MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance

Molecular Basis of Inheritance NCERT Textbook Questions and answers

Question 1.
Group of the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Adenine, Guanosine, Thymine, Uracil and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double stranded DN A has 20 % of cytosine, calculate the percent of adenine in the DNA.
Answer:
According to Chargaff’s rule, the DNA molecule should has an equal ratio
Cytosine = 20% therefore, Guanine = 20%
A+T = 100 – (G + C)
A + T = 100 – 40 since, both Adenine and Thymine are in equal amounts.
Thymine = Adenine = \(\frac { 60 }{ 2 }\) = 30%
So, quantity of Adenine is 30% in DNA helix.

Question 3.
If the sequence of one strand of DNA is written as follows:
5′ -ATG CATGCA TGC ATG CAT GCA TGCATGC – 3′
Write down the sequence of complementary strand in 5′-3’direction.
Answer:
In 3’→ 5′ direction, 3′ –
TACGTACGTACGTACGTACGTACGTACG-5′
In 5’→ 3′ 5′ direction, 5-
GCATGCATGCATGCATGCATGCATGCAT-3′.

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows:
5’ – ATG CAT GCA TGC ATG CAT GCA TGC ATGC – 3’
Write down the sequence of mRNA.
Answer:
5’ – AUGCAUGCAUGC AUGCAUGCAUGCAUG-3’

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesis semiconservative mode of DNA replication? Explain.
Answer:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of parent is conserved while the other complementary strand formed is new.

MP Board Solutions

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acid synthesized from it (DNA or RNA) list the types of nucleic acid polymerases.
Answer:
These 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.

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

  • They grew some bacteriophages on a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.
  • Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA.
  • Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
  • It was found that bacteria which were infected with bacteriophages that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.
  • Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that proteins did not enter the bacteria from the viruses.
    MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 1
    (vi) This was a clear cut proof that DNA is die genetic material that is passed from virus to bacteria.

Question 8.
Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand.
Answer:
(a) Differentiate between Repetitive DNA and Satellite DNA:
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 2
(b) Differentiate between mRNA and fRNA:
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 3
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 4
(c) Differentiate between Template strand and Coding strand:
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 5
Question 9.
List two essential roles of ribosome during translation.
Answer:
Two essential roles of ribosome during translation are:

  1. One of the RNA acts as apeptidyl transferase ribozyme for formation of peptide bonds.
  2. Ribosome provides sites for attachment of mRNA and charged tRNA for polypetide synthesis.

MP Board Solutions

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down sometime after addition of lactose in the medium?
Answer:
Lactose regulates switching on and off of the lac operon. If lactose is provided in the growth medium of the bacteria, it is transported into the cells through the action of permease.

The lactose then induces the operon in the following manner :

  • The repressor of the operon is synthesized all the time from the is gene.
  • It binds the repressor protein which binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.

MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 6

  • In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer.
  • This allows RNA polymerase access to the promoter and transcription proceeds.

Question 11.
Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons.
Answer:
(a) Promoter is an essential component of the transcription unit. It is located at the beginning of 5′ -end.
It provides a site for the attachment of transcription factors and RNA polymerase.
(b) tRNA is a small sized RNA molecule that takes part in transcription. It physically picks up activated amino acids from the cytoplasm and carries (transfers) them to ribosomes, where they join together through peptide bonds and leave the tRNA to fetch more amino acids.
(c) Exons are the coding sequences of DNA that are transcribed and translated.

Question 12.
Why is the human genome project called a mega project ?
Answer:
Human genome project is called a mega project because :

  • Its aim was to determine the nucleotide sequence of complete human genome which was a task of enormous magnitude.
  • A total of 3 x 109 base pairs were to be sequenced and the cost was about 9 billion US dollars.
  • It required bioinformatics data base techniques and other contemporary devices for the analysis, storage and retrieval of information.
  • May countries worked jointly to complete this timed project.

MP Board Solutions

Question 13.
What is DNA Fingerprinting ? Mention its application.
Answer:
DNA Fingerprinting : Every human individual is characterised by unique print at the fingertips. The study of fingers, palm and sole print is called dermatogly phics’.

Like prints of the fingertips, each individual has unique DNA fingerprint. Unlike the prints of finger, the DNA fingerprints can not be altered by surgery. The latter is exactly similar in all the cells and tissues of an individual. It can not be changed by medical treatment. The distinction of individuals on the basis of DNA fingerprint is due to sequence of nucleotides in whole genomic DNA. The technique to identify a person on the basis of his/her DNA specificity is called DNA fingerprinting. This was invented by Sir Alec Jeffreys in 1984 at Leicester University U.K. In India, Dr. V. K. Kashyap and Dr. Lalji Singh started this technique at CCMB, Hyderabad.

DNA fingerprinting involves following steps:

The DNA of the organism to be tested is isolated, it is called host DNA.
Host DNA is cleaved with the help of specific restriction enzymes into several fragments.
Double stranded DNA fragments are denatured to produce single stranded DNA by alkali treatment.
DNA segments are separated by electrophoresis.

MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 7

Question 14.
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics.
Answer:
(a) Transcription : It is the formation of RNA over the template of DNA. It forms single-stranded RNA which has a coded information similar to the sense or coding strand of DNA with the exception that thymine is replaced by uracil. One strand of DNA is used as template strand for the synthesis of a complementary strand of RNA called wRNA.

(b) Polymorphism : Genetic polymorphism means occurrence of genetic material in more than one form. It is of three major types, i. e„ allelic, SNP and RFLP.

Allelic polymorphism : Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur.

SNP or single nucleotide polymorphism : Over 1-4 million single base DNA differences have been observed in human beings. According to SNP, every human being is unique. SNP is very useful for locating alleles, identifying disease-associated sequence and tracing human history.

(c) Translation : It is the process during which the genetic information which is stored in the sequence of nucleotides in an mRNA molecule is converted, following dictations of the genetic code, into the sequence of amino acids in the polypeptide. It takes place in cytoplasm in both eukaryotes and prokaryotes.

(d) Bioinformatics : The science which deals with handling storing of huge information of genomics as databases, analysing, modeling and providing various aspects of biological information, especially the molecules connected with genomics and proteomics is called bioinformatics.

Molecular Basis of Inheritance Other Important Questions

Molecular Basis of Inheritance Objective Type Questions

1. Choose the Correct Answer:

Question 1.
Which component they have free amino acid and carboxylic group:
(a) Glucose
(b) Nucleotide
(c) Amino acid
(d) None of these.
Answer:
(c) Amino acid

Question 2.
Nucleotide which are participate in energy transfer:
(a) NAD
(b) FAD
(c) FMN
(d) ATP.
Answer:
(d) ATP.

Question 3.
What is the unit of protein:
(a) Fatty acid
(b) Monosaccharides
(c) Amino acid
(d) Glycerol.
Answer:
(c) Amino acid

Question 4.
Nucleic acids are polymers of:
(a) Amino acid
(b) Nucleoside
(c) Nucleotide
(d) Globulin.
Answer:
(c) Nucleotide

Question 5.
Peptide bonds are found in:
(a) Protein
(b) Fat
(c) Nucleic acid
(d) Carbohydrate.
Answer:
(a) Protein

MP Board Solutions

Question 6.
Glycosidic bonds are found in:
(a) Nucleic acid
(b) Protein
(c) Polysaccharides
(c) Monosaccharides.
Answer:
(c) Monosaccharides.

Question 7.
By which the control and coordinate of heredity:
(a) By DNA
(b) By RNA
(c) Mostly DNA but some organisms by RNA
(d) None of these.
Answer:
(c) Mostly DNA but some organisms by RNA

Question 8.
It is not a protein:
(a) Myosin
(b)Actin
(c) Haematind
(d) Albumin.
Answer:
(c) Haematind

MP Board Solutions

Question 9.
A source which give immediate energy:
(a) Glucose
(b) NADH
(c) AIP
(d) Pyruvic acid.
Answer:
(c) AIP

Question 10.
Who discovered ATP:
(a) Karl Lohmann
(b)Lipman
(c) Bowman
(d) Blackman.
Answer:
(a) Karl Lohmann

Question 11.
Which nitrogenous base is found in only RNA:
(a) Cytosine
(b) Adenine
(c) Uracil
(d) Guanine.
Answer:
(c) Uracil

Question 12.
Molecule which shows very difference from other molecules in the cell:
(a) Mineral salt
(b) Lipids
(c) Proteins
(d) Carbohydrate.
Answer:
(c) Proteins

Question 13.
Who prescribe the double helix structure of DNA:
(a) Nirenberg
(b) Komberg
(c) Holley and Nirenberg
(d) Watson and Crick.
Answer:
(d) Watson and Crick.

Question 14.
Which is non-essential for the plants:
(a) Ca
(b) Zn
(c) Cu
(b) Na.
Answer:
(b) Na.

Question 15.
Number of nucleotides are found in one whorl of DNA:
(a) 9
(b) 10
(c) 11
(d) 12.
Answer:
(b) 10

Question 16.
Which are microelements:
(a) Ca
(b) N
(c) Mg
(d) Mn.
Answer:
(d) Mn.

Question 17.
Which simillarities is found in DNA and RNA:
(a) Same type of pyrimidine is found in both
(b) Thymine is present in both
(c) Some sugar is found in both
(d) Both are polymers of nucleotide.
Answer:
(d) Both are polymers of nucleotide.

MP Board Solutions

Question 18.
Cholesterol is a:
(a) Simple lipids
(b) Complex lipids
(c) Derivatives lipids
(d) Protein.
Answer:
(c) Derivatives lipids

2. Fill in the Blanks:

  1. Cut up and join of polynucleoid chain is called …………………….
  2. Changes ……………………. done by genetic engineering.
  3. UAA, UAG and UGA are …………………….codon.
  4. …………………….code is universal and non-ambiguous.
  5. Transcription of DNA information is the form of …………………….
  6. ……………………. enzyme is involved in transcription.

Answer:

  1. Genetic engineering
  2. Characters of organisms
  3. Termination
  4. Codon
  5. mRNA
  6. RNA Polymerase.

3. Match the Following:

I.
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 8

Answer:

  1. (c)
  2. (e)
  3. (b)
  4. (a)
  5. (d)
  6. (f)

II.
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 9
Answer:

  1. (c)
  2. (d)
  3. (a)
  4. (b).

III.
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 10
Answer:

  1. (c)
  2. (a)
  3. (b)
  4. (e)
  5. (d).

MP Board Solutions

4. Answer in One Word/Sentence:

  1. Who discovered nucleic acid for the first time ?
  2. Name the Indian-American scientist who is well known for chemical synthesis of gene.
  3. Name the small-RNA which is synthesized before DNA replication.
  4. Which codon is known as the starting codon ?
  5. Name the RNA which function as enzyme.
  6. Who proposed the operon model of gene regulation ?
  7. What chemical is produced by regulator gene of Lac-operon ?
  8. What term is used for those genes which are active in all the cells and tissues of an individual ?
  9. Name the operon which regulate the metabolism of lactose in E. coli.
  10. Name the organism which contain single stranded DNA.

Answer:

  1. Friedrich Miescher
  2. H.G. Khorana
  3. RNA-primer
  4. AUG or GUG
  5. Ribozyme
  6. Jacob and Monod
  7. Repressor
  8. Housekeeping genes
  9. Lac-operon
  10. Φ × 174phage.

Molecular Basis of Inheritance Very Short Answer Type Questions

Question 1.
Structure formed by regulation + structural + operator+ promoter gene.
Answer:
Operon.

Question 2.
What are the animals that have a foreign gene deliberately inserted into their genome ?
Answer:
Transgenic animals.

Question 3.
What are the group of cells or organisms which have same hereditary characters ?
Answer:
Clone.

Question 4.
By which the instructions of our DNA are converted into a functional product ?
Answer:
Gene expression.

MP Board Solutions

Question 5.
Write the name of sugar found in RNA.
Answer:
Ribose sugar.

Question 6.
Which codon is AUG?
Answer:
Anticodon.

Question 7.
Name the enzyme which takes part in transcription.
Answer:
RNA Polymerase.

Question 8.
Who tell that DNA is a heredity material ?
Answer:
Alfred Hershey and Martha Chase. ,

Question 9.
Which bond is make in DNA when join the sugar and phosphoric acid ?
Answer:
Phosphodiester bond.

Question 10.
Name the segment in which any nucleotide sequence within a gene that is removed by RNA splicing during maturation of the final RNA products.
Answer:
Intron.

Question 11.
Who gave the Operon model ?
Answer:
Jacob and Monod.

Question 12.
What do you mean by commaless due to genetic code ?
Answer:
Between two codon has no internal punctuation.

MP Board Solutions

Question 13.
Write the full name of Sn RNP.
Answer:
Small nuclear Ribonucleo Proteins.

Molecular Basis of Inheritance Short Answer Type Questions

Question 1.
What is genetic code ? What do you know about the discovery of genetic code ?
Answer:
Genetic code is that sequence of three nitrogenous bases of mRNA in which genetic information for the synthesis of one amino acid is coded.

The triplet codons of the genetic codes are discovered for the first time by M.W. Nirenberg in 1950. He synthesized a RNA by the use of repetitive sequence of Uracil which is called as polyuracil (UUUUU …………). They added synthesized mRNA to a cell free extract containing protein synthesizing enzymes and ribosome from E. coli together with a mixture of 20 amino acids. The only molecules synthesized in a polypeptide chain were phenyl alanine and their number in chain was one third of the Uracil base on poly-U-m-RNA. This confirmed triplet nature of genetic code.

Question 2.
What are oncogenes ?
Answer:
Genes which are responsible for production of cancer in host by uncontrolled mitotic cell division are called as oncogenes.

Question 3.
What are Okazaki fragments and leadings strands?
Answer:
Okazaki fragments : On second parental DNA template new complementary DNA strands are formed in smaller fragments starting from RNA primer. These short fragments are called Okazaki fragments.
Leading strands : Second strand is formed on 5’→ 3’ strand of parental DNA in a continuous stretch in reverse direction 3’→ 5’ and is called as leading strand.

Question 4.
DNA nucleotide is formed which molecule ?
Answer:
Components are DNA Nucleotides:
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 11
Question 5.
Explain the Watson and Crick model of DNA.
Answer:
The structure of DNA was proposed by Watson and Crick. It is twisted ladder like structure. It has got two coiled polynucleotides which are joined together by nitrogen bases with hydrogen bond in the centre. The longitudinal strands of DNA are made of sugars and phosphates of nucleotides. The horizontally placed nitrogen bases are of two types, purine and pyrimidine. Purines are adenine and guanine whereas pyrimidines are cytosine and thymine.
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 12

Question 6.
What is RNA primer ? Why it is necessary from DNA synthesis ?
Answer:
A primer is a short strand of RNA or DNA. RNA primer serves as a starting point for DNA polymerase, which builds complementary DNA. It is reqiured for DNA replication because the enzymes that catalyze this process. DNA polymerase can only add new nucleotide to an existing strand of DNA.

MP Board Solutions

Question 7.
What is peptide bond ?
Answer:
The bond formed between the carboxylic group (- COOH) of one amino acid and amino group (- NH2) of another amino acid is called as peptide bond. A molecule of water is released during the formation of peptide bond.

MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 13

Question 8.
Define Codon and Anticodon.
Answer:
Codon: A specific sequence of three consecutive nucleotides that is a part of the genetic code and that specifies a paticular amino acid in a protein or starts or stops protein synthesis e.g., AUG codon which is situated on die mRNA, code methionine amino acid.

Anticodon : A sequence of three adjacent nucleotides located on one end of transfer RNA. It bounds to the complementary coding triplet of nucleotides in mRNA during translation phase of protein synthesis.
For example, the anticodon for Glycine is ccc that binds to the codon (which is GGE) of mRNA.

Molecular Basis of Inheritance Long Answer Type Questions

Question 1.
Explain DNA duplication in short
Answer:
Watson and Crick after giving the double helix model of DNA, also postulated the mechanism of DNA duplication, also known as replication. According to them, during duplication, the weak hydrogen bonds between the nitrogenous base of the nucleotides get separated, so that two polynucleotide chains of DNA also separate and uncoil. The chains thus, separated are complementary to one another. These strands act as template and because of the specificity of base pairing each nucleotide of separated chain attracts its complementary nucleotide from the cell cyto-plasm. Once the nucleotides are attached by their hydrogen bonds their sugar radicals write through their Fig. DNA duplication, phosphate components completing the formation of a new polynucleotide chain. This results in the formation of two double helixes of DNA where in each molecule has one old strand contributed by parent DNA and one synthesized new. This method of DNA duplication is known as semi-conservative method.

MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 14

Question 2.
Describe the functions of nucleic acids.
Or
Explain the utility of nucleic acids.
Answer:
Utility of Nucleic acids:

  • Nucleic acids are the hereditary materials of organisms which involve in the transfer of hereditary characters
  • from one generation to the next.
  • DNA controls the synthesis of enzymes which control the various activities of the body.
  • Nucleic acids also control protein synthesis.
  • Nucleic acids form maximum portion of chromatin network.
  • It causes mutation in living beings.
  • They form enzymes.

Question 3.
Explain the structure of RNA.
Answer:
RNA molecules are single stranded nucleic acids composed of nucleotides. Four types of bases are present in RNA. These nitrogenous bases joint in different manner and form the ribonucleoside. Ribonucleoside joins together and make a polyribonucleotide chain.
AH four types of nucleoside and nucleotide are as follows :
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 15

Question 4.
Elaborate the term RNA. Also describe the types and functions of RNA.
Or
Write location and kinds of RNA in the cell.
Answer:
RNA(Ribonucleic acid): RNA located in the nucleus, cytoplasm, ribosomes and in some other cell organelles.
Types of RNA and their functions: RNA are of three types:

1. Messenger RNA (mRNA): It makes a small fraction 5-10%. This RNAdirects the sequence of amino acids in protein synthesis after joining with ribosomes. It carries the genetic information contained in DNA. It is short lived and has rapid turnover. It is formed of 700-1500 nucleotides and has a molecular weight from 5,00,000 to 20,00,000. The sequence of three nitrogenous bases of mRNA forms a codon which is responsible for coding of one amino acid.

2. Ribosomal RNA (rRNA): It makes 80% of total cell RNA. It is the most stable type of RNA and is associated with ribosomes.

3. Transfer RNA (tRNA): It makes a small fraction (10-15%) of RNA. These are smallest molecules formed of 73-93 nucleotides with molecular weight ranging between 25,000 to 30,000. rRNA works as adaptor molecules for carrying amino acids to the mRNA template during protein synthesis.

Question 5.
Describe the structure of nucleotides.
Answer:
Structure of Nucleotides : Nucleotides are the basic unit of nucleic acids and, they are also involved in the energy transfer reactions of the body. A nucleotide consists of a nitrogenous base, a pentose sugar and a phosphate group. A nucleotide possessing two groups of nitrogenous bases :

  1. Purines and
  2. Pyrimidines.

MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 16

Purines are double ringed nitrogenous bases, e.g., Adenine and Guanine. Pyrimidines are single ring nitrogenous bases like cytosine, thymine and uracil.

The pentose sugar of nucleotides is also of two types :

(i) Ribose sugar [CH2OH(CHOH)2.CHOH.CHO].
(ii) Deoxyribose sugar [CH2OH(CHOH2.CH2CHO].

A nucleotide may have one, two or three phosphates to form a nucleotides. A combination of a base and a sugar is called nucleoside and combination of a base, a sugar and phosphate group is known as nucleotide. As there are five bases so, five different kinds of nucleosides and nucleotides are known. These are listed in table given below :
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 17

MP Board Solutions

Question 6.
Give the functions of nucleotides.
Answer:
Functions of Nucleotides :

  • It works as a activated precursors of DNA and RNA.
  • They are perform the storage and conduction of energy to the form of ATP.
  • It required for activation of intermediates in many biosynthetic pathway.
  • It works as Carrier of methyl group in the form of SAM.
  • It’s Components of co-enzyme: NAD, FAD and Co A.
  • Some functions are as a vitamin. ,
  • They are control and coordinates different activities in our body.

Question 7.
Write four features of genetic code.
Answer:
According to Nirenberg, Khorana and Holley, genetic code is that sequence of nitrogenous bases of DNA in which genetic informations for the synthesis of protein are coded.

Characteristic features of genetic code :

1. The code is triplet: The codon is a specific sequence of three nitrogenous bases of mRNA.

2. The code is commaless: The sequence of bases read in blocks of three at a time form a particular position. There is no gap between two subsequent codons.

3. Code is degenerating: Presence of more than one codon for one amino acid is called as degeneracy of codons, e.g., Serine having three codons UCU, UCA, AGU.

4. Codes are universal: Codons are similar in all organisms, e.g., serine is coded by UCU codon in all the living beings.

5. Codes are non-ambiguous : The position of genetic code in cellular medium is nonambiguous because a codon always codes only one amino acid. Sometimes a codon codes more than one amino acid, e.g., in E. coli. UUV codon generally code phenylalanine, after treatment of their ribosome with streptomycin. It can also code isoleucine, leucine and serine.

6. Initiation and termination codon: Codons responsible for the initiation of polypeptide chain are called as initiation codon, e.g., AUG. Likewise, codons responsible for the termination of polypeptide chain are called as chain termination codon, e.g., UAA, UAG, UGA.

Question 8.
Write five characters of gene hypothesis.
Answer:
Sutton, Bridges, Muller and Morgan suggest these theory. The character’s of gene of this theory are as follows :

  • Genes are situated on the chromosome.
  • They make the physiological character of organisms.
  • These are called functional unit of specific characters.
  • Genes have the capacity of self-transcription.
  • They perform mutation.
  • Characters are goes to one generation to other by parents.

Question 9.
What is gene expression ? Explain by different methods of gene expression in animals.
Answer:
The mechanism at molecular level by which a gene is able to express itself in the phenotype of an organism is called gene expression.
Different methods of gene expression in animals are:

1. Transduction : It is the process in which bacteriophages pick up pieces of DNA from one bacterial cell and transfer the same to another on infection.
2. Transformation : It is the process by which DNA isolated from one type of cell when introduced into another, is able to bestow some of the properties of the former to the latter.

MP Board Solutions

Question 10.
What is proof reading and repair of DNA ?
Answer:
Variety of environmental factors such as radiation, chemicals etc. may cause damage in DNA of a cell. The bacterial DNA polymerase III can do proofreading, in the sense that it can go back and remove the wrong base before it proceeds to add new bases in the 5′ → 3′ direction. It is called proof reading. Obviously, the survival of the cell depends on its availability of damages:

(i) Monoadduct: Which involve alterations in a single nitrogenous base.
(ii) Diadducts : They are the alterations involving more than one nitrogenous base. Number of nucleases have been found to be involved in repair replication such as Exonucleases (defined as phosphodiesterases which require a terminus for hydrolysis and cut off terminal nucleotides), Endonucleases (which are also phosphodiesterases which do not require a terminus for hydrolysis and break internal bonds). The endonucleases which act on the damaged DNA and cause repair or correction of this molecule are referred to as correctional nucleases. The following steps are said to be involved in the repair repl’iGatipn i.e., Incision, Excision, Reinsertion and joining of newly formed strands.

Question 11.
Write any four differences between DNA and RNA.
Answer:
Differences between DNA and RNA:
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 18

Question 12.
Write the names of enzymes used in DNA replication.
Ans.
The names of enzymes used in DNA replication are as follows :

  • DNA helicase : For unwinding of two strands.
  • DNA gyrase : For relieving tension.
  • Primase : For formation of primer.
  • DNA polymerase : For DNA synthesis.
  • RNA primer : For initiation of the synthesis of DNA segments.
  • DNA ligase : For joining of DNA Okazaki segments.

Question 13.
What is transcription ? Name the enzyme catalysing it
Answer:
Transcription : Formation of mRNA from DNA in the presence of enzyme is called transcription. It is the first stage of protein synthesis which is catalysed, by RNA polymerase enzyme. The process of transcription involves in the following steps :

1. Exposing of the bases of DNA : The two strands of DNA are separated due to presence of an unwinding protein and thus, their bases are exposed. The exposed chain of DNA functions as template for the synthesis of mRNA in the presence of RNA polymerase enzyme.

2. Base pairing: The ribonucleotides are jointed in a definite fashion on the exposed strand of DNA. G is bonded with ‘C’, ‘C’ bonded with ‘G’, ‘T’ bonded with ‘A’ and ‘A’ bonded with‘‘T’respectively.

3. Synthesis of RNA chain : The new ribonucleotide bonded on DNA template are jointed with the help of RNA polymerase and thus, forming a new chain of RNA. Then this mRNA is separated from DNA and reaches the cytoplasm. Where it combines with ribosomes and thus, initiating the synthesis of protein.

Question 14.
What is translation ? Explain it
Answer:
Translation : The translation step of protein synthesis involves translation of the language of nucleic acids (available in the form of mRNA) into language of protein. The sequence of bases in mRNA, decides the sequence of amino acids in proteins. Each amino acid is programmed by a triplet code. It consists of a sequence of three bases in the DNA and the complementary bases in mRNA. The synthesis of protein occurs in three steps, initiation, elongation and termination. After the final step i.e., termination, the proteins are transported out of the cell or translocated within the cell. Thus, the transformation of nucleotides chain of RNA into polypeptide chain of protein is called as translation.

It is completed in following steps :

  1. Activation of amino acids.
  2. Binding of activated amino acids with rRNA.
  3. Binding of mRNA with smaller unit of ribosome.
  4. Initiation of polypeptide chain.
  5. Elongation of polypeptide chain.
  6. Termination of polypeptide chain.

MP Board Solutions

Question 15.
Describe the evidence given by Griffith in support of DNA as genetic material. Explain it along with suitable diagram.
Answer:
Griffith had done transformation experiments in mice to prove that DNA is the genetic material. He took virulent strain of Diplococcus pneumonae (S-III) which causes pneumonia in mice and injected it into mice which resulted in the production of pneumonia in mice. He also injected a non-virulent strain of that bacteria in the body of mice and found that all the mice were unaffected.

In third experiment he injected heat killed (S-III) strain and non-virulent strain R-II strain together in the body of mice and found that all the mice suffered from pneumonia and became dead. After analysis it was found that these mice contained both the strains of Diplococcus pneumonae. Thus, this experiment proved that any substance of S-III strain is transferred into R-II strain due to which R-II strain become virulent. Later, McLeod, Avery and McCarty observed that DNA molecules are transferred from S-III to R-II strain and make virulent. Thus, it is proved that DNA is the genetic material.
MP Board Class 12th Biology Solutions Chapter 6 Molecular Basis of Inheritance 19

MP Board Class 12th Biology Solutions

Leave a Reply