Molecular basis of inheritance practice · Medium

Option D is correct because splicing is the process of removing introns (non-coding regions) and joining exons (coding regions) in pre-mRNA to form mature mRNA.

Option D (a and c) is correct because rRNA is transcribed by RNA polymerase I and tRNA is transcribed by RNA polymerase III in eukaryotes.

Option B is correct because lactose acts as the inducer of the lac operon.

Option C is correct because tRNA constitutes about 10-20% of the total cellular RNA, playing a crucial role in carrying amino acids to the ribosome during protein synthesis.

Option A is correct because the coding strand has the same sequence as the mRNA (except T is replaced by U) and does not directly code for amino acids; the template strand is used to create the mRNA.

Option C (Topoisomerase) is correct because topoisomerases relieve the torsional stress caused by DNA unwinding during replication by cutting and rejoining DNA strands.

Option B (Protein Capsid of Phage) is correct because sulfur is present in proteins but not in DNA, so ${}^{35}S$ would be incorporated into the protein capsid of the bacteriophage.

Option D (DNA) is correct because Okazaki fragments are short DNA fragments synthesized on the lagging strand during DNA replication.

Option D is correct because ethidium bromide intercalates into DNA and fluoresces bright orange when exposed to UV radiation.

Option B ($32\%$) is correct because in DNA, A=T and G=C. If Adenine is 18%, Thymine is also 18%. The remaining percentage is split equally between Guanine and Cytosine.

Option D is correct because sickle-cell anemia is an autosomal recessive mutation, not an allosomal (sex-linked) one.

Option D (If both Assertion and Reason are false) is correct because a change in the third position of a codon often does NOT change the amino acid due to wobble base pairing, and a codon is read by all three bases, but the third base has more flexibility.

Option B (254) is correct because the calculation involves determining the number of A-T and G-C base pairs and using the number of cytosine molecules to find the number of hydrogen bonds.

Option C ($4.0 \times 10^6 \mathrm{bp}$) is correct because dividing the length of the E.coli DNA by the distance between base pairs gives the number of base pairs, which is approximately 4 million.

Option C (UGG : Non degenerate) is correct because UGG codes only for tryptophan and is therefore non-degenerate.

Option D (Gene therapy) is correct because gene targeting involving gene amplification to treat disease is the definition of gene therapy.

Option D (Nirenberg and Mathaei) is correct because Nirenberg and Mathaei's experiments cracked the genetic code and demonstrated that it is a triplet code.

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 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 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 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 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 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 (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 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.