Genetic cross analysis for dominant trait:

• Normal eyelids are recessive ($pp$), whereas ptosis is caused by dominant allele $P$.
• The affected woman's mother was normal ($pp$), so the woman must have inherited an allele $p$ from her mother; hence her genotype is heterozygous $Pp$.
• She marries a man with normal eyelids ($pp$). Mating: $Pp \times pp$.
• Expected offspring genotypes: $1/2\ Pp$ (affected with ptosis) and $1/2\ pp$ (normal eyelids). Therefore, $1/2$ of her children will have ptosis (Option 1).

Verification of Chargaff's rules for double-stranded DNA:

• Chargaff's rules state that in double-stranded DNA: $A = T$ and $G = C$.
• Since $A = T$, $A/T = 1$; and since $C = G$, $C/G = 1$. Thus, $A/T = C/G = 1$ (Statement C is correct).
• Similarly, $T/A = 1$ and $C/G = 1$, so $T/A = C/G$ (Statement D is correct).
• Total purines equal total pyrimidines: $A + G = C + T$ (Statement E is correct).
• Statements (A) and (B) are incorrect ($A+T \neq G+C$ in general, and $A \neq G$).

Comparison of eukaryotic and prokaryotic mRNA modifications:

• Statement (I) is correct: Eukaryotic mRNAs are co-transcriptionally modified with a 7-methylguanosine cap at their $5'$ terminus ($\text{m}^7\text{G}$ cap), which is absent in prokaryotes (which retain a $5'$ triphosphate).
• Statement (II) is correct: Eukaryotic mature mRNAs receive a post-transcriptional $3'$ polyadenylate tail ($\sim 200\text{ A}$ residues), whereas prokaryotic mRNAs typically lack stable poly-A tails (transient polyadenylation in bacteria targets RNA for degradation).
• Both statements are fundamental and correct.
• Therefore, option 1 (Both Statement (I) and Statement (II) are correct) is the correct answer.