NEET Biology Molecular Basis of Inheritance MCQs Set 20

Download NEET MCQs for NEET Biology: Molecular Basis of Inheritance

Explore reliable objective questions for Molecular Basis of Inheritance tailored for NEET learners. Utilizing these Biology multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Molecular Basis of Inheritance

View or download the dedicated Molecular Basis of Inheritance MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Biology evaluations.

Multiple Choice Questions

Question 1: Choose the wrong statement regarding the observation drawn from Human Genome Project:
(a) Repetitive sequences are stretches of RNA
(b) Less than 2% of the genome codes for protein
(c) Chromosome 'Y' has the fewest number of genes
(d) SNPs help in tracing human history
Show Answer & Explanation

Answer: (a) Repetitive sequences are stretches of RNA

Explanation:
1. Repetitive sequences are stretches of DNA, not RNA.
2. The other statements about the Human Genome Project are true.

Teacher's Note:
1. Less than 2 percent codes for protein.
2. Y has fewest genes.

Question 2: Which of the following biomolecule can acts as both genetic material and a biological catalyst?
(a) DNA
(b) RNA
(c) Protein
(d) Both
Show Answer & Explanation

Answer: (b) RNA

Explanation:
1. RNA can act as genetic material (in many viruses) and as a catalyst (ribozyme).
2. DNA is not a catalyst and proteins are not genetic material.

Teacher's Note:
1. Ribozymes: catalytic RNA.
2. RNA world hypothesis.

Question 3: Match List I (factors or enzymes) with List II (activities).
List I: A. Sigma factor, B. Rho factor, C. DNA polymerase I, D. Aminoacyl synthetase
List II: 1. Termination of transcription, 2. Removal of RNA primer from newly synthesised DNA strand, 3. Correct initiation of transcription, 4. Correct initiation of DNA replication, 5. Attachment of amino acid to tRNA
(a) A-2, B-5, C-4, D-1
(b) A-3, B-1, C-2, D-5
(c) A-2, B-1, C-4, D-5
(d) A-3, B-5, C-2, D-1
Show Answer & Explanation

Answer: (b) A-3, B-1, C-2, D-5

Explanation:
1. Sigma factor helps correct initiation of transcription (3), Rho factor terminates transcription (1), DNA polymerase I removes the RNA primer (2) and aminoacyl synthetase attaches an amino acid to tRNA (5).
2. This gives A-3, B-1, C-2, D-5.

Teacher's Note:
1. Sigma: start.
2. Rho: stop.

Question 4: Read the following statements and select the correct option: (i) Loosely packed and lightly stained region of chromatin is called as heterochromatin. (ii) Densely packed and dark stained region of chromatin is called as euchromatin (iii) A typical nucleosome contains 200 bp of DNA helix.
(a) Statements (i) and (ii) are true, but statement (iii) is false.
(b) Statements (i) and (ii) are false, but statement (iii) is true.
(c) Statements (ii) and (iii) are true, but statement (i) is false
(d) All statements are true.
Show Answer & Explanation

Answer: (b) Statements (i) and (ii) are false, but statement (iii) is true.

Explanation:
1. Euchromatin is loosely packed and light stained; heterochromatin is densely packed and dark stained, so statements (i) and (ii) are false.
2. A nucleosome contains about 200 bp of DNA, so (iii) is true.

Teacher's Note:
1. Euchromatin is active.
2. Heterochromatin is inactive.

Question 5: In the structure of DNA, which nitrogen of the guanine base attaches to C1' of the pentose sugar?
(a) 3rd
(b) 5th
(c) 7th
(d) 9th
Show Answer & Explanation

Answer: (d) 9th

Explanation:
1. In a nucleotide of guanine the N-9 of the purine ring is joined to C-1' of the pentose sugar by an N-glycosidic bond.
2. Pyrimidines use N-1.

Teacher's Note:
1. Purines: N9.
2. Pyrimidines: N1.

Question 6: A molecule that can act as a genetic material must fulfill the following criteria: (i) It should be able to replicate. (ii) It should provide the scope for slow changes (mutation) that are required for evolution. (iii) It should contain adenine, cytosine, guanine and thymine. (iv) It should chemically and structurally be stable. (v) It should be able to express itself. Correctly option would be:
(a) Only (i), (ii) and (iii)
(b) Only (i), (ii), (iii) and (v)
(c) Only (i), (iv) and (v)
(d) Only (i), (ii), (iv) and (v)
Show Answer & Explanation

Answer: (d) Only (i), (ii), (iv) and (v)

Explanation:
1. A genetic material must replicate, allow slow changes (mutation), be chemically and structurally stable and be able to express itself.
2. Containing A, C, G and T is not a requirement since RNA uses U.

Teacher's Note:
1. DNA and RNA fit these criteria.
2. RNA is less stable.

Question 7: Select the correct option for the direction of RNA synthesis and the direction in which the template DNA strand is read.
(a) RNA synthesis 3' to 5'; template read 3' to 5'
(b) RNA synthesis 5' to 3'; template read 3' to 5'
(c) RNA synthesis 3' to 5'; template read 5' to 3'
(d) RNA synthesis 5' to 3'; template read 5' to 3'
Show Answer & Explanation

Answer: (b) RNA synthesis 5' to 3'; template read 3' to 5'

Explanation:
1. RNA is synthesised in the 5' to 3' direction, while the template DNA strand is read in the 3' to 5' direction.
2. This is option (2).

Teacher's Note:
1. Antiparallel strands.
2. Template: 3' to 5'.

Question 8: Sequencing the whole set of genome that contained all the coding and non- coding sequence and later assigning different region in the sequence with functions are referred to as:
(a) Expressed sequence Tags (EST)
(b) Sequence Annotation
(c) SETs
(d) SNP's
Show Answer & Explanation

Answer: (b) Sequence Annotation

Explanation:
1. Assigning functions to different regions of the sequenced genome is called sequence annotation.
2. ESTs are expressed sequence tags used for expressed genes only.

Teacher's Note:
1. Sequencing then annotation.
2. Human Genome Project: both.

Question 9: Study the lists and find the correct match.
List I: a. Exon, b. Capping, c. Tailing, d. Promoter
List II: I. Site for binding of RNA polymerase, II. Coding sequence, III. Lagging strand, IV. Methyl guanosine triphosphate, V. Adenylate residues
(a) a-II, b-IV, c-V, d-I
(b) a-II, b-IV, c-I, d-V
(c) a-III, b-I, c-II, d-IV
(d) a-IV, b-II, c-III, d-I
Show Answer & Explanation

Answer: (a) a-II, b-IV, c-V, d-I

Explanation:
1. Exon is a coding sequence (II), capping adds methyl guanosine triphosphate (IV), tailing adds adenylate residues (V) and the promoter is the binding site of RNA polymerase (I).
2. So a-II, b-IV, c-V, d-I.

Teacher's Note:
1. Capping at 5' end.
2. Tailing at 3' end.

Question 10: Select the incorrect statement:
(a) In the absence of lactose inducer lac operon will be off
(b) Positive regulation of lac operon in E. coli is due to negative effect of glucose for production of cAMP
(c) Positive regulation of lac operon in E. coli is due to positive effect of glucose for production of cAMP
(d) y-gene codes for permease, which helps in entry of -galactoside in E. coli
Show Answer & Explanation

Answer: (c) Positive regulation of lac operon in E. coli is due to positive effect of glucose for production of cAMP

Explanation:
1. Glucose lowers cAMP, so positive regulation of the lac operon occurs when glucose is low, not by a positive effect of glucose on cAMP production.
2. Statement C is incorrect.

Teacher's Note:
1. Low glucose: high cAMP.
2. CAP-cAMP helps transcription.

Question 11: The central dogma is shown as a flowchart: A is DNA to DNA, B is DNA to RNA and C is RNA to DNA. The enzymes used in processes A, B and C are respectively
(a) DNA dependent DNA polymerase; RNA dependent DNA polymerase; DNA dependent RNA polymerase
(b) DNA dependent DNA polymerase; DNA dependent RNA polymerase; RNA dependent DNA polymerase
(c) DNA dependent DNA polymerase; DNA dependent DNA polymerase; RNA dependent DNA polymerase
(d) RNA dependent DNA polymerase; DNA dependent RNA polymerase; RNA dependent RNA polymerase
Show Answer & Explanation

Answer: (b) DNA dependent DNA polymerase; DNA dependent RNA polymerase; RNA dependent DNA polymerase

Explanation:
1. Replication (DNA to DNA) uses DNA dependent DNA polymerase, transcription (DNA to RNA) uses DNA dependent RNA polymerase and reverse transcription (RNA to DNA) uses RNA dependent DNA polymerase.
2. This is option 2.

Teacher's Note:
1. Reverse transcriptase: retroviruses.
2. Central dogma and Temin.

Question 12: On one strand (the template with polarity 3' to 5') replication is A, while on the other (the template with polarity 5' to 3') it is B.
(a) A - Continuous, B - Continuous
(b) A - Discontinuous, B - Discontinuous
(c) A - Continuous, B - Discontinuous
(d) A - Discontinuous, B - Continuous
Show Answer & Explanation

Answer: (c) A - Continuous, B - Discontinuous

Explanation:
1. On the template with 3' to 5' polarity, replication is continuous (leading strand).
2. On the template with 5' to 3' polarity it is discontinuous (lagging strand, Okazaki fragments).

Teacher's Note:
1. Leading: continuous.
2. Lagging: Okazaki fragments.

Question 13: Degeneration of a genetic code is mainly attributed to the:
(a) First member of a codon
(b) Second member of a codon
(c) Entire codon
(d) Third member of a codon
Show Answer & Explanation

Answer: (d) Third member of a codon

Explanation:
1. Degeneracy of the genetic code is mostly due to the third base of the codon (wobble).
2. The first two bases largely decide the amino acid.

Teacher's Note:
1. Wobble hypothesis.
2. Third base changes often silent.

Question 14: What confers additional stability to double helix model of DNA apart from the H-bond?
(a) One base pair provides inductive effect to the other.
(b) One base pair is stacked over the other
(c) One base pair is perpendicular to another
(d) One base pair interacts with another base pair via steric hindrance.
Show Answer & Explanation

Answer: (b) One base pair is stacked over the other

Explanation:
1. Stacking of one base pair over the other adds stability to the double helix apart from hydrogen bonds.
2. The hydrophobic interactions between stacked bases help.

Teacher's Note:
1. Hydrogen bonds plus stacking.
2. Phosphate backbone outside.

Question 15: Go through following statements and select correct option.
(a) Taylor et. al. used radioactive thymidine in Vicia faba (Broad Bean) and proved that Chromosome replicates semiconservatively.
(b) In eukaryotes, replication of DNA takes place in G1-phase of the cell cycle.
(c) A failure in cell division after RNA replication results into polyploidy.
(d) Crick pointed out that DNA replicates semiconservatively but first proof for it came from the experiments of Beadle and Tatum who used 15NH4Cl in E. coli.
Show Answer & Explanation

Answer: (a) Taylor et. al. used radioactive thymidine in Vicia faba (Broad Bean) and proved that Chromosome replicates semiconservatively.

Explanation:
1. Taylor and co-workers used radioactive thymidine in Vicia faba and showed that chromosomes replicate semiconservatively.
2. Replication is in S phase, polyploidy follows failure of cell division after DNA replication, and the 15N experiment was by Meselson and Stahl.

Teacher's Note:
1. Vicia faba: Taylor.
2. E. coli: Meselson and Stahl.

Question 16: What change occurs by changing one base in DNA?
(a) Always a change of one amino acid in protein.
(b) Change in complex sequence of amino acid.
(c) Always a change in property of protein.
(d) Does not necessarily change the phenotype.
Show Answer & Explanation

Answer: (d) Does not necessarily change the phenotype.

Explanation:
1. A single base change may be silent due to degeneracy and need not change the phenotype.
2. It does not always change the amino acid or protein properties.

Teacher's Note:
1. Silent mutation.
2. Sickle cell is a missense case.

Question 17: Which of the following step of translation does consume a ATP?
(a) Peptidyl transferase reaction
(b) Aminoacyl t-RNA binding to A-site
(c) Translocation
(d) Amino acid activation
Show Answer & Explanation

Answer: (d) Amino acid activation

Explanation:
1. Amino acid activation (formation of aminoacyl AMP by aminoacyl tRNA synthetase) uses ATP.
2. Translocation and peptide bond steps use GTP or none.

Teacher's Note:
1. Activation uses ATP.
2. Elongation uses GTP.

Question 18: Repressor in Lac operon system is:
(a) Always protein
(b) May be protein or a nucleic acid
(c) A metabolite
(d) A substrate
Show Answer & Explanation

Answer: (a) Always protein

Explanation:
1. The repressor of the lac operon is a protein encoded by the i gene.
2. It binds the operator and blocks transcription.

Teacher's Note:
1. Lactose is the inducer.
2. Repressor is always protein.

Question 19: Which is not correct regarding DNA finger printing? (a) The basis of DNA finger printing is high degree of polymorphism in sequence (b) Alec Jeffreys used a satellite DNA as probe that show a minimum degree of polymorphism (c) minimum degree of polymorphism called VNTRs (d) VNTRs varies in size from 0.1-20 kb
(a) (a), (b) and (c)
(b) (b) and (c)
(c) (b), (c) and (d)
(d) (a), (c) and (d)
Show Answer & Explanation

Answer: (b) (b) and (c)

Explanation:
1. DNA fingerprinting depends on high polymorphism of VNTRs (variable number tandem repeats), and Jeffreys used a satellite DNA probe showing a high degree of polymorphism.
2. So statements (b) and (c) which say minimum polymorphism are incorrect.

Teacher's Note:
1. VNTR: 0.1 to 20 kb.
2. Used in forensics.

Question 20: Read the following statements are select the incorrect statement:
(a) One codon codes for only one amino acid.
(b) Some amino acids are coded by more than one codon.
(c) The sequence of triplet nitrogenous bases in DNA or mRNA corresponds to the amino acid sequence in the polypeptide chain.
(d) Codes are universal without any exception
Show Answer & Explanation

Answer: (d) Codes are universal without any exception

Explanation:
1. The genetic code is nearly universal, but there are exceptions, for example in mitochondria.
2. The other statements are correct.

Teacher's Note:
1. Code is degenerate.
2. Code is unambiguous.

Question 21: Select the correct sequences of steps in DNA finger printing. I. Hybridisation using labelled VNTR probe II. Isolation of DNA III. Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose membrane or nylon. IV. Detection of hybridised DNA fragments by autoradiography. V. Separation of DNA fragments by electrophoresis. VI. Digestion of DNA by restriction endonucleases.:
(a) I, V, VI, II, III and IV
(b) II, VI, V, III, I and IV
(c) V, I, VI, III, IV and II
(d) II, I, V, VI, IV and III
Show Answer & Explanation

Answer: (b) II, VI, V, III, I and IV

Explanation:
1. The steps are: isolation of DNA (II), digestion with restriction endonucleases (VI), electrophoresis (V), blotting (III), hybridisation with labelled VNTR probe (I) and detection by autoradiography (IV).
2. So II, VI, V, III, I, IV.

Teacher's Note:
1. Southern blotting.
2. Probe is radioactive.

Question 22: Match the columns.
Column I: a. Monoploidy, b. Monosomy, c. Nullisomy, d. Trisomy, e. Tetrasomy
Column II: 1. 2n - 1, 2. 2n + 1, 3. 2n + 2, 4. 2n - 2, 5. x, 6. 3x
(a) a-6, b-5, c-3, d-4, e-2
(b) a-5, b-2, c-4, d-1, e-3
(c) a-5, b-1, c-4, d-2, e-3
(d) a-1, b-5, c-3, d-2, e-4
Show Answer & Explanation

Answer: (c) a-5, b-1, c-4, d-2, e-3

Explanation:
1. Monoploidy is x (5), monosomy is 2n - 1 (1), nullisomy is 2n - 2 (4), trisomy is 2n + 1 (2) and tetrasomy is 2n + 2 (3).
2. So a-5, b-1, c-4, d-2, e-3.

Teacher's Note:
1. Aneuploidy: change in number of a chromosome.
2. Euploidy: whole sets.

Free study material for Biology

Download Chapter MCQs: NEET Biology Molecular Basis of Inheritance

Chapter MCQs with Answers for NEET Biology

Test your conceptual understanding of Molecular Basis of Inheritance with these targeted multiple-choice questions. Designed in alignment with the latest NEET curriculum for NEET Biology, these problem sets build accuracy and prepare students for objective exams.

Expert Practice Material for NEET Biology

Cross-reference your completed choices with comprehensive NCERT solutions for NEET Biology to ensure absolute clarity across all sub-topics in this chapter.

Complete Your Chapter Revision

Follow up your worksheet practice by attempting the interactive online Biology MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest NEET Biology Molecular Basis of Inheritance MCQs Set 20?

You can get most exhaustive NEET Biology Molecular Basis of Inheritance MCQs Set 20 for free on StudiesToday.com. These MCQs for NEET Biology are updated for the 2026-27 academic session as per NEET examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Biology NEET material?

Yes, our NEET Biology Molecular Basis of Inheritance MCQs Set 20 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the NEET paper is now competency-based.

How do practicing Biology MCQs help in scoring full marks in NEET exams?

By solving our NEET Biology Molecular Basis of Inheritance MCQs Set 20, NEET students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Biology.

Do you provide answers and explanations for NEET Biology Molecular Basis of Inheritance MCQs Set 20?

Yes, Biology MCQs for NEET have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused NEET exams.

Can I practice these Biology NEET MCQs online?

Yes, you can also access online interactive tests for NEET Biology Molecular Basis of Inheritance MCQs Set 20 on StudiesToday.com as they provide instant answers and score to help you track your progress in Biology.