At what allelic frequency is the heterozygous genotype (Aa)... | Agrigyan

• 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

• Testcross Phenotypes for a Polyhybrid Genotype:
• Genotype of the individual: $AaBbCcDd$ (heterozygous at $n = 4$ independently assorting gene loci).
• Testcross involves crossing with a homozygous recessive tester: $AaBbCcDd \times aabbccdd$.
• For each independent gene locus under complete dominance, the testcross produces 2 phenotypic classes:
• $Aa \times aa \to 1\,A\_ : 1\,aa$ (2 phenotypes)
• $Bb \times bb \to 1\,B\_ : 1\,bb$ (2 phenotypes)
• $Cc \times cc \to 1\,C\_ : 1\,cc$ (2 phenotypes)
• $Dd \times dd \to 1\,D\_ : 1\,dd$ (2 phenotypes)
• Total number of distinct phenotypic classes in the testcross progeny is:

• Conclusion: 16 distinct phenotypes appear in the testcross progeny (Option 1).

Correct option: (1) 16

• 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