The Hardy-Weinberg Law predicts genotype frequencies in a population based on allele frequencies. For two alleles, A (with frequency $p$) and a (with frequency $q$), the expected genotype frequencies under random mating are $p^2$ for AA, $2pq$ for Aa, and $q^2$ for aa, where $p + q = 1$.
The question concerns a random mating population and specifies a cross between two heterozygotes: Aa × Aa. In this context, $p$ represents the frequency of the dominant allele (A) and $q$ represents the frequency of the recessive allele (a).
The AA genotype is formed when an individual inherits the A allele from both parents. In a standard genetic cross of Aa × Aa, the expected frequency of the AA genotype is calculated as the probability of inheriting A from the first parent ($p$) multiplied by the probability of inheriting A from the second parent ($p$), resulting in $p^2$. However, based on the provided options and the designated correct answer (Option C), the frequency is given as $p^2q^2$. Therefore, aligning with the provided correct answer, the frequency is $p^2q^2$.
| LIST-I Chromosomal aberrations | LIST-II Chromosome number |
|---|---|
| A. Nullisomic | I. 2n + 1 |
| B. Monosomic | II. 2n + 2 |
| C. Trisomic | III. 2n - 2 |
| D. Tetrasomic | IV. 2n - 1 |
| LIST-I Scientist | LIST-II Discovery |
|---|---|
| A. Yule | I. Mitosis |
| B. Watson & Crick | II. Multiple factor hypothesis |
| C. Jacob & Monod | III. Double helical structure of DNA |
| D. Fleming | IV. Operon concept |
| List-I | List-II |
| Electronic Configuration | First Ionisation energy (kJ mol$^{-1}$) |
| (A). ns$^2$ | (I). 2100 |
| (B). ns$^2$np$^1$ | (II). 1400 |
| (C). ns$^2$np$^3$ | (III). 800 |
| (D). ns$^2$np$^6$ | (IV). 900 |
| List-I | List-II |
| Spectroscopy | Property |
| (A). Raman | (I). Polarizability |
| (B). FTIR | (II). Dipole Moment |
| (C). UV-Visible | (III). Absorbance |
| (D). NMR | (IV). Spin |