A homozygous recessive genotype is when an individual carries two copies of the same recessive allele for a trait (written aa or q²). In Hardy-Weinberg problems, it's the only genotype you can count directly from phenotype, which lets you solve for the recessive allele frequency.
A homozygous recessive genotype is when an organism has two identical recessive alleles for a gene, written as aa. Because both copies are recessive, the recessive trait actually shows up in the phenotype. There's no dominant allele around to mask it.
This matters most in Topic 7.5, Hardy-Weinberg Equilibrium. The H-W equation, p² + 2pq + q² = 1, splits a population into three genotypes: homozygous dominant (p²), heterozygous (2pq), and homozygous recessive (q²). Here, q² is the frequency of the homozygous recessive genotype. It's the starting point for almost every H-W calculation because it's the only piece you can measure directly by looking at organisms.
This term lives in Unit 7: Natural Selection, specifically Topic 7.5, and supports learning objective AP Bio 7.5.A (describe the conditions under which allele and genotype frequencies will change in populations). Essential knowledge EK 7.5.A.2 says allele frequencies in a nonevolving population can be calculated from genotype frequencies — and the homozygous recessive genotype (q²) is your way in. Take the square root of q² and you get q, the recessive allele frequency. From there, p = 1 - q, and the whole population falls into place. Without being able to spot q² from phenotype, the H-W equation would be unsolvable on the exam.
Keep studying AP Biology Unit 7
Allele Frequencies (Unit 7)
The homozygous recessive genotype frequency (q²) is the bridge from what you can see to what you can't. Square-root q² to get q, the recessive allele frequency, then p = 1 - q gives you the dominant allele frequency.
Genotype Frequencies (Unit 7)
q² is one of three genotype frequencies in p² + 2pq + q² = 1. The catch: homozygous dominant (p²) and heterozygous (2pq) often look identical in phenotype, so q² is the only one you can count directly.
Recessive Allele (Unit 7)
A homozygous recessive genotype is just two recessive alleles together. One recessive allele paired with a dominant one (heterozygous) hides the recessive trait, so it takes two copies for the trait to appear.
Null Hypothesis (Unit 7)
Hardy-Weinberg is a model of a non-evolving population, so the q² you calculate from it is a baseline prediction. If the real q² differs from the predicted value, something (selection, drift, migration) is changing the population.
Expect H-W math. A classic MCQ gives you a homozygous recessive frequency and asks for the recessive allele frequency. If q² = 0.04, then q = √0.04 = 0.2. Another favorite tests why q² is special: it's the only genotype you can determine directly from phenotypic ratios, because both homozygous dominant and heterozygous individuals show the dominant phenotype. On free response, you may need to calculate allele frequencies from a given genotype count and then argue whether the population is in equilibrium. Always start with the recessive phenotype, set it equal to q², take the square root, and build out p and 2pq from there.
A recessive allele (a) is a single version of a gene. A homozygous recessive genotype (aa) is two of those alleles together in one individual. One is a single building block; the other is the pair. You need two recessive alleles for the recessive trait to actually show up in the phenotype.
A homozygous recessive genotype means two identical recessive alleles (aa), and it's represented by q² in the Hardy-Weinberg equation.
It's the only genotype you can count directly from phenotype, because every individual showing the recessive trait must be homozygous recessive.
Take the square root of q² to get q, the recessive allele frequency, then use p = 1 - q to find everything else.
Homozygous dominant (p²) and heterozygous (2pq) individuals both show the dominant phenotype, so you can't tell them apart by looking.
Hardy-Weinberg assumes a non-evolving population (large size, no migration, no mutation, random mating, no selection), so q² gives a baseline to test against real data.
It's an individual carrying two copies of the same recessive allele for a trait, written aa. In Hardy-Weinberg problems it equals q², the frequency you use to solve for the recessive allele frequency q.
Because the recessive trait only appears when there's no dominant allele to mask it. Homozygous dominant (p²) and heterozygous (2pq) individuals both display the dominant phenotype, so they look the same, but anyone showing the recessive trait must be aa.
The homozygous recessive frequency is q². Take its square root to get q. For example, if q² = 0.04, then q = 0.2, and p = 1 - 0.2 = 0.8.
No. A recessive allele is a single version of a gene (a). A homozygous recessive genotype is two recessive alleles paired up (aa). It takes two for the recessive trait to show.
q² is the frequency of the homozygous recessive genotype in the population. The full equation is p² + 2pq + q² = 1, where p² is homozygous dominant and 2pq is heterozygous.
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