2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance

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Karnataka 2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance

2nd PUC Biology Molecular Basis of Inheritance One Mark Questions and Answers

Question 1.
Name the components of a nucleoside?
Answer:
They are
(a) Pentose sugar
(b) Nitrogen base.

Question 2.
Name the purines of DNA?
Answer:
(a) Adenine
(b) Guanine

Question 3.
Name the pyrimidines of DNA?
Answer:
(a) Cytosine
(b) Thymine

Question 4.
Who proposed the double helix model for DNA?
Answer:
Watson and Crick.

Question 5.
Name the types of RNA.
Answer:

Ribosomal RNA (rRNA)
Messenger RNA (mRNA)
Transfer RNA (tRNA)

Question 6.
Mention the nitrogen base present in RNA but not in DNA.
Answer:
Uracil

Question 7.
Mention the nitrogen base present in DNA but not in RNA.
Answer:
Thymine

Question 8.
Who discovered nucleic acids?
Answer:
Frederick Miescher.

Question 9.
Who discovered the semi conservative nature of DNA replication?
Answer:
Meselson and Stahl.

Question 10.
What is polycistronic mRNA?
Answer:
mRNA carrying genetic information of many genes for synthesis of protein molecules is called polycistronic mRNA.

Question 11.
Who gave the name ‘Nucleic Acid’?
Answer:
Altmann.

Question 12.
What are the two components of adenosine?
Answer:
Ribose sugar and adenine nitrogen base.

Question 13.
Define Gene?
Answer:
Gene is the part of DNA which is responsible for controlling the expression of a character.

Question 14.
Define Cistron?
Answer:
Cistron is the structural unit of gene which contains the genetic information of a character.

Question 15.
Define Muton.
Answer:
It is a small unit of cistron where mutation occurs causing a change in the gene.

Question 16.
Define recon.
Answer:
It is the smallest unit of DNA, capable of genetic recombination.

Question 17.
Define genetic code.
Answer:
The sequence of arrangement of nitrogen bases in the DNA, which determine the sequence of arrangement of amino acids duringprotein synthesis is called ‘genetic code’.

Question 18.
Define Triplet code?
Answer:
A set of three nucleotides that specifies an amino acid constitute a triplet code.

Question 19.
Define Transcription?
Answer:
The process of transfer of genetic message from DNA to mRNA in nucleotide language is called Transcription.

Question 20.
Define Translation?
Answer:
The process by which the protein is synthesised as per the message contained in the codons of mRNA is called Translation.

Question 21.
Who proposed the “Operon concept”?
Answer:
Jacob and Monad.

Question 22.
Name the initiator triplet cordon of protein synthesis?
Answer:
AUG

Question 23.
Define genetic code
Answer:
Relationship between the sequence of nucleotides or bases on mRNA and the sequence of amino acids in the polypeptide.

Question 24.
What is translation.
Answer:
The process of translation of nucleotide language of mRNA into aminoacid language by tRNA is called translation.

Question 25.
Define transcription.
Answer:
Synthesis of messenger RNA on DNA template in the presence of RNA polymerase enzyme is called transcription.

Question 26.
Explain (in one or two lines) the functions of following:
(a) Promoter (b) tRNA (c) Exons.
Answer:
(a) Promoter: It acts as an initiation signal which functions as recognition centre for RNA – polymerase provided the operator gene is switched on.

(b) tRNA: It acts as an carrier molecule which transfers amino acids to ribosomes for synthesis of polypeptides.

Question 27.
Give the initiation codon for protein synthesis. Name the amino acid it codes for?
Answer:
Initiation codon – AUG and it codes for methionine.

Question 28.
In which direction, is the new strand of DNA synthesised during DNA replication.
Answer:
51 to 31

Question 29.
Name the enzyme that joins the short pieces in the lagging strand during synthesis
Answer:
Ligase.

Question 30.
Name the enzyme which helps in the formation of peptide bond?
Answer:
Peptidyl transferase.

Question 31.
Name the three non-sense codons?
Answer:
UAA, UAG, UGA.

Question 32.
Mention the dual functions of AUG?
Answer:
AUG codes for amino acid methionine and also acts as an initiator codon.

Question 33.
Heterochromatin is transcriptionally inactive when compared to euchromatin. Give reason.
Answer:
Heterochromatin is densely packed than eurchromatin (takes more time for uncoiling during transcription)

2nd PUC Biology Molecular Basis of Inheritance Two Marks Questions and Answers

Question 1.
What are “nonsense codons” name them.
Answer:
Out of 64 triplet codons only 61 code for amino acids. Three codons which do not code for amino acids are called nonsense codons.
These are UAA, UAG and UGA. They are also called as ‘terminator codons’.

Question 2.
What is a nucleotide? Give an example.
Answer:
Nucleotide is a monogenic unit of nucleic acid. It consists of a pentose sugar linked with a nitrogen base and a phosphate group. The base and sugar are joined by glycosidic or N-ribosidic bond and the sugar and phosphate are joined by ester bond.
e.g: Nucleoside + phosphate = Nucleotide
Deoxy adenosine + phosphate = Deoxy adenylic acid

Question 3.
Name the different nucleotides of DNA.
Answer:
(a) Deoxyadenylic acid or decoy adenosine monophosphate (d AMP)
(b) Deoxyguanylic acid or deoxyguanosine monophosphate (dGMP)
(c) Deoxycytidylic acid or deoxycytidine monophosphate (d CMP)
(d) Deoxythymidylic acid or deoxythymidine monophosphate (dTMP).

Question 4.
Mention 4 differences between -RNA and DNA.
Answer:

RNA	DNA
1. Single stranded	1. Double stranded.
2. Sugar component is deoxyribose.	2. Sugar component is ribose.
3. Nitrogen bases are adenine, guanine cytosine and uracil	3. Nitrogen bases are adenine, guanine, cytosine and thymine.
4. Genetic material in certain plants and animals and takes part in protein synthesis.	4. Genetic material in all organisms except certain plants and animals.

Question 5.
During translation, if the codon on mRNA is AUG then
1. What is the sequence of anticodon present on corresponding tRNA?
2. Name the amino acid carried by this tRNA.
Answer:
UAC is the anticodon and Methionine is the amino acid.

Question 6.
List any four characters of genetic code?
Answer:

Genetic code is universal
It is triplet in nature
It is non overlapping and highly sequential
It is degenerate.

Question 7.
Give an account of mRNA.
Answer:

mRNA is synthesised on a DNA template by transcription and is complementary to it.
mRNA synthesis is initiated at 51 end and progresses towards 31 end in the presence of RNA polymerase enzyme.
It acts as a messenger to carry the genetic information from DNA to the cytoplasmic site of protein synthesis.
It has a very short lifespan and is broken by ribo nucleases.

Question 8.
Differentiate between sense and antisense strands of DNA.
Answer:

Sense Strands	Antisense Strands
1. It has information of specific polypeptide. if synthesis.	1. It is not involved for synthesis of mRNA.
2. Has promoter region for binding RNA polymerase	2. Does not have a promoter for RNA polymerase
3. Has sequence complementary to mRNA.	3. Sequence of nucleotides similar to mRNA.
4. Also called template.	4. Also called non coding strand.

Question 9.
Mention any two applications of DNA fingerprinting.
Answer:
a. Identification of criminals in cases connected to rape, murder etc.
b. Determination of real father of children, under paternity disputes.

Question 10.
Name the enzyme that catalyses
(a) replication of DNA and
(b) formation of RNA
Answer:
(a) Topoisomerase.
(b) RNA polymerase

Question 11.
Differentiate between Exoiis and Introns.
Answer:
Exons:
(a) They are the coding sequence of DNA /RNA transcript, that form parts of mRNA and code for different regions of the polypeptide.
(b) They are joined together during splicing to make the information continuous.

Introns:
(a) Introns are non coding sequences of DNA/RNA transcript that do not become part of mRNA.
(b) They are removed during splicing.

Question 12.
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil, and Cytosine.
Answer:
Nitrogenous bases-Adenine, thymine, uracil, cytosine. Nucleosides, cytedine, guanosine.

Question 13.
If a double stranded DNA has 20% of cytosines, calculate 4he percent of adenine in the DNA.
Answer:
The % of cytosine = 20 (given)
Therefore the % of guanine = 20.
The % of Thymine +Adenine will be 100 – (20 + 20) = 60.
Therefore the % of adenine will be 60/2 = 30%

Question 14.
If the sequence of one strand of DNA is written as follows: 5’ATGCATGCATGCATGCA TGCATGCA TGC-3’Write down the sequence of complementary strand in 5′ → 3′ direction.
Answer:
Given DNA sequence5′-
ATGCATGCATGCATGCATGCATGCA TGC-3′
Sequence of complementary strand in 3′ → 5’direction3′-
TACG’I ACGTACGTACGI ACGTACGT ACG-5’
These sequence of complementary strand in 5′ → 3′ direction’
will be the reverse of the above sequence-
5′-GCATGCATGCATGCATGCATGCATG CAT-3′.

Question 15.
If the sequence of Coding strand in a transcription unit is written as follows: 5’-ATGCATGCATGCATGCATGCATGC ATGC-3’ Write down the sequence of mRNA.
Answer:
The sequence of coding strand in a transcription unit is S’-
ATGCATGCATGCATGCATGCATGC ATGC-31
The sequence of template strand will be 3′-
3 TACGTACGTACGTACGTACGTACGT ACG-5’
Thus, the sequence of mRNA will be 5′-
AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

Question 16.
Define the terms
(a) Transcription
(b) Translation.
Answer:
(a) The process of transfer of genetic message from DNA to mRNA in nucleotide language is called Transcription.

(b) The process by which the protein is synthesised as per the message contained in the codons of mRNA is called Translation.

2nd PUC Biology Molecular Basis of Inheritance Three Marks Questions and Answers

Question 1.
How is nucleosome formed? Draw a diagram of the nupleosome.
Answer

In eukaryotes, the DNA is wrapped around positively charged histone octamer into structure called nucleosome.
A typical nucleosome consists of 200 bp of DNA helix.
The nucleosomes are the repeating units that form chromatin fibers.
The chromatin fibers condense 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 (NHC) proteins.
In a nucleus, certain regions of the chromatin are loosely packed and they stain lighter than the other regions; these are called euchromatin.
The other regions are tightly packed and they stain darker and are called heterochromatin.
Euchromatin is transcriptionally more active than heterochromatin.

Question 2.
Explain the post trancriptional events in eukaryotes.
Answer:
Transcription in Eukaryotes

In eukaryotes, the structural genes are monocistronic and ‘split’.
They have coding sequences called exons that form part of mRNA and non-coding sequences, called introns, that do not form part of the mRNA and are removed during splicing.
In eukaryotes, there are atleast three different RNA- polymerases in the nucleus, apart from the RNA polymerase in the organelles, which function as follows:

RNA polymerase I transcribes rRNAs (26S, 18S and 5.8S)
RNA poymerase II transcribes the precursor of mRNA (called as heterogenouus nuclear RNA (hnRNA)) and
RNA-polymerase III catalyses transcription of tRNA.

The primary transcript contains both exons and introns and it is subjected to a process, called splicing, where the introns are removed and the exons are joined in a definite order to form mRNA.
The hnRNA undergoes two additional processes called ‘capping’ and ‘tailing’.
In capping, methylguanosine triphosphate is added to the 5’end of hnRNA.
In tailing, adenylate residues (about 200-300) are added at the 3’end of hnRNA.
The fully processed hnRNA is called mRNA and is released from the nucleus into the cytoplasm.

2nd PUC Biology Molecular Basis of Inheritance Five Marks Questions and Answers

Question 1.
Describe the regulation of lac-operon in E.coli.
Answer:

Regulation of gene action is well studied in E.coli bacteria with regard to utilization of lactose by the bacteria.

Utilisation of lactose in E.coli needs three enzymes namely B-galactosidase, B-galactosidase t permease and B – galactoside transacetylase. These are produced by Z,Y and a genes respectively. Enzyme RNA polymerase enzyme initiates the synthesis of these 3 enzymes. The mechanism of lac-operon can be studied under two steps namely,

(a) Switched OFF mechanism: When lactose is absent in the medium, the regulator gene produces repressor protein which binds with operator gene. This prevents the movement of RNA polymerase on the structural genes is blocked so there is no synthesis of mRNA from the structural genes Z,Y and a, so there is no synthesis of enzymes. This is called switched OFF mechanism

(b) Switched ON mechanism: When lactose is added to the culture medium, some of its molecules enter into the bacterial cell and one of them, binds itself with repressor. It induces the repressor protein to undergo structural change and makes it inactive. This inactive repressor becomes detached from the operator region. Now the RNA polymerase moves along the DNA and as a result the structural genes Z, Y and a produce mRNA. This mRNA synthesises 3 enzymes which are necessary for lactose metabolism. This is called switched ON mechanism.

Question 2.
Describe the Semiconservative method of DNA replication.
Answer:
The process of formation of a new copy of DNA from the existing one is called replication of DNA. It takes place by semi conservative method, ie, one strand of parent molecule is conserved in the daughter molecule of DNA. This mechanism was proposed by Watson and crick and experimentally proved by Meselson and Stahl.

Requirements: DNA templates. Unwindase, RNA primer, RNA polymerase, DNA polymerase I and III, DNA lygase and deOxyribonucleotides

Steps:
1. Unwinding of DNA: Due to the activity of REN nicking of DNA strand takes place at a specific point to form orisite. Unwinding of DNA takes place from orisite, with the help of enzyme unwindase or helicase, The two separated strands of DNA look like ‘y’-shape and is called replication fork. The separated DNA strands act as templates for the synthesis of new strand.

2. Synthesis of RNA primer: RNA primer is a short segment of RNA (8-10 nucleotides) synthesized on DNA template with the help of RNA polymerase or primase. The synthesis of RNA primer is required as the enzyme DNA polymerase III can not directly initiate the synthesis of new DNA strand on the template.

3. Formation of leading and lagging strands: Formation of new DNA strands by DNA polymerase enzyme can occur only in 5′ to 3′ direction. Synthesis of new complementary strands takes place on both the templates in opposite directions. New complementary strand that is formed towards the replication fork on Template strand is continuous and is called leading strand or continuous strand. The synthesis of the other new complementary strand from the replication fork occurs discontinuously on parental strand and is called lagging strand or discontinuous strand. This lagging strand will have small segments of RNA primer to which DNA nucleotides are joined, These are called Okazaki fragments.

4. Formation and separation of daughter DNA molecule: In this step RNA primers are removed by the DNA polymerase I and the gaps are filled up by the DNA nucleotides by the same enzyme. Then short segments of DNA are joined by DNA lipase forming a continuous strand of DNA. Thus each daughter DNA molecule will have one old parental strand and one new strand and hence it follows a semi conservative method of DNA. replication

Question 3.
Describe any five properties of genetic code.
Answer:
The properties of genetic code are:
(a) Triplet code: The genetic code is a triplet code. It means that three bases of DNA code for one amino acid.

(b) Genetic code is universal: It means that a particular mRNA codon, codes for the same amino acid in all living organisms. For e.g.: AUG codes for amino acid methionine in all organisms.

(c) Genetic code is degenerate of redundancy : Some of the amino acids are coded by two or more codons. These redundant codons for the same amino acids are called degenerate codons and this property is called degeneracy.

(d) Genetic code is non overlapping: In this property the base of one codon in not shared by the neighbouring codons.

(e) Genetic code is comma less: Genetic code has no punctuation mark inside the message. Once the reading of message begins with the initiator codon AUG and it is continued till the terminator codon is reached.

Question 4.
Explain the experiment of Avery to show DNA acts as genetic material.
Answer:

In 1928 Griffith working on the pathogencity of streptococcus pneumonia discovered a process called transformation. This bacterium is responsible for a form of pneumonia, killing mice. There are two strains of bacteria – smooth strain (S) with a gelatinous coat and a ‘rough strain (R) without the gelatinous coat. Smooth strain bacteria are virulent type and kill the mice while rough strain bacteria are avirulent type and do not kill the mice.

If live rough strain bacteria (R) are injected the mouse does not die. When live smooth bacteria are injected the mouse dies. But if heat killed smooth bacteria (S) are injected, the mouse does not die. If a mixture of heat killed smooth bacteria (S) and live rough bacteria (R) is injected into the mice it dies. Some property of heat killed smooth bacteria had transformed R bacterium to become virulent types. This activity was called transforming principle by Griffith.

In 1944, Avery, MacLeod and Me Carty extended Griffith’s experiment to identify the trans forming principle. They mixed ‘R’ strain with DNA extracted from ‘S’ strain of bacteria. When the enzyme (DNAase) which digests or breaks down DNA was added to this mixtures, there was no transformation of R to S type. But when the enzyme proteases that digest proteins was added to the medium the transformation of R to S type was not prevented. This confirmed that the transforming principle was DNA.

Question 5.
Describe double helical model of DNA
Answer:
Watson and Crick in 1953, proposed a model to explain the structure of DNA called double helix model. According to this model,
(a) DNA is composed to two poly nucleotide chains coiled spirally around a central axis to form a double helix.

(b) The two polynucleotide chains of a double helix are opposite in direction. One chain runs in 5′ → 3′ direction and the other is 3′ → 5′ direction.

(c) Hydrogen bonds are formed between nitrogenous bases. The base-pairing in DNA consists of one Purine and one pyrimidine. Further Adenine (A) always pairs with Thymine (T) and guanine (G) always pairs with cytosine. These base pairing is said to be complementary base pairing. Two hydrogen bonds are formed between A and T and between G and C three hydrogen bonds are formed.

(d) The total amount of A is equal to that of T and the total amount of G is equal to that of C. in DNA molecules. This relationship between base pairs is known as Chargoff’s’s rule. It is represented as A = T, C = G.

(e) Backbone of the DNA helix is made up of sugar and phosphate.

(f) One double helical turn is ‘S’ shaped and contains one major groove and a minor groove. One complete turn of the helix is 34A° in length and it consists of 10 base pairs. Thus the distance between two adjacent bases is 3.4A0. The constant distance between the two strands is 20A°.

Question 6.
Briefly explain the steps involved in DNA Anger printing?
Answer:

This is a technique which helps in the establishment of the identity of a person by detecting and comparing specific nucleotide sequences in DNA of individuals.
This technique was developed by Alec Jeffreys and his associates.
Steps:
a. Collection of sample: Sample like blood, semen, hair, saliva or any other tissue even in the dried state are collected.

b. Extraction of DNA: The material collected is subjected to high speed centrifugation which separates the DNA from the cell.

c. Fragmenting DNA: The DNA in the sample is treated with restriction endonucleases to obtain smaller restriction fragments.

d. Separating DNA fragments: The smaller fragments of DNA are made to run on an agarose gel plate by electrophoresis technique in which depending on molecular size and shape, the nucleotides produce distinct bands in the gel.

e. Extracting DNA from gel: The double stranded DNA fragment in the gel it denatured into single strands by exposing the gel into an alkaline solution (NaOH). Then by Southern blotting technique, these single stranded DNA fragments are transferred onto a nitrocellulose filter or a nylon membrane.

f. Tagging radioactive probes: A probe is radio actively labelled synthetic DNA. The nitrocellulose filter is now incubated with these radioactive probes. The probes act as specific nucleotide detectors and binds or hybridises with the specific nucleotides present in DNA fragments on the filter.

g. Auto radiography: An x-ray film is placed on the filter to obtain autoradiographic prints. The film is then processed to observe visible hydridised pattern of bands. This is the genetic finger print or DNA finger print.

h. Comparison: Comparison of two genetic finger prints with great care can provide information about the degree of similarities or differences.

Question 7.
Enlist the goals and applications of Human Genome project.
Answer:
The objectives of HGP are :
a. To identify the estimated 1,00,000 genes present in the 23 human chromosomes.
b. To determine the sequence of 3 billion base pairs that make up human DNA.
c. To store this information in a Data Base.
d. To improve tools for data analysis.
e. To address ethical, legal and social issues that may arise from the project.

Applications of HGP:
a. It will help in understanding the functioning of genes and genetic regulation.
b. Many genes associated with human diseases like cystic fibrosis, breast cancer, muscle
disorders, cardio vascular diseases, diabetes etc have been identified. This helps in understanding the disease development and formulating new methods of diagnosis and treatment.
c. Helps in medical diagnosis and treatment.

Question 8.
Briefly explain the process of Translation during protein synthesis?
Answer:

Translation is the second step during protein synthesis in which using genetic information on mRNA in nucleotide language, protein synthesis takes place on the ribosome in the amino acid lineage. mRNA carrying coded message for the synthesis of protein will be released into cytoplasm. With the help of this mRNA translation occurs in four steps.
They are

Activation and selection of Amino acids
Chain initiation
Chain elongation
Chain termination.

1. Activation and selection of amino acids: In this step, amino acid combines with ATP in the presence of amino acid synthetase and forms activated amino acid and this process is known as activation. The activated amino acid thus will be picked up by specific +RNA molecule and form amino acid + RNA.
Amino acid + ATP → Activated amino acid.
Activated amino acid + +RNA → Amino acid +RNA

2. Chain initiation: During this process, initially mRNA attaches to smaller ribosomal subunit. Then +RNA carrying methionine attaches to the initiator codon (AUG) of mRNA. Now larger ribosomal subunit attaches to mRNA, such that the +RNA with amino acid coincides at P-Site. Soon after this, another +RNA with a second amino acid attaches to A-site peptide bond is formed, between the first amino acid occupying A-site. Chain initiation needs ATP and proteins called initiation factors.

3. Chain elongation: Soon after chain initiation, ribosomal subunits slide along mRNA. This shifts the +RNA occupying A-site automatically to the P-site. Now to the emptied A-site, +RNA carrying third amino acid gets attached. With this peptide bond is established between second and third amino acids. This results in the growing of polypeptide chain referred to as chain elongation. Chain elongation requires GTP, elongation factors and Mg+2 ions.,

4. Chain termination: As soon as the sliding ribosomal subunits arrive at the terminator codons, the protein synthesis is stopped with the association of releasing factors to the larger subunit of ribosome. This is known as chain termination.

Soon after chain termination, newly formed polypeptide chain undergoes processing and attains specific structure.

Question 9.
Differentiate between continuous and discontinuous synthesis of DNA.
Answer:
Continous synthesis:
(a) One stand of DNA is synthesised as a continuou.s stretch in the 5′ 3′ direction
(b) The template strand of DNA strand is with 3′ → 5′ polarity.
(c) No need of enzyme ligase (fof joining).
(d) There is no need for primers. ‘

Discontinuous synthesis:
(a) Short stretches are synthesised in the 5′ → 3′ direction from replication fork,
(b) The DNA strand with 5′ → 3’ polarity is the template Strand for this.
(c) DNA ligase enzyme is required for joining short stretches.
(d) There is need for primers.

Question 10.
Describe the process Of transcription of mRNA in an Eukaryotic cell.
Answer:
1. In eukaryotes, the structural genes are monocistronic and‘split’.

2. They have coding sequences called exons that form part of mRNA and non-coding sequences, called introns, that do not form part of the mRNA and are removed during splicing.

3. In eukaryotes, there are atleast three different RNA polymerases in the nucleus, apart from the RNA polymerase in the organelles, which function as follows:

RNA polymerase I transcribes rRNAs (26S, 18S and 5.8S)
RNA poymerasell transcribes the precursor of mRNA (called as heterogenouus nuclear RNA (hnRNA)) and
RNA-polymerase HI catalyses transcription of tRNA.

4. The primary transcript contains both exons and introns and it is subjected to a process-, called splicing, where the introns are removed and the exons are joined in a definite order to form mRNA. V

5. The hnRNA undergoes two additional processes called ‘capping’ and ‘tailing’.

6. In capping, methyl guanosine triphosphate is added to the 5 ‘ end of hnRNA.

7. In tailing, adenylate residues (about 200-300) are added at the 3’ end of hnRNA.

8. The fully processed hnRNA is called mRNA and is released from the nucleus into the cytoplasm.

Question 11.
List any four salient features of human genome project.
Answer:

The human genome contains 3164.7 million nucleotide bases.
The average gene consists of 3000 bases.
The largest known human gene being dystrophin at 2.4 million bases.
The total number of gene is estimated at 30,000.
99.9 percent nucleotide base sequences are same in all the people.
The function of 50% genes discovered is still unknown.
Less than 2 percent of the genome, codes for proteins.
Repeated sequences make up very large portion of hurpan genome.
Chromosome I has the most genes (2968) and Y has the fewest (231).
It is identified about 1.4 million locations where single-base DNA differences (SNPs – single nucleotide polymorphism) occur in humans.

Question 12.
Represent diagrammatically hershey-Chase experiment. What did this experiment prove?
Answer:
1. The proof for DNA as the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952), who worked with bacteriophages.

2. The bacteriophage on infection injects only the DNA into the bacterial cell and not the protein coat; the bacterial cell treats the viral DNA as its own and subsequently manufactures more virus particles.

3. They made two different preparations of the phage; in one, the DNA was made radioactive with 32P and in the other, the protein coat was made radioactive with 35S.

4. These two phage preparations were allowed to infect the bacterial cells separately.

5. Soon after infection, the cultures were gently agitated in a blender to separate the adhering protein coats of the virus from the bacterial cells.

6. The culture was also centrifuged to separate the viral coat and the bacterial cells.

7. It was found that when the phage containing radioactive DNA was used to infect the bacteria, its radioactivity was found in the bacterial cells (in the sediment) indicating that the DNA has been injected into the bacterial cell.

8. So, DNA is the genetic material and not proteins.

Question 13.
Draw the schematic structure and explain the different regions of a transcription unit.
Answer:
Transcription Unit
A transcription unit in DNA is defined primarily by the three regions in the DNA:

A Promoter
The Structural gene
A Terminator

There is a convention in defining the two strands of the DNA in the structural gene of a transcription, unit. Since the two strands have opposite polarity and the DNA-dependent RNA polymerase also catalyse the polymerisation in only one direction, that is 5′ → 3′, the strand that has the polarity 3′ → 5′ acts as a template, and is also referred to as template strand.

The other strand which has the polarity ( 5’ → 3′) and the sequence same as RNA (except thymine at the place of uraciil), is displaced through transcription. Strangely, this strand (which does not code for anything) is referred to as coding strand. All the reference point while defining a transcription unit is made with coding strand. To explain the point, a hypothetical sequence from a transcription unit is represented below:
3′ -ATGCATGCATGCATGCATGCATGC-5′ Template Strand
5′ – TACGTACGTACGTACGTACGTACG-3′ Coding Strand
Can you now write the sequence of RNA transribed from the above DNA?

The promoter and terminator flank the structural gene in a transcription unit. The promoter is said to be located towards 5′ – end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand). It is a DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that also defines the template and coding strands.

By switching its position with terminator, the definition of coding and template strands could be reversed. The terminator is located towards 3′ – end (downstream) of the coding strand and it usually defines the end of the process of transcription. (Fig). There are additional regulatory sequences that may be present further upstream or downstream to the promoter.

Question 14.
Explain the different steps involved in translation.
Answer:
The process of translation of nucleotide language of mRNA into amino acid language by tRNA is called translation. This process occurs in the cytoplasm of the cell both in prokaryotes and eukaryotes. The process of translation involves the following steps.

Activation of aminoacids
Initiation of polypeptide chain.
Elongation of polypeptide chain.
Termination of polypeptide chain.

1. Activation of aminoacids: Amino acids are present in the cytoplasm and are in inactive form These are combined with ATP in presence of an enzyme aminoacyl synthetase and Mg++ ions : forming activated aminoacid.

This activated amino acid then combines with 3 end of tRNA forming aminoacyl-tRNA complex , or charged RNA and AMP is released.

2. Initiation of polypeptide chain: In this process, mRNA gets attached to 30s ribosomal subunit. Now the tRNA carrying activated aminoacid, methioninp gets attached to the first codon of mRNA. It takes place in presence of GTP and initiation factors [IF₁, IF₂, IF₃]. This is the initiation complex. Now 50s ribosomal sub-unit combines with 30s ribosomal unit forming 70s ribosome. The first tRNA-aminoacid complex is now at ‘P’ site of ribosome.

3. Elongation of polypeptide chain: The ‘A’ site of 50s ribosome is now free and so the second tRNA with second activated aminoacid is get attached at ‘A’ site. A peptide bond is formed between 1 st and 2^nd aminoacid in presence of peptidyl transferase at ‘P’ site. Now the ribosome moves a distance of one codon so that the ‘A’site become vacant. Meanwhile the 3rd tRNA with 3^rd aminoacid gets attached to ‘A’ site. Again the ribosome moves on mRNA by one codon and bonding takes place between 2^nd and 3^rd aminoacid at ‘P’ site. 1st tRNA is released through ‘E’ site. This process continues till all the codons on mRNA are completely read. Thus polypeptide chain elongates along the mRNA.

4. Termination and release of polypeptide chain: The process of protein synthesis continues till the arrival of terminator codon on mRNA at.‘A’ site. The terminator codon may be UAA, UAG and UGA. These codons do not code for any aminoacids. During this process the enzyme peptidyl synthetase catalyses the cleavage of polypeptide chain from the last tRNA. Here releasing factors RF₁, RF₂, and RF₃ are involved. Thus the polypeptide chain is finally released.

The speed of transcription is 30 nucleotides per second in bacteria.
The speed of translation 20 aminoacids per second in bacteria.

Question 15.
What is operon? With a diagram describe how is lac operon switched off and switched on.
Answer:
Regulation of gene action is well studied in E.coli bacteria with regard to utilization of lactose by the bacteria. This is an example for inductive operon because, the gene expression is turned on by the addition of lactose [Inducer] into the medium.

Utilization of lactose in E.coli needs three enzymes namely P-galactosidase, p-galactoside permease and P-galactoside transacetylase. These are produced by z, y and a structural genes respectively in presence of RNA polymerase enzyme. This enzyme is bound to the promoter region of DNA. This enzyme has to move along the structural genes to initiate the synthesis of mRNA for these 3 enzymes.

The mechanism of Lac Operon can be studied under two steps namely.

Switched off mechanism and
Switched on mechanism.

1. Switched off Mechanism: When lactose is absent in the medium, the regulator gene produces repressor protein which binds with the operator gene. The repressor protein is an allosteric protein with 2 specific sites. One is active site with which it binds to operator gene and another is effecter site at which the lactose molecule can bind. Thus movement of RNA polymerase on the structural genes is blocked. So there is no synthesis of mRNA from the structural genes z,y and a. So there is no synthesis of enzymes. This is called Switched OFF mechanism.

2. Switched on Mechanism: When lactose is added to the culture medium some of its molecules enter into the bacterial cell and one of them binds itself with repressor. It induces the repressor protein to undergo structural change and makes it inactive. This inactive repressor becomes detached from the operator region. Now the RNA polymerase moves along the