Molecular Basis of Inheritance NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Molecular Basis of Inheritance question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Molecular Basis of Inheritance previous year questions
87 questions. Tap an option to check your answer.
Q1NEET 2021
Complete the flow chart on central dogma. (a) ↻ DNA —(b)→ mRNA —(c)→ (d)
Not quite — the answer is C.
In the central dogma DNA copies itself by replication, is transcribed into mRNA, and mRNA is translated into protein. Transduction is transfer of DNA by viruses and is not a step of the central dogma.
Q2NEET 2021
Identify the correct statement.
Not quite — the answer is B.
In bacteria, RNA polymerase associates transiently with the rho factor to terminate transcription. Capping adds methyl guanosine triphosphate to the 5′ end, the template strand (not the coding strand) is copied, and split genes are typical of eukaryotes.
Q3NEET 2021
What is the role of RNA polymerase III in the process of transcription in eukaryotes ?
Not quite — the answer is B.
In eukaryotes RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs. RNA polymerase I makes the 28S, 18S and 5.8S rRNAs and RNA polymerase II makes hnRNA, the precursor of mRNA.
Q4NEET 2021
DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as :
Not quite — the answer is B.
DNA fingerprinting looks for differences in regions of repetitive DNA, where short stretches are repeated many times and the number of repeats differs between people. Satellite DNA is one type of repetitive DNA, so repetitive DNA is the broader, textbook answer.
Q5NEET 2021
Which of the following RNAs is not required for the synthesis of protein ?
Not quite — the answer is D.
mRNA carries the code, tRNA brings amino acids and rRNA forms the ribosome, so all three are needed to make protein. siRNA works in RNA interference to silence genes, not in protein synthesis.
Q6NEET 2021
Which is the "Only enzyme" that has "Capability" to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes ?
Not quite — the answer is B.
In bacteria a single DNA-dependent RNA polymerase catalyses initiation (with the sigma factor), elongation and termination (with the rho factor) of transcription. DNA-dependent DNA polymerase acts in replication, not transcription.
Q7NEET 2021
If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it ?
Not quite — the answer is C.
By Chargaff's rule A = T and G = C, so if A is 30% then T is also 30%. The remaining 40% is shared equally, giving G = 20% and C = 20%.
Q8NEET 2021
Which one of the following statements about Histones is wrong ?
Not quite — the answer is B.
Histones are rich in the basic amino acids lysine and arginine, so they carry positive side-chain charges and are basic, not slightly acidic. Eight histone molecules form the histone octamer of the nucleosome.
Q9NEET 2021
Statement I : The codon 'AUG' codes for methionine and phenylalanine. Statement II : 'AAA' and 'AAG' both codons code for the amino acid lysine. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is D.
AUG codes only for methionine (it is also the start codon); phenylalanine is coded by UUU and UUC, so Statement I is false. AAA and AAG both code for lysine, showing degeneracy of the code, so Statement II is true.
Q10NEET 2020 · Q67
Name the enzyme that facilitates opening of DNA helix during transcription.
Not quite — the answer is D.
In transcription RNA polymerase itself opens the DNA helix and elongates the RNA, with no separate helicase needed. DNA helicase opens the helix in replication, and DNA ligase and DNA polymerase are replication enzymes.
Q11NEET 2020 · Q74
If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6 × 10⁹ bp, then the length of the DNA is approximately:
Not quite — the answer is C.
Length = number of base pairs × distance per base pair = 6.6 × 10⁹ × 0.34 × 10⁻⁹ m = 2.244 m, which is about 2.2 m. The values 2.0, 2.5 and 2.7 m do not match the product.
Q12NEET 2020 · Q77
Which of the following statements is correct?
Not quite — the answer is A.
Adenine pairs with thymine through two hydrogen bonds, while guanine pairs with cytosine through three. A-T pairing is the basis of Chargaff's rule, so it never pairs through one or three bonds and does pair with thymine.
Q13NEET 2019 · Q115
Purines found both in DNA and RNA are :
Not quite — the answer is B.
The purines adenine and guanine occur in both DNA and RNA. Cytosine, thymine and uracil are pyrimidines, and thymine is found only in DNA.
Q14NEET 2019 · Q154
Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology ?
Not quite — the answer is C.
Because the genetic code is nearly universal, a bacterium reads the codons of the human insulin gene the same way a human cell does and makes the same polypeptide. Redundancy (degeneracy) and specificity are real features but do not explain expression across species.
Q15NEET 2019 · Q158
Expressed Sequence Tags (ESTs) refers to :
Not quite — the answer is A.
In the Human Genome Project, ESTs identified all the genes that are expressed as RNA. Sequence annotation, the other approach, sequenced the whole genome including non-coding parts.
Q16NEET 2019 · Q166
Match the following genes of the Lac operon with their respective products : (a) i gene — (i) β-galactosidase (b) z gene — (ii) Permease (c) a gene — (iii) Repressor (d) y gene — (iv) Transacetylase Select the correct option.
Not quite — the answer is C.
The i gene codes for the repressor, z for β-galactosidase, y for permease and a for transacetylase. Swapping the products of the y and a genes is the common error.
Q17NEET 2019 · Q172
Under which of the following conditions will there be no change in the reading frame of following mRNA ? 5′ AACAGCGGUGCUAUU 3′
Not quite — the answer is D.
The mRNA reads AAC AGC GGU GCU AUU; positions 7–9 form the whole codon GGU, so deleting them removes one amino acid and the downstream codons GCU AUU stay in frame. Inserting or deleting one or two bases shifts every codon after that point (frameshift).
Q18NEET 2018 · Q102
Select the correct match:
Not quite — the answer is C.
A ribozyme is a catalytic RNA, such as the peptidyl transferase activity of rRNA, so it is a nucleic acid. Morgan is linked with linkage, a cross of the F₂ (or F₁) with the recessive parent is a test cross, and transformation was discovered by Griffith, not Mendel.
Q19NEET 2018 · Q106
The experimental proof for semiconservative replication of DNA was first shown in a
Not quite — the answer is B.
Meselson and Stahl (1958) grew Escherichia coli in ¹⁵N medium and then ¹⁴N medium and showed hybrid and light DNA by density gradient centrifugation, the first proof of semiconservative replication. Taylor later showed the same in Vicia faba, a plant, so the plant option is the later, not the first, proof.
Q20NEET 2018 · Q109
Select the correct match:
Not quite — the answer is D.
Jacob and Monod proposed the lac operon model of gene regulation in E. coli. Meselson and Stahl worked on E. coli, Hershey and Chase used bacteriophage T2, and Alec Jeffreys developed DNA fingerprinting; Streptococcus pneumoniae was Griffith's organism.
Q21NEET 2018 · Q158
All of the following are part of an operon except
Not quite — the answer is A.
An operon such as the lac operon consists of a promoter, an operator and structural genes (with a regulator gene nearby). Enhancers are regulatory DNA sequences of eukaryotic genes that can act from a distance and are not parts of prokaryotic operons.
Q22NEET 2018 · Q161
AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?
Not quite — the answer is C.
The mRNA has the same sequence and polarity as the coding strand, except that uracil replaces thymine, so AGGTATCGCAT becomes AGGUAUCGCAU. The complementary sequence UCCAUAGCGUA would be obtained by copying the coding strand itself, which is the trap.
Q23NEET 2017 · Q65
If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered?
Not quite — the answer is C.
Base 901 is the first base of codon 301 (bases 901–903). Deleting it shifts the reading frame from codon 301 to the end, so codons 301 to 333, that is 333 − 300 = 33 codons, are altered while the first 300 stay the same. Choosing 333 wrongly assumes the whole message is misread.
Q24NEET 2017 · Q103
Spliceosomes are not found in cells of:
Not quite — the answer is D.
Spliceosomes remove introns from eukaryotic hnRNA. Bacterial genes are not split into exons and introns, so bacteria have no spliceosomes, while plants, fungi and animals, all eukaryotes, do.
Q25NEET 2017 · Q117
DNA replication in bacteria occurs:
Not quite — the answer is C.
Bacteria have no defined cell cycle phases or nucleolus; they replicate their circular DNA before the cell divides by binary fission. The S phase belongs to the eukaryotic cell cycle.
Q26NEET 2017 · Q120
Which of the following RNAs should be abundant in animal cell?
Not quite — the answer is A.
rRNA is the most abundant RNA in any cell, about 80% of total RNA, because it is the structural and catalytic part of the very large number of ribosomes. mRNA forms only a few percent, as it is short-lived and varies from gene to gene.
Q27NEET 2017 · Q128
During DNA replication, Okazaki fragments are used to elongate:
Not quite — the answer is D.
DNA polymerase works only 5′→3′, so on the lagging strand it makes short Okazaki fragments in the direction away from the moving replication fork, and ligase joins them. The leading strand is made continuously towards the fork and needs no fragments.
Q28NEET 2017 (Gujarati paper) · Q68
Consider the following statements and choose the correct option: (a) Six codons do not code for any amino-acid. (b) Codon is read in m-RNA in a contiguous fashion. (c) Three codons function as stop codons. (d) The initiator codon AUG codes for methionine.
Not quite — the answer is D.
Only three codons (UAA, UAG, UGA) do not code for any amino acid, so the claim of six is the only false statement. The code is read contiguously without punctuation, the three stop codons terminate translation, and AUG both initiates and codes for methionine.
Q29NEET 2017 (Gujarati paper) · Q87
If the ratio of (T + C)/(A + G) in one strand of DNA is 1.43, same ratio in the complementary strand is:
Not quite — the answer is B.
Because A pairs with T and G with C, (T + C) of the complementary strand equals (A + G) of the given strand, so the ratio inverts: 1/1.43 ≈ 0.70. Keeping the ratio at 1.43 would apply only to (A + T)/(G + C), which stays the same in both strands.
Q30NEET 2017 (Gujarati paper) · Q128
The technique of DNA fingerprinting is superior to conventional fingerprinting because it can:
Not quite — the answer is C.
DNA fingerprinting compares repetitive, highly polymorphic DNA sequences (VNTRs), which differ between individuals and are inherited from parents. It does not compare whole genomes, and it is slower and more expensive than taking conventional fingerprints.
Q31NEET 2017 (Gujarati paper) · Q132
A couple claimed in court that a child belonged to them. Their claim can be true if the DNA fingerprint pattern of the child shows:
Not quite — the answer is D.
A child gets one set of chromosomes from each parent, so about half of the bands in its DNA fingerprint match the father and half match the mother. A complete match with either parent, or with both, is impossible for a biological child.
Q32NEET 2016 Phase 2 · Q120
Taylor conducted the experiments to prove semiconservative mode of chromosome replication on
Not quite — the answer is B.
Taylor and colleagues in 1958 used radioactive thymidine on root tips of Vicia faba (faba bean) to show that chromosomes replicate semiconservatively. The E. coli experiment with heavy nitrogen was done by Meselson and Stahl and dealt with DNA, not whole chromosomes.
Q33NEET 2016 Phase 2 · Q122
The equivalent of a structural gene is
Not quite — the answer is B.
A cistron is the segment of DNA coding for one polypeptide, which is what a structural gene is. A muton and a recon are the smallest units of mutation and recombination, and an operon is a group of structural genes with their regulatory elements.
Q34NEET 2016 Phase 2 · Q124
Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria?
Not quite — the answer is C.
The 23S rRNA of the bacterial 50S subunit forms part of the ribosome's structure and also catalyses peptide bond formation as peptidyl transferase. 18S and 5.8S rRNAs are found in eukaryotic ribosomes, and 5S rRNA is structural only.
Q35NEET 2016 Phase 1 · Q107
Which of the following is required as inducer(s) for the expression of Lac operon?
Not quite — the answer is C.
Lactose (strictly its isomer allolactose) binds the lac repressor and inactivates it, so the structural genes are transcribed. Glucose actually represses the operon and galactose is not the inducer.
Q36NEET 2016 Phase 1 · Q110
A complex of ribosomes attached to a single strand of RNA is known as:
Not quite — the answer is A.
Several ribosomes translating the same mRNA at the same time form a polysome (polyribosome). A polypeptide is the product of translation, and Okazaki fragments are short DNA pieces of the lagging strand.
Q37NEET 2016 Phase 1 · Q141
Which of the following is not required for any of the techniques of DNA fingerprinting available at present?
Not quite — the answer is B.
DNA fingerprinting uses restriction digestion, PCR amplification and hybridisation with labelled VNTR probes (Southern blotting). Zinc finger analysis is not a part of any fingerprinting technique; zinc fingers are DNA-binding protein motifs.
Q38NEET 2016 Phase 1 · Q158
Which one of the following is the starter codon?
Not quite — the answer is A.
AUG, coding for methionine, is the initiation (start) codon. UGA, UAA and UAG are stop codons that terminate translation.
Q39AIPMT 2015
Gene regulation governing lactose operon of E.coli that involves the lac I gene product is:
Not quite — the answer is C.
The lac I gene codes for a repressor that binds the operator and blocks transcription, so control is negative. Lactose (allolactose) inactivates the repressor and switches the operon on, so it is inducible, not repressible.
Q40AIPMT 2015
In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are:
Not quite — the answer is D.
By Chargaff's rule G = C = 17%, so G + C = 34% and A + T = 100 - 34 = 66%. Since A = T, each is 33%. Options with unequal A and T break the base-pairing rule.
Q41AIPMT 2014
Select the correct option (Direction of RNA synthesis ; Direction of reading of the template DNA strand):
Not quite — the answer is D.
RNA polymerase can add nucleotides only to the 3′ end, so RNA grows 5′ → 3′; because the template is antiparallel, it is read 3′ → 5′. Any combination that has RNA growing 3′ → 5′ or the template read in the same direction as synthesis is wrong.
Q42AIPMT 2014
Commonly used vectors for human genome sequencing are:
Not quite — the answer is A.
In the Human Genome Project the DNA was broken into large fragments and cloned in BAC (bacterial artificial chromosome) and YAC (yeast artificial chromosome) vectors, which accept very large inserts. Expression vectors, T/A cloning vectors and T-DNA are not the hosts used for genome sequencing.
Q43AIPMT 2014
An analysis of chromosomal DNA using the Southern hybridization technique does not use:
Not quite — the answer is C.
Southern hybridisation involves restriction digestion of DNA, gel electrophoresis, blotting onto a membrane, hybridisation with a labelled probe and detection by autoradiography. PCR is not a step of this technique.
Q44AIPMT 2014
Transformation was discovered by:
Not quite — the answer is B.
Frederick Griffith (1928), working with Streptococcus pneumoniae and mice, found that heat-killed virulent bacteria could transform live non-virulent ones: transformation. Hershey and Chase proved DNA is the genetic material, Watson and Crick gave the double helix, and Meselson and Stahl proved semiconservative replication.
Q45AIPMT 2014
Which one of the following is wrongly matched?
Not quite — the answer is D.
Transcription copies genetic information from DNA into RNA, chiefly mRNA, so 'writing information from DNA to t-RNA' is the wrong match. Translation, the repressor binding the operator, and an operon made of structural genes, operator and promoter are all correctly described.
Q46NEET 2013
Which enzyme/s will be produced in a cell in which there is a nonsense mutation in the lac Y gene?
Not quite — the answer is A.
The lac operon genes are transcribed in the order z, y, a. A nonsense mutation in lacY stops translation of permease, but β-galactosidase from lacZ, which lies upstream, is still made. Permease cannot be produced, and transacetylase downstream is lost or greatly reduced.
Q47AIPMT 2012 Mains
What is it that forms the basis of DNA fingerprinting?
Not quite — the answer is D.
DNA fingerprinting compares repetitive DNA, satellite DNA such as VNTRs whose copy number differs between people. The purine to pyrimidine ratio is the same in everyone, and the technique has nothing to do with skin ridges.
Q48AIPMT 2012 Prelims
Removal of introns and joining of exons in a defined order during transcription is called
Not quite — the answer is A.
Splicing is the removal of introns from the primary transcript (hnRNA) and joining of exons in a defined order. Slicing is a distractor that sounds similar but is not the term for this process.
Q49AIPMT 2012 Prelims
PCR and Restriction Fragment Length Polymorphism are the methods for
Not quite — the answer is A.
DNA (genetic) fingerprinting uses PCR to amplify DNA and RFLP (restriction digestion and hybridisation with probes) to compare banding patterns between individuals. These methods are not used to study enzymes or to transform cells.
Q50AIPMT 2012 Prelims
Which one of the following is not a part of a transcription unit in DNA ?
Not quite — the answer is B.
A transcription unit consists of a promoter, a structural gene and a terminator. An inducer (such as lactose in the lac operon) is a small molecule that regulates expression; it is not a stretch of DNA.
Q51AIPMT 2012 Prelims
Removal of RNA polymerase III from nucleoplasm will affect the synthesis of
Not quite — the answer is B.
In eukaryotes, RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs, so removing it stops tRNA synthesis. RNA polymerase I makes most rRNA, and RNA polymerase II makes hnRNA (the precursor of mRNA).
Q52AIPMT 2011 Mains
The unequivocal proof of DNA as the genetic material came from the studies on a
Not quite — the answer is B.
Hershey and Chase grew bacteriophages (bacterial viruses) with radioactive phosphorus or sulphur and found that only the labelled DNA entered the bacteria, giving unequivocal proof that DNA is the genetic material. Griffith's and Avery's work on the bacterium Streptococcus was suggestive but not unequivocal.
Q53AIPMT 2011 Mains
In the history of biology, the human genome project led to the development of
Not quite — the answer is A.
The Human Genome Project produced enormous amounts of sequence data, and storing, retrieving and analysing it led to the rapid growth of bioinformatics. Biotechnology existed long before the project and is not its product.
Q54AIPMT 2011 Prelims
Which one of the following also acts as a catalyst in a bacterial cell?
Not quite — the answer is A.
In bacteria, the 23S rRNA of the large ribosomal subunit acts as a ribozyme, catalysing peptide bond formation (peptidyl transferase activity). snRNA and hnRNA are eukaryotic nuclear RNAs, and 5S rRNA is structural.
Q55AIPMT 2011 Prelims
What are those structures that appear as 'beads-on-string' in the chromosomes when viewed under electron microscope?
Not quite — the answer is D.
DNA wrapped around histone octamers forms nucleosomes, which repeat along the chromatin fibre and look like beads on a string under the electron microscope. Base pairs, nucleotides and genes are far too small or not visible as separate beads.
Q56AIPMT 2010 Mains · Q4
Which one of the following statements about the particular entity is true?
Not quite — the answer is B.
Every somatic cell carries the full genome, so the insulin gene is present in all body cells even though it is expressed only in pancreatic beta cells. Asters are formed by centrioles (centrosome), not by the centromere. A nucleosome is DNA wrapped around a histone octamer, so it is the nucleosome (not DNA) that has a core of eight histones.
Q57AIPMT 2010 Mains · Q24
In eukaryotic cell transcription, RNA splicing and RNA capping take place inside the
Not quite — the answer is B.
In eukaryotes the DNA is in the nucleus, so the primary transcript (hnRNA) is made there and is also capped, tailed and spliced there before the mature mRNA leaves for the cytoplasm. Ribosomes are where translation happens, after these steps.
Q58AIPMT 2010 Mains · Q52
The lac operon consists of
Not quite — the answer is B.
The lac operon has one regulatory gene (i, coding the repressor) and three structural genes: z (β-galactosidase), y (permease) and a (transacetylase), along with promoter and operator sites. The promoter and operator are DNA sites, not genes, so it does not have two or three regulatory genes.
Q59AIPMT 2010 Mains · Q60
The 3′-5′ phosphodiester linkages inside a polynucleotide chain serve to join
Not quite — the answer is C.
In a polynucleotide, the phosphate of one nucleotide links the 3′-carbon of one sugar to the 5′-carbon of the next, so a 3′-5′ phosphodiester bond joins successive nucleotides. The two DNA strands are held together by hydrogen bonds, and a base is joined to its sugar by an N-glycosidic bond.
Q60AIPMT 2010 Prelims · Q121
The one aspect which is not a salient feature of genetic code, is its being
Not quite — the answer is C.
The genetic code is unambiguous and specific: each codon codes for only one amino acid. It is degenerate (several codons per amino acid) and nearly universal, so 'ambiguous' is the one property it lacks.
Q61AIPMT 2010 Prelims · Q122
Satellite DNA is useful tool in
Not quite — the answer is D.
Satellite DNA repeats such as VNTRs are highly polymorphic between individuals, so they form the basis of DNA fingerprinting used in forensic identification and paternity testing.
Q62AIPMT 2010 Prelims · Q123
Which one of the following does not follow the central dogma of molecular biology?
Not quite — the answer is A.
HIV is a retrovirus: it copies its RNA genome into DNA with reverse transcriptase, which reverses the usual DNA → RNA direction of the central dogma. Pea, Mucor and Chlamydomonas are cellular organisms with DNA genomes that follow DNA → RNA → protein.
Q63AIPMT 2010 Prelims · Q126
Select the two correct statements out of the four (a–d) given below about lac operon. (a) Glucose or galactose may bind with the repressor and inactivate it (b) In the absence of lactose the repressor binds with the operator region (c) The z-gene codes for permease (d) This was elucidated by Francois Jacob and Jacque Monod The correct statements are
Not quite — the answer is D.
Without lactose, the repressor binds the operator and blocks transcription, and the lac operon was described by Jacob and Monod. The inducer that inactivates the repressor is lactose (allolactose), not glucose or galactose. The z gene codes for β-galactosidase, while permease comes from the y gene.
Q64AIPMT 2009 Prelims · Q89
What is not true for genetic code?
Not quite — the answer is D.
The genetic code is read in a contiguous way, codon after codon without punctuation, so a 'non-contiguous' reading is false. It is indeed nearly universal, degenerate (several codons per amino acid) and unambiguous (one codon codes for one amino acid).
Q65AIPMT 2009 Prelims · Q96
Removal of introns and joining the exons in a defined order in a transcription unit is called:
Not quite — the answer is D.
In eukaryotes the primary transcript (hnRNA) has introns, which are removed and the exons joined in order by splicing. Capping adds methyl guanosine at the 5′ end and tailing adds a poly-A tail at the 3′ end; neither removes introns.
Q66AIPMT 2009 Prelims · Q99
Semiconservative replication of DNA was first demonstrated in:
Not quite — the answer is A.
Meselson and Stahl (1958) grew E. coli in ¹⁵N and then ¹⁴N medium and showed by CsCl density gradient centrifugation that DNA replicates semiconservatively. Streptococcus pneumoniae was used by Griffith for transformation, not replication.
Q67AIPMT 2009 Prelims · Q142
Whose experiments cracked the DNA and discovered unequivocally that a genetic code is a triplet?
Not quite — the answer is D.
Nirenberg and Matthaei used cell-free protein synthesis with synthetic RNA (poly-U giving polyphenylalanine), which helped decode the codons and confirm the triplet code. Hershey and Chase proved DNA is the genetic material, not the nature of the code.
Q68AIPMT 2008 Prelims · Q113
Polysome is formed by
Not quite — the answer is B.
A polysome (polyribosome) is a chain of several ribosomes translating the same mRNA at the same time. The ribosomes are linked through the mRNA, not directly to one another, and ribosomes bound to ER form rough ER rather than a polysome.
Q69AIPMT 2008 Prelims · Q121
Which one of the following pairs of nitrogenous bases of nucleic acids, is wrongly matched with the category mentioned against it?
Not quite — the answer is A.
Adenine is a purine but thymine is a pyrimidine, so grouping them together as purines is wrong. Adenine and guanine are purines, and cytosine, thymine and uracil are pyrimidines, so the other pairs are correct.
Q70AIPMT 2008 Prelims · Q123
In the DNA molecules
Not quite — the answer is A.
The two polynucleotide chains of the double helix run antiparallel, one 5'→3' and the other 3'→5'. By Chargaff's rule, purines always equal pyrimidines and A always equals T in double-stranded DNA, so those options are false.
Q71AIPMT 2008 Prelims · Q189
Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino acid?
Not quite — the answer is C.
UAA, UAG and UGA are the three stop codons. In the other options only one codon of each pair fits: UCA codes serine, GUU codes valine and ACG codes threonine.
Q72AIPMT 2007 Prelims · Q107
The Okazaki fragments in DNA chain growth
Not quite — the answer is C.
DNA polymerase can add nucleotides only in the 5'->3' direction, so on the lagging strand short Okazaki fragments are made 5'->3' and later joined by DNA ligase. This explains how that strand is built overall in the 3'->5' direction. No polymerase works 3'->5', and semi-conservative replication was proved by Meselson and Stahl, not by Okazaki fragments.
Q73AIPMT 2007 Prelims · Q117
The length of DNA molecule greatly exceeds the dimensions of the nucleus in eukaryotic cells. How is this DNA accommodated?
Not quite — the answer is A.
Negatively charged DNA wraps around positively charged histone octamers to form nucleosomes, which coil further into chromatin fibres, packing about 2.2 m of DNA into a tiny nucleus. Cells do not digest DNA or delete genes or repetitive sequences to save space.
Q74AIPMT 2007 Prelims · Q122
During transcription, RNA polymerase holoenzyme binds to a gene promoter and assumes a saddle-like structure. What is its DNA-binding sequence?
Not quite — the answer is C.
RNA polymerase is positioned at the promoter by the TATA box (Pribnow box in prokaryotes), an AT-rich consensus sequence TATAAT/TATAAAA about 10-30 bases upstream of the start site. AATT and TTAA are not the recognised promoter consensus.
Q75AIPMT 2007 Prelims · Q148
Differentiation of organs and tissues in a developing organism, is associated with
Not quite — the answer is A.
All cells of an organism carry the same genes; they differentiate into different tissues because different sets of genes are switched on or off in each. Mutations and gene deletions are abnormal events, not the basis of normal development.
Q76AIPMT 2007 Prelims · Q156
One gene-one enzyme relationship was established for the first time in
Not quite — the answer is D.
Beadle and Tatum studied nutritional mutants of the mould Neurospora crassa and showed each gene controls one enzyme. Diplococcus pneumoniae was used by Griffith for transformation, and E. coli for lac operon and replication studies.
Q77AIPMT 2007 Prelims · Q170
Telomere repetitive DNA sequences control the function of eukaryotic chromosomes because they
Not quite — the answer is C.
Telomeres are repetitive caps at chromosome ends that stop the ends from being degraded, fused or shortened at each replication, so they keep chromosomes stable and prevent their loss. Replication origins (replicons) and promoters are different sequences.
Q78AIPMT 2007 Prelims · Q190
Molecular basis of organ differentiation depends on the modulation in transcription by
Not quite — the answer is B.
Which genes a cell transcribes, and therefore what it becomes, is controlled by regulatory proteins (transcription factors) that bind promoters and enhancers to switch genes on or off. RNA polymerase carries out transcription in every cell but by itself does not decide which genes are expressed; ribosomes and anticodons act in translation.
Q79AIPMT 2007 Prelims · Q192
The two polynucleotide chains in DNA are
Not quite — the answer is B.
In the double helix one strand runs 5'->3' and its partner 3'->5', so the chains are antiparallel. 'Semiconservative' describes how DNA replicates, not how its strands are arranged.
Q80AIPMT 2006 Prelims · Q135
Amino acid sequence, in protein synthesis is decided by the sequence of :
Not quite — the answer is B.
The order of codons in mRNA decides the order of amino acids in the polypeptide. tRNA only brings amino acids by reading those codons; it does not decide the order.
Q81AIPMT 2006 Prelims · Q169
One gene – one enzyme hypothesis was postulated by :
Not quite — the answer is D.
Beadle and Tatum, working with Neurospora mutants, proposed that one gene controls one enzyme. Garrod earlier described inborn errors of metabolism but did not state this hypothesis.
Q82AIPMT 2006 Prelims · Q170
One turn of the helix in a B-form DNA is approximately :
Not quite — the answer is C.
B-DNA has about 10 base pairs per turn with 0.34 nm between pairs, so one turn (pitch) is about 3.4 nm. 0.34 nm is the distance between adjacent base pairs, and 2 nm is the helix diameter.
Q83AIPMT 2006 Prelims · Q172
Antiparallel strands of a DNA molecule means that :
Not quite — the answer is C.
The two strands of DNA run in opposite directions, one 5'→3' and the other 3'→5', so the phosphate end (5') of each strand lies at opposite ends of the molecule. Both strands turn the same way in a right-handed helix.
Q84AIPMT 2005 Prelims · Q114
Telomerase is an enzyme which is a :
Not quite — the answer is D.
Telomerase has a protein reverse-transcriptase part plus its own RNA, which serves as the template for adding telomere repeats. That makes it a ribonucleoprotein, not a simple protein and not RNA alone.
Q85AIPMT 2005 Prelims · Q115
During transcription holoenzyme RNA polymerase binds to a DNA sequence and the DNA assumes a saddle like structure at that point. What is that sequence called ?
Not quite — the answer is D.
The promoter's TATA box (Pribnow box in prokaryotes) is the AT-rich site where RNA polymerase binds. The DNA bends and opens easily there to start transcription. The CAAT box lies further upstream and mainly regulates how often transcription starts.
Q86AIPMT 2005 Prelims · Q129
E. coli cells with a mutated Z gene of the lac operon cannot grow in medium containing only lactose as the source of energy because :
Not quite — the answer is D.
The z gene codes for β-galactosidase, which splits lactose into glucose and galactose. Without a working enzyme, lactose cannot be used for energy. Lactose transport depends on permease, the y gene product, so the transport option describes a y mutation instead.
Q87AIPMT 2005 Prelims · Q177
Which one of the following makes use of RNA as a template to synthesize DNA ?
Not quite — the answer is A.
Reverse transcriptase, found in retroviruses such as HIV, is an RNA-dependent DNA polymerase that copies RNA into DNA. DNA polymerase copies DNA into DNA, and RNA polymerase copies DNA into RNA.
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