrium, allele frequencies are stable due to the absence of evolutionary pressures.

Flashcard 21: What is the allele frequency of $A$ if genotype frequencies are $f(AA)=0.36$, $f(Aa)=0.48$, $f(aa)=0.16$?

Answer: $p = 0.60$. Allele frequency $p$ is calculated as the proportion of $A$ alleles from genotype frequencies: $p = f(AA) + 0.5 f(Aa)$.

Flashcard 22: If $q = 0.20$ under Hardy–Weinberg, what is the expected frequency of affected recessive genotype $aa$?

Answer: $q^2 = 0.04$. For a recessive genotype, the frequency is the square of the recessive allele frequency under equilibrium.

Flashcard 23: If $q^2 = 0.09$ for an autosomal recessive disease in Hardy–Weinberg, what is $q$?

Answer: $q = 0.30$. The recessive allele frequency $q$ is the square root of the homozygous recessive genotype frequency in equilibrium.

Flashcard 24: If $q^2 = 0.09$ in Hardy–Weinberg, what is the carrier frequency $2pq$?

Answer: $2pq = 0.42$. Carrier frequency is $2pq$, where $p = 1 - q$ and $q = \sqrt{q^2}$, assuming equilibrium.