Molecular basis of inheritance practice · Medium

Option D (All of these) is correct because X-rays, UV rays, and γ-rays are all known mutagens that can induce mutations.

Option A (Different density and do not resemble with their parent DNA) is correct because, after one generation, each DNA molecule consists of one heavy $^{15}N$ strand and one light $^{14}N$ strand, resulting in a hybrid density and not resembling the original fully $^{15}N$ DNA.

Option B is correct because both statements are true, but the ability of some RNAs to act as catalysts (ribozymes) doesn't explain why DNA is chemically less reactive.

Option D is correct because the ratio of (G+C)/(A+T) is generally constant for a given species, reflecting the stability of the DNA molecule due to the specific base pairing.

Option D (Radioactive sulphur in bacterial cells) is correct because if DNA contained sulfur instead of phosphorus, the radioactive sulfur would be found in the bacterial cells after infection.

Option D (It is the largest of the RNAs) is correct because tRNA is the smallest of the RNAs, not the largest.

Option B (Purine/pyrimidine+sugar) is correct because a nucleoside consists of a nitrogenous base (purine or pyrimidine) attached to a sugar molecule (ribose or deoxyribose).

Option D (5’AUGUACCGUUUAUAGGUAAGU3’) is correct because DNA dependent RNA polymerase reads the DNA template in the 3' to 5' direction and synthesizes the RNA transcript in the 5' to 3' direction, using complementary base pairing (A with U, T with A, C with G, and G with C).

Option A (0.4) is correct because in double-stranded DNA, the amount of guanine (G) equals cytosine (C), and adenine (A) equals thymine (T). If G is 30%, then C is 30%, leaving 40% for A and T combined, so A = T = 20%. Thus, the percentage of adenine and thymine base pairs is 0.4.

Option D (94 amino acids) is incorrect. The correct answer is 93 amino acids because the mRNA consists of 282 nucleotides, and each codon consists of 3 nucleotides. Therefore, 282 / 3 = 94 codons. However, one codon is a stop codon, so 94 - 1 = 93 amino acids.

Option A is correct because release factors recognize stop codons in the mRNA and trigger the release of the polypeptide chain from the ribosome, terminating translation.

Option B (Leading strand) is correct because the leading strand is synthesized continuously in the 5' to 3' direction during DNA replication.

Option C (An intestinal cell possesses a full set of genes) is correct because the ability of the intestinal cell nucleus to direct the development of a normal tadpole demonstrates that it contains all the genetic information necessary for development.

Option C (Genetic code is nearly universal) is correct because the near-universality of the genetic code allows bacteria to produce human proteins.

Option A (Mutations) is correct because polymorphism arises primarily due to mutations in the DNA sequence.

Option B ($20{A^0}$) is correct because the width of the DNA molecule is approximately 20 Angstroms.

Option B (Both statements are incorrect) is correct because the structural genes of the Lac operon are transcribed only when lactose is available and glucose is absent or at low levels, and the mRNA of the Lac Operon has multiple terminating codons, one for each structural gene.

Option A (A 24 amino acid polypeptide is formed) is correct because the mutation to UAA introduces a stop codon at position 25, terminating translation prematurely.

Option C (Lower density than parental types) is correct because after 40 minutes (two generations), all DNA molecules will contain at least one strand of the normal 14N, resulting in a lower overall density compared to the original 15N DNA.

Option A (Promoter) is correct because the promoter is the site on the DNA molecule where RNA polymerase binds during transcription.

Option A (Two) is correct because ribosomes have two binding sites for tRNA molecules: the A site and the P site.

Option A is correct because DNA synthesis requires RNA primers to initiate the process at the origin of replication.

Option A (0%) is correct because after two generations in N14 medium, all DNA molecules will contain at least one strand of N14, thus no heavy DNA will remain.

Option D (UACAAAUGACCA) is correct because the mRNA sequence is the same as the sense strand (coding strand) with Thymine (T) replaced by Uracil (U).

Option A is correct because gamma rays are a form of ionizing radiation that can induce mutations in plant cells by damaging DNA.

Option C is correct because the Y gene transcribes the permease protein in the lac operon.

Option B (siRNA) is correct because siRNA is involved in gene silencing and RNA interference, but it is not directly required for the synthesis of proteins from mRNA.

Option B (0.35) is correct because in DNA, A=T and G=C. If Adenine is 15%, Thymine is also 15%. The remaining 70% is split equally between Guanine and Cytosine, resulting in 35% each.

Option B (RNA polymerase) is correct because RNA polymerase is responsible for unwinding the DNA helix during transcription.

Option D (Four) is correct because the core nucleosome is formed by two copies each of four histone proteins: H2A, H2B, H3, and H4.