Molecular basis of inheritance practice · Medium

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.

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

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

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

Option C (3164.7 Million BP) is correct because the human genome contains approximately 3.1647 billion base pairs, which is 3164.7 million base pairs.

Option A (Mutated gene does not appear on a photographic film as the probe has no complementary with it) is correct because a mutated gene will not hybridize with a complementary probe, thus not appearing on the film.

Option D (Watson and Crick) is correct because they proposed the double helix structure of DNA and its copying mechanism.

Option D (${\mathrm{S}^{35}}$) is the correct answer because sulfur is not a component of DNA, while hydrogen, phosphorus, and nitrogen are.

Option A (A polypeptides of 24 amino acids will be formed) is correct because the UAA codon is a stop codon, and translation will terminate prematurely at the 25th codon position.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why the fragment lengths differ after restriction enzyme digestion.

Option C is correct because discontinuous synthesis occurs on the template with 5' to 3' polarity, not 3' to 5'.

Option D (Ochoa) is correct because Severo Ochoa received the Nobel Prize in 1959 for his work on the enzymatic synthesis of RNA.

Option D (All of these) is correct because the synthesis of DNA from RNA is described as reverse transcription, which is a reversal of the central dogma, and is also known as Teminism.

Option D ($3.3 \times {10^9}bp \times 3.4\\,{A^0}$) is correct because it represents the approximate length of the haploid content of human DNA.

Option A (met - Val - trp – Thr) is correct because the mRNA sequence 5' AUGGUGUGGACC 3' translates to Met-Val-Trp-Thr.

Option A (T:30; G:20; C:20) is correct because in DNA, A=T and G=C. If Adenine is 30%, Thymine is also 30%. The remaining 40% is split equally between Guanine and Cytosine.

Option D (Nucleotide of RNA) is correct because adenosine monophosphate (AMP) is a nucleotide containing adenine, ribose (sugar found in RNA), and a phosphate group.

Option B is correct because translation in eukaryotes occurs on the ribosomes.

Option D ((A) 5 (B) 7) is correct because under normal conditions, 5 amino acids are inserted, and with the deletion, 7 amino acids are inserted.

Option C (AUG) is correct because AUG is the start codon, while UAA, UAG, and UGA are stop codons.

Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion. Different polypeptide chains can be formed from a single mRNA due to alternative splicing and the presence of multiple start codons, not just terminator codons.

Option D (Watson and Crick) is correct because James Watson and Francis Crick received the Nobel Prize for their discovery of the structure of DNA.

Option B (Exons) is correct because exons are the coding regions of DNA that are translated into proteins.