Population Genetics and Hardy–Weinberg (1C) - MCAT Biological and Biochemical Foundations of Living Systems
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Identify the expected genotype frequencies under Hardy–Weinberg in terms of $p$ and $q$.
Identify the expected genotype frequencies under Hardy–Weinberg in terms of $p$ and $q$.
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$AA: p^2$, $Aa: 2pq$, $aa: q^2$. These frequencies represent the expected proportions of homozygous dominant, heterozygous, and homozygous recessive genotypes under equilibrium.
$AA: p^2$, $Aa: 2pq$, $aa: q^2$. These frequencies represent the expected proportions of homozygous dominant, heterozygous, and homozygous recessive genotypes under equilibrium.
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If $p = 0.70$ in Hardy–Weinberg, what is the expected heterozygote frequency?
If $p = 0.70$ in Hardy–Weinberg, what is the expected heterozygote frequency?
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$2pq = 0.42$. Heterozygote frequency is $2pq$, with $q = 1 - p$ under Hardy-Weinberg conditions.
$2pq = 0.42$. Heterozygote frequency is $2pq$, with $q = 1 - p$ under Hardy-Weinberg conditions.
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If $p = 0.80$ under Hardy–Weinberg, what are the expected frequencies of $AA$ and $aa$?
If $p = 0.80$ under Hardy–Weinberg, what are the expected frequencies of $AA$ and $aa$?
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$AA: p^2 = 0.64$; $aa: q^2 = 0.04$. Homozygous frequencies are squares of their respective allele frequencies in equilibrium.
$AA: p^2 = 0.64$; $aa: q^2 = 0.04$. Homozygous frequencies are squares of their respective allele frequencies in equilibrium.
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Identify the formula for allele frequency $p$ from genotype counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$.
Identify the formula for allele frequency $p$ from genotype counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$.
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$p = \frac{2N_{AA}+N_{Aa}}{2(N_{AA}+N_{Aa}+N_{aa})}$. This formula counts $A$ alleles from homozygotes and heterozygotes, divided by total alleles in the population.
$p = \frac{2N_{AA}+N_{Aa}}{2(N_{AA}+N_{Aa}+N_{aa})}$. This formula counts $A$ alleles from homozygotes and heterozygotes, divided by total alleles in the population.
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Identify the formula for allele frequency $q$ from genotype counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$.
Identify the formula for allele frequency $q$ from genotype counts $N_{AA}$, $N_{Aa}$, and $N_{aa}$.
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$q = \frac{2N_{aa}+N_{Aa}}{2(N_{AA}+N_{Aa}+N_{aa})}$. This formula tallies $a$ alleles from homozygotes and heterozygotes, normalized by total alleles.
$q = \frac{2N_{aa}+N_{Aa}}{2(N_{AA}+N_{Aa}+N_{aa})}$. This formula tallies $a$ alleles from homozygotes and heterozygotes, normalized by total alleles.
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Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of random mating?
Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of random mating?
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Nonrandom mating (assortative mating or inbreeding). Nonrandom mating disrupts random union of gametes, altering genotype frequencies from Hardy-Weinberg expectations.
Nonrandom mating (assortative mating or inbreeding). Nonrandom mating disrupts random union of gametes, altering genotype frequencies from Hardy-Weinberg expectations.
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Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no migration?
Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no migration?
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Gene flow. Gene flow introduces alleles from other populations, changing allele frequencies and violating isolation.
Gene flow. Gene flow introduces alleles from other populations, changing allele frequencies and violating isolation.
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Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no mutation?
Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no mutation?
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Mutation. Mutation alters allele sequences, directly changing allele frequencies in the population.
Mutation. Mutation alters allele sequences, directly changing allele frequencies in the population.
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Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no selection?
Which evolutionary mechanism directly violates the Hardy–Weinberg assumption of no selection?
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Natural selection. Natural selection favors certain genotypes, shifting allele frequencies away from equilibrium.
Natural selection. Natural selection favors certain genotypes, shifting allele frequencies away from equilibrium.
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If observed genotype frequencies differ significantly from $p^2:2pq:q^2$, what is the correct conclusion?
If observed genotype frequencies differ significantly from $p^2:2pq:q^2$, what is the correct conclusion?
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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.
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.
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If a recessive allele is lethal in homozygotes, which genotype is selected against most strongly?
If a recessive allele is lethal in homozygotes, which genotype is selected against most strongly?
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$aa$. Homozygous recessives express the lethal phenotype, experiencing complete selection against viability.
$aa$. Homozygous recessives express the lethal phenotype, experiencing complete selection against viability.
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In a bottleneck event, which population genetics force is primarily responsible for allele frequency changes?
In a bottleneck event, which population genetics force is primarily responsible for allele frequency changes?
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Genetic drift. Bottlenecks reduce population size, amplifying random sampling effects of genetic drift on allele frequencies.
Genetic drift. Bottlenecks reduce population size, amplifying random sampling effects of genetic drift on allele frequencies.
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Which genotype frequency equals the heterozygote frequency under Hardy–Weinberg?
Which genotype frequency equals the heterozygote frequency under Hardy–Weinberg?
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$2pq$. Heterozygote frequency is calculated as twice the product of allele frequencies, accounting for both allele combinations.
$2pq$. Heterozygote frequency is calculated as twice the product of allele frequencies, accounting for both allele combinations.
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Which evolutionary mechanism most directly violates the Hardy–Weinberg assumption of a very large population?
Which evolutionary mechanism most directly violates the Hardy–Weinberg assumption of a very large population?
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Genetic drift (including founder and bottleneck effects). Genetic drift causes random allele frequency fluctuations in finite populations, especially small ones.
Genetic drift (including founder and bottleneck effects). Genetic drift causes random allele frequency fluctuations in finite populations, especially small ones.
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Under inbreeding (nonrandom mating), which Hardy–Weinberg genotype class typically decreases relative to expectation?
Under inbreeding (nonrandom mating), which Hardy–Weinberg genotype class typically decreases relative to expectation?
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Heterozygotes decrease (homozygotes increase). Inbreeding increases homozygosity by promoting mating between relatives, reducing heterozygote proportions.
Heterozygotes decrease (homozygotes increase). Inbreeding increases homozygosity by promoting mating between relatives, reducing heterozygote proportions.
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What do $p$ and $q$ represent in Hardy–Weinberg for a two-allele locus?
What do $p$ and $q$ represent in Hardy–Weinberg for a two-allele locus?
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$p$ 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.
$p$ 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.
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State the Hardy–Weinberg allele frequency equation for two alleles.
State the Hardy–Weinberg allele frequency equation for two alleles.
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$p + q = 1$. This equation ensures the sum of allele frequencies for two alleles at a locus equals unity in Hardy-Weinberg equilibrium.
$p + q = 1$. This equation ensures the sum of allele frequencies for two alleles at a locus equals unity in Hardy-Weinberg equilibrium.
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State the Hardy–Weinberg genotype frequency equation for two alleles.
State the Hardy–Weinberg genotype frequency equation for two alleles.
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$p^2 + 2pq + q^2 = 1$. This equation describes the expected distribution of genotypes in a population under Hardy-Weinberg equilibrium for a locus with two alleles.
$p^2 + 2pq + q^2 = 1$. This equation describes the expected distribution of genotypes in a population under Hardy-Weinberg equilibrium for a locus with two alleles.
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What are the five assumptions required for Hardy–Weinberg equilibrium to hold?
What are the five assumptions required for Hardy–Weinberg equilibrium to hold?
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Large population, random mating, no mutation, no migration, no selection. These assumptions prevent evolutionary forces from altering allele frequencies, maintaining equilibrium across generations.
Large population, random mating, no mutation, no migration, no selection. These assumptions prevent evolutionary forces from altering allele frequencies, maintaining equilibrium across generations.
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If a population is in Hardy–Weinberg equilibrium, how do allele frequencies change over generations?
If a population is in Hardy–Weinberg equilibrium, how do allele frequencies change over generations?
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They remain constant across generations. Under Hardy-Weinberg equilibrium, allele frequencies are stable due to the absence of evolutionary pressures.
They remain constant across generations. Under Hardy-Weinberg equilibrium, allele frequencies are stable due to the absence of evolutionary pressures.
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What is the allele frequency of $A$ if genotype frequencies are $f(AA)=0.36$, $f(Aa)=0.48$, $f(aa)=0.16$?
What is the allele frequency of $A$ if genotype frequencies are $f(AA)=0.36$, $f(Aa)=0.48$, $f(aa)=0.16$?
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$p = 0.60$. Allele frequency $p$ is calculated as the proportion of $A$ alleles from genotype frequencies: $p = f(AA) + 0.5 f(Aa)$.
$p = 0.60$. Allele frequency $p$ is calculated as the proportion of $A$ alleles from genotype frequencies: $p = f(AA) + 0.5 f(Aa)$.
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If $q = 0.20$ under Hardy–Weinberg, what is the expected frequency of affected recessive genotype $aa$?
If $q = 0.20$ under Hardy–Weinberg, what is the expected frequency of affected recessive genotype $aa$?
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$q^2 = 0.04$. For a recessive genotype, the frequency is the square of the recessive allele frequency under equilibrium.
$q^2 = 0.04$. For a recessive genotype, the frequency is the square of the recessive allele frequency under equilibrium.
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If $q^2 = 0.09$ for an autosomal recessive disease in Hardy–Weinberg, what is $q$?
If $q^2 = 0.09$ for an autosomal recessive disease in Hardy–Weinberg, what is $q$?
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$q = 0.30$. The recessive allele frequency $q$ is the square root of the homozygous recessive genotype frequency in equilibrium.
$q = 0.30$. The recessive allele frequency $q$ is the square root of the homozygous recessive genotype frequency in equilibrium.
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If $q^2 = 0.09$ in Hardy–Weinberg, what is the carrier frequency $2pq$?
If $q^2 = 0.09$ in Hardy–Weinberg, what is the carrier frequency $2pq$?
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$2pq = 0.42$. Carrier frequency is $2pq$, where $p = 1 - q$ and $q = \sqrt{q^2}$, assuming equilibrium.
$2pq = 0.42$. Carrier frequency is $2pq$, where $p = 1 - q$ and $q = \sqrt{q^2}$, assuming equilibrium.
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