Molecular Basis of Inheritance — Practice Quiz

30%

Step 1: Chargaff's rule: In dsDNA, %A = %T and %G = %C. Step 2: Given %G = 20%, therefore %C = 20%. Step 3: Total of G + C = 20% + 20% = 40%. Step 4: Remaining percentage for A + T = 100% - 40% = 60%. Step 5: Since %A = %T, each = 60%/2 = 30%. Therefore, %Adenine = 30%. Option A (20%) is wrong — that is the percentage of G or C. Option C (40%) is wrong — that is the combined G+C content. Option D (60%) is wrong — that is the combined A+T content.

Proteins contain sulfur-containing amino acids (methionine and cysteine), whereas DNA contains no sulfur; so ³⁵S specifically labels protein and not DNA.

Step 1: The Hershey-Chase experiment needed to separately label protein and DNA of the bacteriophage. Step 2: DNA contains phosphorus (in the phosphate groups) but no sulfur. Step 3: Proteins contain amino acids; two amino acids — methionine (Met) and cysteine (Cys) — contain sulfur in their side chains. Step 4: Therefore, growing phages in ³⁵S medium specifically incorporates radioactivity into proteins only, not into DNA. Step 5: Similarly, ³²P was used to label DNA specifically because DNA has phosphate groups and proteins generally lack phosphorus. This differential labelling allowed resea

DNA polymerase can only add nucleotides in the 5'→3' direction, but at the replication fork one template strand runs 5'→3' (same direction as fork movement), forcing discontinuous synthesis in short fragments.

Step 1: DNA polymerase always synthesises the new strand in the 5'→3' direction — it adds nucleotides to the 3'-OH end of the growing chain. Step 2: At the replication fork, the two parental strands run antiparallel. Step 3: One template strand (3'→5') allows continuous synthesis in the 5'→3' direction as the fork opens — this is the leading strand. Step 4: The other template strand runs 5'→3' in the direction of fork movement. Polymerase cannot synthesise in the 3'→5' direction on this strand, so it must repeatedly start new fragments (Okazaki fragments) moving away from the fork — this is th

Addition of an unusual methylated guanosine triphosphate (7-methylguanosine cap) at the 5'-end of hnRNA in a template-independent manner.

Step 1: eukaryotic hnRNA undergoes three major post-transcriptional modifications: capping, tailing, and splicing. Step 2: Capping involves adding an unusual nucleotide — 7-methylguanosine triphosphate (methyl guanosine triphosphate) — to the 5'-end of the hnRNA. Step 3: This addition is template-independent (it is not directed by the DNA template). Step 4: The cap protects mRNA from degradation by exonucleases and helps in ribosome recognition and initiation of translation. Step 5: Option A describes tailing (poly-A tail at 3'-end), not capping. Option C describes splicing. Option D describes