Molecular Basis of Inheritance Question Answers: NCERT Class 12 Biology

Nitrogenous bases present in the list are adenine, thymine, uracil, and cytosine.

Nucleosides present in the list are cytidine and guanosine.

Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolize lactose into glucose and galactose.

In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolized into glucose and galactose.

After sometime, when the level of inducer decreases as it is completely metabolized by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.

(a) Promoter

Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.

(b) tRNA

tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

(c) Exons

Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

Human genome project was considered to be a mega project because it had a specific goal to sequence every base pair present in the human genome. It took around 13 years for its completion and got accomplished in year 2003. It was a large scale project, which aimed at developing new technology and generating new information in the field of genomic studies. As a result of it, several new areas and avenues have opened up in the field of genetics, biotechnology, and medical sciences. It provided clues regarding the understanding of human biology.

DNA fingerprinting is a technique used to identify and analyze the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.

Application

(1) It is used in forensic science to identify potential crime suspects.

(2) It is used to establish paternity and family relationships.

(3) It is used to identify and protect the commercial varieties of crops and livestock.

(4) It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

(a) Transcription

Transcription is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promoter region of the template DNA and terminates at the terminator region. The segment of DNA between these two regions is known as transcription unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of ribonucleotides, and certain cofactors such as Mg2+.

The three important events that occur during the process of transcription are as follows.

(i) Initiation

(ii) Elongation

(iii) Termination

The DNA-dependent RNA polymerase and certain initiation factors (σ) bind at the double stranded DNA at the promoter region of the template strand and initiate the process of transcription. RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex into two separate strands. Then, one of the strands, called sense strand, acts as template for mRNA synthesis. The enzyme, RNA polymerase, utilizes nucleoside triphosphates (dNTPs) as raw material and polymerizes them to form mRNA according to the complementary bases present on the template DNA. This process of opening of helix and elongation of polynucleotide chain continues until the enzyme reaches the terminator region. As RNA polymerase reaches the terminator region, the newly synthesized mRNA transcripted along with enzyme is released. Another factor called terminator factor (ρ) is required for the termination of the transcription.

(b) Polymorphism

Polymorphism is a form of genetic variation in which distinct nucleotide sequence can exist at different sites in a DNA molecule. This heritable mutation is observed at a high frequency in a population. It arises due to mutation either in somatic cell or in the germ cells. The germ cell mutation can be transmitted from parents to their offsprings. This results in accumulation of various mutations in a population, leading to variation and polymorphism in the population. This plays a very important role in the process of evolution and speciation.

(c) Translation

Translation is the process of polymerizing amino acid to form a polypeptide chain a ribosome by reading mRNA molecule. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain.

The process of translation involves three steps.

(i) Initiation

(ii) Elongation

(iii) Termination

During the initiation of the translation, tRNA gets charged when the amino acid binds to it using ATP. The start (initiation) codon (AUG) present on mRNA is recognized only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in a large subunit for the attachment of subsequent amino acids. The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave the space for binding of another charged tRNA. The amino acid brought by tRNA gets linked with the previous amino acid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (UAA, UAG, and UGA), the process of translation gets terminated. The polypeptide chain is released and the ribosomes get detached from mRNA.

(d) Bioinformatics

Bioinformatics is the application of computational and statistical techniques to the field of molecular biology. It solves the practical problems arising from the management and analysis of biological data. The field of bioinformatics developed after the completion of human genome project (HGP). This is because enormous amount of data has been generated during the process of HGP that has to be managed and stored for easy access and interpretation for future use by various scientists. Hence, bioinformatics involves the creation of biological databases that store the vast information of biology.

It develops certain tools for easy and efficient access to the information and its utilization. Bioinformatics has developed new algorithms and statistical methods to find out the relationship between the data, to predict protein structure and their functions, and to cluster the protein sequences into their related families.

According to Chargaff’s rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.

% A = % T and % G = % C

If dsDNA has 20% of cytosine, then according to the law, it would have 20% of guanine.

Thus, percentage of G + C content = 40%

The remaining 60% represents both A + T molecule. Since adenine and guanine are always present in equal numbers, the percentage of adenine molecule is 30%.

The DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is

5'- ATGCATGCATGCATGCATGCATGCATGC − 3’

Then, the sequence of complementary strand in direction will be

3'- TACGTACGTACGTACGTACGTACGTACG − 5’

Therefore, the sequence of nucleotides on DNA polypeptide in direction is

5'- GCATGCATGCATGCATGCATGCATGCAT− 3’

If the coding strand in a transcription unit is

5’− ATGCATGCATGCATGCATGCATGCATGC-3’

Then, it is known that the sequence of mRNA is same as the coding strand of DNA.

However, in RNA, thymine is replaced by uracil.

Hence, the sequence of mRNA will be

5’ − AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’

Watson and Crick observed that the two strands of DNA are anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each daughter DNA molecule would have one parental strand and a newly synthesized daughter strand.

Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication.

There are two different types of nucleic acid polymerases.

(1) DNA-dependent DNA polymerases

(2) DNA-dependent RNA polymerases

The DNA-dependent DNA polymerases use a DNA template for synthesizing a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesizing RNA.

Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage.

They grew some bacteriophages on a medium containing radioactive phosphorus (32P) to identify DNA and some on a medium containing radioactive sulphur (35S) to identify protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated from the bacterial cell by blending and then subjected to the process of centrifugation.

Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. In case I, supematatn was found radioactive. Which shows protein did not enter in bacterial cell during infection. While in case II, Bacterial cells in pelld were radioactive as they have radioactive DNA.

Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.

(a) Repetitive DNA and satellite DNA

(b) mRNA and tRNA

(c) Template strand and coding strand

The important functions of ribosome during translation are as follows.

(a) Ribosome acts as the site where protein synthesis takes place from individual amino acids. It is made up of two subunits.

The smaller subunit comes in contact with mRNA and forms a protein synthesizing complex whereas the larger subunit acts as an amino acid binding site.

(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s r-RNA in bacteria acts as a ribozyme.

Hershey and Chase conducted experiments using viruses called bacteriophage which infects bacteria to prove that DNA is the genetic material.

1. Some bacteriophage virus were grown on a medium which contained radioactive phosphorus (³²P) and some were grown in another medium with radioactive Sulphur (³⁵S).
2. Viruses grown in the presence of radioactive phosphorus (³²P) contained radioactive DNA.
3. Similar viruses were been grown in the presence of radioactive Sulphur (³⁵S) that also contained this radioactive protein.
4. Both the radioactive virus types were allowed to infect E.coli separately.
5. Soon after the infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria.
6. The culture was also centrifuged to separate the viral particle from the bacterial cell.

Observation and conclusions:

1. Only radioactive phosphorus (³²P) was found to be associated with the bacterial cell, whereas radioactive Sulphur (³⁵S) was only found in surrounding medium and not in the bacterial cell.
2. This indicates that only DNA and not protein coat entered the bacterial cell.
3. This proves that DNA is the genetic material which is passed from virus to bacteria and not protein.

If both DNA and proteins contained Phosphorus and Sulphur, the result might change.

RNA is more liable and prone to degradation like owing to the presence of 2’OH group in its ribose. Hence, heat killed S- strain may not have regain its ability to transform the R- strain into the virulent form if it was having RNA in the form of genetic material.

Use of ¹⁵N will be inappropriate because method of detection of ³²P and ¹⁵N different (³²P being a radioactive isotope while ¹⁵N is non- radioactive but is the heavier isotope of nitrogen).

Even if the ¹⁵N was radioactive isotope then its presence would have been detected, both inside the cell (¹⁵N incorporated as nitrogenous base in DNA) as well as in the supernatant, because ¹⁵N would also get incorporated in amino acids in the proteins of amino group. Hence, the use of ¹⁵N would not give any conclusive results.

Some amino acids are coded by more than one codon which is known as degeneracy of codons. Hence, on deducing a nucleotide sequence from an amino acid sequence, multiples nucleotide sequence will be obtained, example; Ile (isoleucine) has three codons AUU, AUC, AUA. Hence, a dipeptide Met- Ile can have the following nucleotide sequences.

1. AUG-AUU
2. AUG-AUC
3. AUG-AUA

And if, we deduce amino sequence from the above nucleotide sequences then all the three sequences will code for Methionine and Isoleucine.

The statement is correct. Because of degeneracy of codons, mutation at third base of codon, usually does not result into any change is phenotype. This is called silent mutations.

On the other hand, if codon is changed in any way that now it specifies another amino acid, it may do other protein functions as it happens in case of β- globulin of haemoglobin protein. Whereas substitution of valine instead of glutamic acid causes change in its structure and function and resulting into sickle cell trait.  

In complete absence of expression of lac operon, permease will not be synthesized which is essential for the transport of lactose from medium into the cells. And, lactose acts as inducer when it cannot be transported into the cell. Hence, cannot relieve the lac operon from its repressed state.   

The sequencing of human genome in enhancing the basic understanding of genetics and immunity to various disorders. Various genes that cause genetic disorders were identified with the help of this project. It was found that more than 1200 genes are responsible for common human cardiovascular diseases, endocrine diseases, neurological disorders and cancers and many more. These diseases can be treated easily by knowing the particular gene mutation that is responsible for the particular disease.

The total number of genes is estimated at 25000 much lower than previous estimates of 140000 that had based on extrapolation from gene- rich areas as opposed to a composite of gene rich and gene poor areas.

Almost all 99.9% nucleotide bases are exactly the same in all people. Functions for over 50% discovered genes are not known yet. Scientists have identified about 1.4 million locations where single base DNA difference occur in humans. This information provides to revolutionize the processes of finding chromosomal locations which is for disease associated sequence and tracing human history.    

Human genome helps to find out the complete genome sequence of the human. It has many advantages and disadvantages.

Some important advantages:

It enhances the basic understanding of human genetics. Reveals the genes responsible for diseases such as cardiovascular ailments, Alzheimer’s diseases, cancer etc. Provides information that will help in the prevention of inherited diseases, Leads to the treatment of genetic disorders through gene therapy etc. are the advantages.

Some important disadvantages:

People might discover and untreatable genetic disease. People may not take serious about the knowledge obtained from the HGP. Problem can occur for the ownership of the genetic test result or from mutation and the patenting of human genes and DNA.  

Bacteriophage does not contains repetitive sequences like Variable Number Tandem Repeats (VNTRs) in its genome, as its genome is very small and have all the coding sequence. DNA finger printing is not applied for phages.

Further Polymerization would not occur, as the 3’OH on sugar is not there to add a new nucleotide for forming ester bond.

A molecule that can act as a genetic material must fulfill the following requirements:

1. It should be able to generate its replica also known as replication.
2. It should chemically and structurally be stable.
3. It should provide the scope for slow changes or we can mutations that are required for evolution.
4. It should be able to represent itself in the form of Mendelian.

Biochemical differences between DNA and RNA:

1. Both nucleic acid (DNA and RNA) are able to direct their duplication proteins fails for the first criteria.
2. RNA is reactive, it also acts are catalyst, hence DNA is less reactive and structurally more stable than RNA.
3. Presence of thymine at the place of uracil also confers additional stability to DNA.

Watson and Crick had the following information which helped them to develop a model of DNA.

1. Chargaff’s law suggesting A= T and C= G
2. Wilkins and Rosalind Franklin’s work on DNA crystal’s X- ray diffraction studies about DNA physical structure.

Watson and Crick proposed

1. Pattern of complementary base pairs
2. Semi- conservative replication
3. Mutation through tautomerism

1. Methylated guanine cap helps in binding of mRNA to smaller ribosomal sub-unit in the initiation of translation.
2. Poly- A tail provides longevity to mRNA's life. Tail length and longevity of mRNA are positively correlated.   

Functional mRNA of structural genes need not always include all of its exons. This alternate splicing of exons of sex- specific, tissue specific and even developmental stage- specific. By such alternate splicing of exons, a single gene may encode for several isoproteins.

In the absence of such kind of splicing, there should have been new variety of genes for every protein/ isoprotein. This extravagancy has and should been avoided in natural phenomena by way of alternate splicing.   

Tandemness in repeats provides many copies of the sequence for finger- printing and variability in nitrogen base sequences present in them. Being individual specific, this proves to be useful in the process of DNA fingerprinting.  

Post transcriptional Modifications:

The primary transcripts are non- functional that contains both the coding region, exon and non- coding region, intron in RNA and are called heterogeneous RNA or hnRNA. In eukaryotes, three types of RNA polymerases are found in the nucleus:

1. RNA polymerase I transcribes rRNA (28S and 5.8S)
2. RNA polymerase II transcribes the precursor of mRNA
3. RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs.

The hnRNA undergoes two additional or main processes called capping and tailing.

In capping, an unusual nucleotide, methyl guanosine triphosphate is added to the 5’ end of the hnRNA. In tailing, adenylate residues about 200- 300 are added at the 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 known as Splicing.    

There are three stages of protein synthesis:

1. Initiation:

Assembly of ribosomes on mRNA In prokaryotes, initiation needs the large and small ribosome subunits, the mRNA, initiation tRNA and three initiation factors (IFs); IF₁, IF₂ and IF₃.

Activation of Amino Acid These amino acids have become activated by binding with aminoacyl tRNA synthetase enzyme in the presence of ATP.

Transfer of Amino Acid to tRNA The AA-AMP- Enzyme complex formed reacts with specific tRNA to form aminoacyl tRNA complex.

AA-AMP- Enzyme complex + tRNA                                     AAtRNA + AMP + Enzyme

The cap region of the mRNA binds to the smaller subunit of ribosome on that site.

The ribosome has two sites, A- site and P- site.

The smaller subunit first binds the initiator tRNA then and then binds to the larger subunit so, that initiation codon (AUG) lies on the P- site.

The initiation tRNA, i.e., methionyl tRNA then binds to the P- site.

2. Elongation: Another charged aminoacyl tRNA complex binds to the A- site of the ribosome.

Peptide bond formation and movement along with the mRNA is defined as translocation. A peptide bond is formed between carboxyl group (-COOH) of amino acid at the P- site and amino group (-NH) of amino acid at A- site by the enzyme peptidyl transferase. The ribosome slides over mRNA from codon to codon in the 5’ to 3’ direction.

According to the codon sequence, amino acids are attached to one another with the peptide bonds and then a polypeptide chain is formed.

3. Termination: When the A- site of ribosome reaches a termination codon which does not code for any amino acid, no charged tRNA binds to the A- site. Dissociation of polypeptide from the ribosome takes place which is catalyzed by a release factor. There are three termination codon:

UGA, UAG and UAA  

The concept of operon was first proposed by Jacob and Monad, in 1961. An operon is a unit of prokaryotic gene expression which includes coordinately regulated or structural genes and control elements which are recognized by regulatory gene product.

Components of an operon:

1. Structural gene When lactose is added to the E.coli, the structural genes produces mRNAs, which in turn, synthesize specific polypeptides on the ribosomes (translation).
2. Promoter The sequence of DNA where RNA polymerase binds and initiates transcription of structural genes is called promoter.
3. Operator The sequence of DNA adjacent to promoter where specific repressor protein binds is called operator.
4. Regulator gene The gene that codes for the repressor protein which gets bind to the operator and suppresses its activity as a final result by which transcription will get switched off.
5. Inducer It is a chemical substrate which inactivates the repressor by the latter’s structure and many other processes are in turn induces in terms of this. It is a chemical of diverse of diverse nature like metabolite, hormone substrate etc.

Inducible operon system:

An inducible operon system is a regulatory unit of genetic material which remains switched on in response to the presence of a chemical signal, e.g., the lactose or lac operon of E.coli.

The lactose operon: the lac Z, Y and A genes are transcribed from a lac transcription unit under the control of a single promoter. They encode enzyme required for the use of lactose as a carbon source. The lac I gene product and the lac repressor are expressed from a separated transcription unit that is upstream from the operator.

Lac operon consists of three structural genes (Z, Y and A), operator, promoter and a separate regulatory gene. The three structural genes (Z, Y and A) transcribe polycistronic mRNA.

1. Gene z codes for β- galactosidase enzyme which breaks lactose into galactose and glucose
2. Gene y codes for permease, which increases the permeability of the cell to lactose.
3. Gene a codes for enzyme transacetylase that catalyses the transacetylation of lactose in the active form.   

1. When lactose is absent

• When lactose is absent, I gene regulates and produces repressor mRNA which translate repression.
• The repressor protein binds to the operator region of the operon and as a result prevent RNA polymerase to bind the operon.
• The operon is switched off.

2. When lactose is present

• Lactose acts as an inducer which binds to the repressor and forms an inactive repressor.
• The repressor fails to bind the operator region.
• The RNA polymerase binds to the operator and transcript lac mRNA.
• lac mRNA is polycistronic, i.e. produces all three enzymes, β-galactosidase, permease and transacetylase.
• The lac operon is switched on. 

DNA fingerprinting is defined as the technique which is used in solving the paternity dispute for a child. This is a technique of determining nucleotide sequences of certain areas of DNA which are unique in each individual.

The DNA fingerprinting is based or works on the DNA polymorphism. Although the DNA from different individuals is more alike than different, there are many regions of the human chromosomes that exhibits diversity in its form. These types of variable sequences are known as polymorphic.

A special type of polymorphism known as VNTR (Variable Number of Tandem Repeats) which is composed of repeated copies of a DNA sequence and that lie adjacent to one another on the chromosome. Since, polymorphism is the basis of genetic mapping in human beings.  

Human genome Sequencing has made it possible to understand the link between various genes and their functions. If there are any gene found defected, is expressed as disorders or increases the susceptibility of an individual to a disease then specific gene therapies can be applied or used.  

Methodologies of human genome sequencing:

1. Expressed Sequence Tags (ESTs): This method focusses in identifying all the genes that are expressed as RNA.
2. Sequence annotation: It is an process of simply sequencing the whole set of genome which contains all the coding and non- coding sequences and later assigning different regions in the sequence with the specific functions. 

For sequencing, firstly the total DNA from cell, i.e., isolated and broken down in relatively small sizes as fragments. Then, DNA fragments are cloned in suitable host using suitable vectors. When bacteria is used as vector, they are called Bacterial Artificial Chromosome (BAC) and when yeast is used as vector, they are known as Yeast Artificial Chromosome (YACs).

Frederick Sanger described a principle according to which, the DNA fragments are sequenced by automated DNA sequences. On the basis of overlapping regions on fragmented DNA, these sequences are arranged accordingly. For alignment of these sequences, specialized computer based programs were developed.

Finally, the genetic and physical maps of the genome were constructed by collecting information or we can say on the base of data collection, certain repetitive DNA sequences and DNA polymorphism, based on endonuclease recognition sites.

Dr. Alec Jeffrey defined the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases. DNA fingerprinting uses short nucleotide repeats called Variable Number Tandem Repeats as the suitable markers. VNTRs vary from person to person and are inherited from one generation to another. Only closely individuals have similar VNTRs.  

In above case which is shown in the question, as E.coli bacteria is a mutant for DNA ligase (which helps in binding) enzyme, it will lead to no further joining of Okazaki fragments on lagging strand.

This will results into the formation of both high molecular weight fragments on the leading strand and low molecular weight fragments on the lagging strand. Hence, only the graph (a) could be the appropriate result after centrifugation.