Molecular basis of inheritance practice · Easy

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

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

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

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

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 A (AUG) is correct because AUG is the start codon, while the others are stop codons.

Option A (Kornberg) is correct because Arthur Kornberg discovered DNA polymerase in 1956.

Option A (chromosome 1) is correct because chromosome 1 was the last human chromosome to be completely sequenced.

Option C (Thymine) is correct because in RNA, Uracil replaces Thymine as one of the four nitrogenous bases.

Option B (Polymorphism) is correct because DNA fingerprinting exploits the high degree of polymorphism in satellite DNA sequences.

Option D is correct. If guanine constitutes 20% of the bases, cytosine also constitutes 20%. Therefore, adenine and thymine together make up 60%, and adenine is 30%.

Option B (Chromosome 1) is correct because the sequence of Chromosome 1 was completed in May 2006 as part of the Human Genome Project.

Option D (Uracil + ribose + phosphate) is correct because RNA contains uracil as a nitrogenous base, ribose as the sugar, and phosphate group.

Option C (231) is correct because the Y-chromosome in humans has the least number of genes, approximately 231.

Option A (Frederick Griffith - Discovered the phenomenon of transformation) is correct because Frederick Griffith's experiment demonstrated the transformation of bacteria.

Option A (Point mutation) is correct because sickle cell anemia results from a single base substitution, which is the definition of a point mutation.

Option D (5’ CAU 3’) is correct because the anticodon loop of tRNA is complementary and antiparallel to the codon AUG, hence it is 5' CAU 3'.

Option B ($rRNA$) is correct because ribosomal RNA (rRNA) forms the structural framework of ribosomes and also catalyzes peptide bond formation during translation.

Option A (AUG, UGA, UUU) is correct because AUG is a start codon, UGA is a stop codon, and UUU codes for an amino acid (phenylalanine), representing a possible transcript.

Option C (Y-Chromosome) is correct because the Y-chromosome has the fewest number of genes compared to other human chromosomes.

Option A is correct because recombinant DNA technology allows for the manipulation and insertion of genes into living organisms, enabling genetic engineering.