Molecular basis of inheritance practice · Hard

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because DNA fingerprinting identifies differences in repetitive DNA sequences, which exhibit high polymorphism.

Option D (0.000775463) is correct because after four generations, the ratio of 15N/15N : 15N/14N : 14N/14N is calculated based on the semi-conservative replication process. There are 2 molecules with 15N/14N and the rest are 14N/14N. The total number of molecules is 2⁴ = 16. So the ratio is 0:2:14, which simplifies to 0:1/8:7/8. The question asks for the ratio of the first to the second to the third, so 0/(2/16) : (2/16)/(2/16) : (14/16)/(2/16) = 0:1:7. If we divide each by the total (16), we get 0:0.125:0.875. If we divide each by 16², we get 0: 2/256: 224/256. The ratio is 0:2:224.

Option B (Ribosomal RNA does not act as a catalyst for peptide bond formation) is correct because ribosomal RNA (rRNA) acts as a ribozyme and catalyzes peptide bond formation during translation.

Option C (A site) is correct because the aminoacyl-tRNA (charged tRNA) initially binds to the A site of the ribosome during translation.

Option A is correct because RNA is not the primary genetic material due to its instability, which leads to a higher mutation rate.

Option A (GGU and CUU) is correct because GGU codes for glycine and CUU codes for leucine.

Option B (mRNA) is correct because mRNA carries the genetic instructions for polypeptide chain formation from DNA to the ribosome.

Option A (Redundancy of genetic code) is correct because the redundancy of the genetic code means that more than one codon can specify the same amino acid.

Option D is correct because Chargaff's rule applies to double-stranded DNA, not RNA, and RNA is typically single-stranded.

Option A (AUG) is correct because AUG is the start codon that initiates translation and codes for methionine.

Option C ((A) is true but (R) is false) is correct because the gene I codes for the repressor, but the y-gene codes for permease, which increases permeability of the cell to $\beta$-galactosidase, not $\beta$-galactosidase itself.

Option D (Only nucleic acids enter the cell) is correct because the Hershey-Chase experiment demonstrated that only the DNA of the bacteriophage enters the E. coli cell during infection.

Option D (z, y, a genes will not be translated) is correct because if the repressor cannot bind the inducer (lactose), it will remain bound to the operator, preventing transcription and therefore translation of the structural genes.

Option A is incorrect. In the formation of a polynucleotide chain using DNA, there are 4 different types of nucleotides that participate: Adenine, Guanine, Cytosine, and Thymine.

Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because the uptake of DNA during transformation requires specific enzymatic machinery and energy expenditure, making the reason a valid explanation for the assertion.

Option C (Adenylate, 3'-end) is correct because during tailing, adenylate residues (poly(A) tail) are added to the 3'-end of hnRNA.

Option A is correct because the Meselson and Stahl experiment showed that DNA replication is semi-conservative, resulting in 50% hybrid (radioactive and non-radioactive) DNA molecules after one generation in a non-radioactive medium.

Option B ((a) & (b)) is correct because allolactose binds to and inactivates the repressor, and in the absence of lactose, the repressor binds to the operator region.

Option A is correct because the repressor protein attaches to the operator region in the lac operon.

Option B (G – 17%, A – 33%, T – 33%) is correct because in double-stranded DNA, G=C and A=T. Since Cytosine is 17%, Guanine is also 17%. The remaining percentage is split equally between Adenine and Thymine.

Option A is correct. A 10 bp ds DNA molecule has 20 nucleotides. Each nucleotide is linked by a phosphodiester bond, except for the terminal nucleotides on each strand, so there are 18 phosphodiester bonds.

Option B (Termination gene message for polypeptide synthesis) is correct because nonsense codons (stop codons) signal the termination of translation.

Option C (CGU) is correct because the mRNA codon is GCA (complement of CGT), and the tRNA anticodon is CGU (complement of GCA, with T replaced by U).

Option D (Transcription of tRNA, 5S rRNA and snRNA) is correct because RNA polymerase III is responsible for transcribing tRNA, 5S rRNA, and snRNA genes in eukaryotes.

Option C (Only (a) and (c)) is correct because less than 2% of the genome codes for protein and chromosome Y has 231 genes are correct statements about the human genome project.

Option C ($\mathrm{S}^{35}$ in capsid of bacteriophage) is correct because Hershey and Chase used radioactive sulfur to label the protein capsid of the bacteriophage, and this radioactivity was found in the medium after infection.

Option B (tRNA) is correct because tRNA is the smallest RNA molecule, typically around 75-95 nucleotides long.

Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because while both statements are true, Sanger sequencing doesn't explain why BAC and YAC are used.