NEET Biology Molecular Basis of Inheritance Important Questions Set 01

These Class 12 NEET Biology important questions on Molecular Basis of Inheritance give 45 questions with step-by-step answers, arranged as very short, short and long answer types, as asked in school and board examinations. They help NEET aspirants revise the definitions, structures, differences and key terms of the chapter.

Very Short Answer Type Questions

Question 1. What is the number of N-glycosidic linkages in φ × 174 bacteriophage?
Answer: φ × 174 bacteriophage – 5386 nucleotides, therefore 5386 N-glycosidic linkages.

Question 2. Write down the components of cytidine and uridine.
Answer: Cytidine → Cytosine + Sugar (ribose) Uridine → Uracil + Sugar

Question 3. Two nucleotide molecules are linked by which bond?
Answer: Phosphodiester bond

Question 4. Which biomolecule acts as messenger, structural as well as adapter?
Answer: RNA

Question 5. Calculate the length of DNA in human diploid cell.
Answer: 6.6 × 10⁹ bp × 0.34 × 10–9 m/bp = 2.2 meter

Question 6. Which structure can be observed as "beads-on-string" when chromatin is viewed under electron microscope?
Answer: Nucleosome

Question 7. What determines the order and sequence of amino acids in proteins?
Answer: Sequences of bases in mRNA

Question 8. Who gave experimental proof of semiconservative mode of DNA replication in prokaryotes?
Answer: Meselson and Stahl (1958)

Question 9. Write down the sequence in mRNA if the sequence of nucleotides in DNA is : 3′-ATGCATTGCCAT-5′
Answer: 5′-UACGUAACGGUA-3′

Question 10. What is coded by i-gene and y-gene of lac operon?
Answer: i-gene → Repressor protein y-gene → Permease

Short Answer Type Questions

Question 11. Explain degeneracy of codons. Give the examples of non-degenerate codons.
Answer: Some amino acids are coded by more than one codon, hence the code is degenerate. Non-degenerate codons are AUG and UGG.

Question 12. Write down the name of any three non-human model organisms, genome of which have been sequenced.
Answer: (Any three)
(i) Bacteria
(ii) Yeast
(iii) Caenorhabditis elegans
(iv) Drosophila
(v) Rice
(vi) Arabidopsis

Question 13. How many nucleosomes do you imagine are present in a mammalian cell?
Answer: One nucleosome has about 200 bp. 6.6 ×109 = 3.3 ×107 nucleosomes / 33 × 10⁶ nucleosomes 200

Question 14. Which features of DNA confer the stability of the helical structure?
Answer: (i) The plane of one base pair stacks over the other base pair.
(ii) H-bonds in base pairs of two strands.

Question 15. Identify the following components (a to c) in the given diagram.
(b) 3′ 5′
U C A (a) A G U (c) 5′ 3′
Answer: (a) Anticodon
(b) Serine (Ser)
(c) Codon

Question 16. Mention the chemical differences between DNA and RNA. (Any three)

Answer:

Question 17. Write down the evidences that justify the statement "RNA was the first genetic material".
Answer: Essential life processes, such as metabolism translation, splicing etc., evolved around RNA. It used to act as genetic material as well as catalyst. There are some important biochemical reactions in living systems that are catalysed by RNA catalysts (ribozyme) and not by protein enzymes. But, RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable.

Question 18. What is the significance of release factors in the process of translation?
Answer: At the end of translation, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.

Question 19. How many types of DNA polymerase are found in bacteria? Mention the rate of polymerisation and direction of exonuclease activity of main polymerising enzyme.
Answer: Three types of DNA polymerase in bacteria are : DNA polymerase I - Exonuclease activity in 5′ → 3′ and 3′ → 5′ direction. DNA polymerase II - Exonuclease activity in 3′ → 5′ direction. DNA polymerase III–Rate of polymerisation is 2000 bp per second and exonuclease activity in 3′ → 5′ direction.

Question 20. In which respect polynucleotide phosphorylase is different from RNA polymerase?
Answer: Polynucleotide phosphorylase (Severo Ochoa enzyme) polymerises RNA with defined sequences in a template independent manner.

Question 21. (a) What do you mean by tailing of RNA?
(b) Where does it occur in cell?

Answer:

(a) During this process, adenylate residues (200–300) are added at 3'-end of RNA in a template-independent manner.

(b) It occurs in the nucleus.

Question 22. (a) Define chromatin.
(b) Write the composition of its repeating unit.

Answer:

(a) Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin. In nucleus, it is visible in the form of thread-like stained bodies.

(b) Nucleosome consists of histone octamer around which negatively charged DNA is wrapped which contain 200 bp.

Question 23. Consider the given strand of DNA duplex.
3′-TACCATTGAGACCGACTC-5′
(a) Draw the complementary DNA polynucleotide chain.
(b) Write down the mRNA sequence transcribed by it.

Answer:

(a) 5′-ATGGTAACTCTGGCTGAG-3′

(b) 5′-AUGGUAACUGUGGCUGAG-3′

Question 24. Frame shift mutations form the genetic basis of proof that codon is a triplet and it is read in a contiguous manner. Explain it.
Answer: Insertion or deletion of three or its multiple bases insert or delete one or multiple codon hence one or multiple amino acids and reading frame remains unaltered from that point onwards.

Question 25.

Given below sequence represent steps involved in eukaryotic transcription. Fill up the blanks A, B and C shown in the sequence.

7-mG

Answer:

A - RNA polymerase-III

B - hnRNA

C - Poly A-tail

Question 26. Write down the significance of operator in operon.
Answer: The accessibility of promoter regions of prokaryotic DNA is regulated by the interaction of proteins with operators.

Question 27. Expand VNTR, SSR, SNPs, RFLP.

Answer:

VNTRs – Variable Number of Tandem Repeats

SSRs – Single Sequence Repeats

SNPs – Single Nucleotide Polymorphism

RFLP – Restriction Fragment Length Polymorphism

Question 28. (a) What is contribution of Friedrich Meischer in molecular biology?
(b) Which experimental material was used by Griffith for finding transforming principle?

Answer:

(a) DNA as an acidic substance present in nucleus was first identified by Meischer in 1869 and he named it as 'Nuclein'.

(b) Streptococcus pneumoniae

Question 29. Differentiate between codon and anticodon.
Answer: S. No. Codon Anticodon
i. It is a sequence of three Sequence of three nitrogenous successive nitrogenous bases bases which is complementary which specifies an amino acid. to a codon.
ii. Present on mRNA. Present on tRNA.

Question 30. (a) Draw a clover-leaf secondary structure of tRNA.
(b) How does charging of tRNA take place?

Answer:

(a)

A.

(b) Attachment of activated amino acids with specific tRNA. Thus, amino acid is transferred to tRNA. Enzyme and AMP are released. It is also called as aminoacylation of tRNA.

AA₁ - AMP - E₁complex + tRNA₁ → AA₁ - tRNA₁ + AMP + E₁

Long Answer Type Questions

Question 31. (i) Which feature in DNA generates approximately uniform distance between the two strands?
(ii) Explain Chargaff's rule.

Answer:

(i) Adenine forms hydrogen bonds with thymine from opposite strand and vice versa. Similarly, guanine is bonded with cytosine. It means, always a purine comes opposite to a pyrimidine or vice versa. This generates approximately uniform distance between the two strands of the helix.

(ii) Chargaff's rule explains that :

(a) Purine and pyrimidines occur in equal amounts, i.e., purine adenine is equimolar with pyrimidine thymine, purine guanine is equimolar with pyrimidine cytosine.

A + G
(b) A + G = T + C or = 1 in all species. T + C

A + T
(c) Base ratio is specific for a species. C + G
(d) Sugar deoxyribose and phosphate residues occurs in equal number.

Question 32. How did Hershey and Chase prove that DNA is the genetic material?

Answer:

Hershey and Chase performed experiment in three steps:

1. Infection

2. Blending

3. Centrifugation

Hershey and Chase raised T₂ bacteriophage over two different bacterial colonies. One having radioactive phosphorus ³²P and the other having radioactive sulphur ³⁵S. Radioactive sulphur got incorporated in the capsid proteins of bacteriophage. Radioactive phosphorus ³²P became component of DNA of bacteriophage. The two types of bacteriophages were introduced to infect different bacterial colonies of E. coli. The two cultures were taken out and independently agitated on a blender. The empty phage capsid got separated from the bacterial cells. The virus particles were separated from the bacteria by spinning them in a centrifuge.

Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating the DNA was the material that passed from virus to the bacteria. Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicates that protein did not enter the bacteria from the viruses. DNA is therefore the genetic material.

Question 33. Mention the salient features of DNA.

Answer:

The salient features of DNA are as follows:

(i) It is made of two polynucleotide chains, where the backbone is constituted by sugar-phosphate, and the bases project inside.
(ii) The two chains have anti-parallel polarity i.e., one chain has the polarity 5′ → 3′ and the other has 3′ → 5′.
(iii) The bases in two strands are paired through hydrogen bonds i.e., A = T(by two H-bonds) and G ≡ C (by three H-bonds)

(iv) The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm and there are roughly 10 bp in each turn. The distance between a bp in a helix is approximately equal to 0.34 nm.

(v) The plane of one base pair stacks over the other in double helix. This, in addition to H-bonds, confers stability to the helical structure.

Question 34. (a) What do you mean by reverse central dogma?
(b) What is nu-body? What will be the length of DNA of a nucleosome?

Answer:

(i) Reverse central dogma is the flow of information in reverse direction i.e., from RNA to DNA in some viruses e.g., retrovirus. Transcription Translation DNA RNA Protein Reverse transcription
(ii) nu-body is the central octamer histone protein (two copies of each– H₂A, H₂B, H₃ and H₄) of nucleosome.

DNA coils around the histone octamer has 200 bp. The distance between one bp is approximately 0.34 nm. Thus, length of DNA with 200 bp = 200 × 0.34 = 68 nm.

Question 35. Discuss the role of NHC proteins. Differentiate between euchromatin and heterochromatin.

Answer:

The packaging of chromatin at higher level requires additional set of proteins that collectively are referred to as NHC (Non-histone chromosomal) proteins.

Question 36. Transforming principle was biochemically characterised by Avery et. al. Explain their work.
Answer: Avery, MacLeod and McCarty worked to determine the biochemical nature of "transforming principle" in Griffith's experiment. They purified biochemicals (proteins, DNA, RNA etc.) from the heat killed S-cells to see which ones could transform live R-cells into S-cells. They discovered that DNA alone from S-bacteria caused R-bacteria to become transformed. They also discovered that protein-digesting enzymes i.e., proteases and RNA digesting enzymes i.e., RNases did not affect transformation, so the transforming substance was not a protein or RNA. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation. Thus, they concluded that DNA is the hereditary material.

Question 37. (a) Regulation of gene expression considering polypeptide formation in eukaryotes could be exerted at how many levels? Write down these levels.
(b) Which is the main level of DNA regulation in bacteria?
Answer: (a) Regulation of gene expression refers to a very broad term that may occur at various levels. Considering that gene expression results in the formation of a polypeptide, it can be regulated at several levels. In eukaryotes, the regulation could be exerted at:
(i) Transcriptional level i.e., formation of primary transcript.
(ii) Processing level i.e., regulation of splicing.
(iii) Transport of mRNA from nucleus to the cytoplasm.
(iv) Translational level.
(b) In bacteria, the main level of DNA regulation occurs at transcription level.

Question 38. Explain semiconservative mode of replication. How was it experimentally proved and by whom?

Answer:

In semiconservative mode of DNA replication, the two strands separate and act as template for the synthesis of new complementary strands. After the completion of replication, each DNA molecule would have one parental and one newly synthesised strand. It was experimentally proved by Meselson and Stahl in E. coli. by cesium chloride density gradient centrifugation technique. They grew E. coli in a medium containing 15NH₄Cl (¹⁵N is the heavy isotope of nitrogen) as the only nitrogen source for many generations. The result was that 15N was incorporated into newly synthesised DNA (as well as other nitrogen-containing compounds). This heavy DNA molecule could be distinguished from the normal DNA by centrifugation in a cesium chloride (CsCl) density gradient. Then they transferred the cells into a medium with normal 14NH₄Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double-stranded helices. The various samples were separated independently on CsCl gradients to measure the densities of DNA. Generation I Generation II 15N-DNA 14N-DNA 14 N-DNA 15N-DNA 15N-DNA 20 min 40 min 14N-DNA Gravitational Force 14N-DNA

15 15 14N15 14 14 14 15 HeavyN N HybridN LightN N HybridN N (Separation of DNA by centrifugation)

Messelson and Stahl's Experiment

Thus, the DNA that was extracted from the culture one generation after the transfer from 15N to 14N medium [that is after 20 minutes; E.coli divides in 20 minutes] had a hybrid or intermediate density. DNA extracted from the culture after another generation [that is after 40 minutes, II generation] was composed of equal amounts of this hybrid DNA and of light DNA.

Question 39. Label a, b and c and identify the figure. Write the name of main enzyme of this process.
(b)
(a) 3′ (c)

Answer:

A well-labelled diagram should show these parts: 5′ 3′; 5′; 3′ 5′; 5′ 3′; (a) Continuous DNA synthesis; (b) Template DNA (parental strands); (c) Discontinuous DNA synthesis.

It is figure of replicating fork. Main enzyme of this process is DNA-dependent DNA polymerase.

Question 40. (a) How many types of RNA polymerases are found in eukaryotes? Mention their functions also.
(b) Name the specific amino acids coded by GAG, UUC, AUG and UGG.

Answer:

Three types of RNA polymerase in eukaryotes are :

(a) RNA polymerase I – 5.8S, 18S, 28S rRNA synthesis

RNA polymerase II – HnRNA, mRNA

RNA polymerase III – tRNA, ScRNA, 5S rRNA, SnRNA

(b) GAG – Glutamic acid

UUC – Phenylalanine

AUG – Methionine

UGG – Tryptophan

Question 41. Explain the transcription unit and its components with the help of diagram.

Answer:

The segment of the DNA which takes part in transcription is called transcription unit. It has a promoter region in the beginning, a terminator region in the end and a structural gene in between promoter and terminator region. Transcription start site Promoter Terminator Structural gene Template strand

Coding strand Schematic structure of a transcription unit

Question 42. Write the salient features of genetic code.

Answer:

Salient features of genetic code:

(i) Triplet nature

(ii) Non-overlapping nature

(iii) Non-ambiguous nature

(iv) Genetic code is commaless

(v) Initiation codons : They function as start signals for the synthesis of polypeptides e.g., AUG

(vi) Termination codon : They are stop signals which cause termination of polypeptide synthesis, e.g., UAA, UAG, UGA

(vii) Degeneracy of codons : More than one codon specify one amino acid.

Question 43. Study the schematic representation of the genes involved in the lac operon given below and answer the following questions.
p i p o z y a
(i) Identity and name the regulatory gene in this operon. Explain its role in "switching off" the operon.
(ii) Identity and name the structural genes in this operon and trace the events that lead to the transcription of these structural genes.
(iii) Why the regulation of lac operon by repressor is referred to as negative control?
(iv) Name the inducer molecule and the product of 'y' gene. Write the function of this gene product.

Answer:

(i) Regulator gene is i-gene, it codes for repressor protein. Repressor protein covers the operator gene (O) so that RNA polymerase is unable to pass over and reach the structural genes for transcription.

(ii) z, y and a are structural genes. Presence of lactose or galactoside activates the operon. It binds to repressor and removes it from operator gene. RNA polymerase is now able to reach over the structural genes and transcribe them. Thus, switching on the operon.

(iii) Negative regulation is the one in which presence of a substrate or a factor inhibits a process. Here, presence of the product of regulatory gene acts as a repressor of lac operon.

(iv) Lactose or allolactose is the inducer of lac operon. Gene y codes for galactoside permease which permits entry of lactose into the cell.

Question 44. (a) Give an outline of the steps involved in technique that was initially developed by Alec Jeffreys.
(b) Mention the causes of DNA polymorphism.

Answer:

(a) Steps involved in DNA fingerprinting are :

(i) Extraction of DNA from source i.e., WBCs, semen, saliva, vaginal swabs, skin cells etc.

(ii) DNA amplification by PCR

(iii) Fragmentation of DNA by restriction enzymes

(iv) Separation of VNTRs, by gel electrophoresis

(v) Denaturation of VNTRs by alkali or high temperature treatment.

(vi) Southern blotting : Transfer of DNA from gel to synthetic membrane, i.e., nitrocellulose or nylon.

(vii) DNA probes synthesis

(viii) Hybridisation between VNTR and DNA probe

(ix) Exposure to X-ray film to recognise hybridised DNA segments.

(b) The cause of DNA polymorphism is

(i) Allelic polymorphism i.e., occurrence of multiple alleles of a gene.

(ii) Single nucleotide polymorphism : Over 1.4 million single base difference occurs in DNA.

(iii) RFLP (Restriction Fragment Length Polymorphism) : Differences appear in the restriction maps of individuals.

Question 45. Briefly describe the following :
(i) Template strand
(ii) VNTR
(iii) RFLP
(iv) Exons
(v) ESTs

Answer:

(i) Template strand : It is a strand of DNA which takes part in transcription. It shows 3′ → 5′ polarity. Its nucleotide sequence is complementary to one present in mRNA.

(ii) VNTR (Variable Number Tandem Repeats) : It is the part of satellite DNA or it is minisatellite sequence. The numbers of repeat show very high degree of polymorphism. As a result the size of VNTR varies in size from 0.1 to 20 Kb. It is the basis of DNA fingerprinting.

(iii) RFLP (Restriction Fragment Length Polymorphism) : The length of minisatellites and position of restriction sites is different for each person. Therefore, when the genomes of two people are cut using the same restriction enzyme, the length and number of fragments obtained are different for both. This is called RFLP.

(iv) Exons : Coding sequences of DNA that specify functional polypeptides.

(v) ESTs : Expressed sequence tags. Identity all the genes that are expressed as RNAs and sequence the same.