• Central Dogma of Molecular Biology:
• Articulated by Francis Crick (1958, 1970), the central dogma describes the sequential flow of detailed genetic information within a biological system:

$$\text{DNA} \xrightarrow{\text{Transcription}} \text{RNA} \xrightarrow{\text{Translation}} \text{Protein}$$

• Genetic information stored in DNA is transcribed into messenger RNA (mRNA) by RNA polymerase, which is subsequently translated into polypeptides/proteins by ribosomes.
• Conclusion: DNA is transcribed into RNA, which is translated into proteins (Option 1).

Correct option: (1) DNAis transcribed into RNA, which is translated into proteins

• Definition of Homology in Computational Biology:
• In bioinformatics and evolutionary genetics, homology strictly means sharing common evolutionary ancestry between genes, proteins, or morphological structures.
• Homology is a qualitative attribute (sequences are either homologous or not homologous; they cannot be '80% homologous').
• Homologous genes are subdivided into:
• Orthologs: Homologs separated by a speciation event (often preserving function).
• Paralogs: Homologs arising from gene duplication events within a genome.
• Conclusion: Homology refers to Evolutionarily related genes or proteins (Option 3).

Correct option: (3) Evolutionarily related genes or proteins

• Genotypic Frequency in Dihybrid Selfing:
• Self-pollination cross: $AaBb \times AaBb$ with unlinked, independently assorting loci.
• At locus A ($Aa \times Aa$):

$$P(AA) = 1/4, \quad P(Aa) = 1/2, \quad P(aa) = 1/4$$

• At locus B ($Bb \times Bb$):

$$P(BB) = 1/4, \quad P(Bb) = 1/2, \quad P(bb) = 1/4$$

• By independent assortment, the joint probability of obtaining the homozygous dominant genotype $AABB$ is the product of the individual marginal probabilities:

$$P(AABB) = P(AA) \times P(BB) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$$

• Conclusion: The probability of getting $AABB$ genotype is 1/16 (Option 1).

Correct option: (1) 112

• 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