CBSE Class 12 Biology Molecular Basis of Inheritance Question Bank

Long Answer Questions

Question. How is hnRNA processed to form mRNA? 
Answer. The hnRNA undergoes the following processes to form mRNA:
(i) Capping: Addition of methyl guanosine triphosphate at 5’-end.
(ii) Tailing: Addition of 200-300 adenylate residues at 3’-end.
(iii) Splicing: Removal of introns and rejoining of exons.

Question. What is hnRNA? Explain the changes hnRNA undergoes during its processing to form mRNA.
Answer. hnRNA is the precursor of mRNA that is transcribed by RNA ploymerase II and is called heterogenous nuclear RNA.
Changes:
• The hnRNA undergoes two additional processes called capping and tailing.
• In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5’-end of hnRNA.
• In tailing, adenylate residues (about 200–300) are added at 3’-end in a template independent manner.
• Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.
• Fragments on one of the template strands.

Question. (a) Identify the polarity from a to a′ in the given diagram and mention how many more amino acids are expected to be added to this polypeptide chain.
(b) Mention the DNA sequence coding for serine and the anticodon of tRNA for the same amino acid.
(c) Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA?

Answer. (a) Polarity from a to a’ is 5’→ 3′.
No more amino acid will be added to this polypeptide chain.
(b) TCA; anticodon is UCA.
(c) The untranslated sequence of bases are required for efficient translation process. They are present before the start codon at the 5′-end and after the stop codon at 3′-end.

Question. One of the codons on mRNA is AUG. Draw the structure of tRNA adapter molecule for this codon. Explain the uniqueness of this tRNA.
Answer. This tRNA is specific for amino acid Methionine and it also acts as initiator codon (initiator tRNA) 

Question. Explain the process of translation in a bacterium.
Answer. • Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
•  A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide,a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.

Question. Unambiguous, universal and degenerate are some of the terms used for the genetic code.
Explain the salient features of each one of them. 
Answer. Unambiguous code means that one codon codes for only one amino acid, e.g., AUG codes for only methionine.
Universal code means that codon and its corresponding amino acid are the same in all organisms, e.g., from bacteria to human, UUU codes for phenylalanine.
Degenerate code means that same amino acids are coded by more than one codon, e.g., UUU and UUC code for phenylalanine.

Question. State the conditions when ‘genetic code’ is said to be
(i) degenerate,
(ii) unambiguous and specific,
(iii) universal. 
Answer. (i) Degenerate—When some amino acids are coded by more than one amino acids.
(ii) Unambiguous and speci—ficWhen one codon codes for only one specific amino acid.
(iii) A particular codon codes for same amino acid in all organisms except in mitochondria and few protozoa.

Question. 

Study the mRNA segment given above which is complete to be translated into a polypeptide chain.
(i) Write the codons ‘a’ and ‘b’.
(ii) What do they code for?
(iii) How is peptide bond formed between two amino acids in the ribosome? 
Answer. (i) a is AUG and b is UAA/UAG/UGA
(ii) AUG codes for methionine (initiation codon).
UAA/UAG/UGA do not code for any amino acid, i.e., stop or terminating codons.
(iii) There are two sites (P-site and A-site) in the large subunit of ribosome, where subsequent amino acids bind to and thus are close enough to form peptide bond by peptidyl transferase enzyme. The ribosome also acts as a catalyst for the formation of peptide bond.

Question. “The codon is a triplet and is read in a contiguous manner without punctuations.” Provide the genetic basis for the statement.
Answer. Since there are only four bases which code for twenty amino acids, the code should be made up of three bases, i.e., (4 × 4 × 4) = 64 codons; a number more than the required.
If the codon consists of four letters, only (4 × 4), only sixteen codons are possible, which is less than the required. Hence the codon is a triplet.
As the ribosome moves on mRNA, continuously without break, the codons are read in a contiguous manner.

Question. Identify giving reasons, the salient features of genetic code by studying the following nucleotide sequence of mRNA strand and the polypeptide translated from it.
AUG UUU UCU UUU UUU UCU UAG
Met – Phe – Ser – Phe – Phe – Ser
Answer. 

Question. (a) Explain the observations of Meselson and Stahl when
(i) they cultured E. coli in a medium containing 15NH4Cl for a few generations and centrifuged the content.
(ii) they transferred one such bacterium to the normal medium of NH4Cl and cultured for 2 generations?
(b) What does the above experiment prove?
(c) Which is the first genetic material identified?
Answer. (a) (i) Meselson and Stahl observed that in the E. coli bacterium the DNA becomes completely labelled with 15N medium by centrifugation for few generations.
(ii) After two generations, density changed and showed equal amount of light DNA (14N) and dark hybrid DNA (15N–14N).
(b) They concluded that DNA replicates semi-conservatively.
(c) Ribonucleic acid (RNA) was the first genetic material.

Question. (a) State the ‘Central dogma’ as proposed by Francis Crick. Are there any exceptions to it?Support your answer with a reason and an example.
(b) Explain how the biochemical characterisation (nature) of ‘Transforming Principle’ was determined, which was not defined from Griffith’s experiments.
Answer. (a) Francis Crick proposed the central dogma of molecular biology which states that genetic information flows from DNA to mRNA (transcription) and then from mRNA to protein (translation) always unidirectionally (except bidirectionally in some viruses and the process is called reverse transcription).

Yes, there are some exceptions to it. In some viruses flow of information is in reverse direction (reverse transcription).

Question. (a) Name the molecule ‘X’ synthesised by ‘i’ gene. How does this molecule get inactivated?
(b) Which one of the structural genes codes for b-galactosidase?
(c) When will the transcription of this gene stop?

Answer (a) The molecule ‘X’ is repressor. It gets inactivated when lactose (inducer) binds with the repressor molecule.
(b) z gene codes for b-galactosidase.
(c) Transcription of the gene stops when lactose is absent and thus repressor is free to bind with the operator.

Question. Draw a schematic diagram of lac operon in its ‘switched off’ position. Label the following:
(i) The structural genes (ii) Repressor bound to its correct position
(iii) Promoter gene (iv) Regulatory gene. 
Answer. (i) z, y and a are structural genes.
(iii) p is the promoter sequence.
(iv) i is the regulatory gene.

Question. (a) Explain the process of aminoacylation of tRNA. Mention its role in translation.
(b) How do ribosomes in the cells act as factories for protein synthesis?
(c) Describe ‘initiation’ and ‘termination’ phases of protein synthesis.
Answer. (a) Aminoacylation is the process by which amino acids become activated by binding with its aminoacyl tRNA synthetase in the presence of ATP.
If two charged tRNAs come close during translation process, the formation of peptide bond between them in energetically favourable.
(b) The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists
as two subunits: a large subunit and a small subunit. When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins. There are two sites in
the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst (23S rRNA in
bacteria is the enzyme ribozyme) for the formation of peptide bond.
(c) • Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
• A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide, a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.

Question. (a) Name the scientist who postulated the presence of an adapter molecule that can assist in protein synthesis.
(b) Describe its structure with the help of a diagram. Mention its role in protein synthesis.
Answer. (a) Francis Crick
(b) Structure 

Question. In a maternity clinic, for some reasons the authorities are not able to hand over the two newborns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter. 
Answer. The technique is DNA fingerprinting. It includes the following steps:
Methodology and Technique
(i) DNA is isolated and extracted from the cell or tissue by centrifugation.
(ii) By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification. 
(iii) DNA is cut into small fragments by treating with restriction endonucleases.
(iv) DNA fragments are separated by agarose gel electrophoresis.
(v) The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye.
(vi) The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting.
(vii) VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation.

Question. Describe the elongation process of transcription in bacteria. 
Answer. After initiation, RNA polymerase loses the s factor but continues the polymerisation of ribonucleotides to form RNA. It uses nucleoside triphosphates as substrate and polymerises in a template-dependent fashion, following the rule of complementarity. For diagram,

Question. Describe the termination process of transcription in bacteria. 
Answer. Once the RNA polymerase reaches the termination region of DNA, the RNA polymerase is separated from DNA–RNA hybrid, as a result nascent RNA separates. This process is facilitated by a termination factor r (rho). In prokaryotes, mRNA does not require any processing, so transcription and translation both occur in the cytosol.
For diagram,

Question. Write any six salient features of the human genome as drawn from the human genome project. 
Answer. (i) The human genome contains 3164.7 million nucleotide bases.
(ii) The average gene consists of 3000 bases; the largest known human gene being dystrophin at 2.4 million bases.
(iii) The total number of genes is estimated to be 30,000 and 99.9 per cent nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50 per cent of the discovered genes.
(v) Less than 2 per cent of the genome codes for proteins.
(vi) The human genome contains large repeated sequences, repeated 100 to 1000 times.
(vii) The repeated sequence is thought to have no direct coding functions but they throw light on chromosome structures, dynamics and evolution.
(viii) Chromosome 1 has most genes (2968) and the Y has the fewest genes (231).
(ix) Scientists have identified about 1.4 million locations where single base DNA sequence differences called SNPs or single nucleotide polymorphism (pronounced as ‘snips’) occur in humans.This information promises to revolutionise the processes of finding chromosomal locations for disease—associated sequences and tracing human history.

Question. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and write the procedure that would help in the identification of kinship. 
                                                                          OR
A number of passengers were severely burnt beyond recognition during a train accident.
Name and describe a modern technique that can help to hand over the dead to their relatives.
Answer. DNA fingerprinting can help in identification of kinship.
For procedure, • Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases.
• Human genome has 3 × 109 bp. 99.9% of base sequences among humans are the same, which makes every individual unique in phenotype.

Question. (a) Draw a schematic representation of the structure of a transcription unit and show the following in it:
(i) Direction in which the transcription occurs
(ii) Polarity of the two strands involved
(iii) Template strand
(iv) Terminator gene
(b) Mention the function of promoter gene in transcription.
Answer. (a) (i) Transcription occurs in 5′→3′.

(b) Promotor gene has DNA sequence that provide binding site for RNA polymerase.

Question. Describe the initiation process of transcription in bacteria. 
Answer. In bacteria, the transcription of all the three types of RNA (mRNA, tRNA, rRNA) is catalysed by single DNA-dependent enzyme called the RNA polymerase. The RNA polymerase has cofactors that catalyse the process. During initiation, s (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription.
For diagram,

Question. Observe the representation of genes involved in the lac operon given below: 
(a) Identify the region where the repressor protein will attach normally.
(b) Under certain conditions repressor is unable to attach at this site. Explain.
(c) If repressor fails to attach to the said site what products will be formed by z, y and a?
(d) Analyse why this kind of regulation is called negative regulation.
Answer. (a) The repressor protein will attach to operator region, o.
(b) In presence of an inducer, lactose, repressor is unable to attach.
(c) z—b galactosidase.
y—Permease
a—Transacetylase
(d) It is called negative regulation as it involves constitutive (all the time) repressor. The operon is always in off position due to presence of repressor and is switched on only in presence of an inducer. Inducer Lactose or allolactose interacts with repressor making it inactive.

Question. Name the transcriptionally active region of chromatin in a nucleus. 
Answer. Euchromatin is the transcriptionally active region of chromatin in a nuclecus.

Question. Name the negatively charged and positively charged components of a nucleosome.
Answer.The negatively charged and positively charged components of a nucleosome are DNA and histones respectively.

Question. Name the specifc components and the linkages between them that form deoxyguanosine.
Answer. The specific components that form deoxyguanosine are guanine (nitrogenous base) and deoxyribose (pentose sugar) linked together by glycosidic bond.

Question. Name the two basic amino acids that provide positive charges to histone proteins.
Answer. Lysine and arginine are the two basic amino acids that provide positive charge to histone proteins.

Question. If the base adenine constitutes 31 percent of an isolated DNA fragment, then what is the expected percentage of the base cytosine in it?
Answer. According to Chargaff ’s rule, [A] + [G] = [C] + [T] = 50%
Therefore, if [A] = 31%, then [T] = 31% [C] + [T] = 50%
Therefore, [C] = 50% – 31% = 19%

Question. Name the positively charge proteins around which the negatively charged DNA is wrapped.
Answer. Histones are the positively charged proteins around which negatively charged DNA is wrapped.

Question. How is the length of DNA usually calculated?
Answer. Length of DNA segment is usually calculated by finding the number of base pairs and multiplying it by the distance between adjoining base pairs.

Question. What is Central dogma ? Who proposed it?
Answer. Concept of central dogma was proposed by Crick in 1958. It refers to the flow of information from DNA to mRNA (transcription) and then decoding the information present in mRNA in the formation of polypeptide chain or protein (translation). 

Question. State the central dogma in molecular biology.Who proposed it? Is it universally applicable?
Explain.
Answer. Central dogma in molecular biology states that :
– The DNA replicates its information in a process called replication that involves many enzymes and proteins.
– The DNA codes for the production of messenger RNA (mRNA) during transcription.
– Messenger RNA carries coded information to ribosomes. The ribosomes use this information for protein synthesis. This process is called translation.
Francis Crick (1958) proposed central dogma in molecular biology. It is not universally applicable as retroviruses with the help of enzyme reverse transcriptase perform the reverse transcription of its genome from RNA into DNA.

Question.(a) A DNA segment has a total of 1000 nucleotides, out of which 240 are adenine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer. (a) According to Chargaffs rule,
[A] + [G] = [C] + [T]
Also [A] = [T] and [C] = [G]
As [A] = [T], therefore [T] = 240
[A] + [T] = 240 + 240 = 480
As total number of nucleotides = 1000,
therefore [G] + [C] = 1000 – 480 = 520
[G] = [C]
therefore, [G] = [C] = 520/2
=260
Thus, total number of pyrimidines.
i.e. [C] + [T] = 260 + 240 = 500

Question. (a) A DNA segment has a total of 1,500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this segment possesses?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer. a) Cytosine and thymine are pyrimidines.
According to Chargaff’s Rule, purines and pyrimidine
base pairs are in equal amount, therefore
Total nucleotides = 1500
[A + G + C + T] = 1500
[A] = [T] and [G] = [C]
Guanine = 410
Therefore, A + 410 + 410 + T = 1500
A + T + 820 = 1500
A + T = 1500 – 820
A + T = 680
T = 680/2
= 340
Therefore, total pyrimidine,
C + T = [410 + 340] = 750

Question. (a) A DNA segment has a total of 2,000 nucleotides, out of which 520 are adenine containing nucleotides. How many purine bases this DNA segment possesses?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer. (a) According to Chargaff’s rule,
[A] + [G] = [C] + [T]
Also [A] = [T] and [G] = [C]
As [A] = [T], therefore [A] = [T] = 520
[A] + [T] = 520 + 520 = 1040
As total number of nucleotides = 2000
therefore, [G] + [C] = 2000 – 1040 = 960
[G] = [C] =
960/2 = 480
Thus, total number of purines i.e.
[A] + [G] = 520 + 480 = 1000

Question. The base sequence in one of the strands of DNA is TAGCATGAT.
(a) Give the base sequence of its complementary strand.
(b) How are these base pairs held together in a DNA molecule?
(c) Explain the base complementarity rules.
Name the scientist who framed this rule
Answer.
(a) The base sequence of the complementary strand is ATCGTACTA.
(b) The base pairs in a DNA molecule are held together by hydrogen bonds.
(c) Base complementarity rules or Chargaff’s rules are the important generalisations made by Chargaff (1950) on the bases and other components of DNA. These rules are as follows:
(i) Purine and pyrimidine base pairs are in equal amount, that is,
adenine + guanine = thymine + cytosine.
[A + G] = [T + C], i.e., [A+G]/[T+ C] = 1
(ii) Molar amount of adenine is always equal to the molar amount of thymine. Similarly, molar concentration of guanine is equalled by molar concentration of cytosine.
[A] = [T], i.e., [A]/[T] = 1 ; [G] = [C], i.e., [G ]/[C]= 1
(iii) Sugar deoxyribose and phosphate occur in equimolar proportions.
(iv) A–T base pairs are rarely equal to C–G base pairs.
(v) The ratio of [A+T]/[G +C] is variable but constant for a species. It can be used to identify the source of DNA. The ratio is low in primitive organism and
higher in advanced ones.

Question. The length of a DNA molecule in a typical mammalian cell is calculated to be approximately 2.2 meters. How is the packaging of this long molecule done to accommodate it within the nucleus of the cell?
Answer. There is a set of positively charged, basic proteins called histones (rich in the basic amino acid residues lysines and arginines) used for DNA packing. The histone proteins are of 5 types (H1, H2A, H2B, H3, H4). Four of them (except H1) occur in pair to produce octamer or core particle. The DNA wrap around this
octamer by 1 3/4 turns and are plugged by H protein.
This is called nucleosome. A typical nucleosome contains approximately 200 bp of DNA. Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin. The nucleosomes in chromatin are seen as ‘beads -on-string’ structure when viewed under electron microscope. The beads-on-string structure in chromatin is packaged to form chromatin fibres that are further coiled and condensed at metaphase stage of cell division to form chromosomes. The packaging of chromatin at higher level requires additional set of proteins called non-histone chromosomal proteins.

Question. (a) How are the following formed and involved in DNA packaging in a nucleus of a cell?
(i) Histone octamer
(ii) Nucleosome
(iii) Chromatin
(b) Differentiate between euchromatin and heterochromatin. 
Answer. (a) (i) Histone octamer : Histones are positively charged proteins, rich in basic amino acid residues lysines and arginines. These amino acids carry positive charges on their side chains. There are five types of histone proteins : H1, H2A, H2B, H3 and H4. Four of them (H2A, H2B, H3 and H4) are organised in pairs to from a unit of eight molecules called histone octamer, nu body or core of nucleosome. Negatively charged DNA wraps around this octamer to form nucleosome.
(ii) Nucleosome : It is the compaction unit. The positively charged ends of histone octamer attract the negatively charged strands of DNA. The DNA is thus wrapped around the positively charged histone octamer to form a structure called nucleosome. Around 200 bp of DNA is wrapped around the nu body or histone octamer for 13/4 turns. DNA connecting two adjacent nucleosomes is called linker DNA which bears H1 histone protein. Nucleosome and linker DNA together constitute chromatosome. Nucleosome chain gives a bead on string appearance under electron microscope.
(iii) Chromatin : The nucleosomal organisation has approximately 10 nm thickness, which further gets condensed and coiled to produce a solenoid (having 6 nucleosomes per turn) of 30 nm diameter. This solenoid structure further undergoes coiling to produce a chromatin fibre of 30-80 nm thickness. These chromatin fibres are further coiled and condensed to form chromatid which further forms chromosome at metaphase stage of cell division.
The packaging can be summarised as follows :
DNA       →  Nucleosome → Solenoid → Chromatin fibre
(2nm           (10 nm           (30 nm         (30-80 nm
diameter)     diameter)       diameter)         diameter)
                                                                 ↓
                                                                 Chromatid
                                                                 (700 nm diameter)
                                                                 ↓
                                                                 Chromosome
                                                                 (1400 nm diameter)
(b) The differences between euchromatin and heterochromatin are as follows :

Question. (a) Explain the chemical structure of a single stranded polynucleotide chain.
(b) Describe the salient features of the doublehelix structure of DNA molecule.
Answer. (a) A polynucleotide chain is formed by the end to end polymerisation of a large number of nucleotides. A nucleotide is a condensation product of three chemicals – a pentose sugar, phosphoric acid and a nitrogenous base.
The nitrogen base combines with the sugar molecule at its carbon atom 1′ in a glycosidic bond (C–N–C) by one of its nitrogen atoms (usually 1 in pyrimidines and 9 in purines).
The phosphate group is connected to carbon 5′ of the sugar residue of its own nucleotide and carbon 3′ of the sugar residue of the next nucleotide by phosphodiester bonds. –H of phosphate and –OH of sugar are eliminated as H2O during each ester connecting two adjacent nucleosomes is called linker formation.

At the end of the polynucleotide chain, last sugar has its 5–C free while at the other end 3–C of first sugar is free. They are respectively called 5′ and 3′ ends.
(b) Watson and Crick proposed the double helix model for structure of DNA in 1953. Its salient features are as follows:
– In a DNA double helix, two polynucleotide chains are coiled to form a helix. Sugarphosphate forms backbone of this helix while bases project inwards towards each other.
– Complementary bases, pair with each other through hydrogen bonding. Purines (adenine, guanine) always pair with their corresponding pyrimidines (thymine, cytosine). Adenine pairs with thymine through two hydrogen bonds while guanine pairs with cytosine through three hydrogen bonds.
– The helix is right-handed.
– The plane of one base pair stacks over the other in a double helix. This provides stability to the helix along with hydrogen bonding.
– The two chains of DNA have antiparallel polarity, 5′ → 3′ in one and 3′ → 5′ in other.
– The pitch of helix is 3.4 nm (34 Å) with roughly 10 base pairs in each turn. The average distance between two adjacent base pairs comes to about 0.34 nm (0.34 × 10–9 m or 3.4 Å).
– DNA is acidic. For its compaction, it requires basic (histone) proteins. The histone proteins are positively charged and occupy the major grooves of DNA at an angle of 30° to helix axis.

Question. (a) Mention the contributions of the following scientists:
(i) Maurice Wilkins and Rosalind Franklin
(ii) Erwin Chargaff
(b) Draw a double-stranded dinucleotide chain with all the four nitrogen bases. Label the polarity and the components of the dinucleotide.
Answer. (a) (i) Maurice Wilkins and Rosalind Franklin (1953) carried out X-ray diffraction studies to study the structure of DNA molecule. The fine X-ray photographs of DNA taken by them showed that DNA was a helix with a width of 2.0 nm. One turn of the helix was 3.4 nm with 10 layers of bases stacked in it. Waston and Crick (1953) worked out the first correct double helix model from the X-ray photographs of Wilkins and Franklin.
(ii) Erwin Chargaff (1950) proposed generalisations called Chargaff ’s rules (or base complimentarity rules) about DNA. These generalisations are as follows:
– Purine and pyrimidine base pairs are in equal amount, that is,
adenine + guanine = thymine + cytosine.

[A + G] = [T + C], i.e., [A+ G]/[T+ C] = 1 
Molar amount of adenine is always equal to the molar amount of thymine. Similarly, molar concentration of guanine is equalled by molar concentration of cytosine.
[A] = [T], i.e., [A]/[T]
=1;[G]=[C], . ., [G]/[C] i e =1
– Sugar deoxyribose and phosphate occur in equimolar proportions.
– A – T base pairs are rarely equal to C – G base pairs.
– The ratio of [A+T]/[G+ C] is variable but constant for a species. 

Question. Why is RNA more reactive in comparison to DNA?
Answer. RNA is more reactive in comparison to DNA because:
– 2′ –OH group present in ribose sugar of every nucleotide of RNA is a reactive group. It makes RNA highly reactive, labile and easily degradable.
– RNA functions as an enzyme, therefore is reactive and unstable.

Question. Why is DNA considered a better genetic material?
Answer. (i) DNA is chemically less reactive and structurally more stable as its nucleotides are not exposed except when they are to express their effect.
(ii) Presence of thymine in DNA instead of uracil in RNA, provides stability to DNA.
(iii) Hydrogen bonding between purines and pyrimidines and their stacking make DNA more stable for storage of genetic information.
(iv) DNA is capable of undergoing slow mutations. (v) It has power of repairing.
Thus, DNA which is stable enough not to change with different stages of life cycle, age or with change in metabolism of the organism, is a better material for the storage of genetic information 

Question. Why is DNA a better genetic material when compared to RNA? 
Answer. The criteria which makes DNA a better genetic material than RNA are as follows :
(i) DNA is chemically less reactive and structurally more stable than RNA as its nucleotides are not exposed except when they are to express their effect whereas 2′ –OH group in ribose sugar of every nucleotide of RNA makes it more reactive. RNA also functions as an enzyme and is therefore more reactive and unstable.
(ii) Presence of thymine in DNA instead of uracil in RNA provides stability to DNA.
(iii) Hydrogen bonding between purines and pyrimidines and their stacking make DNA more stable for storage of genetic information than RNA. (iv) DNA has power of repairing and there is no such repairing mechanism in RNA.

Question. Frederich Grifth claimed that R-strain Streptococcus pneumoniae had been transformed by heat-killed S-strain bacteria. Explain the findings. 
Answer. Transformation is the phenomenon by which the DNA isolated from one type of cell, when introduced (artificially or naturally) into another type, is able to bestow some of the properties of the former to the latter.
Griffith observed transformation in Streptococcus pneumoniae (bacterium responsible for causing pneumonia). He grew bacteria on a culture plate, some produced smooth shiny colonies (S) while others produced rough colonies (R). Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia.
S strain → Injected into mice → Mice die
R strain → Injected into mice → Mice live
Griffith observed that heat-killed S strain bacteria, when injected into the mice, did not kill them.
S strain (heat-killed) → Inject into mice → Mice live When he injected a mixture of heat-killed S strain and live R bacteria, the mice died. Moreover, he recovered living S strain bacteria from the dead mice.
S strain (heat - killed) + R strain (live)
→ Injected into mice → Mice die From the experiment Griffith concluded that the R strain bacteria had been transformed by the heat-killed S strain bacteria as some ‘transforming principle’ transferred from heat killed S strain enabled the R strain to become virulent.

Question. Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase.
Answer. Hershey and Chase used following two types of culture media :
(i) Medium that contained radioactive phosphorus (P32).
(ii) Medium that contained radioactive sulphur (S35).
They used two different kinds of culture media to detect whether the genetic material is DNA or protein.
Viruses grown in the medium with radioactive phosphorus contained radioactive DNA but not radioactive protein as DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur medium contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

Question. List any four properties of a molecule to be able to act as a genetic material. 
Answer. A molecule that can act as genetic material must fulfill the following criteria :
(a) It should be able to generate its replica.
(b) It should chemically and structurally be stable.
(c) It should provide the scope for slow changes that are required for evolution.
(d) It should be able to express itself in the form of ‘Mendelian characters’.

Question. Answer the following questions based on
Hershey and Chase experiments :
(a) Name the kind of virus they worked with and why?
(b) Why did they use two types of culture media to grow viruses in? Explain.
(c) What was the need for using a blender and later a centrifuge during their experiments?
(d) State the conclusion drawn by them after the experiments.
Answer. (a) Hershey and Chase worked with virus T2 bacteriophage. T2 bacteriophage is a bacterial virus which has ability to infect Escherichia coli and it possess linear double stranded DNA (deoxyribose nucleic acid) as genetic material, therefore they used this bacteriophage or virus for their work.
(b) Hershey and Chase used following two types of culture media :
(i) Medium that contained radioactive phosphorus (P32).
(ii) Medium that contained radioactive sulphur (S35).
They used two different kinds of culture media to detect whether the genetic material is DNA or protein.
Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
(c) Blender was used in experiment to remove the empty phage capsids (or ghosts) sticking to the surface of bacteria.
Centrifuge was used to separate virus particle from the bacteria. The bacterial cultures were centrifuged. Both the supernatant and the pellets were checked for radioactivity. In culture with radioactive S35 it was found that phage with labelled protein did not make bacteria labelled. Instead, radioactivity was restricted to supernatant which was found to contain only capsid. On the other hand, in the second culture with P32 it was found that supernatant containing capsid was not radioactive instead bacteria become labelled proving that only DNA of the phage entered the bacteria.
(d) Hershey and Chase from this experiment concluded that genetic material is DNA and not the protein.

Question. (a) Describe the series of experiments of F. Grifth. Comment on the significance of the results obtained.
(b) State the contribution of MacLeod, Mc Carty and Avery. 
Answer. (b) Avery, Macleod and McCarty (1944) performed the biochemical characterisation of the ‘transforming principle’ of Griffith’s experiment. They separated the extract of smooth, virulent bacteria into protein, DNA and carbohydrate fractions. Each fraction was separately added to a culture medium containing live rough bacteria. Only the culture that received the DNA fraction of the extract from virulent bacteria produced smooth bacteria. This proved that DNA was the transforming agent. When DNA fraction was treated with deoxyribonuclease (an enzyme that digests DNA), it became inactive and incapable of transforming the rough strain into the smooth strain. This confirmed that DNA is the genetic material.
The following table represent the result of Avery’s experiment.

Question. How did Hershey and Chase established that DNA is transferred from virus to bacteria?
Answer. Alfred D. Hershey and Martha Chase, chose T2 bacteriophage as their experimental material. They decided to see which of the bacteriophage components-protein or DNA-entered bacterial cells and directed reproduction of the virus.
Hershey and Chase experiment is based on the fact that DNA but not the protein contains phosphorus, and similarly sulphur is present in proteins (cysteine and methionine) but not in DNA. They incorporated radioactive isotope of phosphorus (32P) into phage DNA and that of sulphur (35S) into proteins of separate phage cultures. These phage types were
used independently to infect the bacterium Escherichia coli. After sometime, the cultures were agitated in a blender to separate the empty phage capsids from the surface of bacterial cells and the two were separated by centrifugation. Hershey and Chase showed that in bacterial cells, infected with virus containing radioactive phosphorus (32P), radioactivity was associated with bacterial cells and also, appeared in the progeny phage. However, in bacterial culture where radioactive sulphur (35S) was used, all radioactive material was limited to phage
‘ghosts’ (empty viral protein coats).
These results indicated that the DNA of the bacteriophage and not the protein enters the host, where viral replication takes place. Therefore, DNA is the genetic material of T2 bacteriophage. It directs protein coat synthesis and allows replication to occur. 

Question. (a) Write the conclusion drawn by Griffth at the end of his experiment with Streptococcus
pneumoniae.
(b) How did O. Avery, C. MacLeod and M. McCarty prove that DNA was the genetic material? Explain. 
Answer. a) From his experiment Griffth concluded that R strain bacteria has been transformed by heat killed S strain bacteria as some ‘transforming principle’ from it enabled R strain bacteria to become virulent.

Question. Name the scientists who proved experimentally that DNA is the genetic material. Describe their experiment. (Delhi 2012)
Answer. The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952).
They decided to see which of the bacteriophage components (protein or DNA) entered bacterial cells and directed reproduction of the virus.
Their experiment is based on the fact that DNA but not the protein contains phosphorus, and similarly sulphur is present in proteins but not in DNA. They incorporated 32P (radioactive isotope of phosphorus) into phage DNA and 35S (radioactive isotope of sulphur) into proteins of a separate phage culture. These phage types were independently used to infect the bacterium Escherichia coli.
After sometime, this mixture was agitated in a blender. And the two were separated by centrifugation. Harshey and Chase observed that when 32P was used, all radioactivity was associated with bacterial cells and if followed, appeared in the progeny phage. However, when 35S was used, all radioactive material was limited to the protein coats.
These results confirmed that the DNA is the genetic material. 

Question. Describe the structure of a RNA polynucleotide chain having four different types of nucleotides.
Answer. RNA is single stranded or chain (ds RNA is reported in wound tumour virus, Rice Dwarf virus) which is formed by end to end polymerisation of a number of ribonucleotides or ribotides. Four types of ribonucleotides occur in RNA. They are adenosine monophosphate, guanosine monophosphate, uridine monophosphate and cytidine monophosphate. A ribonucleotide is formed of ribose sugar, phosphoric acid and a nitrogen base. The four nitrogen bases present in RNA are adenine, guanine (purines), cytosine and uracil (pyrimidines). The union of nitrogen base is with carbon 1′ of ribose sugar by glycosidic bond through 3 N or 9 N region. Phosphate combines with carbon 5′ of its sugar and carbon 3′ of next sugar by phosphodiester bond. 

Question. It is established that RNA is the first genetic material. Explain giving three reasons.
Answer. The first genetic material was RNA. It can be explained as :
(a) Metabolism, splicing and translation evolved around RNA.
(b) The first biocatalysts were RNAs. Even now some enzymes are made of RNAs, e.g., ribozyme
(c) RNAs worked well in early unstable environmental conditions. As the environment become stable, RNAs were replaced in two of its functions (i) By small chemical modifications RNA gave rise to DNA as genetic material. (ii) For biocatalysis, RNA was replaced by protein enzymes. The latter were more stable, more efficient and more varied.

Question. (a) Name the enzyme that catalyses the transcription of hnRNA.
(b) Why does the hnRNA need to undergo changes? List the changes hnRNA undergoes and where in the cell such changes take place.
Answer. (a) RNA polymerase II transcribes hnRNA.
(b) Post transcription processing of hnRNA is required to convert primary transcript of all types of RNA into functional RNAs. The eukaryotic transcription involves certain complexity, one of the complexity is that the primary transcript contain both the exons and the introns (which are non-functional). Hence, it is subjected to a process called splicing where the introns are removed and exons are joined in a defned order. The transcribed eukaryotic RNA or heterogenous nuclear RNA (hnRNA), undergoes additional processing called as capping and tailing.
In capping, an unusual nucleotide (methyl guanosine triphosphate) is added to the 5’ - end of hnRNA. In tailing, adenylate residues (200 - 300) are added at the 3′-end in a template independent manner. It is the fully processed hnRNA, now called mRNA. This process takes place in the nucleus of a cell and then mRNA is transported out of the nucleus for translation.

Question. Monocistronic structural genes in eukaryotes have interrupted coding sequences. Explain. How are they different in prokaryotes?
Answer. In eukaryotes, monocistronic structural genes could be said as interrupted, as both introns and exons are present. 

Question. How does a degenerate code differ from an unambiguous one? 
Answer. The  difference  between  unambiguous   and degenerate codons is :

Question. Mention the role of the codons AUG and UGA during protein synthesis. 
Answer. AUG  has  dual   functions.   It   functions   as initiation codon during protein synthesis and also codes for methionine.  UGA does not  specify any amino acid hence it functions as terminator codon.

Question. Following are the features of genetic codes.What does each one indicate?
Stop codon; Unambiguous codon; Degenerate codon; Universal codon. 
Answer. (a) Stop codon : Codons that do not code for any amino acids and signal polypeptide chain termination. E.g., UAA, UAG, UGA.
(b) Unambiguous codon : Codons that specify only one amino acid and not any other. E.g., AUG codes for methionine.
(c) Degenerate codon : More than one codons codes for a single amino acid. In degenerate codons,generally the first two nitrogen bases are similar while the third one is different. E.g., UUU and UUC codes for phenylalanine.
(d) Universal codon : A codon that is applicable universally i.e., specifies the same amino acid from a virus to a tree or human being.

Question. (a) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(b) Explain the basis on which he arrived at this conclusion. 
Answer. (a) George Gamow suggested that the genetic code should be made up of three nucleotides.
(b) He proposed that since there are only 4 bases and if they have to code for 20 amino acids, the code should constitute a combination of bases made up of three nucleotides. Combination of 43 (4 × 4 × 4) would generate 64 codons; generating many more codons than required.

Question. Write the two specific codons that a translational unit of mRNA is flanked by one on either sides.
Answer. The two specific codons are initiation  codon on one side (AUG or GUG) and termination codon (UAA, UAG or UGA) on the other side of mRNA.

Question.Which one of the two sub-units of ribosome encounters an mRNA? 
Answer. Smaller sub-unit of ribosome encounters mRNA during initiation of protein synthesis.

Question. What is aminoacylation? State its significance.
Answer. Aminoacylation or charging of the tRNA is the process during which the amino-acyl-adenylate- enzyme complex reacts with tRNA specific for the amino acid to form aminoacyl-tRNA complex. Enzyme and AMP are released. tRNA complexed with amino acid is sometimes called charged tRNA. The amino acid is linked to 3-OH-end of tRNA through its -COOH group,
AA ∼ AMP — E + tRNA → AA -tRNA + AMP + E
Aminoacyl adenylate enzyme
The aminoacyl-tRNA complex specific for the initiation codon reaches the P-site to initiate the process of protein synthesis.

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