• 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

• Probability with Combinations:
• Composition of eligible officers:
• Production department: 3
• Purchase department: 4
• Sales department: 2
• Chartered Accountant (CA): 1
• Total pool size: $N = 3 + 4 + 2 + 1 = 10$ individuals
• Committee size: $k = 4$ individuals
• Direct Symmetry Method:
• Since each of the 10 individuals has an equal chance of being chosen in a committee of 4, the probability that any specific individual (the CA) is included is simply:

• Combinatorial Verification:
• Total number of possible committees: $\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
• Favorable committees (including the 1 CA, remaining 3 chosen from the other 9): $\binom{1}{1} \times \binom{9}{3} = 1 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$
• Probability: $P = \frac{84}{210} = \frac{2}{5}$
• Conclusion: The probability is 2/5 (Option 4).

Correct option: (4) 215