Top 100+ Molecular Basis of Inheritance - Biology Questions and Answers For NEET 2024 Exam Preparation

Q1. Which of these do not follow independent assortment ?

1. Genes on homologous chromosomes
2. Genes on non-homologous chromosomes and absence of linkage
3. Linked genes on same chromosome
4. Unlinked genes on same chromosome

Answer: (c)

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Q2. Which enzymes will be produced in a cell in which there is a nonsense mutation in the lac Y gene?

1. Transacetylase
2. Lactose permease
3. Lactose permease and transacetylase
4. b- galactosidase

Answer: (d)

A nonsense mutation is the one which stops polypeptide synthesis due to formation of a terminating or non sense codon. e.g. ATT(UAA), ATC (UAG), ACT(UGA) . The lactose or lac operon of Esherichia coli contains structural genes (Z, Y, A). If Y codes for termination of polypeptide chain then only the product of ‘Z’ gene transcribe to form β galactosidase.

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Q3. Change in the sequence of nucleotide in DNA is called as

1. mutation
2. mutagen
3. recombination
4. translation

Answer: (a)

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Q4. Lightly stained part of chromatin which remains loosely packed and is transcriptionally active named as

1. heterochromatin
2. euchromatin
3. chromatosome
4. chromonemata

Answer: (b)

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Q5. The lac operon consists of

1. one regulatory gene and three structural genes
2. four regulatory genes only
3. two regulatory genes and two structural genes
4. three regulatory genes and three structural genes

Answer: (a)

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Q6. A sequential expression of a set of human genes

1. DNA sequence
2. messenger RNA
3. ribosome
4. transfer RNA

Answer: (a)

A sequential expression of a set of human genes is the DNA sequence. Because gene is the functional part of DNA sequence.

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Q7. In a nucleosome, histones are organized in which structure?

1. Hexamer
2. Octamer
3. Decamer
4. Septamer

Answer: (b)

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Q8. In protein synthesis, the polymerisation of amino acids involves three steps. Which one of the following is not involved in the polymerisation of protein ?

1. Initiation
2. Termination
3. Elongation
4. Transcription

Answer: (d)

Transcription is the mechanism of copying the message of DNA on RNA with the help of enzyme RNA polymerase. It is meant for taking the coded information from DNA to the site where it is required for protein synthesis.

Translation or protein synthesis is a complicated process involving several steps such as – activation of amino acid, transfer of amino acid to tRNA, initiation of polypeptide synthesis, elongation of polypeptide chain and termination of polypeptide chain.

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Q9. The Okazaki fragments in DNA chain growth

1. polymerize in the 3' to 5' direction and forms replication fork
2. result in transcription
3. prove semi-conservative nature of DNA replication
4. polymerize in the 5' to 3' direction and explain 3' to 5' DNA replication

Answer: (a)

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Q10. The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to C.

1. A - Translation, B - Extension, C - Rosalind Franklin
2. A - Transcription, B -Translation, C - Francis Crick
3. A - Transcription, B - Replication, C - James Watson
4. A - Translation, B - Transcription, C - Ervin Chargaff

Answer: (b)

The expression of the genetic material occurs normally through the production of proteins. This involves two consecutive steps. These are transcription and translation. The DNA codes for the production of messenger RNA (mRNA) during transcription. Messenger RNA carries coded information to ribosomes.

The ribosomes read this information and use it for protein synthesis. This process is called translation. F.H.C. Crick described this unidirectional flow of information in 1958 as the 'central dogma of molecular biology'.

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Q11. Which of the following option is true for Human Genome Project (HGP)?

1. Total estimated cost of the project would be 9 billion US dollars
2. It was launched in the year 1990 and was called mega project
3. It aims to identify all 20000-25000 genes in human DNA
4. All of the above

Answer: (d)

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Q12. In terms of DNA and RNA structure, what is a nucleotide ?

1. A nucleotide is a sugar molecule covalently bonded to a heterocyclic base.
2. A nucleotide is a heterocyclic base.
3. A nucleotide is a sugar molecule bonded to phosphate group and a heterocyclic base.
4. A nucleotide is a heterocyclic base bonded to phosphate group.

Answer: (c)

A nucleoside is made up of a sugar molecules and a heterocyclic base while a nucleotide is made up of a sugar molecule, phosphate group and a heterocyclic base.

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Q13. In eukaryotes, RNA polymerase III is responsible for synthesis of

1. tRNA, hnRNA, rRNA
2. 28S RNA, 18S RNA and 5.8S RNA
3. tRNA, 5sRNA, snRNA
4. hnRNA, tRNA, rRNA

Answer: (c)

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Q14. Transformation experiments in search of genetic material were performed by

1. T. H. Morgan
2. G. J. Mendel
3. F. Griffith
4. F. H. Crick

Answer: (c)

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Q15. In split genes, the coding sequences are called

1. cistrons
2. exons
3. introns
4. operons

Answer: (b)

Split genes are those genes that consist of continuous sequence of nucleotide (coding sequence) interrupted by intervening sequences. Most eukaryotic genes are split as are genes of some animal viruses.

The continuous coding sequences are called exons and the intervening non-coding sequence are called introns. These introns are not represented in mRNA transcribed from the gene and are not utilised for the synthesis of proteins.

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Q16. What would happen if in a gene encoding a polypeptide of 50 amino acids, the 25th codon (UAU) is mutated to UAA?

1. A polypeptide of 24 amino acids will be formed
2. A polypeptide of 25 amino acids will be formed
3. Two polypeptides of 24 and 25 amino acids will be formed
4. A polypeptide of 49 amino acids will be formed

Answer: (a)

UAA is the stop codon. Therefore at 25th amino acid the synthesis of polypeptide stops. So, a polypeptide of 24Amino acid is formed.

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Q17. What does 'c' represent in the figure?

1. Continuous synthesis
2. Template DNA
3. Discontinuous synthesis
4. Newly synthesized strands

Answer: (b)

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Q18. The RNA that pick up specific amino acid from amino acid pool in the cytoplasm to ribosome during protein synthesis is called

1. RNA
2. rRNA
3. mRNA
4. tRNA.

Answer: (d)

Transfer RNA or tRNA help in transfer of amino acids to ribosomes mRNA complex to form the polypeptide chain. It has four key regions a carrier and recognition end, enzyme site and ribosome site.

This recognition end has three anticodons with the help of which amino acids are identified. rRNA forms 67% of 70S ribosomes and 50% of 80S ribosomes. mRNA carries the coded information from DNA for the synthesis of proteins.

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Q19. The first genetic material could be

1. carbohydrates
2. protein
3. DNA
4. RNA

Answer: (d)

The first genetic material could be RNA. We know that RNA is present as a genetic material in some viruses, and it also works as a catalyst (there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein enzymes).

But, RNA being a catalyst is reactive and hence unstable. Therefore, it is considered that DNA has evolved from RNA thereby making RNA the first genetic material.

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Q20. What are the components required for the initiation of transcription in bacteria?

1. Promoter, RNA polymerase and ρ–factor
2. RNA polymerase only
3. ρ factor only
4. σ factor only

Answer: (a)

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Q21. Which one of the following statements about the particular entity is true?

1. The gene for producing insulin is present in everyone's cell.
2. Centromere is found in animal cells, which produces aster during cell division.
3. Nucleosome is formed of nucleotides.
4. DNA consists of core of eight histones.

Answer: (a)

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Q22. Erwin Chargaff found that

1. Ratios of adenine and thymine and guanine and cytosine are constant.
2. Ratios of adenine and cytosine and thymine and guanine are constant.
3. Ratios of adenine and guanine and thymine and cytosine are constant.
4. Ratios of adenine – guanine and thymine – cytosine are constant.

Answer: (a)

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Q23. The whole structure indicated in the figure is of ________.

1. Nucleolus
2. Chromosome
3. Nucleosome
4. Proteasome

Answer: (c)

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Q24. Gene regulation governing lactose operon of E. coli that involves the lacI gene product is

1. Negative and inducible because repressor protein prevents transcription.
2. Positive and inducible because it can be induced by lactose.
3. Negative and repressible because repressor protein prevents transcription.
4. Feedback inhibition because excess of β galactosidase can switch off transcription.

Answer: (a)

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Q25. Assertion (A): Replication and transcription occur in the nucleus, but translation takes place in the cytoplasm.

Reason (R): mRNA is transferred from the nucleus into cytoplasm where ribosomes and amino acids are available for protein synthesis.

1. If both A and R are true, but R is not the correct explanation of A
2. If both A and R are true and R is the correct explanation of A
3. If A is true, but R is false
4. If A is false, but R is true

Answer: (b)

Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

In eukaryotes, replication and transcription take place in the nucleus.

The fully processed hnRNA now called mRNA, is transferred from the nucleus into the cytoplasm, where translation occurs.

This is because all the amino acids, tRNA and ribosomes required for translation are present in the cytoplasm.

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Q26. Name the nucleotide added to 5′ end of hnRNA in capping.

1. Ethyl guanosine triphosphate
2. Ethyl cytosine triphosphate
3. Methyl guanosine triphosphate
4. Methyl cytosine triphosphate

Answer: (c)

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Q27. Assertion (A): Polycistronic mRNA is capable of forming a number of different polypeptide chains

Reason (R): Polycistronic mRNA has terminator codons.

1. If both A and R are true, but R is not the correct explanation of A
2. If both A and R are true and R is the correct explanation of A
3. If A is true, but R is false
4. If A is false, but R is true

Answer: (a)

Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

Polycistronic mRNA, commonly found in prokaryotes, two or more coding regions are present and can specify a number of polypeptide chains or proteins. It further contains multiple (open) reading frames to enable the formation of two or more proteins.

Both polycistronic and monocistronic mRNAs have a 5′ leader sequence, the coding region (containing initiation codon and termination codon) and a non-translated 3′ trailer sequence.

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Q28. In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids?

1. 20
2. 60
3. 64
4. 61

Answer: (d)

Out of a total of 64 codons, 3 codons do not make any sense. Hence only 61 codons are used in the formation of the 20 essential amino acids (polypeptides).

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Q29. Which is true about the structure of heterochromatin?

1. Loosely packed; Stain dark
2. Loosely packed; Stain light
3. Densely packed; Stain light
4. Densely packed; Stain dark

Answer: (d)

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Q30. What sequence on the template strand of DNA corresponds to the first amino acid inserted into a protein ?

1. UAC
2. TAC
3. UAG
4. AUG

Answer: (b)

The first mRNA codon to specify an amino acid is always AUG. A DNA strand with the sequence TAC will corresponds to the first amino acid i.e., AUG. On DNA strand A always pairs with T while on RNA strand A always pairs with U.

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Q31. VNTRs are

1. Very Narrow Tandem Repeats.
2. Variable Number of Tandem Repeats.
3. Variable Non-cistronic Transposon Repeats.
4. Valuable Non-cistronic Transposon Regions.

Answer: (b)

In human genome, there are about 200,000 satellite loci. These simple tandem repeats of short sequences are called ‘Variable Number Tandem Repeats’ (VNTRs).

These repeats are inherited from the parents, and are used as genetic markers in a personal identity test.

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Q32. 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:

1. they cannot synthesize functional beta-galactosidase
2. the lac operon is constitutively active in these cells
3. in the presence of glucose, E.coli cells do not utilize lactose
4. they cannot transport lactose from the medium into the cell

Answer: (a)

Operons are segments of genetic material which function as regulated unit or units that can be switched on and switched off. An operon consists of one to several structural genes. (Three in lac operon).

These are genes which produce mRNAs for forming polypeptides / proteins / enzymes. Z (produces enzyme β-galactosidase for splitting lactose into glucose and galactose).

Y (produces enzyme galactoside permease required in entry of lactose). A (produces enzyme thiogalactoside trans-acetylase). The three structural genes of the operon produce a single polycistronic mRNA.

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Q33. Promoter and terminator flanks the

1. structural gene
2. house-keeping gene
3. recon
4. transcription unit

Answer: (a)

Promoter is present at the 5′-site of structural gene and terminator is present at the 3′-site of structural genes. Thus, we can say that promoter and terminator flanks the structural genes.

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Q34. Nucleoside differs from a nucleotide. It lacks the :

1. sugar
2. base
3. Phosphate group
4. Hydroxyl group

Answer: (c)

A nitrogenous base is attached to the pentose sugar by an N- glycosidic linkage to form a nucleoside, i.e., Nucleoside = Nitrogen base + Pentose sugar.

When a phosphate group is attached to the 5’-OH of a nucleoside through phosphor diester linkage, a nucleotide is formed, i.e. Nucleotide = Nitrogen base + Pentose sugar +phosphate ($PO_{4}$) So, a nucleoside differs from a nucleotide as it lacks the phosphate group.

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Q35. Removal of introns and joining of exons in a defined order during transcription is called

1. inducing
2. looping
3. slicing
4. splicing

Answer: (d)

Introns, which occur principally in eukaryotes, are transcribed into messenger RNA (mRNA) but are subsequently removed from the transcription before translation. In certain cases, removal of the introns is an autocatalytic process (self-splicing) whereby the RNA itself has the properties of an enzyme.

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Q36. Chemically, RNA is (i) reactive and (ii) stable as compared to DNA.

1. (i) more, (ii) less
2. (i) less, (ii) more
3. (i) equally, (ii) equally
4. (i) more, (ii) equally

Answer: (a)

2'–OH group present in ribose sugar of every nucleotide of RNA is a reactive group. Uracil present in RNA is less stable as compared to thymine (= methyl uracil) of DNA.

Being unstable, RNA mutates at a much faster rate and there is no repairing system. That is why RNA viruses have shorter life span. They mutate and evolve very fast.

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Q37. What is indicated by 'r' in the figure?

1. Operator
2. Suppresor
3. Repressor
4. Inducer

Answer: (c)

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Q38. The lac operon is an example of

1. overlapping genes
2. repressible operon
3. arabinose operon
4. inducible operon

Answer: (d)

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Q39. β - galactosidase is synthesized by E. coli to catalyze hydrolysis of ________ into ________ and glucose.

1. Galactose, glucose
2. Galactose, lactose
3. Lactose, galactose
4. Maltose, galactose

Answer: (c)

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Q40. The commonly used vectors for human genome sequencing are

1. BAC and YAC
2. T-DNA
3. Expression vectors
4. T/Cloning vectors

Answer: (a)

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Q41. Both husband and wife have normal vision though their fathers were colourblind. The probability of their daughter becoming colourblind is

1. 25%
2. 0%
3. 50%
4. 75%

Answer: (b)

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Q42. Genes are packaged into a bacterial chromosome by

1. basic protein
2. histones
3. acidic protein
4. actin

Answer: (a)

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Q43. Determination of one amino acid by more than one codon is due to

1. continuous nature of genetic code.
2. redundancy of genetic code.
3. punctuation in genetic code.
4. universal nature of genetic code.

Answer: (b)

Degeneracy of codons is the redundancy of the genetic code. A single amino acid may be specified by many codon i.e., called degeneracy. Degeneracy is due to the last base in codon (which is known as wobble base).

Thus, first two codon are more important to determine the amino acid and third one differ without affecting the coding i.e., known wobble hypothesis, proposed by Crick which establishes an economy of tRNA molecule.

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Q44. Escherichia coli fully labelled with N15 is allowed to grow in N14 medium. The two strands of DNA molecule of the first generation bacteria have

1. different density but resemble parent DNA
2. different density and do not resemble parent DNA
3. same density and resemble parent DNA
4. same density but do not resemble parent DNA

Answer: (a)

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Q45. Who proved that DNA is basic genetic material?

1. Watson
2. Griffith
3. Boveri and Sutton
4. Hershey and Chase

Answer: (d)

Hershey &Chase (1962) discovered that DNA is the genetic material of bacteriophage. They experimented with $T_{2}$ phage which attacks the bacterium E. coli. Some virus made to grow on culture containing radioactive sulphur and some on radioactive phosphorus. Findings indicated that protein did not enter the bacteria from the viruses but DNA from the virus particle enters bacteria as genetic material.

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Q46. Lac operon' in E. coli, is induced by

1. promoter gene
2. 'I' gene
3. β-galactosidase
4. lactose

Answer: (c)

Lac operon in E. coli is induced by β-galactosidase an enzyme meant for hydrolysis of lactose in glucose and galactose.

$Lactose ⟶ ↖{\text"β - galactosidase"} Glucose + Galactose$

These enzymes are called as inducible enzymes, because the synthesis of such enzymes are induced by adding substrate such as lactose by 10, 000 times.

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Q47. What does "lac" refer to in what we call the lac operon?

1. Lactase
2. Lactose
3. Lac insect
4. The number 1,00,000

Answer: (b)

In lac operon, lac refers to lactose. The lac operator is a part of the structural genes (lac Z, lac Y, lac A and lac I). It is responsible for the uptake and initial catabolism of lactose.

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Q48. Choose the incorrect option.

1. HGP will help in developing new ways to diagnose, treat and some day prevent disorders affecting humans
2. HGP is closely associated with bioinformatics
3. Fragment sequenced during HGP are done by method developed by Frederick Sanger
4. Repetitive DNA sequences are stretches of DNA repeated 2-3 times in a DNA sequence

Answer: (d)

This option is incorrect and can be corrected as Repetitive DNA sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. Rest of the options are correct.

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Q49. The promoter site and the terminator site for transcription are located at

1. 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit
2. 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit
3. the 5′ (upstream) end
4. the 3′ (downstream) end

Answer: (a)

The promoter is located towards 5′ end (upstream) of the structural gene of coding strand and provides the binding site for RNA polymerase to initiate transcription. The terminator region is where the transcription stops and it is present at the 3′ end (downstream).

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Q50. Which of the following is required as inducer(s) for the expression of Lac operon? [NEET - I, 2016]

1. Galactose
2. Glucose
3. Lactose
4. Lactose and galactose

Answer: (c)

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Q51. In the genetic code dictionary, how many codons are used to code for all the 20 essential amino acids?

1. 64
2. 20
3. 61
4. 60

Answer: (c)

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Q52. The order and sequences of amino acids are defined by the sequences of the bases in

1. mRNA
2. rRNA
3. tRNA
4. All of these

Answer: (a)

According to the sequences present on the mRNA, amino acids are produced. Thus, the order and the sequence of the amino acids are defined by mRNA.

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Q53. Which one of the following does not follow the central dogma of molecular biology?

1. Mucor
2. Pea
3. Chlamydomonas
4. HIV

Answer: (d)

This one-way flow of information from DNA to mRNA and then to protein is called the central dogma of molecular biology by F.H.C. Crick (1958). But later on two American workers H. Temin and D. Baltimore reported that DNA is also formed from RNA in retroviruses, e.g., HIV. This is called reverse transcription or Teminism, i.e.,

This reverse transcription occurs under the influence of reverse transcriptase enzyme. So, HIV viruses does not follow central dogma.

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Q54. Identify the incorrect statement for lac operon model.

1. RNA polymerase stay away from promoter in the presence of repressor
2. Lactose acts as inducer which inactivates repressor
3. Regulation of lac operon by repressor is referred to as negative regulation
4. The repressor of the operon is synthesised during specific periods from r-gene

Answer: (d)

The incorrect statement can be corrected as

The repressor of the operon is synthesised (all-the-timeconstitutively) from the i gene. The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon. Rest of the statements are correct.

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Q55. Semi-conservative replication of DNA was first demonstrated in:

1. Streptococcus pneumoniae
2. Escherichia coli
3. Salmonella typhimurium
4. Drosophila melanogaster

Answer: (b)

Semi conservative replication of DNA was first demonstrated in Escherichia coli. E. coli is a common type of bacteria that can get into food, like beef and vegetables. The strange thing about these bacteria is that they are not always harmful to you. E. coli normally lives inside your intestines, where it helps your body breakdown and digest the food you eat.

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Q56. In some viruses, DNA is synthesized by using RNA as template. Such a DNA is called

1. B-DNA
2. A - DNA
3. c DNA
4. r DNA

Answer: (c)

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Q57. DNA is a polymer of nucleotides which are linked to each other by 3' to 5' phosphodiester bond. To prevent the polymerization of nucleotides, which of the following modifications would you choose?

1. Remove/replace 3' OH group in deoxyribose.
2. Replace purine with pyrimidines.
3. Remove/replace 2' OH group with some other group in deoxyribose.
4. Both 'a' and 'c'.

Answer: (a)

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Q58. Additional post-transcriptional processing like capping and tailing is characteristic to

1. hnRNA
2. rRNA
3. snRNA
4. tRNA

Answer: (a)

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Q59. Sequence of which of the following is used to know the phylogeny ?

1. rRNA
2. mRNA
3. tRNA
4. DNA

Answer: (a)

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Q60. Degeneration of a genetic code is attributed to the

1. second member of codon
2. first member of a codon
3. entire codon
4. third member of a codon

Answer: (d)

In a triplet, for a particular amino acid more than one word (synonyms) can be used. This phenomenon is described by saying that the code is degenerate. A degenerate code would be one where there is one to one relation between amino acids and the codons that 44 codons out of 64 will be useless or nonsense codons.

A code is degenerate because of the third base of the codon. It has been shown that the same tRNA can recognize more than one codons differing only at the third position. For example GCU, GCC and GCA all code for alanine amino acids.

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Q61. Nucleotide arrangement in DNA can be seen by

1. electron microscope
2. X-ray crystallography
3. ultracentrifuge
4. light microscope

Answer: (b)

X-ray crystallography is the study of molecular structure by examining diffraction patterns made by X-rays beamed through a crystalline form of the molecules. It is widely used in biochemistry to examine the molecular structure of molecules such as proteins and DNA.

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Q62. How many pair of nucleotides are present in one helix of B- DNA?

1. 5
2. 12
3. 10
4. 6

Answer: (c)

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Q63. What is indicated by 'a' in figure?

1. Ionic bond
2. H-bond
3. Covalent bond
4. Van der Waals bond

Answer: (a)

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Q64. Identify A, B and C of a nucleosome. ABC Core of histone molecules

1. A – H1 histone; B – DNA; C – Histone octamer
2. A – DNA; B – H1 histone; C – Histone octamer
3. A – Histone octamer; B – RNA; C – H1 histone
4. A – RNA; B – H1 histone; C – Histone octamer

Answer: (b)

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Q65. Khorana first deciphered the triplet codons of

1. threonine and histidine
2. serine and isoleucine
3. tyrosine and tryptophan
4. phenylalanine and methionine

Answer: (a)

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Q66. Typically a nucleosome consists of how many base pairs?

1. 200
2. 190
3. 300
4. 310

Answer: (a)

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Q67. In tertiary structure of DNA, what is a histone octamer ?

1. A complex consisting of eight negatively charged histone proteins
   (two of each $H_{2}A, H_{2}B, H_{3}$ and $H_{4}$) that aid in the packaging of DNA.
2. A complex consisting of eight positively charged histone proteins
   (two of each $H_{2}A, H_{2}B, H_{3}$ and $H_{4}$) that aid in the packaging of DNA.
3. A complex consisting of nine positively charged histone proteins
   (H1 and two of each $H_{2}A, H_{2}B, H_{3}$ and $H_{4}$ ) that aid in the packaging DNA.
4. A complex consisting of nine negatively charged histone proteins
   ($H_{1}$ and two of each $H_{2}A, H_{2}B, H_{3}$ and $H_{4}$) that aid in the packaging of DNA.

Answer: (b)

A histone octamer is a complex of eight positively charged histone proteins (two of each $H_{2}A, H_{2}B, H_{3}$ and $H_{4}$) that aid in the packaging of DNA. Negatively charged DNA wraps around these histone octamers to form the nucleosome. The DNA is held there by ionic bonds. Linker histone H1 binds to each nucleosome where the DNA enters and exits and this draws a string of nucleosomes closer together to form the 10 nm fibre. The nucleosomes in chromatin are seen as beads-on string structure when viewed under electron microscope.

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Q68. Gene regulation governing lactose operon of E.coli that involves the lac I gene product is :

1. Negative and repressible because repressor protein prevents transcription
2. Negative and inducible because repressor protein prevents transcription
3. Feedback inhibition because excess of galactosidase can switch off transcription
4. Positive and inducible because it can be induced by lactose

Answer: (b)

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Q69. Control of gene expression takes place at the level of

1. Transcription
2. DNA-replication
3. Translation
4. None of the above

Answer: (a)

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Q70. Satellite DNA is important because it :

1. does not code for proteins and is same in all members of the population
2. shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable from parents to children.
3. codes for enzymes needed for DNA replication
4. codes for proteins needed in cell cycle.

Answer: (b)

Satellite DNA displays high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is inherited from parents to children(offsprings).

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Q71. The polytene chromosomes were discovered for the first time in

1. Chironomus
2. Drosophila
3. Musca nebulo
4. Musca domestica

Answer: (a)

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Q72. Choose the incorrect option.

1. Erwin Chargaff said, the ratio between A and T and G and C of dsDNA are constant and equals one
2. Friedrich Miescher in 1869 identified DNA as an acidic substance and named it nuclein
3. The two strands of dsDNA are complementary to each other
4. None of the above

Answer: (d)

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Q73. Tandem Repeat DNA

1. is classified as microsatellites and minisatellites.
2. normally does not code for any protein.
3. shows polymorphism.
4. is used in fingerprinting.

Choose the correct option to complete the statement.

1. I, II and III
2. I and III
3. I, III and IV
4. I, II, III and IV

Answer: (d)

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Q74. BAC, a vector used in biotechnology for cloning, stands for

1. Bacterial Artificial Chromosome
2. Bacterial Actual Chromosome
3. Bacterial Annoted Chromosome
4. None of these

Answer: (a)

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Q75. Heavy DNA can be differentiated from normal DNA by which centrifugation technique?

1. $CaSO_4$ density gradient
2. AgCl density gradient
3. CsCl density gradient
4. KCl density gradient

Answer: (c)

Heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient.

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Q76. In transgenics, expression of transgene in target tissue is determined by

1. transgene
2. enhancer
3. promoter
4. reporter

Answer: (a)

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Q77. The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was

1. protein
2. DNA
3. mRNA
4. polysaccharide

Answer: (b)

The transforming chemical discovered by Griffith in his experiments with Pneumococcus, was confirmed as DNA by Avery, McLeod and McCarty.

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Q78. In mutational event, when adenine is replaced by guanine, it is a case of

1. transcription
2. frame shift mutation
3. transition
4. transversion.

Answer: (c)

Transition mutant is one in which a purine is substituted by a different purine, or a pyrimidine by a different pyrimidine. Such a change involves a base pair change between a G–C pair and an A–T pair in the DNA whereas transversion results when one nitrogen base is replaced by another different type e.g., C-G and A-T. Transcription is the formation of mRNA on DNA templete.

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Q79. How genetic and physical maps were generated in HGP?

1. By using RNase
2. By using DNase
3. By using restriction endonuclease
4. By using automated DNA sequences

Answer: (c)

For generating genetic and physical maps, restriction endonuclease was used. For sequencing the human DNA, the entire DNA from a cell is isolated and converted into random fragments of relatively smaller size by using restriction endonuclease enzyme and cloned in suitable host using specialised vectors.

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Q80. A DNA strand with the sequence AACGTAACG is transcribed. What is the sequence of the mRNA molecule synthesized ?

1. UUGCAUUGC
2. AACGTAACG
3. AACGUAACG
4. TTGCATTGC

Answer: (a)

When a DNA strand with the sequence AACGTAACG is transcribed, the resultant sequence of the mRNA molecule synthesized is UUGCAUUGC. This is based on the pairing of nitrogenous bases - adenine pairs with thymine (in DNA) and uracil (in RNA) and guanine with cytosine.

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Q81. What is indicated by 'b' in the figure?

1. 3' hydroxyl
2. 5' phosphate
3. Ribose sugar
4. Nitrogen base

Answer: (a)

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Q82. Lactose operon produces enzymes

1. b-galactosidase, permease and transacetylase.
2. b-galactosidase, permease and glycogen synthetase.
3. Permease, glycogen synthetase and transacetylase.
4. b-galactosidase, permease and phosphoglucose isomerase.

Answer: (a)

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Q83. Apart from histones, chromatin at higher level of coiling requires which proteins?

1. None-histone chromatic proteins
2. Neo histone complex proteins
3. Non-histone chromosomal proteins
4. None of these

Answer: (c)

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Q84. Choose the incorrect option for tRNA molecule.

1. It has an amino acid acceptor end to which it binds to amino acids
2. It has an anticodon loop that has bases complementary to the code
3. tRNA are not specific for each amino acid
4. tRNA looks like a clover leaf

Answer: (c)

Option (c) is incorrect and can be corrected as tRNA are specific for each amino acid, e.g. for initiation there is a specific initiator tRNA. Rest of the options are correct for tRNA.

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Q85. Select the correct match.

1. $F_2$ × Recessive parent- Dihybrid cross
2. Ribozyme - Nucleic acid
3. T.H. Morgan - Transduction
4. G. Mendel - Transformation

Answer: (b)

Ribozymes (ribonucleic acid enzymes) are RNA molecules that are capable of catalysing specific biochemical reactions similar to the action of protein enzymes.

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Q86. The chromosomes in which centromere is situated close to one end are:

1. Telocentric
2. Acrocentric
3. Sub-metacentric
4. Metacentric

Answer: (b)

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Q87. What effect would you expect if gene expression of the lac operon were completely repressed ?

1. Allolactose would accumulate within the cell and become toxic.
2. The cell would be more efficient without 'wasting' the energy required for the low level of Lac Z, LacY, and Lac A gene expression.
3. Lactose would not be converted into the inducer and the operon could not be induced.
4. All of the above

Answer: (c)

A low level of Lac Z expression is required for conversion of lactose to the inducer, allolactose.

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Q88. DNA replication is

1. semi-conservative and semi-discontinuous
2. conservative and discontinuous
3. semi-conservative and discontinuous
4. conservative.

Answer: (a)

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Q89. Transformation was discovered by

1. Hershey and Chase
2. Meselson and Stahl
3. Griffith
4. Watson and Crick.

Answer: (c)

Transformation was first studied by S.F. Griffith in 1928 while studying Streptococcus pneumoniae. He found that R-Type non virulent bacteria pick up virulence from heat killed virulent S-type bacteria and transform into virulent forms. It was this experiment which indicated presence of a 'transforming principle' which was later found out to be DNA, by Avery et al.

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Q90. In HGP, the chromosome ________ was completed in May 2006.

1. 1
2. 2
3. 3
4. 4

Answer: (a)

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Q91. DNA fingerprinting using Variable Number Tandem Repeats (VNTRs) is based on the observation that

1. the DNA of VNTR loci is more stable than that of loci which code for proteins.
2. every individual has unique alleles at each VNTR locus.
3. VNTR sequences show little variability.
4. VNTR loci are highly polymorphic.

Answer: (d)

The technique of DNA fingerprinting was initially developed by Alec Jeffrey's. He used a satellite DNA as probe that shows high degree of polymorphisms. DNA fingerprinting using variable number tandem repeats is based on the observation that VNTR loci are highly polymorphic.

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Q92. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called.

1. B-DNA
2. A-DNA
3. cDNA
4. rDNA

Answer: (c)

In viruses, like retroviruses (e.g., HIV), an enzyme called reverse transcriptase is used to generate complementary DNA (cDNA) from and RNA template. This process is termed as reverse transcription.

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Q93. Satellite DNA is important because it

1. shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which is heritable from parents to children
2. codes for proteins needed in cell cycle
3. does not code for proteins and is same in all members of the population
4. codes for enzymes needed for DNA replication

Answer: (a)

Sattelite DNA is important because it shows high degree of polymorphism (variation at genetic level) in population and also the same degree of polymorphism in an individual, which is heritable from parents to children.

Also satellite DNA does not code for proteins or enzymes but are different in all the members of a population.

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Q94. The primary purpose of Griffith’s experiments on the Streptococcus pneumoniae bacterium was to

1. prevent cancers caused by exposure to ultraviolet light.
2. find a cure for pneumonia in humans.
3. determine if DNA is the hereditary material.
4. discover the molecular structure of DNA.

Answer: (b)

Griffith was a physician trying to find a way to effectively treat or cure pneumonia. Nothing much was known about the structure or function of DNA at the time he did his work.

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Q95. 5 - end of ribose sugar bears

1. Deoxyribose
2. N-base
3. Phosphate group
4. Hydroxyl group

Answer: (c)

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Q96. Which of the following is initiation codon?

1. AUC
2. UAG
3. AUG
4. CCU

Answer: (c)

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Q97. Which form of RNA has a structure resembling clover leaf?

1. m RNA
2. hn RNA
3. rRNA
4. t RNA

Answer: (d)

rRNA occurs inside ribosomes. m RNA brings information from DNA to polypeptides. hnRNA are heterogenous nuclear RNA.

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Q98. DNA-dependent DNA polymerases catalyses polymerisation in which direction?

1. 5′⟶ 2′
2. 3′⟶ 5′
3. 5′⟶ 3′
4. 2′⟶ 5′

Answer: (c)

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Q99. The given diagram of the lac operon showing an operon of inducible enzymes. Identify components and enzymes (A, B, C, D and E).

1. A– Galactosidase, B– Permease, C– Transacetylase, D– Inducer (lactose), E– Repressor protein
2. A– Galactosidase, B– Permease, C– Transacetylase, D– Repressor protein, E– Inducer (lactose)
3. A– Galactosidase, B– Transacetylase, C– Permease, D– Repressor protein, E– Inducer (lactose)
4. A– Permease, B– Transacetylase, C– Galactosidase, D– Repressor protein, E– Inducer (lactose)

Answer: (b)

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Q100. Select the two correct statements out of the four (i – iv) statements given below about lac operon

1. Glucose or galactose may bind with the repressor and inactivate it
2. In the absence of lactose, the repressor binds with the operator region
3. The z-gene codes for permease
4. This was elucidated by Francois Jacob and Jacques Monod.

The correct statements are

1. (i) and (iii)
2. (ii) and (iii)
3. (ii) and (iv)
4. (i) and (ii)

Answer: (c)

The two French scientists, Jacob and Monod proposed the lac operon of E. coli. The lac operon (an inducible operon) contains a promoter, an operator, a regulator gene and three structural genes z, y, and a, coding for the enzymes β-galactosidase, β-galactoside permease and β-galactoside transacetylase, respectively. β-galactoside permease "pumps" lactose into the cell, where β-galactosidase cleaves it into glucose and galactose.

The function of the transacetylase is still not clear. The lac regulator gene, designated the i gene, codes for a repressor. In the absence of the inducer (i.e., lactose, actually allolactose), the repressor binds to the lac operator sequence, preventing RNA polymerase from binding to the promoter and transcribing the structural genes.

The inducer of the operon, allolactose, is derived from lactose in a reaction that is catalysed by β-galactosidase. Once formed, allolactose binds to the repressor, causing it to be released from the operator; in doing so, it induces transcription of the z, y and a structural genes. CAP is activator called catabolic activator protein. It exerts a positive control in lac operon because in its absence RNA polymerase is unable to recognise promotor gene.

CAP activates lac genes only when glucose is absent. Such enzymes whose synthesis can be induced by adding the substrate are known as inducible enzymes and the genetic systems responsible for the synthesis of such an enzyme are known as inducible operons.

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