• Hardy-Weinberg Equilibrium Genotype Frequencies:
• For a diallelic gene with allele frequencies $p$ (allele $A$) and $q$ (allele $a$) such that $p + q = 1$:
• Heterozygote frequency: $f(Aa) = 2pq$
• Recessive homozygote frequency: $f(aa) = q^2$
• Given condition: The heterozygous genotype is twice as frequent as the homozygous recessive genotype:

• Dividing both sides by $2q$ (for $q \neq 0$):

• Since $p + q = 1$ and $p = q$:

• Conclusion: The required allelic frequencies are p = 0.5; q = 0.5 (Option 4).

Correct option: (4) p=05;q=05

• Correlation vs. Independence in Standard Normal Variables:
• Given $X \sim N(0, 1)$ and $Y = X^2$:
• The Pearson correlation coefficient is given by:

• For a standard normal distribution:
• $E[X] = 0$
• $E[XY] = E[X \cdot X^2] = E[X^3]$
• Since the standard normal distribution is symmetric about 0, all its odd central moments are zero:

• Therefore, the covariance is:

• Consequently, $\rho(X, Y) = 0$.
• Key Insight: A correlation coefficient of 0 indicates absence of linear association, even though $X$ and $Y$ have a perfect non-linear functional dependence ($Y = X^2$) (Option 2).

Correct option: (2) 0

• ANOVA Relationship in Completely Randomized Design (CRD):
• Given parameters:
• Degrees of freedom for Treatment Sum of Squares ($df_{\text{treatment}}$) = 4
• Mean Sum of Squares due to Error ($MSE$) = 2
• Observed $F$-ratio ($F_{\text{cal}}$) = 12.5
• In ANOVA, the $F$-statistic is defined as:

• Rearranging gives the Treatment Mean Sum of Squares ($MST$):

• The Treatment Sum of Squares ($SST$) is the product of $MST$ and its degrees of freedom:

• Conclusion: The Treatment Sum of Squares is 100 (Option 4).

Correct option: (4) 100

• Orthogonal Latin Squares (Graeco-Latin Squares):
• Latin square of order $n$ is an $n \times n$ square array containing $n$ distinct symbols such that each symbol appears exactly once in each row and once in each column.
• Orthogonality: Two Latin squares of the same order $n$ are said to be orthogonal if, when one square is superimposed over the other, every ordered pair of symbols $(A_i, B_j)$ occurs exactly once across all $n^2$ cells.
• pair of orthogonal Latin squares forms a Graeco-Latin square, which allows simultaneous elimination of two extraneous sources of variation in experimental design.
• Other Terms:
• Conjugate Latin square: Formed by transposing rows, columns, or symbols.
• Reduced (Standardized) Latin square: One where both the first row and first column are arranged in natural order.
• Conclusion: The two squares are Orthogonal (Option 3).

Correct option: (3) Orthogonal