Molecular basis of inheritance practice · Mixed difficulty

Option C (It can decode AUG but not AUC, AUU and AUA) is correct because according to the wobble hypothesis, U in the anticodon can pair with A or G in the codon. Therefore, UAC can only decode AUG.

Option A (It binds with an amino acid at its. 3’ end) is correct because tRNA binds to an amino acid at its 3' end, which is crucial for its role in translation.

Option A is correct because continuous replication occurs on the template strand with 3' to 5' polarity, allowing for 5' to 3' synthesis of the new strand.

Option C (2 amino acids) is correct because the original sequence codes for 4 amino acids, and the insertion changes the reading frame such that the sequence now codes for 2 amino acids before encountering a stop codon.

Option D is correct because Jacob and Monod are known for their work on the Lac operon and mRNA.

Option B (78.4Å) is the closest to the correct answer because the distance between two base pairs in DNA is approximately 3.4Å and 23 * 3.4 is 78.2.

Option C ((A) is true but (R) is false) is correct because the presence of introns is considered an ancient feature, but the split gene arrangement is not necessarily an advanced feature.

Option D is correct. The ratio of A+T/G+C in prokaryotic cells can vary and is not necessarily greater than 1, less than 1, or equal to 1. It depends on the specific organism.

Option C (2.2 meters) is correct because it is the approximate length calculated by multiplying the number of base pairs by the distance between them.

Option B (There are no tRNAs for stop codons) is correct because stop codons are recognized by release factors, not tRNAs.

Option B (Guanine) is correct because guanine is a purine, which has a 9-membered double-ring structure.

Option C (4) is correct because the mRNA sequence is read in codons (groups of three nucleotides), and the sequence AUGCUGAGGUAAUGA has four codons before the stop codon (UAA, UGA, or UAG). AUG-CUG-AGG-UAA-UGA. The stop codon is UAA, so there are 3 codons that code for amino acids.

Option B (b → c → a → d) is correct because gene regulation in eukaryotes occurs at the transcriptional level, processing level, mRNA transport level, and translational level, in that order.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because the lac operator specifically interacts with the lac repressor, and this specificity is a general feature of operons.

Option B (Eight histones of four different types) is correct because a histone octamer consists of two copies each of the four core histone proteins: H2A, H2B, H3, and H4.

Option B (Ribozymes) is correct because ribozymes are RNA molecules that act as enzymes, and therefore are non-proteinaceous.

Option A (RNA polymerase) is correct because the sigma factor is a subunit of RNA polymerase that helps in promoter recognition during transcription.

Option A (II and III only) is correct because reverse transcriptase and DNA polymerase are enzymes involved in DNA synthesis.

Option D (Poly A polymerase) is correct because Poly A polymerase is involved in adding a poly(A) tail to mRNA during RNA processing, not DNA replication.

Option C is correct. Antiparallel strands in DNA mean the phosphate groups at the start of the two strands are in opposite positions (poles), indicating opposite 5' to 3' directions.

Option A (Replication of DNA) is correct because Okazaki fragments are formed during the discontinuous synthesis of the lagging strand in DNA replication.

Option A is correct because splicing is a process unique to eukaryotes and supports the idea that RNA played a more central role in early life forms.

Option D (All of these) is correct because Topoisomerase, Helicase, and RNA polymerase are all involved in DNA synthesis.

Option B (Glycosidic bond) is correct because a glycosidic bond connects the nitrogenous base to the deoxyribose sugar in a DNA nucleotide.

Option D (60 cells) is correct because after 60 minutes (3 generations), the number of cells is 10 * 2³ = 80. Only the original 20 strands will have any N15, so 60 cells will be free of N15.

Option B is correct because the Meselson and Stahl experiment on E.coli provided evidence for the semi-conservative nature of DNA replication.

Option C ($30\%$) is correct because in DNA, A=T and G=C. If Thymine is 20%, Adenine is also 20%. The remaining 60% is split equally between Guanine and Cytosine.

Option B is correct because alternating C and U residues in the coding strand (CU CU CU...) would correspond to alternating G and A residues in the template strand (GA GA GA...). This would result in codons CUC and UCU, which code for leucine and serine, respectively.

Option C (Both (1) and (2)) is correct because mutations that alter nucleotide sequences within a gene can be either frameshift mutations or base pair substitutions.

Option B is correct because both the assertion and reason are true, but the reason doesn't directly explain why tRNA is an adapter molecule.