Skip to main content
Chapter 5 of 5
Important Questions

Molecular Basis of Inheritance — Important Questions

Punjab Board · Class 12 · Biology

45 important questions from Molecular Basis of Inheritance for Punjab Board Class 12 Biology, with answers. Includes multiple choice questions.

45 questions33 flashcards5 concepts

Interactive on Super Tutor

Studying Molecular Basis of Inheritance? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for important questions and more.

Free trial, no card needed.

A detailed labeled diagram showing the components of a DNA nucleotide: a deoxyribose sugar, a phosphate group, and one of the four nitrogenous bases (Adenine, Guanine, Cytosine, or Thymine). Emphasize
Super Tutor

One of 22 illustrations for Molecular Basis of Inheritance in Super Tutor — alongside flashcards, concept maps and practice questions.

45 Questions·
multiple choice

Important Questions from Molecular Basis of Inheritance

1multiple choice
1 marks

Chargaff's rule states that in a double-stranded DNA molecule, A=T and G=C. If a DNA molecule has 20% Guanine, what percentage of the total nucleotides is Adenine?

Show answer

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.

2multiple choice
1 marks

The Hershey-Chase experiment conclusively proved that DNA is the genetic material. Which of the following best explains why radioactive sulfur (³⁵S) was used to label the protein component of the bacteriophage?

Show answer

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

3multiple choice
1 marks

During DNA replication in E. coli, one strand is synthesised continuously while the other is synthesised discontinuously. What is the molecular reason for discontinuous synthesis on one strand?

Show answer

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

4multiple choice
1 marks

In eukaryotic transcription, the primary transcript (hnRNA) undergoes processing before becoming functional mRNA. Which of the following correctly describes the process of CAPPING?

Show answer

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

+41 more questions on Molecular Basis of Inheritance (Punjab Board Class 12 Biology)

Practise All

Frequently Asked Questions

What are the important topics in Molecular Basis of Inheritance for Punjab Board Class 12 Biology?
Key topics in Molecular Basis of Inheritance include Structure of DNA, Packaging of DNA in the Nucleus, Search for Genetic Material – Key Experiments, Properties of Genetic Material and RNA World. Study these first, then practise questions on each for the Punjab Board Class 12 board exam.
How many important questions are there in Molecular Basis of Inheritance?
Super Tutor has 45 practice questions for Molecular Basis of Inheritance, including multiple choice questions. A sample with answers is on this page.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

For serious students

Get the full Molecular Basis of Inheritance chapter — start free.

Quizzes, flashcards, an AI doubt solver and a study plan for Punjab Board Class 12 Biology. Free to start, no card needed.