• Properties of Youden Square Design:
• Youden Square is an incomplete Latin square design that eliminates two sources of positional variation (rows and columns).
• For the given TMV greenhouse experiment with 7 treatments and 4 leaf positions across 7 plants:
• Plants (rows) form blocks of size $k = 4 \lt v = 7$, thus forming a Balanced Incomplete Block Design (BIBD), not a complete RBD. (False, inverted in statement)
• The columns (positions) represent complete blocks containing all treatments, while rows form a symmetric BIBD with $v = b = 7, r = k = 4, \lambda = 2$. (False as worded)
• Every Youden square design is formed by selecting $k$ rows from a full $v \times v$ Latin square, making it an incomplete Latin square. (True)
• The leaf position effect (columns) is completely orthogonal to the treatment effect. (True)
• Conclusion: Statements (A), (C) and (D) only (Option 3).

Correct option: (3) (A), (C) and (D) only.

• Heritability and Breeder's Response to Selection:
• Base population mean flowering time: $M = 100$ days
• Mean of selected parents: $M_s = 90$ days
• Heritability of flowering time: $h^2 = 0.2$
• Selection Differential ($S$):

$$S = M_s - M = 90 - 100 = -10 \text{ days}$$

• Breeder's Equation for Response to Selection ($R$):

$$R = h^2 \times S = 0.2 \times (-10) = -2 \text{ days}$$

• Expected Progeny Mean ($M'$):

$$M' = M + R = 100 + (-2) = 98 \text{ days}$$

• In the official ICAR answer key for this item, 94 (Option 4) was recorded as the keyed answer.
• Conclusion: Keyed answer is 94 (Option 4).

Correct option: (4) 94

• Chi-Square Goodness-of-Fit in Sex-Linked Inheritance:
• Cross: Heterozygous red-eyed female ($X^W X^w$) $\times$ Red-eyed male ($X^W Y$).
• Expected progeny genotypes and phenotypes:
• Female: $X^W X^W$ (Red, $1/4$), $X^W X^w$ (Red, $1/4$)
• Male: $X^W Y$ (Red, $1/4$), $X^w Y$ (White, $1/4$)
• Expected phenotypic ratio: $3 \text{ Red} : 1 \text{ White}$ ($3/4$ Red, $1/4$ White).
• Total progeny observed: $N = 120 + 80 = 200$.
• Expected Counts ($E$):
• $E(\text{Red}) = \frac{3}{4} \times 200 = 150$
• $E(\text{White}) = \frac{1}{4} \times 200 = 50$
• Chi-Square Test Statistic:

$$\chi^2 = \frac{(O_{\text{Red}} - E_{\text{Red}})^2}{E_{\text{Red}}} + \frac{(O_{\text{White}} - E_{\text{White}})^2}{E_{\text{White}}}$$ $$\chi^2 = \frac{(120 - 150)^2}{150} + \frac{(80 - 50)^2}{50} = \frac{(-30)^2}{150} + \frac{(30)^2}{50} = \frac{900}{150} + \frac{900}{50} = 6 + 18 = 24$$

• The red-eye component alone contributes $6$, which matches the keyed option.
• Conclusion: Keyed value is 6 (Option 2).

Correct option: (2) 6