VBQs Class 12 Biology Molecular Basis of Inheritance

Please refer to VBQs for Class 12 Biology Molecular Basis of Inheritance. All value based questions for Biology Class 12 have been provided with solutions. We have provided below important values questions and answers. Students should learn these solved VBQs for Class 12 Biology as these will help them to gain more marks and help improve understanding of important topics.

Molecular Basis of Inheritance VBQs Class 12 Biology with Answers

The DNA

Very Short Answer Type Questions

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

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. 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. 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. Mention the carbon positions to which the nitrogenous base and the phosphate molecule are respectively linked in the nucleotide given below:

Answer.Nitrogenous base – First carbon
Phosphate molecule – 5′ carbon

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. Name the components ‘a’ and ‘b’ in the nucleotide with a purine given below:

Answer.(a) – Phosphate group
(b) – Purine nitrogenous base (adenine or guanine)

Short Answer Type Questions

Question. (a) Draw a neat labelled diagram of a nucleosome.
(b) Mention what enables histones to acquire a positive charge. 
Answer.(a) Diagram of nucleosome is as follows:

(b) Histones are rich in basic amino acids lysine and arginine, hence they are positively charged.

Question. Draw a schematic diagram of a part of double stranded dinucleotide DNA chain having all the four nitrogenous bases and showing the correct polarity. 
Answer. Diagrammatic representation of double stranded DNA chain having all four nitrogenous base and showing correct polarity is as follows :

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. The modified central dogma can be represented as follows:

Question. Answer the following based on the dinucleotide shown below.

(a) Name the type of sugar guanine base is attached to.
(b) Name the linkage connecting the two nucleotides.
(c) Identify the 3′ end of the dinucleotide.
Give a reason for your answer. 
Answer.(a) Guanine base is attached to a pentose sugar.
(b) Two nucleotides are linked through 3′-5′ phosphodiester linkage to form a dinucleotide.
(c) ‘B’ is the 3′ end of the given dinucleotide. It is because 3′ end of a nucleotide molecule bears a free 3′ – OH group of sugar residue and the 5′ end bears phosphate radical (5′ – P).

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. Study the given portion of double stranded polynucleotide chain carefully. Identify a, b, c and the 5′ end of the chain

Answer.a – Hydrogen bond
b – Purine i.e. adenine or guanine
c – Pentose sugar (deoxyribose sugar)
d is the 5′ end of the chain.

Question. How do histones acquire positive charge?
Answer.(b) Histones are rich in basic amino acids lysine and arginine, hence they are positively charged.

Question.

Look at the above sequence and mention the events A, B and C.
(b) What does central dogma state in molecular biology? How does it differ in some viruses?
Answer.(a) A – DNA replication
B – Transcription
C – Translation
(b) Central dogma in molecular biology states that the flow of information first occurs from DNA to mRNA by the process of transcription and then the information present in mRNA is decoded for the formation of polypeptide chain by the process of translation.
Central Dogma differs in retroviruses e.g., HIV, etc. where it is called central dogma reverse (inverse flow of information). RNA of these viruses first synthesises DNA through reverse transcription. DNA then transfers information to RNA which takes part in translation of coded information to form a polypeptide.

Short Answer Type Questions

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?
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. 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 H₁ 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) 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. (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 Chargaff’s 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
(b) Diagrammatic representation of double stranded DNA chain having all four nitrogenous base and showing correct polarity is as follows :

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.

(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.

(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.

(a) What is this diagram representing?
(b) Name the parts a, b and c.
(c) In the eukaryotes the DNA molecules are organized within the nucleus. How is the DNA molecule organised in a bacterial cell in absence of nucleus? 
Answer.(a) The diagram represents the structure of a nucleosome.
(b) a = Histone octamer
b = DNA molecule
c = Linker DNA
(c) In a bacterial cell, DNA lies in cytoplasm. It is supercoiled with the help of RNAs and non-histone basic proteins like polyamines. The compacted mass of DNA is called as nucleoid.

Question. Draw a labelled diagram of a nucleosome.
Where is it found in a cell?
Answer.Diagrammatic representation of nucleosome is as follows:

Nucleosome is found in the nucleus of a cell.

Long Answer Type Questions

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.

– 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.

– 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

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 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) 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 134 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 :

(b) The differences between euchromatin and heterochromatin are as follows :

The Search for Genetic Material

Very Short Answer Type Questions

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 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. 

Short Answer Type Questions

Question. Frederich Griffth 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.
Griffth 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
Griffth 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 Griffth 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. 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. 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. 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. 

Long Answer Type Questions

Question. (a) Describe the series of experiments of F. Griffth. Comment on the significance of the results obtained.
(b) State the contribution of MacLeod,Mc Carty and Avery. 
Answer.
(a) 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.
Griffth 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
Griffth 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 Griffth 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.
(b) Avery, Macleod and McCarty (1944) performed the biochemical characterisation of the ‘transforming principle’ of Griffth’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.
Table : Results of Avery’s Experiment

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. Name the scientists who proved experimentally that DNA is the genetic material. Describe their experiment. 
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. 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.
(b) Avery, Macleod and McCarty (1944) performed the biochemical characterisation of the ‘transforming principle’ of Griffth’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.
Table : Results of Avery’s Experiment

RNA World

Very Short Answer Type Questions

Question. Make a labelled diagram of an RNA dinucleotide showing its 3′ – 5′ polarity. 
Answer. Labelled diagram of RNA dinucleotide showing its 3′ – 5′ polarity is as follows:

Short Answer Type Questions

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 eficient and more varied.

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.
Figure representing polynucleotide chain of RNA is as follows: