25 questions in 25 minutes on the NTA computer-based test interface, all from Molecular Basis of Inheritance. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Molecular Basis of Inheritance, of which 18 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
Revising first? Read the Molecular Basis of Inheritance chapter notes, then come back and take this test to check it stuck.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Molecular Basis of Inheritance mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 18 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Molecular Basis of Inheritance?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Botany
- Q1The codons causing chain termination are:
- AGT, TAG, UGA
- TAG, TAA, TGA
- GAT, AAT, AGT
- UAA, UAG, UGA
Answer: (D) UAA, UAG, UGA
The three stop (termination) codons on mRNA are UAA (ochre), UAG (amber) and UGA (opal). They code for no amino acid and end translation. (Options with T are DNA-form, not mRNA codons.)
- Q2During transcription, if the nucleotide sequence of the DNA strand being coded (template) is ATACG, then the nucleotide sequence in the mRNA would be:
- UATGG
- TATGC
- UAUGC
- TCTGG
Answer: (C) UAUGC
The mRNA is complementary to the template with U for T: A->U, T->A, A->U, C->G, G->C = UAUGC.
- Q3Jacob and Monod studied lactose metabolism in E. coli and proposed the Operon concept, which is applicable for:
- all prokaryotes and all eukaryotes
- all prokaryotes and some protozoans
- all prokaryotes
- all prokaryotes and some eukaryotes
Answer: (D) all prokaryotes and some eukaryotes
The operon model is fundamentally a prokaryotic mechanism but has some value/application in explaining certain eukaryotic gene expression, so it is best stated as applying to all prokaryotes and some eukaryotes.
- Q4During replication of DNA, its two strands separate and each serves as a template for the formation of new strands. Such type of replication is called:
- flexible
- conservative
- non-conservative
- semiconservative
Answer: (D) semiconservative
Each daughter DNA molecule retains one parental strand and one newly synthesised strand. This is semiconservative replication (Meselson and Stahl, 1958).
- Q5Telomerase is an enzyme which is a:
- ribonucleoprotein
- simple protein
- RNA
- repetitive DNA
Answer: (A) ribonucleoprotein
Telomerase is a ribonucleoprotein (it has both a protein catalytic part and an internal RNA template) that synthesises the repetitive telomeric DNA at the ends of eukaryotic chromosomes.
- Q6What will be the sequence of mRNA produced from a template strand 3′-ATGCATGCATGCATG-5′?
- 3′-AUGCAUGCAUGCAUG-5′
- 5′-UACGUACGUACGUAC-3′
- 5′-AUGCAUGCAUGCAUG-3′
- 3′-UACGUACGUACGUAC-5′
Answer: (B) 5′-UACGUACGUACGUAC-3′
mRNA is complementary and antiparallel to the template. Template 3′-ATG…-5′ gives mRNA 5′-UAC…-3′ (A->U, T->A, G->C, C->G), i.e. 5′-UACGUACGUACGUAC-3′.
- Q7The anticodon is an unpaired triplet of bases in an exposed position of:
- snRNA
- rRNA
- tRNA
- mRNA
Answer: (C) tRNA
The anticodon is located on the tRNA, in the anticodon loop; this exposed triplet base-pairs temporarily with the mRNA codon during translation.
- Q8The number of possible base substitutions in the sense (amino-acid coding) codons is:
- 535
- 261
- 284
- 549
Answer: (D) 549
There are 61 sense codons, each with 3 bases, and each base can be changed to one of 3 other bases: 61 x 3 x 3 = 549 possible single-base substitutions.
- Q9Assertion (A): The lac operon is described as an inducible operon under negative control. Reason (R): In the absence of lactose, the repressor binds the operator and switches the operon off.
- A is true but R is false
- Both A and R are true and R correctly explains A
- A is false but R is true
- Both A and R are true but R does not explain A
Answer: (B) Both A and R are true and R correctly explains A
It is negative control because a repressor normally turns the operon off; it is inducible because lactose (inducer) switches it on by removing the repressor. R explains A.
- Q10What are the structures called that give the appearance of ‘beads on a string’ in chromosomes when viewed under the electron microscope?
- Nucleotides
- Nucleosomes
- Base pairs
- Genes
Answer: (B) Nucleosomes
Nucleosomes (DNA wrapped around a histone octamer) give the repeating ‘beads-on-string’ appearance, the beads being the nucleosomes and the string the connecting linker DNA.
- Q11The Okazaki fragments in DNA chain growth:
- polymerise in the 3′->5′ direction and form the replication fork
- prove the conservative nature of replication
- polymerise in the 5′->3′ direction and explain the 3′->5′ DNA replication (lagging-strand synthesis)
- result in transcription
Answer: (C) polymerise in the 5′->3′ direction and explain the 3′->5′ DNA replication (lagging-strand synthesis)
Okazaki fragments are short pieces made by DNA polymerase in the 5′->3′ direction on the lagging strand. They are how the lagging strand (overall 3′->5′ template direction) is copied discontinuously, then joined by ligase.
- Q12The Hershey and Chase experiment using bacteriophages conclusively proved that:
- protein is the genetic material
- DNA is the genetic material
- both DNA and protein are genetic material
- RNA is the genetic material
Answer: (B) DNA is the genetic material
Radioactive ³²P labels DNA and ³⁵S labels protein. Only ³²P (DNA) entered the bacteria and appeared in progeny phages, proving DNA is the genetic material.
- Q13The protein that helps in opening the DNA double helix in front of the replication fork is:
- topoisomerase
- DNA gyrase
- DNA polymerase-I
- DNA ligase
Answer: (B) DNA gyrase
DNA gyrase (a type-II topoisomerase) relieves the supercoiling tension ahead of the replication fork, helping to open/unwind the double helix so replication can proceed.
- Q14In the genetic code dictionary, how many codons are used to code for all 20 essential amino acids?
- 64
- 60
- 20
- 61
Answer: (D) 61
Of the 64 codons, 3 are stop codons (UAA, UAG, UGA) which code no amino acid. The remaining 64 – 3 = 61 codons code for the 20 amino acids (the code is degenerate).
- Q15Which enzyme(s) will be produced in a cell in which there is a nonsense mutation in the lac Y gene?
- beta-galactosidase
- Transacetylase
- Lactose permease
- Lactose permease and transacetylase
Answer: (A) beta-galactosidase
The genes are transcribed/translated in order z, y, a. A nonsense (stop) mutation in y truncates translation, so y (permease) and the downstream a (transacetylase) are not made, but the upstream z (beta-galactosidase) is still produced.
- Q16DNA fingerprinting refers to:
- analysis of DNA samples using an imprinting device
- techniques used for molecular analysis of different specimens of DNA
- molecular analysis or profiles of DNA samples
- techniques used for identification of finger-prints of individuals
Answer: (C) molecular analysis or profiles of DNA samples
DNA fingerprinting is the molecular analysis/profiling of DNA samples (comparing VNTR patterns), developed by Alec Jeffreys (1985). It has nothing to do with skin fingerprints.
- Q17The transfer RNA (tRNA) molecule in 3D appears:
- L-shaped
- E-shaped
- Y-shaped
- S-shaped
Answer: (A) L-shaped
While the 2D (secondary) structure of tRNA is a clover-leaf, its 3D (tertiary) structure is L-shaped (Kim et al., 1973).
- Q18A double-stranded DNA molecule contains 120 adenine bases. The total number of hydrogen bonds in the A=T pairs of this molecule is:
- 360
- 480
- 240
- 120
Answer: (C) 240
Each A pairs with one T via 2 hydrogen bonds. With 120 A (and hence 120 A=T pairs), total H-bonds from A=T pairs = 120 × 2 = 240.
- Q19According to the findings of the Human Genome Project, the approximate fraction of the human genome that actually codes for proteins is:
- less than 2%
- about 25%
- about 50%
- about 98%
Answer: (A) less than 2%
The HGP revealed that less than 2% of the ~3 billion bp human genome codes for proteins; the rest is non-coding (regulatory, repetitive, etc.).
- Q20What is the role of RNA polymerase-III in the process of transcription in eukaryotes?
- Transcribes the precursor of mRNA (hnRNA)
- Transcribes rRNAs (28S, 18S and 5.8S)
- Transcribes tRNA, 5S rRNA and snRNAs
- Transcribes only snRNAs
Answer: (C) Transcribes tRNA, 5S rRNA and snRNAs
RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs. Pol I makes 28S/18S/5.8S rRNA; Pol II makes hnRNA (mRNA precursor).
- Q21DNA replication in bacteria occurs:
- just before transcription
- just prior to fission
- within the nucleolus
- during the S phase
Answer: (B) just prior to fission
Bacteria have no nucleus and no distinct S phase. DNA replication occurs during the C period of the bacterial cell cycle, just prior to binary fission (cell division).
- Q22If the coding (sense) strand of a gene reads 5′-ATGCGA-3′, the mRNA sequence produced will be:
- 5′-UACGCU-3′
- 3′-AUGCGA-5′
- 5′-TACGCT-3′
- 5′-AUGCGA-3′
Answer: (D) 5′-AUGCGA-3′
The mRNA has the same sequence as the coding strand except that T is replaced by U. So 5′-ATGCGA-3′ becomes 5′-AUGCGA-3′.
- Q23During transcription, the DNA site at which RNA polymerase binds is called the:
- enhancer
- regulator
- promoter
- receptor
Answer: (C) promoter
The promoter is the DNA sequence located towards the 5′ end of the coding strand where RNA polymerase binds to initiate transcription.
- Q24DNA fingerprinting relies on differences between individuals in regions of DNA called:
- promoters
- start codons
- VNTRs (variable number tandem repeats)
- exons
Answer: (C) VNTRs (variable number tandem repeats)
DNA fingerprinting compares VNTRs (satellite/minisatellite repeats) whose copy number varies greatly between individuals, giving a unique pattern.
- Q25During DNA replication, the lagging strand is synthesized:
- discontinuously as Okazaki fragments later joined by ligase
- continuously in the 5’→3′ direction without primers
- only after the leading strand is complete
- continuously in the 3’→5′ direction
Answer: (A) discontinuously as Okazaki fragments later joined by ligase
Because DNA polymerase works only 5’→3′ and the strands are antiparallel, the lagging strand is made in short Okazaki fragments which are then joined by DNA ligase.