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Molecular Basis of Inheritance

Punjab Board · Class 12 · Biology

Practice quiz for Molecular Basis of Inheritance — Punjab Board Class 12 Biology. MCQs and questions with answers to test your preparation.

45 questions33 flashcards5 concepts

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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
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Quick Quiz: Molecular Basis of Inheritance

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1

In the Meselson-Stahl experiment, E. coli was grown in ¹⁵N medium for several generations and then shifted to ¹⁴N medium. After THREE generations of growth in ¹⁴N medium, what will be the ratio of light (¹⁴N/¹⁴N) DNA to hybrid (¹⁵N/¹⁴N) DNA?

2

Which of the following correctly explains why only ONE strand of DNA (template strand) is used during transcription, and not both strands simultaneously?

3

A segment of the coding strand of a structural gene reads: 5'-ATGAAATTCGGG-3'. What will be the anticodon of the tRNA that recognises the SECOND codon in the resulting mRNA?

4

In the lac operon, even in the absence of lactose, a very low level of expression of the operon is always maintained. What is the biological significance of this?

45 Questions·
multiple choice

Sample Questions

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?

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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

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What are the important topics in Molecular Basis of Inheritance for Punjab Board Class 12 Biology?
Key topics in Molecular Basis of Inheritance include Central Dogma and Gene Expression Pathway, Molecular Basis of Inheritance – Chapter Overview, Molecular Basis of Inheritance — Complete Chapter Overview. These are the concepts Punjab Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Molecular Basis of Inheritance — Punjab Board Class 12 Biology?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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