NCERT Solutions Class 12 Biology Chapter 6

Q.1 Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine

Ans-

Explanation: A nucleotide has three components: a nitrogenous base, a pentose sugar (ribose in case of RNA and deoxyribose in case of DNA) and a phosphate group. A nitrogenous base is linked to the pentose sugar through N-glycosidic linkage to form a nucleoside.

Q.2 If the double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.

Ans-

If the double-stranded DNA has 20 percent of cytosine, the percent of adenine in the DNA would be 30%.

Explanation: As a rule no. of A= no. of T and no. of G = no. of C. Also, A+T=G+C. The bases in the two strands are paired through hydrogen bond (H-bonds) forming base pairs. Cytosine forms three H-bonds with Guanine that means if Cytosine is 20%, Guanine is also 20%. This leaves 60% for Adenine and Thymine which are paired through two H-bonds. Thus, Adenine would be 30%.

Q.3 If the sequence of one strand of DNA is written as follows:
5′ -ATGCATGCATGCATGCATGCATGCATGC-3′
Write down the sequence of complementary strand in 5’→3′ direction.

Ans-

The sequence of the complementary strand in 5’to 3’direction:

5’-GCATGCATGCATGCATGCATGCATGCAT -3’

Explanation: The two complementary (A pairs with T and G pairs with C) strands of double-stranded DNA have anti-parallel polarity. It means, if one strand is 5’to 3’, the other will be 3’to 5’. Thus, the complementary strand will be

5’- ATGCATGCATGCATGCATGCATGCATGC – 3’

3’- TACGTACGTACGTACGTACGTACG TACG -5’

Q.4 If the sequence of the coding strand in a transcription unit is written as follows:
5’-ATGCATGCATGCATGCATGCATGCATGC–3’.
Write down the sequence of mRNA.

Ans-

mRNA sequence:

5’- AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’

Explanation: The process of converting genetic information from one strand of the DNA into RNA is termed as transcription. It is the template strand is for mRNA synthesis. The coding strand in a transcription unit does not code for any mRNA but has a sequence similar to mRNA except that uracil replaces thymine.

Q.5 Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.

Ans-

The specific base pair property of nucleotides (Adenosine pairing with Thymine and Cytosine pairing with Guanine) in DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication. They proposed that the two strands of a DNA molecule separate during replication (the process of duplication of DNA). After separation, each strand acts as a template for the synthesis of the new strand, following the base-pairing rule. Thus, the new DNA double helix that is created has one parental strand and one new strand. This scheme is referred to as semi-conservative DNA replication.

Q.6 Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.

Ans-

The various types of nucleic acid polymerases involved in the following processes are:

a. Replication process (DNA to DNA): DNA-dependent DNA Polymerases

b. Transcription process (DNA to RNA): DNA-dependent RNA polymerase

Q.7 How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

Ans-

Hershey and Chase grew some bacteriophages on a medium containing radioactive phosphorus and some others on medium containing sulphur. Viruses grown on the radioactive phosphorus contained radioactive DNA since phosphorus is part of DNA however, their protein was normal. On the other hand, viruses grown on the radioactive sulphur had radioactive protein, not the DNA since sulphur is the part of the protein.

Then, they went ahead with these radiolabelled bacteriophages (cultures in which either the protein capsule is labelled with radioactive sulphur or the DNA core is labelled with radioactive phosphorus) to infect E.coli host in two separate set of experiments followed by agitation in a blender to dislodge phage particles from bacterial cells and centrifugation to concentrate cells, separating them from the phage particles left in the supernatant.

Result: They checked for the presence of radioactivity in these two sets of experiments and the following observations were made:

a. Radioactivity was detected in bacteria, which were infected with viruses containing radioactive phosphorus and therefore, radioactive DNA. No radioactivity was detected in the supernatant which contained the protein head. New generation of radioactive and infective phages could be isolated from this culture. This suggested that radioactive DNA has entered the bacterial cell.

b. The other set of bacteria which were infected with bacteriophages labelled with radioactive sulphur had no radioactivity in them. Instead, the entire radioactive sulphur was detected in the supernatant. This indicated that protein did not enter the bacterial cell.

Therefore, this experiment clearly showed that it is the DNA, not protein that acts as genetic material and passes the information from one generation to the next.

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

c. Template strand and Coding strand

Q.9 List two essential roles of ribosome during translation.

Ans-

The translation is the process of making proteins using cellular machinery with the help of specialised units called ribosomes. During translation, the information carried by messenger RNA (mRNA) produced by transcription is translated into a polypeptide using the ribosome complex. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. The amino acids are joined by a bond called peptide bond and this process requires energy. In the first phase of the process, the amino acids are activated in the presence of ATP and linked to their cognate tRNA, which acts as an adaptor molecule. When two such charged tRNAs come in close proximity, the formation of a peptide bond occurs between the two loaded amino acids. A catalyst is needed for this entire process. This is where the machinery of ribosomes comes into action in the following ways:

a. The ribosome consists of structural RNAs and about 80 different proteins. There are two subunits, which remain separate in its inactive state however when the smaller subunit encounters an mRNA, the process of translation begins. The larger subunit has two sites for the activated tRNAs to bind, thus bringing amino acids in close proximity for the formation of a peptide bond.

b. It also acts as a catalyst for the formation of the peptide bond.

Q.10 In the medium where E.coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?

Ans-

The lac operon consists of one regulatory gene (the i gene – the term i is derived from the word inhibitor) and three structural genes (z, y and a). The i gene codes for the repressor of the operon. The z gene codes for beta-galactosidase (β-gal), which is responsible for the hydrolysis of the sugar lactose into monomeric units, galactose and glucose. The y gene codes for permease, which increases cell permeability to β -galactosidases. The a gene encodes a transacetylase. Hence, all three genes of the operon are involved in lactose metabolism.

Lactose which is the substrate for the enzyme β –galactosidases, also acts as an inducer of the operon by switching on and off of the operon. When lactose is provided in the medium as a carbon source, it is transported into the cells by the action of a little amount of permease present in the cell. This lactose then induces the operon by inactivating the repressor. When lactose was absent, the repressor produced by the i gene remains bound to the operator region and prevents RNA polymerase from transcribing the operon. Once the repressor is inactivated in the presence of lactose, RNA polymerase initiates the transcription process.

However, after sometime, the lac operon shuts down because of catabolite repression of the lac operon. The level of lactose decreases and that of glucose rises. If lactose and glucose are present, the cell will use glucose before the lac operon is turned on. This type of control is termed catabolite repression. To prevent lactose metabolism, the second level of control of gene expression exists. The promoter of the lac operon has two binding sites. One site is where RNA polymerase binds. The second location is the binding site for a complex between the catabolite activator protein (CAP) and cyclic AMP (cAMP). The binding of the CAP-cAMP complex to the promoter site is required for transcription of the lac operon. The presence of this complex is closely associated with the presence of glucose in the cell. With the increase in the concentration of glucose, the amount of cAMP decreases. Thus, the amount of complex decreases. This decrease in the complex inactivates the promoter, and the lac operon is turned off. Because the CAP-cAMP complex is needed for transcription, the complex exerts positive control over the expression of the lac operon.

Q.11 Explain (in one or two lines) the function of the following:
(a) Promoter
(b) tRNA
(c) Exons

Ans-

(a) Promoter: A promoter is the part of DNA from where transcription of a particular gene is initiated and is usually located towards 5’-end of the structural gene on the coding strand, close of the gene it works for. It contains specific sequences and response elements that provide the binding site for RNA polymerase and other transcription factors that help in recruiting the RNA polymerase on the promoter.

(b) tRNA: tRNA is the adaptor molecule responsible for reading the genetic information on the mRNA and bringing in specific amino acid, thus translating the information on mRNA into a polypeptide. tRNA has an anticodon loop that has bases complementary to the code on mRNA and the other end, has an amino acid acceptor end to which amino acids bind.

(c) Exons: Exons are parts of DNA that are converted into mature messenger RNA (mRNA). These are the DNA sequences of a eukaryotic split gene that code for proteins. The exons are interrupted by introns.

Q.12 Why is the Human Genome project called a mega project?

Ans-

The quest to find differences between two individuals at a gross level due to differences at the DNA level lead to a very ambitious project in the 1990s. This mega project was called the Human Genome Project (HGP). The goal was to discover the genetic make-up of individuals to the DNA level. With the establishment of precise genetic engineering methods, it became possible to clone any sequence of DNA and availability of simple and fast DNA sequencing techniques made HGP a possibility.

The goals of the Human Genome Project were:
a. To identify all the genes in human DNA which were estimated to be around 20,000-25,000.
b. To determine the sequence of 3 billion base pairs which make up the entire human genome.
c. To store this massive information in properly catalogued databases.
d. To improve data analysis tools to analyse the generated data.
e. To address the ethical, legal and social issues arising out of the project.

The human genome is said to have approximately 3 X 10⁹ base pairs. With the cost of sequencing each base way back in the 1990s to be around US $ 3 per base pair, the total estimated cost of the entire HGP was calculated to be around 9 billion US dollars. Further, it was calculated that around 3300 books (of 1000 pages with 1000 letters typed on each page) would be required to store the data generated by sequencing of a single human cell. This enormous data necessitated the need for high-speed computational devices for data storage, retrieval and analysis.
Human Genome Project took 13 long years to finish. It was coordinated by the U.S. Department of Energy and the National Institute of Health. During the early years of the HGP, the Wellcome Trust (U.K.) became a major partner with support coming to Japan, France, Germany, China and others. The project was completed in 2003.
These were the reasons to term Human Genome Project as a mega project.

Q.13 What is DNA fingerprinting? Mention its application.

Ans-

DNA fingerprinting is a quick way to compare the DNA sequences of any two individuals. 99.9% of base sequence among individuals is similar; it is only the remaining 0.1% which gives distinguishing characteristics to each individual. Now, if one needs to find out genetic differences between two individuals in a population, sequencing 3X10⁹ bases every time would be a very time consuming and expensive affair. DNA fingerprinting is a way to identify differences in some specific regions in the DNA known as VNTR (Variable number of tandem repeats) repetitive DNA or more specifically, microsatellites, to make identification much easier. Microsatellites are short pieces of DNA which repeat many times in a given person’s DNA. By comparing a number of microsatellites in a given area, one can identify a person easily.

Applications of DNA fingerprinting:

1. In paternity issues since polymorphisms are inherited from parents to children.

2. In forensic crime analysis (using blood, hair-follicle, skin, saliva, semen, etc)

3. In population genetics to analyse variation within the population.

4. In conservation biology to study the genetic variability of endangered species.

5. In evolutionary biology to compare DNA extracted from fossils to present-day animals or humans.

Q.14 Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics

(a) Transcription: The process of copying genetic information from one strand of the DNA into RNA is termed as transcription. The segment of DNA, which is transcribed during the process of transcription constitutes the transcription unit and is comprised of a promoter, the structural gene and a terminator. All three types of RNA, namely, mRNA, tRNA, and rRNA are synthesised using the process of transcription.

DNA-dependent RNA polymerase catalyses the reaction of transcription using the template strand which has 3’ to 5’polarity. It binds to the promoter and initiates the transcription process. This process follows the base complementarity rule except for that adenosine pairs up with uracil instead of thymine. This process is also called elongation. Once the polymerase reaches the terminator sequence, the RNA polymerase along with the nascent RNA falls off from the template. The process of falling fo nascent RNA is called termination.

RNA polymerase by itself is capable of the only elongation. It takes the help of the initiation factor and termination factor to initiate and terminate transcription. mRNA generated during the transcription process in bacteria doesn’t require further processing, thus, translation can begin at the same time. However, in eukaryotes, there are two additional complexities associated with transcription:

1. Different kinds of RNA polymerases catalyse the formation of different kinds of RNAs.
2. The primary transcript, which contains both exons and introns, and is non-functional, is subjected to the process of splicing, capping and tailing.

(b) Polymorphism: It is a form of genetic variation in which a DNA molecule consists of the distinct nucleotide sequence at a particular site. It is a heritable mutation and can be in a population at a high frequency. It can arise due to mutation in somatic or germ cells. The mutations in germ cells are transmitted from parents to offsprings. This causes accumulation of various mutations in a population leading to variation and polymorphism.

(c) Translation: The process of polymerisation of amino acids to form a polypeptide chain using the information defined by the sequence of bases in the mRNA is called translation. This is an energy-driven process. The amino acids are first activated in the presence of ATP and linked to their cognate t-RNA (aminoacylation of tRNA). When two such charged tRNAs come in close proximity to each other, the formation of the peptide bond is favoured. Also, a catalyst required for this bond formation enhances the reaction. These two processes are carried out by very specialised cellular machinery called ribosomes. The ribosome consists of structural RNAs and about 80 different proteins. There are two subunits, which remain separated in their inactive state however when the smaller subunit encounters an mRNA, the process of translation begins. The larger subunit has two sites for the activated tRNAs to bind to, thus bringing amino acids in close proximity for the formation of the peptide bond. The ribosome also acts as a catalyst for the formation of the peptide bond.

A translational unit in mRNA is the sequence of RNA that is flanked by the start codon (AUG) and the stop codon and codes for polypeptides. mRNA has some characteristic sequences called untranslated regions (UTRs) for the efficient translation process, but they are not translated.

For initiation, the ribosome binds to the mRNA at the start codon (AUG) followed by the elongation phase of protein synthesis. During this stage, charged tRNAs sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA adding amino acids one by one till it reaches the stop codon. Release factor binding to the stop codon results in termination and release of the complete polypeptide from the ribosome.

(d) Bioinformatics: The application of computer technology to manage biological information is called bioinformatics. This is a field that combines the knowledge of computer science, statistics, mathematics, and engineering to process biological data. Computers gather, store, analyse and then integrate the information which is then applied to various method developments in biological systems. A major activity in bioinformatics is to develop software tools to generate useful biological knowledge out of existing data.

The branch of bioinformatics came into existence with the progress of the Human Genome Project when a large amount of data became publicly available. The main aim of this project was to determine the sequence of the entire human genome (approximately three billion base pairs). This systematic approach of deciphering information using computers helped drastically in gene discovery programs across the world using biological information.

With the use of Next Generation sequencing technologies, where massive data is being generated across the world for many animals and plants, the meaning and usefulness of bioinformatics have become all the relevant.