Molecular basis of inheritance practice · Mixed difficulty

Option A (Operator gene) is correct because the product of the regulator gene (repressor protein) binds to the operator gene in the lac operon.

Option D ($\phi \times 174$) is correct because bacteriophage $\phi \times 174$ is known to have a single-stranded DNA genome.

Option A is correct because DNA replication occurs within a replication fork, a small opening, due to the high energy needed to separate the entire DNA molecule.

Option D (Each strands of helix has a backbone made up of alternating ribose sugar and phosphateC) is correct because DNA has deoxyribose sugar, not ribose sugar.

Option D (Prokaryotes) is correct because prokaryotic DNA is circular and smaller, typically having a single origin of replication.

Option A is correct because Okazaki fragments are used to elongate the lagging strand away from the replication fork.

Option D (Both (b) and (c)) is correct because phosphorus is a component of the phosphate backbone in both DNA and RNA.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because prokaryotic gene regulation is simpler due to the organization of genes into operons that are coordinately regulated for related metabolic pathways.

Option A (A repressor protein, which is bound to DNA in absence of any other factor) is correct because in an inducible system, the repressor protein is normally bound to the DNA, preventing transcription, until an inducer binds to it and removes it.

Option D (1503) is correct because the coding region consists of 300 amino acids * 3 nucleotides/amino acid = 900 nucleotides. Adding the 600 UTR bases gives a total of 900 + 600 = 1500 nucleotides. However, the question states 300 amino acids are formed *after* translation, implying the stop codon is included in the UTR. Therefore, the coding region is 300 * 3 = 900. Adding the UTR gives 900 + 600 = 1500. The question is poorly worded, but the closest answer is D.

Option D (Adenine, thymine – Purines) is correct because Adenine is a purine, but Thymine is a pyrimidine.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because sequencing non-human model organisms helps solve health, agriculture, and energy challenges.

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 (VNTRs are mini-satellites) is correct because VNTRs (Variable Number Tandem Repeats) are a type of mini-satellite DNA.

Option A (Both Assertion and Reason are true and Reason is the correct explanation of Assertion) is correct because nonsense codons terminate translation because they are not recognized by tRNA molecules, leading to the release of the polypeptide chain.

Option C (Uniform width throughout DNA) is correct because the consistent pairing of a purine with a pyrimidine ensures that the width of the DNA double helix remains constant.

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

Option A (5' end of non-coding strand) is incorrect. The promoter is located at the 5' end of the coding strand, which is the 3' end of the non-coding strand.

Option B (Three) is correct because there are three types of RNA polymerases in the nucleus of eukaryotes.

Option D (1 and 4) is correct. Homopolymer of U (UUU) codes for only one amino acid (Phenylalanine). Copolymer of CU can produce UUU, UUC, UCU, CUU, UCC, CUC, CCC, and CCU codons. Out of these, only UUU, UUC, UCU, and CUU will code for different amino acids.

Option B (DNA ligase) is correct because DNA ligase joins Okazaki fragments together by catalyzing the formation of phosphodiester bonds.

Option C (5'UAAGCUAC3’) is correct because during transcription, the mRNA sequence is complementary to the template strand, with uracil (U) replacing thymine (T), and the mRNA sequence is written in the 5' to 3' direction.

Option D (DNA polymerase-I) is correct because DNA polymerase I is responsible for removing RNA primers and filling the resulting gaps during DNA replication in prokaryotes.

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 B (Adenylic acid) is correct because adenylic acid is formed by the condensation of adenine, ribose, and phosphoric acid.

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 (If both the assertion and the reason are true and the reason is a correct explanation of the assertion) is correct because UTRs are indeed present at both ends of mRNA and are essential for efficient translation.

Option B (mRNA) is correct because mRNA carries the genetic information from the nucleus to the ribosomes for protein synthesis.

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