case study questions class 12 biology molecular basis of inheritance

Class 12 Biology Case Study of Chapter 6 Molecular Basis of Inheritance

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Downloads of CBSE Class 12 Biology Chapter 6 Molecular Basis of Inheritance Case Study and Passage Based Questions with Answers were Prepared Based on the Latest Exam Pattern. Students can solve Class 12 Biology Case Study Questions Molecular Basis of Inheritance  to know their preparation level.

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In CBSE Class 12 Biology Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Molecular Basis of Inheritance Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Biology  Chapter 6 Molecular Basis of Inheritance

Case Study/Passage-Based Questions

Case Study 1: In prokaryotes, DNA is circular and present in the cytoplasm but in eukaryotes, DNA is linear and mainly confined to the nucleus. DNA or deoxyribonucleic acid is a long polymer of nucleotides. In 1953, the first correct double helical structure of DNA was worked out by Watson and Crick. Based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin. It is composed of three components, i.e., A phosphate group, deoxyribose sugar, and a nitrogenous base. Different forms of DNA are B-DNA, Z-DNA, A-DNA, C-DNA, and D-DNA.

(i) Name the linkage present between the nitrogen base and pentose sugar in DNA.

Answer: (b) Glycosidic bond

(ii) The double helix structure of DNA was proposed by

Answer: (a) James Watson and Francis Crick

(iii) The double chain of B-DNA is coiled in a helical fashion. The spiral twisting of the B-DNA duplex produces

Answer: (b) major and minor grooves

(iv)  Assertion:  The two strands of DNA helix have a uniform distance between them. Reason: A large-sized purine is always paired opposite to a small-sized pyrimidine.

Answer: (a) Both assertion and reason are true and reason is the correct explanation of assertion

 (v) Which of the following describes the structure of B-DNA?

Answer: (b)

Case Study 2: The process of translation requires the transfer of genetic information from a polymer of nucleotides to synthesize a polymer of amino acids. The relationship between the sequence of amino acids in a polypeptide and the nucleotide sequence of DNA or mRNA is called genetic code. George Gamow suggested that in order to code for all the 20 amino acids, code should be made up of three nucleotides.

What is the process by which a polymer of nucleotides is used to synthesize a polymer of amino acids? A) Replication B) Transcription C) Translation D) Mutation

What is the relationship between the sequence of amino acids in a polypeptide and the nucleotide sequence of DNA or mRNA known as? A) Genetic Translation B) Genetic Mutation C) Genetic Replication D) Genetic Code

Who suggested that the genetic code should be made up of three nucleotides? A) James Watson B) Francis Crick C) George Gamow D) Rosalind Franklin

How many nucleotides are required to code for one amino acid in the genetic code? A) One B) Two C) Three D) Four

What is the main reason that three nucleotides are used to code for each amino acid? A) To ensure replication accuracy B) To allow for more combinations and code for all 20 amino acids C) To facilitate faster transcription D) To prevent mutations

Which of the following molecules carries the genetic information from DNA to the ribosome for translation? A) tRNA B) rRNA C) mRNA D) DNA

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CBSE 12th Standard Biology Subject Molecular Basic of Inheritance Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 Biology, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

12th Standard CBSE

Final Semester - June 2015

Case Study Questions

Read the following and answer any four questions from (i) to (v) given below: In prokaryotes, DNA is circular and present in the cytoplasm but in eukaryotes, DNA is linear and mainly confined to the nucleus. DNA or deoxyribonucleic acid is a long polymer of nucleotides. In 1953, the first correct double helical structure of DNA was worked out by Watson and Crick. Based on the X-ray diffraction data produced by Maurice Wilkins and Rosalind Franklin. It is composed of three components, i.e., A phosphate group, a deoxyribose sugar and a nitrogenous base. Different forms of DNA are B-DNA, Z-DNA, A-DNA, C-DNA and D-DNA. (i) Name the linkage present between the nitrogen base and pentose sugar in DNA.

(ii) The double helix structure of DNA was proposed by

(iii) The double chain of B-DNA is coiled in a helical fashion. The spiral twisting of B-DNA duplex produces

(iv) Assertion: The two strands of DNA helix have uniform distance between them. Reason: A large sized purine always paired opposite to a small sized pyrimidine.

 (v) Which of the following describes the structure of B-DNA?

Read the following and answer any four questions from (i) to (v) given below: The process of translation requires transfer of genetic information from a polymer of nucleotides to synthesise a polymer of amino acids. The relationship between the sequence of amino acids in a polypeptide and nucleotide sequence of DNA or mRNA is called genetic code. George Gamow suggested that in order to code for all the 20 amino acids, code should be made up of three nucleotides. (i) What is a codon?

(ii) Three consecutive bases in the DNA molecule provide the code for each amino acid in a protein molecule. What is the maximum number of different triplets that could occurs ?

(iii) Listed below are some amino acids and their corresponding mRNA triplets.

Which DNA sequence would be needed to produce the following polypeptide sequence? Alanine- Arginine- Lysine- Phenylalanine

(iv) Identify the non-sense codons among the following.

(v) A polypeptide is made using synthetic mRNA molecules as shown.

What are the DNA codes for the amino acids phenylalanine and lysine?

*****************************************

Cbse 12th standard biology subject molecular basic of inheritance case study questions with solution 2021 answer keys.

(i) Replication of DNA occurs in small replication forks, because DNA is such a long molecule that the separation of the two strands along its entire length requires a very high amount of energy. (ii) a - Continuous synthesis. b - Discontinuous synthesis (iii) A - 5' B-3'.

(a) 'Termination of transcription. (b) A - Template strand of DNA. B - Coding strand of DNA. C - RNA synthesised D - RNA-polymerase E - rho (p) factor.

(a) 11 amino acids will be coded, as the last codon is a termination codon that does not code for any amino acid. (b) Dual functions of AVG: (i) It acts as the initiation codon for translation. ·(ii) It codes for the amino acid, methionine.

(i) (b): In DNA the nitrogenous base and a pentose sugar joins to form nucleoside with the help of bond called glycosidic bond or N-glycosidic linkage. (ii) (a): The correct structure of DNA was first worked out by James Watson and Francis Crick in 1953. Their double-helix model of DNA structure was based on two major investigations, i.e., Chargaff's rules of base i'airing and study of X-ray diffraction pattern of DNA produced by Maurice Wilkins and Rosalind Franklin which helped Watson and Crick to design the 3-dimensional structure of DNA. (iii) (b): Due to spiral twisting, the B-DNA duplex comes to have two types of alternate grooves, i.e., majo: (length 22 Â) and minor (length 12 Â). (iv) (a)  (v) (b): The double helical chains of B-DNA are bound to each other via hydrogen bonds in an antiparallel fashion, i.e., 5'-3' in one and 3'-5' in other. The pitch of helix per turn is 3.4 nm with 10 base pairs in each turn.

(i) (d): Codon is complementary to a triplet of templet strand. It is found in mRNA. Anticodon is complementary to a codon it occurs in tRNA. (ii) (d) : The triplet code consists of three of the four nucleotide bases - A, C, G or T. Thus the maximum number of codon is 4 3 = 64. (iii) (b): The complementary bases of GCA-CGAAAG- UUU are CGT-GCT-TTC-AAA on the DNA strand. (iv) (c): AUG and GUG are initiation codon which codes for methionine and valine respectively: UGG codes for tryptophan. UAA (ochre) is a termination codon. (v) (a): The triplet codon of phenylalanine (UUU) ")Willbase pair with AAA in the DNA molecule and that oflysine (AAA) will base pair with TTT.

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Case Study and Passage Based Questions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

• Last modified on: 2 years ago
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Case Study/Passage Based Questions:

Question 1:

Given below is the diagram of a tRNA molecule.

Answer the questions based on the above diagram: (i) Why is charging of tRNA essential in translation? (ii) Where does peptide bond formation occur in a bacterial ribosome? (iii) Name the scientist who called tRNA an adaptor molecule.

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Total Papers :

CBSE Class 12 Biology Syllabus

• Reproduction in Organisms
• Sexual Reproduction in Flowering Plants
• Human Reproduction
• Reproductive Health
• Principles of Inheritance and Variation
• Molecular Basis of Inheritance
• Human Health and Diseases
• Strategies for Enhancement in Food Production
• Microbes in Human Welfare
• Biotechnology - Principles and Processes
• Biotechnology and its Application
• Organisms and Populations
• Biodiversity and its Conservation
• Environmental Issues

Course Syllabus Details

Unit vi. reproduction.

Chapter 1: Reproduction in Organisms

• A characteristic feature of all organisms for continuation of species
• Asexual reproduction
• Sexual reproduction
• Modes-Binary fission
• Sporulation
• Fragmentation
• vegetative propagation in plants

Chapter 2: Sexual Reproduction in Flowering Plants

• Flower structure
• Development of male and female gametophytes
• Outbreeding devices
• Pollen-Pistil interaction
• Double fertilization
• Post fertilization Events-Development of endosperm and embryo
• Development of seed and formation of fruit
• Parthenocarpy
• Polyembryony
• Significance of seed and fruit formation

Chapter 3: Human Reproduction

• Male and female reproductive systems
• Microscopic anatomy of testis and ovary
• Spermatogenesis
• Menstrual cycle
• Fertilisation embryo development up to blastocyst formation, implantation pregnancy and placenta formation (Elementary idea)
• Parturition (Elementary idea)
• Lactation (Elementary idea)

Chapter 4: Reproductive Health

• Need for reproductive health and prevention of sexually transmitted diseases (STD)
• Need and Methods
• Contraception
• Medical Termination of Pregnancy (MTP)
• Amniocentesis
• GIFT (Elementary idea for general awareness)

Unit VII. Genetics and Evolution

Chapter 5: Principles of Inheritance and Variation

• Mendelian Inheritance
• Incomplete dominance
• Co-dominance
• Multiple alleles
• Inheritance of blood groups
• Elementary idea of polygenic inheritance
• Chromosome theory of inheritance
• Chromosomes and genes
• Linkage and crossing over
• Haemophilia
• Colour blindness
• Thalassemia
• Down's syndrome
• Klinefelter's syndromes

Chapter 6: Molecular Basis of Inheritance

• Search for genetic material and DNA as genetic material
• Structure of DNA and RNA
• DNA packaging
• DNA replication
• Central dogma
• Transcription, genetic code, translation
• Genome and human ganeome project
• DNA fingerprinting

Chapter 7: Evolution

• Origin of life
• Biological evolution and evidences for biological evolution (Paleontological, comparative anatomy, embryology and molecular evidence)
• Darwin's contribution
• Modern Synthetic theory of Evolution
• Variation (Mutation and Recombination)
• Natural Selection with examples
• Types of natural selection
• Gene flow and genetic drift
• Hardy - Weinberg's principle
• Adaptive Radiation
• Human evolution

Unit VIII. Biology and Human Welfare

Chapter 8: Human Health and Diseases

• Common cold
• Adolescence, drug and alcohol abuse

Chapter 9: Strategies for Enhancement in Food Production

• Plant breeding
• Tissue culture
• Single cell protein
• Biofortification
• Apiculature
• Animal husbandry

Chapter-10: Microbes in Human Welfare

• In household food processing
• Industrial production
• Sewage treatment
• Energy generation and as biocontrol agents
• Biofertilizers
• Production and judicious use

Unit IX. Biotechnology and Its Applications

Chapter 11: Biotechnology - Principles and Processes

• Genetic engineering (Recombinant DNA technology).

Chapter 12: Biotechnology and its Application

• Human insulin and vaccine production, gene therapy
• Genetically modified organisms - Bt crops
• Transgenic Animals; biosafety issues, biopiracy and patents

Unit X. Ecology and Environment

Chapter 13: Organisms and Populations

• Ecological adaptations
• Competition
• Age distribution

Chapter 14: Ecosystem

• Productivity
• Decomposition
• Energy flow
• Pyramids of number, biomass, energy
• Nutrient cycles (carbon and phosphorous)
• Ecological succession
• Carbon fixation
• Pollination
• Seed dispersal
• Oxygen release (in brief)

Chapter-15: Biodiversity and its Conservation

• Concept of biodiversity
• Patterns of biodiversity
• Importance of biodiversity
• Loss of biodiversity
• Endangered organisms
• Red data book
• Biosphere reserves
• National parks
• Sanctuaries
• Ramsar sites

Chapter-16: Environmental Issues

• Air pollution and its control
• Water pollution and its control
• Agrochemicals and their effects
• Solid waste management
• Radioactive waste management
• Greenhouse effect and climate change
• Ozone layer depletion
• Deforestation
• Any one case study as success story addressing environmental issue(s).

Practical Works

Part A: List of Experiments

• Study pollen germination on a slide.
• Collect and study soil from at least two different sites and study them for texture, moisture content, pH and water holding capacity. Correlate with the kinds of plants found in them.
• Collect water from two different water bodies around you and study them for pH, clarity and presence of any living organisms.
• Study the presence of suspended particulate matter in air at two widely different sites.
• Study of plant population density by quadrat method.
• Study of plant population frequency by quadrat method.
• Prepare a temporary mount of onion root tip to study mitosis.
• Study the effect of different temperatures and three different pH on the activity of salivary amylase on starch.
• Isolation of DNA from available plant material such as spinach, green pea seeds, papaya, etc.

Part B: Study/observation of the following (Spotting)

• Flowers adapted to pollination by different agencies (wind, insect, bird).
• Pollen germination on stigma through a permanent slide.
• Identification of stages of gamete development, i.e., T.S. of testis and T.S. of ovary through permanent slides (from grasshopper/mice).
• Meiosis in onion bud cell or grasshopper testis through permanent slides.
• T.S. of blastula through permanent slides.
• Mendelian inheritance using seeds of different colour/sizes of any plant.
• Prepared pedigree charts of any one of the genetic traits such as rolling of tongue, blood groups, ear lobes, widow's peak and colour blindness.
• Controlled pollination - emasculation, tagging and bagging.
• Common disease causing organisms like Ascaris, Entamoeba, Plasmodium, Roundworm through permanent slides or specimens. Comment on symptoms of disease that they cause.
• Two plants and two animals (models/virtual images) found in xeric conditions. Comment upon their morphological adaptations.
• Two plants and two animals (models/virtual images) found in aquatic conditions. Comment upon their morphological adaptations.

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There are many ways to ascertain whether a student has understood the important points and topics of a particular chapter and is he or she well prepared for exams and tests of that particular chapter. Apart from reference books and notes, Question Banks are very effective study materials for exam preparation. When a student tries to attempt and solve all the important questions of any particular subject , it becomes very easy to gauge how much well the topics have been understood and what kind of questions are asked in exams related to that chapter.. Some of the other advantaging factors of Question Banks are as follows

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• CBSE Question Banks are great tools to help in analysis for Exams. As it has a collection of important questions that were asked previously in exams thereby it covers every question from most of the important topics. Thus solving questions from the question bank helps students in analysing their preparation levels for the exam. However the practice should be done in a way that first the set of questions on any particular chapter are solved and then solutions should be consulted to get an analysis of their strong and weak points. This ensures that they are more clear about what to answer and what can be avoided on the day of the exam.
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CBSE Class 12 Biology –Chapter 6 Molecular Basis of Inheritance- Study Materials

NCERT Solutions Class 12 All Subjects Sample Papers Past Years Papers

Molecular basis of Inheritance : Notes and Study Materials -pdf

Notes and study materials.

• Concepts of Molecular basis of Inheritance
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• Molecular basis of Inheritance : Practice Paper 1
• Molecular basis of Inheritance : Practice Paper 2
• Molecular basis of Inheritance : Practice Paper 3

6. MOLECULAR BASIS OF INHERITANCE

·      Nucleic acids (DNA & RNA) are the building blocks of genetic material.

·      DNA is the genetic material in most of the organisms.

·      RNA is the genetic material in some viruses. RNA mostly functions as messengers.

STRUCTURE OF POLYNUCLEOTIDE CHAIN

Polynucleotides are the polymer of nucleotides. DNA & RNA are polynucleotides. A nucleotide has 3 components:

1.     A nitrogenous base.

2.     A pentose sugar (ribose in RNA & deoxyribose in DNA).

3.     A phosphate group.

Nitrogen bases are 2 types:

} Purines: It includes Adenine (A) and Guanine (G).

} Pyrimidines: It includes Cytosine (C), Thymine (T) & Uracil (U). Thymine (5-methyl Uracil) present only in DNA and Uracil only in RNA.

A nitrogenous base is linked to the OH of 1′ C pentose sugar through an N-glycosidic linkage to form nucleoside.

A phosphate group is linked to OH of 5′ C of a nucleoside through phosphoester linkage to form nucleotide (or deoxynucleotide).

In RNA, each nucleotide has an additional –OH group at 2 ‘ C of the ribose (2’- OH).

2 nucleotides are linked through 3’-5’ phosphodiester bond to form dinucleotide.

When more nucleotides are linked, it forms polynucleotide.

STRUCTURE OF THE DNA

} Friedrich Meischer (1869): Identified DNA and named it as ‘Nuclein’.

} James Watson & Francis Crick (1953) proposed double helix model of DNA. It was based on X-ray diffraction data produced by Maurice Wilkins & Rosalind Franklin.

} DNA is made of 2 polynucleotide chains coiled in a right-handed fashion. Its backbone is formed of sugar & phosphates. The bases project inside.

} The 2 chains have anti-parallel polarity , i.e. one chain has the polarity 5’→3’ and the other has 3’→5’.

} The bases in 2 strands are paired through H-bonds forming base pairs (bp).

 A=T (2 hydrogen bonds)      C≡G (3 hydrogen bonds)

} Purine comes opposite to a pyrimidine. This generates uniform distance between the 2 strands.

} Erwin Chargaff’s rule: In DNA, the proportion of A is equal to T and the proportion of G is equal to C.

∴    [A] + [G] = [T] + [C]      or     [A] + [G] / [T] + [C] =1

v Ф 174 (a bacteriophage) has 5386 nucleotides.

v Bacteriophage lambda has 48502 base pairs (bp).

v E. coli has 4.6×10 6 bp.

v Haploid content of human DNA is 3.3×10 9 bp.

Length of DNA = number of base pairs X distance between two adjacent base pairs.

Number of base pairs in human    = 6.6 x 10 9

Hence, the length of DNA            = 6.6 x10 9 x 0.34x 10 -9

                                                    = 2.2 m

In E. coli , length of DNA          =1.36 mm (1.36 x 10 -3 m)

∴ Therefore the number of base pairs 

PACKAGING OF DNA HELIX

§ In prokaryotes (E.g. E. coli ), the DNA is not scattered throughout the cell. DNA is negatively charged. So it is held with some positively charged proteins to form nucleoid.

§ In eukaryotes, there is a set of positively charged, basic proteins called histones.

§ Histones are rich in positively charged basic amino acid residues lysines and arginines.

§ 8 histones form histone octamer.

§ Negatively charged DNA is wrapped around histone octamer to give nucleosome.

§ A typical nucleosome contains 200 bp.

Therefore, total number of nucleosomes in human =

6.6 x 10 9 bp 200 = 3.3x 10 7

§ Nucleosomes constitute the repeating unit to form chromatin. Chromatin is the thread-like stained bodies.

§ Nucleosomes in chromatin = ‘beads-on-string’.

§ Chromatin is packaged → chromatin fibres → coiled and condensed at metaphase stage → chromosomes.

§ Higher level packaging of chromatin requires non-histone chromosomal (NHC) proteins.

§ Chromatin has 2 forms:

·    Euchromatin: Loosely packed and transcriptionally active region of chromatin. It stains light.

·    Heterochromatin: Densely packed and inactive region of chromatin. It stains dark.

THE SEARCH FOR GENETIC MATERIAL

Frederick Griffith used mice & Streptococcus pneumoniae .

Streptococcus pneumoniae has 2 strains :

◦    Smooth (S) strain (Virulent): Has polysaccharide mucus coat. Cause pneumonia.

◦    Rough (R) strain (Non-virulent): No mucus coat. Do not cause Pneumonia.

Experiment:

• S-strain → Inject into mice → Mice die
• R-strain → Inject into mice → Mice live
• S-strain (Heat killed) → Inject into mice → Mice live
• S-strain (Heat killed) + R-strain (live) → Inject into mice → Mice die

He concluded that some ‘transforming principle’ transferred from heat-killed S-strain to R-strain. It enabled R-strain to synthesize smooth polysaccharide coat and become virulent. This must be due to the transfer of genetic material.

2. Biochemical characterization of transforming principle

–    Oswald Avery, Colin MacLeod & Maclyn McCarty worked to determine the biochemical nature of ‘transforming principle’ in Griffith’s experiment.

–   They purified biochemicals (proteins, DNA, RNA etc.) from heat killed S cells using suitable enzymes.

–   They discovered that

• Digestion of protein and RNA (using Proteases and RNases ) did not affect transformation. It means that the transforming substance was not a protein or RNA.
• Digestion of DNA with DNase inhibited transformation. It means that DNA caused transformation of R cells to S cells. It proves that DNA was the transforming principle.

3. Hershey-Chase Experiment (Blender Experiment)-1952

• Hershey & Chase grew some bacteriophage viruses on a medium containing radioactive phosphorus ( P 32 ) and some others on medium containing radioactive sulphur ( S 35 ) .
• Viruses grown in P 32 got radioactive DNA because only DNA contains phosphorus. Viruses grown in S 35 got radioactive protein because protein contains sulphur.
• These preparations were used separately to infect E. coli.
• After infection, the E. coli cells were gently agitated in a blender to remove the virus particles from the bacteria.
• Then the culture was centrifuged to separate lighter virus particles from heavier bacterial cells.
• Bacteria infected with viruses having radioactive DNA were radioactive. i.e., DNA had passed from the virus to bacteria. Bacteria infected with viruses having radioactive proteins were not radioactive. i.e., proteins did not enter the bacteria from the viruses. This proves that DNA is the genetic material .

PROPERTIES OF GENETIC MATERIAL (DNA v/s RNA)

A genetic material must have the following properties:

• Ability to generate its replica (Replication).
• Chemical and structural stability.
• Provide the mutations that are required for evolution.
• Ability to express as Mendelian Characters.

Reasons for stability (less reactivity) of DNA Reasons for mutability (high reactivity) of RNA Double stranded Single stranded Presence of thymine Presence of Uracil Absence of 2’-OH in sugar Presence of 2’-OH in sugar

–     RNA is unstable. So, RNA viruses (E.g. Q.B bacteriophage, Tobacco Mosaic Virus etc.) mutate and evolve faster.

–     DNA strands are complementary. On heating, they separate. In appropriate conditions, they come together. In Griffith’s experiment, some properties of DNA of the heat killed bacteria did not destroy. It indicates the stability of DNA.

–     For the storage of genetic information, DNA is better due to its stability. But for the transmission of genetic information, RNA is better.

–   RNA can directly code for the protein synthesis, hence can easily express the characters. DNA is dependent on RNA for protein synthesis.

• RNA was the first genetic material.
• It acts as genetic material and catalyst.
• Essential life processes (metabolism, translation, splicing etc.) evolved around RNA.
• DNA evolved from RNA for stability.

CENTRAL DOGMA OF MOLECULAR BIOLOGY

· It is proposed by Francis Crick. It states that the genetic information flows from DNA → RNA → Protein.

· In some viruses, flow of information is in reverse direction (from RNA to DNA). It is called reverse transcription.

DNA REPLICATION

· Replication is the copying of DNA from parental DNA.

· Watson & Crick proposed Semi-conservative model of replication. It suggests that the parental DNA strands act as template for the synthesis of new complementary strands. After replication, each DNA molecule would have one parental and one new strand.

· Matthew Messelson & Franklin Stahl (1958) experimentally proved Semi-conservative model.

Messelson & Stahl’s Experiment

} They grew E. coli in 15 NH 4 Cl medium ( 15 N = heavy isotope of nitrogen) as the only nitrogen source. As a result, 15 N was incorporated into newly synthesised DNA (heavy DNA or 15 N DNA).

} Heavy DNA can be distinguished from normal DNA (light DNA or 14 N DNA) by centrifugation in a cesium chloride (CsCl) density gradient.

} E. coli cells from 15 N medium were transferred to 14 NH 4 Cl medium. After one generation (i.e. after 20 minutes), they isolated and centrifuged the DNA. Its density was intermediate (hybrid) between 15 N DNA and 14 N DNA. This shows that in newly formed DNA, one strand is old ( 15 N type) and one strand is new ( 14 N type). This confirms semi-conservative replication.

} After II generation (i.e. after 40 minutes), there was equal amounts of hybrid DNA and light DNA.

Taylor & colleagues (1958) performed similar experiments on Vicia faba (faba beans) using radioactive thymidine to detect distribution of newly synthesized DNA in the chromosomes. It proved that the DNA in chromosomes also replicate semi-conservatively.

The Machinery and Enzymes for Replication

· DNA replication starts at a point called origin ( ori ) .

· A unit of replication with one origin is called a replicon.

· During replication, the 2 strands unwind and separate by breaking H-bonds in presence of an enzyme, Helicase.

· Unwinding of the DNA molecule at a point forms a ‘Y’-shaped structure called replication fork.

· The separated strands act as templates for the synthesis of new strands.

· DNA replicates in the 5’→3’ direction.

· Deoxyribonucleoside triphosphates (dATP, dGTP, dCTP & dTTP) act as substrate and provide energy for polymerization.

· Firstly, a small RNA primer is synthesized in presence of an enzyme, primase .

· In presence of an enzyme, DNA dependent DNA polymerase , many nucleotides join with one another to primer strand and form a polynucleotide chain (new strand).

· During replication, one strand is formed as a continuous stretch in 5’ → 3’ direction (Continuous synthesis). This strand is called leading strand.

· The other strand is formed in small stretches (Okazaki fragments) in 5’ → 3’ direction (Discontinuous synthesis).

· The Okazaki fragments are then joined together to form a new strand by an enzyme, DNA ligase . This new strand is called lagging strand.

· If a wrong base is introduced in the new strand, DNA polymerase can do proof reading.

· E. coli completes replication within 18 minutes. i.e. 2000 bp per second.

· In eukaryotes, the replication of DNA takes place at S-phase of the cell cycle. Failure in cell division after DNA replication results in polyploidy.

TRANSCRIPTION

– It is the process of copying genetic information from one strand of the DNA into RNA.

– Here, adenine pairs with uracil instead of thymine.

– The DNA- dependent RNA polymerase catalyzes the polymerization only in 5’→3’direction.

– 3’→5’ acts as template strand . RNA is built from this.

– 5’→3’ acts as coding strand. This is copied to RNA.

3’-ATGCATGCATGCATGCATGCATGC-5’ template strand.

5’-TACGTACGTACGTACGTACGTACG-3’ coding strand.

– During transcription, both strands are not copied because

◦ The code for proteins is different in both strands. This complicates the translation.

◦ If 2 RNA molecules are produced simultaneously, this would be complimentary to each other. It forms a double stranded RNA and prevents translation.

Transcription Unit

– It is the segment of DNA between the sites of initiation and termination of transcription. It consists of 3 regions:

◦ A promoter: Binding site for RNA polymerase. Located towards 5′-end (upstream).

◦ Structural gene: The region between promoter and terminator where transcription takes place.

◦ A terminator: The site where transcription stops. Located towards 3′-end (downstream).

Transcription unit and gene

Gene is a functional unit of inheritance. It is the DNA sequence coding for an RNA (mRNA, rRNA or tRNA).

Cistron is a segment of DNA coding for a polypeptide during protein synthesis. It is the largest element of a gene.

Structural gene in a transcription unit is 2 types:

} Monocistronic structural genes (split genes) : It is seen in eukaryotes. Here, coding sequences (exons or expressed sequences) are interrupted by introns (intervening sequences).

Exons appear in processed mRNA.

Introns do not appear in processed mRNA.

} Polycistronic structural genes : It is seen in prokaryotes. Here, there are no split genes.

Transcription in prokaryotes

In bacteria (Prokaryotes), synthesis of all types of RNA are catalysed by a single RNA polymerase. It has 3 steps:

} Initiation: Here, the enzyme RNA polymerase binds at the promoter site of DNA. This causes the local unwinding of the DNA double helix. An initiation factor ( σ factor) present in RNA polymerase initiates the RNA synthesis.

} Elongation: RNA chain is synthesized in 5’-3’ direction. In this process, activated ribonucleoside triphosphates (ATP, GTP, UTP & CTP) are added. This is complementary to the base sequence in the DNA template.

} Termination: A termination factor ( ρ factor) binds to the RNA polymerase and terminates the transcription.

In bacteria, transcription and translation can be coupled ( translation begins before mRNA is fully transcribed) because

· mRNA requires no processing to become active.

· Transcription and translation take place in the same compartment (no separation of cytosol and nucleus).

Transcription in eukaryotes

In eukaryotes, there are 2 additional complexities:

1. There are 3 RNA polymerases:

· RNA polymerase I: Transcribes rRNAs (28S, 18S & 5.8S).

· RNA polymerase II: Transcribes the heterogeneous nuclear RNA (hnRNA). It is the precursor of mRNA.

· RNA polymerase III: Transcribes tRNA, 5S rRNA and snRNAs (small nuclear RNAs).

2. The primary transcripts (hnRNA) contain exons and introns and are non-functional. Hence introns must be removed. For this, it undergoes the following processes:

· Splicing: From hnRNA, introns are removed (by the spliceosome) and exons are spliced (joined) together.

· Capping: Here, a nucleotide methyl guanosine triphosphate (cap) is added to the 5’ end of hnRNA.

· Tailing (Polyadenylation): Here, adenylate residues (200-300) are added at 3’-end.

Now, it is the fully processed hnRNA, called mRNA.

GENETIC CODE

§ It is the sequence of nucleotides (nitrogen bases) in mRNA that contains information for protein synthesis (translation).

§ The sequence of 3 bases determining a single amino acid is called codon.

§ George Gamow suggested that for coding 20 amino acids, the code should be made up of 3 nucleotides. Thus, there are 64 codons (4 3 = 4 x 4 x 4).

§ Har Gobind Khorana developed the chemical method in synthesizing RNA molecules with defined combinations of bases (homopolymers & copolymers).

§ Marshall Nirenberg developed cell-free system for protein synthesis.

§ Severo Ochoa ( polynucleotide phosphorylase ) enzyme is

used to polymerize RNA with defined sequences in a template independent manner.

20 types of amino acids involved in translation

Alanine (Ala) Arginine (Arg) Asparagine (Asn) Aspartic acid (Asp) Cystein (Cys) Glutamine (Gln) Glutamic acid (Glu) Glycine (Gly) Histidine (His) Isoleucine (Ile) Leucine (Leu) Lysine (Lys) Methionine (Met) Phenyl alanine (Phe) Proline (Pro) Serine (Ser) Threonine (Thr) Tryptophan (Trp) Tyrosine (Tyr) Valine (Val)

The codons for various amino acids

Salient features of genetic code

· Codon is triplet (three-letter code).

· 61 codons code for amino acids. 3 codons ( UAA, UAG & UGA) do not code for any amino acids. They act as stop codons ( Termination codons or non-sense codons) .

· Genetic code is universal. E.g. From bacteria to human UUU codes for Phenylalanine. Some exceptions are found in mitochondrial codons, and in some protozoans.

· No punctuations b/w adjacent codons (comma less code). The codon is read in mRNA in a contiguous fashion.

· Genetic code is non-overlapping.

· An amino acid is coded by more than one codon (except AUG for methionine & UGG for tryptophan). Such codons are called degenerate codons.

· Genetic code is unambiguous and specific. i.e. one codon specifies only one amino acid.

· AUG has dual functions. It codes for Methionine and acts as initiator codon. In eukaryotes, methionine is the first amino acid and formyl methionine in prokaryotes.

Mutations and Genetic Code

– Relationship between genes & DNA are best understood by mutation studies. Deletions & rearrangements in a DNA may cause loss or gain of a gene and so a function.

– Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion. It is called frame-shift insertion or deletion mutations.

– Insertion/ deletion of three or its multiple bases insert or delete one or multiple codon. The reading frame remains unaltered from that point onwards. Hence one or multiple amino acids are inserted /deleted.

– It proves that codon is a triplet and is read contiguously.

TYPES OF RNA

· mRNA (messenger RNA): Provide template for translation (protein synthesis).

· rRNA (ribosomal RNA): Structural & catalytic role during translation. E.g. 23S rRNA in bacteria acts as ribozyme.

· tRNA (transfer RNA or sRNA or soluble RNA): Brings amino acids for protein synthesis and reads the genetic code.

Francis Crick postulated presence of an adapter molecule that can read the code and to link with amino acids.

tRNA is called adapter molecule because it has

· An Anticodon (NODOC) loop that has bases complementary to the codon.

· An amino acid acceptor end to which amino acid binds.

· Ribosome binding loop.

· Enzyme binding loop.

– For initiation, there is another tRNA called initiator tRNA.

– There are no tRNAs for stop codons.

– Secondary (2-D) structure of tRNA looks like a clover-leaf. 3-D structure looks like inverted ‘L’.

TRANSLATION (PROTEIN SYNTHESIS)

– It is the process of polymerisation of amino acids to form a polypeptide based on the sequence of codons in mRNA.

– It takes place in ribosomes. Ribosome consists of structural RNAs and about 80 types of proteins.

– Ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond ( peptidyl transferase enzyme in large subunit of ribosome).

– Translation includes 4 steps:

• Charging of tRNA
• Termination

1. Charging (aminoacylation) of tRNA

· Formation of peptide bond needs energy obtained from ATP.

· For this, amino acids are activated (amino acid + ATP) and linked to their cognate tRNA in presence of aminoacyl tRNA synthetase. Thus, the tRNA becomes charged.

2. Initiation

· In this, small subunit of ribosome binds to mRNA at the start codon (AUG).

· Now large subunit binds to small subunit to form initiation complex.

· Large subunit consists of aminoacyl tRNA binding site (A site) and peptidyl site (P site).

· The initiator tRNA (which carries methionine) binds on P site. Its anticodon (UAC) recognises start codon AUG.

3. Elongation

· Second aminoacyl tRNA binds to the A site of ribosome. Its anticodon binds to the second codon on the mRNA and a peptide bond is formed between first and second amino acids in presence of peptidyl transferase.

· First amino acid and its tRNA are broken. This tRNA is removed from P site and second tRNA from A site is pulled to P site along with mRNA. This is called translocation.

· These processes are repeated for other codons in mRNA.

· During translation, ribosome moves from codon to codon.

4. Termination

· When a release factor binds to stop codon, the translation terminates.

· The polypeptide and tRNA are released from the ribosomes.

· The ribosome dissociates into large and small subunits.

A group of ribosomes associated with a single mRNA for translation is called a polyribosome (polysomes).

An mRNA has additional sequences that are not translated (untranslated regions or UTR). UTRs are present at both 5’-end (before start codon) and 3’-end (after stop codon). They are required for efficient translation process.

REGULATION OF GENE EXPRESSION

In eukaryotes, gene expression occurs by following levels:

1. Transcriptional level (formation of primary transcript).

2. Processing level (splicing, capping etc.).

3. Transport of mRNA from nucleus to the cytoplasm.

4. Translational level (formation of a polypeptide).

The metabolic, physiological and environmental conditions regulate gene expression. E.g.

ú In E. coli, the beta-galactosidase enzyme hydrolyses lactose into galactose & glucose. In the absence of lactose, the synthesis of beta-galactosidase stops.

ú The development and differentiation of embryo into adult are a result of the expression of several set of genes.

If a substrate is added to growth medium of bacteria, a set of genes is switched on to metabolize it. It is called induction.

When a metabolite (product) is added, the genes to produce it are turned off. This is called repression.

OPERON CONCEPT

§ “Each metabolic reaction is controlled by a set of genes”

§ All the genes regulating a metabolic reaction constitute an Operon. E.g. lac operon, trp operon, ara operon, his operon, val operon etc.

Lac Operon in E. coli

– The operon controlling lactose metabolism.

– It is proposed by Francois Jacob & Jacque Monod.

It consists of

a) A regulatory or inhibitor (i) gene: Codes for repressor protein.

b) 3 structural genes:

i. z gene: Codes for b galactosidase. It hydrolyses lactose to galactose and glucose.

ii. y gene: Codes for permease. It increases permeability of the cell to b -galactosides ( lactose).

iii. a gene: Codes for a transacetylase.

– Genes in the operon function together in the same or related metabolic pathway.

– If there is no lactose (inducer), lac operon remains switched off. The regulator gene synthesizes mRNA to produce repressor protein. This protein binds to the operator region and blocks RNA polymerase movement. So the structural genes are not expressed.

– If lactose or allolactose is provided in the growth medium , it is transported into E. coli cells by the action of permease. Lactose (inducer) binds with repressor protein. So repressor protein cannot bind to operator region. The operator region becomes free and induces the RNA polymerase to bind with promoter. Then transcription starts.

– Regulation of lac operon by repressor is called negative regulation.

HUMAN GENOME PROJECT (HGP)

·    The entire DNA in the haploid set of chromosomes of an organism is called a Genome .

·    In Human genome, DNA is packed in 23 chromosomes.

·    Human genome contains about 3×10 9 bp.

·    Human Genome Project (1990-2003) was the first mega project for the sequencing of nucleotides and mapping of all the genes in human genome.

·    HGP was coordinated by U.S. Department of Energy and the National Institute of Health.

Goals of HGP

a.     Identify all the estimated genes in human DNA.

b.     Sequencing of 3 billion chemical base pairs of human DNA.

c.     Store this information in databases.

d.     Improve tools for data analysis.

e.     Transfer related technologies to other sectors.

f.      Address the ethical, legal and social issues (ELSI) that may arise from the project.

Methodologies of HGP: 2 major approaches.

• Expressed Sequence Tags (ESTs): Focused on identifying all the genes that are expressed as RNA.
• Sequence annotation: Sequencing whole set of genome containing all the coding & non-coding sequence and later assigning different regions in the sequence with functions.

Procedure of sequencing:

Isolate DNA from a cell → Convert into random fragments → Clone in a host (bacteria & yeast) using vectors (e.g. BAC & YAC) for amplification → Sequencing of fragments using Automated DNA sequencers (Frederick Sanger method) → Arrange the sequences based on overlapping regions→ Alignment of sequences using computer programs.

BAC= Bacterial Artificial Chromosomes

YAC= Yeast Artificial Chromosomes

• Sanger   has also developed method for sequencing of amino acids in proteins.
• DNA is converted to fragments as there are technical limitations in sequencing very long pieces of DNA.
• HGP was closely associated with Bioinformatics.
• Bioinformatics: Application of computer science and information technology to the field of biology & medicine.
• Of the 24 chromosomes (22 autosomes and X & Y), the last sequenced one is chromosome 1 (May 2006).
• DNA sequencing also have been done in bacteria, yeast, Caenorhabditis elegans (a free living non-pathogenic nematode), Drosophila, plants (rice & Arabidopsis ), etc.

Salient features of Human Genome

a.      Human genome contains 3164.7 million nucleotide bases.

b.     Total number of genes= about 30,000.

c.     Average gene consists of 3000 bases, but sizes vary. Largest known human gene ( dystrophin on X-chromosome) contains 2.4 million bases.

d.     99.9% nucleotide bases are same in all people. Only 0.1% (3×10 6 bp) difference makes every individual unique.

e.      Functions of over 50% of discovered genes are unknown.

f.      Chromosome I has most genes (2968) and Y has the fewest (231).

g.     Less than 2% of the genome codes for proteins.

h.     Very large portion of human genome is made of Repeated (repetitive) sequences. These are stretches of DNA sequences that are repeated many times. They have no direct coding functions. They shed light on chromosome structure, dynamics and evolution.

i.     About 1.4 million locations have single-base DNA differences. They are called SNPs (Single nucleotide polymorphism or ‘snips’). This helps to find chromosomal locations for disease-associated sequences and tracing human history.

DNA FINGERPRINTING (DNA PROFILING)

• It is the technique to identify the similarities and differences of the DNA fragments of 2 individuals.
• It is developed by Alec Jeffreys (1985).

Basis of DNA fingerprinting

• DNA carries some non-coding repetitive sequences.
• Repetitive DNA can be separated from bulk genomic DNA as different peaks during density gradient centrifugation.
• The bulk DNA forms a major peak and the small peaks are called satellite DNA.
• Satellite DNA is classified as micro-satellites, mini-satellites etc. based on base composition (A:T rich or G:C rich), length of segment and number of repetitive units.
• A DNA sequence which is tandemly repeated in many copy numbers is called variable number tandem repeats (VNTR). It belongs to mini-satellite DNA.
• In a person, copy number varies in each chromosome.
• The two alleles (paternal and maternal) of a chromosome also contain different copy numbers of VNTR.
• VNTR is specific from person to person.
• The size of VNTR varies from 0.1 to 20 kb.
• Any difference in the nucleotide sequence (inheritable mutation) observed in a population is called DNA polymorphism (variation at genetic level).
• Polymorphism is higher in non-coding DNA sequence because mutations in these sequences may not affect an individual’s reproductive ability. These mutations accumulate generation to generation causing polymorphism.
• Polymorphisms have great role in evolution & speciation.

Steps of DNA fingerprinting (Southern Blotting Technique)

• Isolation of DNA (from any cells or blood stains, semen stains, saliva, hair roots, bone, skin etc.).
• Digestion of DNA by restriction endonucleases.
• Separation of DNA fragments by gel electrophoresis.
• Transferring (blotting) DNA fragments to synthetic membranes such as nitrocellulose or nylon.
• Hybridization using radioactive labelled VNTR probe.
• Detection of hybridized DNA by autoradiography.

The autoradiogram gives an image in the form of dark & light bands. It is called DNA fingerprint.

• DNA fingerprint differs in everyone except in monozygotic (identical) twins.
• The sensitivity of the technique can be increased by use of polymerase chain reaction (PCR). Therefore, DNA from a single cell is enough for DNA fingerprinting. 
• Forensic tool to solve paternity, rape, murder etc.
• For the diagnosis of genetic diseases.
• To determine phylogenetic status of animals.
• To determine population and genetic diversities.

CBSE Class 12 Biology Important Questions Chapter 6 – Molecular Basis of Inheritance

1 mark questions.

Chapter 6 Molecular Basis of Inheritance 

1 Marks Questions 1. Name the factors for RNA polymerase enzyme which recognises the start and termination signals on DNA for transcription process in Bacteria. Ans. Sigma (s) factor and Rho(p) factor)

2. Mention the function of non-histone protein. Ans. Packaging of chromatin

3. During translation what role is performed by tRNA Ans.   (i)  Structural role (ii)  Transfer of amino acid.

4. RNA viruses mutate and evolve faster than other viruses. Why? Ans.  -OH group is present on RNA, which is a reactive group so it is unstable and mutate faster.

6. Mention the dual functions of AUG. Ans. (i)  Acts as initiation codon for protein synthesis (ii)  It codes for methionine.

7. Write the segment of RNA transcribed from the given DNA – 3´ -A T G C A G T A C G T C G T A ‘5´- Template Strand 5´ – T A C G T C A T G C A G C A T ‘3´ – Coding Strand. Ans.  5’- U A C G U C A U G C A G C A U – 3’ (In RNA ‘T’ is replaced by‘U’)

8.Name the process in which unwanted mRNA regions are removed & wanted regions are joined. Ans. RNA splicing.

9.Give the initiation codon for protein synthesis. Name the amino acid it codes for? Ans. Initiation codon – AUG & it code for methionine.

10.In which direction, the new strand of DNA synthesised during DNA replication. Ans. 5’ → →  3

11.What is the function of amino acyl tRNAsynthetase. Ans. Amino acyl tRNAsynthetasecatalyses activation of amino and attachment of activated amino acids to the 3-end of specific tRNA molecule.

12.What is point mutation? Ans. Mutation due to change in a single base pair in a DNA sequence is called point mutation.

13.Name the enzyme that joins the short pieces in the lagging strand during synthesis of DNA? Ans. Ligase.

14.Name the enzyme which helps in formation of peptide bond? Ans. Peptidyltransferase

15.Who experimentally prove that DNA replication is semi conservative. Ans. Messelson&stahl.

16.What is a codon? Ans. Triplet sequence of bases which codes for a single amino is called a codon.

17.Name the three non-sense codons? Ans. UAA, UAG, UGA

18.What is the base pairing pattern of DNA? Ans. In DNA, adenine always binds with thymine & cytosine always binds with Guanine.

19.Mention the dual functions of AUG? Ans. AUG codes for amino acid methionine & also acts as an initiator codon.

2 Mark Questions

2. Complete the blanks a, b, c and d on the basis of Frederick Griffith Experiment. S Strain  → →  inject into mice  → →  (a) R strain  → →  inject into mice  → →  (b) S strain (heat killed)  → →  inject into mice  → →  (c) S strain (heat killed) + R strain (live)  → →  inject into mice  → →  (d) Ans.(a)  Mice die (b)  mice live (c)  mice live (d)  mice die

3. Give two reasons why both the strands of DNA are not copied during transcription. Ans. (a)  If both the strands of DNA are copied, two different RNAs(complementary to each other) and hence two different polypeptideswill produce; If a segment of DNA produces two polypeptides, thegenetic information machinery becomes complicated. (b)  The two complementary RNA molecules (produced simultaneously)would form a doublestranded RNA rather than getting translated intopolypeptides. (c)  RNA polymerase carries out polymerisation in 53’direction andhence the DNA strand with 35’ polarity acts as the template strand.(Any two)

4. Mention any two applications of DNA fingerprinting. Ans.(i)  To identify criminals in the forensic laboratory. (ii)  To determine the real or biological parents in case of disputes. (iii)  To identify racial groups to rewrite the biological evolution. (Any two)

5. State the 4 criteria which a molecule must fulfill to act as a genetic material. Ans.(i)  It should be able to generate its replica. (ii)  Should be chemically and structurally stable. (iii)  Should be able to express itself in the form of Mendelian characters. (iv)  Should provide the scope for slow changes (mutations) that are necessary for evolution.

6.“DNA polymerase plays a dual function during DNA replication” comment on statement? Ans. DNA polymerase plays a dual function –it helps in synthesis of new strand & also helps in proof reading i.e replacement of RNA strands lay DNA fragments.

7.Three codons on mRNA are not recognised by tRNA what are they? What is the general term used for them what is their significance in protein synthesis? Ans. UAG UAA & UGA are the three codons that are not recognised by tRNA these are known as stop codon or non-sense codon. Since these three codons are not recognised by any tRNA they help in termination of protein chain during translation.

8.Give two reasons why both the strands of DNA are not copied during DNA transcription? Ans. I)If both the strands code for RNA two different RNA molecules & two different proteins wouldbe formed hence genetic machinery would become complicated II) Since the two RNA molecules would be complementary to each other, they would wind togetherto form dsRNA without carrying out translation which means process of transcription would befutile

9.Why is it essential that tRNA binds to both amino acids & mRNA codon during protein synthesis? Ans. It is essential that tRNA binds to both amino acids & mRNA codon because tRNA acts as an adapter molecule with picks up a specific activated aminoacid from the cytoplasm & transferred it to the ribosomal in the cytoplasm where proteins are synthesized. It attracts itself to ribosome with the sequence specified by mRNA & finally it transmits its amino acid to new polypeptide chain.

11.Explain what happens in frameshift mutation? Name one disease caused by the disorder? Ans. Frameshift mutation is a type of mutation where addition or deletion of one or two bases changes the reading from the site of mutation, resulting in protein with different set of amino acid.

12.What do you mean by “Central Dogma of Molecular genetics?” Ans. The central dogma of molecular genetics is the flow of genetic information from DNA to DNA through replication, DNA to mRNA through transcription & mRNA to proteins through translation. ReplicationDNA  → →  mRNA  → →  proteins. transcription translation

13.Give two reasons why both the strands are not copied during transcription? Ans. i) If both the strands codes for RNA, two different RNA molecules & two different proteins areformed hence genetic machinery would be complicated. ii)Since two RNA molecules produced would be complementary to each other, they would wind together to form ds-RNA.

14.Why is human Genome project considered as mega project? Ans. Human Genome project was called mega project for the following facts.

• The human genome has approximately 3.3 x 109bp, if the cost of sequencing is US  3 p e r b p , t h e a p p r o x i m a t e c o s t i s a b o u t U S 3perbp,theapproximatecostisaboutUS  g billion.
• If the sequence obtained were to be stored in a typed form in books & if each page contained1000 letters & each book contained 1000 page than 3300 such books would be needed to store complete in formation
• The enormous quantity of data expected to be generated also necessitates the use of high speedcomputational devices for data storage, retrieval & analysis.

15.Why is DNA & not RNA is the genetic material in majority of organisms? Ans. The -OH group in the nucleotides of RNA is much more reactive & makes RNA labile & easily degradable thus, DNA and not RNA acts as genetic material in majority of organisms.

16.Mention any four important characteristics of genetic code. Ans. Genetic codon has following imp-features :-

• Each codon is a triplet consisting of three bases.
• Each codon codes for only one amino acid i.e. – unambiguous.
• Some amino acids are coded lay more than one codon  ∴ ∴  said to be degenerative.
• Codons are read in a continuous manner in direction & have no punctuation.

17.Why it is that transcription & translation could be coupled in prokaryotic cell but not in eukaryotic cell? Ans. In prokaryotes the mRNA synthesised does not require any processing to become active &both transcription & translation occurs in the same cytosol but In Eukaryotes, primary transcriptcontains both exon & intron & is subjected to a process called splicing where introns are removed &exons are joined in a definite order to form mRNA.

3 Mark Questions

3 Marks Questions 1. Give six points of difference between DNA and RNA in their structure/chemistry and function. Ans.

2. Explain how does the hnRNA becomes the mRNA. OR Explain the process of splicing, capping and tailing which occur during transcription in Eukaryotes. Ans. hnRNA is precursor of mRNA. It undergoes (i) Splicing : Introns are removed and exons are joined together. (ii) Capping : an unusual nucleotide (methyl guanosine triphosphate isadded to the 5´ end of hnRNA. (iii) Adenylate residues (200-300) are added at 3´ end of hnRNA.

3. Name the three major types of RNAs, specifying the function of each inthe synthesis of polypeptide. Ans. (i)  mRNA-(Messenger RNA) : decides the sequence of amino acids. (ii)  tRNA-(Transfer RNA) : (a) Recognises the codon on mRNA (b) transport the aminoacid to the site of protein synthesis. (iii)  rRNA (Ribosomal RNA) : Plays the structural and catalytic role during translation.

4. Enlist the goals of Human genome project. Ans.  The Human Genome Project (HGP) is an international scientific research project with the goal of determining the sequence of chemical base pairs which make up human DNA, and of identifying and mapping all of the genes of the human genome from both a physical and functional standpoint

5. A tRNA is charged with the amino acid methionine. (i) Give the anti-codon of this tRNA. (ii) Write the Codon for methionine. (iii) Name the enzyme responsible for binding of amino acid to tRNA. Ans.   (a)  UAC  (b)  AUG  (c)  Amino-acyltRNAsynthetase.

6. Illustrate schematically the process of initiation, elongation and termination during transcription of a gene in a bacterium. Ans. In bacteria, the mRNA provides the template, tRNA brings aminoacids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription (Initiation) It somehow also facilitates opening of the helix and continues elongation Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination

7.What is transformation? Describe Grifith’s experiment to show transformation? What did he prove from his experiment? Ans. Transformation means change in genetic makeup of an individual. Fredrick Grifith conducted aseries of experiments on streptococcus pneumoniae. He observed two strains of this bacterium –one forming smooth colonies with capsule (s-type) & other forming rough colonies without capsule (R-type) (i) when live s-type cells are infected into mice, they produced pneumonia & mice dies. (ii) When live R-type cells are infected into mice, disease was not produced did not appear. (iii) When heat – killed S-type cells were infected into mice, the disease did not appear. (iv) When heat killed S-type cells were mixed with live R-cells & infected into mice, the mice died. He concluded that R-strain bacteria had somehow been transformed by heat –killed S-strain bacteria which must be due to transfer of genetic material

8.The base sequence on one strand of DNA is ATGTCTATA (i) Give the base sequence of its complementary strand. (ii) If an RNA strand is transcribed from this strand what would be the base sequence of RNA? (iii) What holds these base pairs together? Ans. (i)  TACAGATAT. (ii)  UACAGAUAU (iii)  Hydrogen bonds hold these base pairs together. Adenine & thymine are bounded by twohydrogen bonds & cytosine & Guanine are bonded by three hydrogen bonds.

9.Two claimant fathers filed a case against a lady claiming to be the father of her only daughter. How could this case be settled identifying the real biological father? Ans. This case to identify the real biological father could lee settled lay DNA – finger printingtechnique. In this technique :-

• first of all, DNA of the two claimants who has to be tested is isolated.
• Isolated DNA is then digested with suitable restriction enzyme & digest is subjected to gelelectrophoresis.
• The fragments of ds DNA are denatured to produce ss DNA by alkali treatment.
• The electrophoresed DNA is then transferred from get into a nitrocellulose filter paper where itis fixed.
• A known sequence of DNA is prepared called probe – DNA & is labelled with radioactive esotope32p & then probe is added to nitrocellulose paper.
• The nitrocellulose paper is photographed on X –ray film through auto radiography. The film isanalysed to determine the presence of hybrid nucleic acid.

Then, the DNA fingerprints of the two claimants is compared with the DNA fingerprint of the lady &her daughter, whosoever matches with each other would be declared as biological father of herdaughter.

11.A tRNA is charged with amino acid methionine. i) At what site in the ribosome will the tRNA bind? ii) Give the anticodon of this tRNA? iii) What is the mRNA codon for methionine? iv) Name the enzyme responsible for this binding? Ans. (i)  P- site (ii)  UAC (iii)  AUG (iv)  Amino acyl tRNASynthetase

13.What are the three types of RNA & Mention their role in protein Synthesis? Ans. There are three types of RNA :

• Messenger RNA (mRNA) :- It is a single – stranded RNA which brings the genetic information ofDNA transcribed on it for protein synthesis.
• Transfer RNA (tRNA) :- It has a clover leaf like structure which acts as an adapter moleculewhich contains an “anticodon loop” on one end that reads the code on one hand &” an amino acid acceptor end which binds to the specific amino acid on other hand.
• Ribosomal RNA (rRNA) :- Ribosomes provides the site for synthesis of protein &catalyse theformation of peptide bond.

14. Define bacterial transformation? Who proved it experimentally & how? Ans.  The transformation is a mode of exchange or transfer of genetic information betweenorganism or from one organism to another. Fredrick Griffith tested the virulence of two strains of Diplococci to show transformation in thefollowing steps :-

• When S-III strains of bacteria are injected into mice. It developed pneumonia & died.
• When R-II strains are infected into mice, they did not develop pneumonia & survive.
• When heat – killed S-III strains of bacteria are injected into mice, No symptoms of pneumoniadevelops& mice remain healthy.
• When a mixture of heat – killed S-III strain & lives R-II strain is injected into mice, theydeveloped pneumonia & died.

From these results, Griffith concluded that the presence of heat – killed S-III bacteria must convertliving R-II type bacteria to type S-III so as to restore them the capacity for capsule formation. Thiswas called “BACTERIAL TRANSFORM ATION” S strain  → →  Inject into mice  → →  Mice die R strain  → → Injct into mice  → →  Mice live S strain (heat-killed) → → Inject into mice  → →  Mice live S strain (heat-killed) + R strain (live)  → →  Inject into mice  → → Mice die

5 Marks Questions

5 Marks Questions 1. What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally? Ans. Meselson and Stahl, performed an experiment using E.coli to prove that DNA replication is semi conservative.

• They grew E.coli in a medium containing  15 N H 4 C l 15NH4Cl .
• Then separated heavy DNA from normal (14N) by centrifugation in CsCl density gradient.
• The DNA extracted, after one generation of transfer from 15N medium to 14N medium, had an intermediate density.

-The DNA extracted after two generations consisted of equal amounts of light and hybrid DNA. -They proved that DNA replicates in a semiconservative manner.

2. What does the lac operon consist of? How is the operator switch turned on and off in the expression of genes in this operon? Explain. Ans. Lac Operon consists of the following :

• Structural genes : z, y, a which transcribe a polycistronic mRNA.
• gene ‘z’ codes for b-galactosidase
• gene‘y’ codes for permease.
• gene‘a’ codes for transacetylase.
• Promotor : The site where RNA polymerase binds for transcription.
• Operator : acts as a switch for the operon
• Repressor : It binds to the operator and prevents the RNAPolymerase from transcribing.
• Inducer : Lactose is the inducer that inactivates the repressor by binding to it.
• Allows an access for the RNA polymerase to the structural gene andtranscription.

Unit 6: The Molecular Basis Of Inheritance

About this unit, discovery of dna as the genetic material.

• DNA as the "transforming principle" (Opens a modal)
• Hershey and Chase: DNA is the genetic material (Opens a modal)
• Classic experiments: DNA as the genetic material (Opens a modal)
• Discovery of DNA 4 questions Practice

Structure of DNA and RNA

• Introduction to nucleic acids and nucleotides (Opens a modal)
• The discovery of the double helix structure of DNA (Opens a modal)
• Discovery of the structure of DNA (Opens a modal)
• DNA (Opens a modal)
• Molecular structure of DNA (Opens a modal)
• Antiparallel structure of DNA strands (Opens a modal)
• Molecular structure of RNA (Opens a modal)
• Nucleic acids (Opens a modal)
• DNA packaging 4 questions Practice
• DNA and RNA structure 4 questions Practice
• Nucleic acids 4 questions Practice

DNA Replication

• Semi conservative replication (Opens a modal)
• Mode of DNA replication: Meselson-Stahl experiment (Opens a modal)
• Leading and lagging strands in DNA replication (Opens a modal)
• Speed and precision of DNA replication (Opens a modal)
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Transcription and RNA processing

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• Central dogma of molecular biology (Opens a modal)
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• Overview of transcription (Opens a modal)
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Translation

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• Translation (mRNA to protein) (Opens a modal)
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• Impact of mutations on translation into amino acids (Opens a modal)
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Regulation of Gene Expression

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• The lac operon (Opens a modal)
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Human genome project and DNA fingerprinting

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• Molecular Basis of Inheritance Class 12 Notes CBSE Biology Chapter 6 (Free PDF Download)
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Revision Notes for CBSE Class 12 Biology Chapter 6 (Molecular Basis of Inheritance) - Free PDF Download

DNA or Deoxyribonucleic acid is the molecule that contains the genetic code of living organisms. DNA is present in each cell of the organism and gives information to living cells about what type of protein to generate. In chapter 6 of class 12, students are going to learn about DNA and molecular basis of inheritance, thoroughly. Students will be able to revise the chapter precisely.

Beside studying textbooks, students can always refer to class 12 biology chapter 6 revision notes PDF offered by Vedantu.

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Molecular Basis of Inheritance Class 12 Notes Biology - Basic Subjective Questions

Section–a (1 mark questions) 1. name the process in which unwanted mrna regions are removed & wanted regions are joined. ans. rna splicing is the process in which unwanted mrna regions (introns) are removed & wanted regions (exons) are joined. 2. mention the dual functions of aug ans. aug has dual functions. it codes for methionine (met), and it also acts as initiator codon. 3. in which direction, the new strand of dna synthesized during dna replication. ans. the dna-dependent dna polymerases catalyse polymerisation only in one direction, that is 5'→3'. 4. list three components of the transcription unit. ans. a transcription unit in dna is defined primarily by the three regions in the dna:  (i) a promoter  (ii) the structural gene  (iii) a terminator 5. what is a replication fork ans. for long dna molecules, since the two strands of dna cannot be separated in its entire length, the replication occurs within a small opening of the dna helix, referred to as replication fork., section–b (2 marks questions).

6. Give two reasons why both the strands of DNA are not copied during transcription.

Ans. Two reasons why both the strands of DNA are not copied during transcription are:

If both the strands code for RNA two different RNA molecules & two different proteins would be formed hence genetic machinery would become complicated. 

Since the two RNA molecules would be complementary to each other, they would wind together to form dsRNA without carrying out translation which means the process of transcription would be futile.

7. What do you mean by “Central Dogma of Molecular Biology?”

Ans. The central dogma of molecular genetics is the flow of genetic information from DNA to DNA through replication, DNA to mRNA through transcription & mRNA to proteins through translation. It was proposed by Francis Crick.

8. “The genetic material in the majority of organisms is DNA and not RNA.” Comment.

Ans. The following characteristics of DNA contributes to that fact that it is the genetic material in the majority of organisms:

2'-OH group present at every nucleotide in RNA is a reactive group and makes RNA labile and easily degradable.

DNA chemically is less reactive and structurally more stable when compared to RNA. The presence of thymine at the place of uracil also confers additional stability to DNA.

9. Which three codons on mRNA are not recognized by tRNA? What is the general term used for them and what is their significance in protein synthesis?

Ans. UAA, UAG & UGA are the three codons that are not recognized by tRNA. These are known as stop codons or nonsense codons. Since these three codonsare not recognized by any tRNA they help in the termination of the protein chain during translation.

 10. State the criterias which a molecule must fulfill to act as genetic material.

Ans. A molecule that can act as a genetic material must fulfill the following criteria: 

(i) It should be able to generate its replica (Replication). 

(ii) It should be stable chemically and structurally. 

(iii) It should provide the scope for slow changes (mutation) that are required for evolution. 

(iv) It should be able to express itself in the form of 'Mendelian Characters’.

11. Mention any three applications of DNA fingerprinting.

Ans. Three applications of DNA fingerprinting are:

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

(ii) It is used to determine the real or biological parents in case of disputes. 

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

Access class 12 Biology Chapter 6- Molecular basis of inheritance Notes in 30 Minutes 

The dna .

DNA (deoxyribonucleic acid) is a double helix structure that was cracked by Watson and Crick based on the results of X-ray crystallography. Each strand of a DNA helix consists of repeating nucleotide units. The nucleotide consists of 3 components: ribose or deoxyribose sugar, nitrogen-based. which is either Purines or pyrimidines and phosphate.

Structure of Nucleotide 

There are two types of purines which are known as adenine, and guanine. There are three types of pyrimidines: thymine, cytosine, and uracil. 

All nucleotides are common in DNA and RNA, but uracil is found in RNA, and thymine is only found in DNA. 

DNA is negatively charged due to the presence of negatively charged phosphate groups. A nitrogenous base binds to the pentose sugar via the glycosidic bond. 

Two nucleotides join by a 3'5 'phosphodiester bond to form a dinucleotide. A polymer so formed has a free phosphate group at the 5 'end of the ribose sugar known as the 5' end of the polynucleotide chain. The other end of the polymer has a free 3'OH group of ribose sugar. 

This is known as the 3 'end of the polynucleotide chain. The bond between sugars and phosphates forms the backbone of a polynucleotide chain. The nitrogenous bases are bound to the sugar content and protrude from the backbone.

The salient feature of the double helix structure of DNA is as follows- 

Two polynucleotides chains that are present, wrap around each other. Here, the backbone is constituted by sugar-phosphate, and bases project inside. 

The two DNA chains are always antiparallel to each other. Antiparallel means that if one chain has the polarity 5'-3', the other has 3'-5'. 

The bases which are present in the two strands are paired through hydrogen bonding forming base pairs. Adenine form two hydrogen bonds with thymine whereas cytosine forms three hydrogen bonds with guanine. 

 The two strands are coiled in the right-handed pattern.  

The plane of one of the base pair stacks over the other in a double helix. This, in addition to H-bonds, confers the stability of the helical structure. 

Packaging of DNA helix 

Positively charged basic proteins that surround the DNA are known as histones. Histones are found to be rich in basic amino acids such as lysine and arginine. Histones molecules are organized in a specific manner to form a unit of eight molecules called histone octamer. The DNA is negatively charged in nature and is packaged by wrapping around the positively charged histone octamer. This forms a structure called a nucleosome. A nucleosome was found to be containing 200 base pairs of DNA helix. Nucleosomes form a specific repeating unit of a structure which is called chromatin in the nucleus. Chromatin is known to be a thread-like stained body seen in the nucleus. The nucleosomes in chromatin appear as a ‘beads-on-string’ structure when viewed under an electron microscope (EM). The beads on string structure in chromatin are packaged to form chromatin fibers that are further coiled and condensed at the metaphase stage of cell division to form chromosomes. At the high levels, chromatin packaging requires additional proteins. These proteins are the Non-Histone Chromosomal (NHC) proteins. In a typical nucleus, a loosely packed region of chromatin stains light and is referred to as euchromatin. The dense chromatin stains dark is called Heterochromatin. The euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive. 

Packaging of DNA Helix 

DNA as a Genetic Material 

Griffith performed an experiment which is commonly known as the transforming experiment. He used the two different strains of Pneumococcus. These two different strains were used to infect the mice. The two strains which were used are type III-S (smooth), that contains outer capsule made up of polysaccharide and type II-R (rough) strain do not contain capsule. The capsule protects the bacteria from the host immune system. 

Griffith Experiment

The experiment of Griffith is explained below- 

The rough strain of Pneumococcus is injected into the mouse. The mouse is alive. 

The smooth strain of Pneumococcus is injected into the mouse. The mouse dies

When the heat-killed smooth strain of Pneumococcus is injected into the mouse, the mouse is alive.

In the last set of experiments, rough strain, and heat-killed smooth strain are injected into the mouse. The mouse dies. 

This proves that there is some transforming substance present in the heat-killed S strain that is converting or transforming the rough strain into a virulent strain that is responsible for the death of the mouse. The substance that was transformed. later found to be DNA.  

The Genetic Material is DNA 

Alfred Hershey and Martha Chase in 1952 performed an experiment to prove that DNA is the genetic material. They worked on bacteriophages, which are viruses that infect the bacteria. It was found that when the bacteriophage attaches to the bacteria, its genetic material enters into the bacterial cell. The viral genetic material uses the bacterial cell to synthesize more viral particles. Hershey and Chase grew some viruses on a medium that contained radioactive phosphorus and another medium that contained radioactive sulfur. When phosphorus which was radioactive was present in the medium the viruses contained radioactive DNA but not radioactive protein. This is due to the fact that DNA contains phosphorus, but protein does not. Similarly,  when the growth medium contained radioactive sulfur the viruses contained radioactive protein but not radioactive DNA.  This is because DNA does not contain sulfur.  

Hershey and Chase Experiment

These bacteria were only found to be radioactive when infected with viruses that contained radioactive DNA. This indicates that DNA was the material that was transferred from the virus to the bacteria. Bacteria which were infected with viruses containing radioactive proteins were not radioactive. This indicates that the proteins did not get into the bacteria from the virus. So DNA is the genetic material that is transferred from viruses to bacteria. This experiment shows that DNA is the genetic material.

The Central Dogma of Molecular Biology 

It is an explanation of how genetic information transfer in a biological system. It tells us how the DNA replicates and then transcribes into messenger RNA (mRNA). This mRNA will now be translated to make proteins. 

DNA Replication 

DNA replication is the process by which two identical copies of DNA are made from a single DNA molecule. As we know, DNA is a double helix in which two strands are complementary to each other. These two strands of a helix separate at replication time to form two new DNA molecules. Of the two DNA strands formed, one is identical to one of the strands and the other is complementary to the original strand. This form of replication is defined as semi-conservative replication. goes into mitosis, DNA replicates in the S phase of the interface DNA polymerase is the most important enzyme involved in DNA replication is an energy-dependent process During the replication process, the two DNA chains do not separate completely, replication occurs within the small opening in the DNA helix known as the replication fork.DNA polymerase catalyzes the reaction in 5 'to 3' so that in one strand (the template with 3'5 'polarity) the replication is continuous, while in the other (the template with 5'3' polarity) the enzyme DNA ligase then binds to the discontinuously synthesized fragments. The continuously synthesized strand is referred to as the main strand, while the discontinuously synthesized strand is referred to as the lag strand. Replication begins at a specific location in DNA known as the origin of replication.

DNA Replication

Transcription 

It is a process of making RNA, like messenger RNA, from DNA before gene expression or protein synthesis takes place. During transcription, one of the DNA strands acts as a template for mRNA formation. The synthesis of mRNA is carried out by the enzyme RNA polymerase. It generally occurs for a specific stretch of DNA that is most needed for gene expression. In addition to messenger RNA, other forms of RNA such as ribosomal RNA, microRNA, small nuclear RNA can be transcribed in a similar way. It consists of the following three regions: a promoter, a structural gene, and a terminator DNA-dependent RNA polymerase catalyzes polymerase in the 5'3 'direction. The promoter is the region to which the RNA polymerase binds. The terminator defines the end of the transcription.

Process of Transcription

Transcription consists of a total of three steps- initiation, elongation, and termination.  

Initiation is the step that involves the binding of RNA polymerase to the promoter. A single type of DNA-dependent RNA polymerase catalyzes the transcription of all types of RNA in bacteria. 

Elongation is the process of adding nucleotides to form RNA. 

The termination factor helps in the termination of the transcription. The RNA synthesized after transcription is called the primary transcript. The primary transcript undergoes modifications such as splicing, coating, tailing, etc.

The primary transcript consists of the introns and exons. Introns are known as splicing. The addition of the polyA tail to the 3 'end of the RNA is referred to as the tail. When capping, an unusual nucleotide (methylguanosine triphosphate) is attached to the 5 'end of the RNA. 

Some viruses have the property of reverse transcription. You can convert RNA templates into DNA. The enzyme used is known as reverse transcriptase. 

For example, the human immunodeficiency virus that causes AIDS."

Translation 

This is the process of gene expression or protein synthesis, that occurs in the cytosol. Ribosomes are known to be the cellular organelles that participate in protein synthesis. Messenger RNA, made by the transcription process, is deciphered by ribosomes to form a composite polypeptide. Of amino acids. Messenger RNA is made up of a polymer of nucleotides, or codons. Each codon consists of 3 nucleotides that code for a single amino acid. There are several important components involved in the synthesis of ribosome proteins, messenger RNA, and transfer RNA (tRNA). Transfer RNA is involved in the physical binding of mRNA and the amino acid sequence of proteins.

Translation

It involves 4 main steps- 

Activation of amino acids- amino acids bind to specific tRNA molecules. 

Initiation of the polypeptide synthesis- In the process of capping an unusual nucleotide (methyl guanosine triphosphate) is added to  the 5'-end of RNA 

Elongation of polypeptide synthesis- It involves the addition of amino acids to the growing polypeptide chains 

Termination of polypeptide synthesis- It involves the end of the translation of protein synthesis. 

Genetic code 

The set of different rules according to which the information encoded in genetic material is translated into proteins in living cells. The outstanding features of the genetic code are: 

The codon consists of 3 nucleotides. 61 codons code for 20 different types of amino acids. 

1 codon codes for a single amino acid. 

1 amino acid can be encoded by more than one codon.

Genetic Code

Regulation of Gene Expression 

All the genes in the living cells are not active all the time. They become active when needed. Expression is controlled by genes are known as regulatory genes. Regulation in the eukaryotes can occur at the following different levels- 

Transcriptional level. 

Processing level. 

Transportation of mRNA from the nucleus to the cytoplasm. 

Translational level. 

Lac Operon or Lactose Operon 

An operon consists of structural genes, operator genes, promoter genes, promoter genes, regulator genes, and repressors.  Lac operon consist of lac Z, lac Y and lac A genes. Lac Z codes for galactosidase, lac Y codes for permease, and lac A codes for transacetylase. When repressor molecules are bound to the operator, genes are not transcribed. When the repressor does not bind the operator and instead inducer binds, transcription is switched on. In the case of the lac operon, lactose is an inducer. So, binding lactose to the operator, switch on the transcription. 

Lactose Operon

Human Genome Project 

The most important and basic features of the Human Genome Project are: 

The human genome contains 3,164.7 million nucleotide bases. 

The average gene is 3000 bases, but the size varies. 

Humans are said to have around 30,000 genes. 

The functions of more than 50 percent of the discovered genes are unknown. 

Less than 2 percent of the genome code for proteins. 

The human genome consists to a large extent of repetitive sequences. 

Chromosome 1 has the most genes (2968) and Y has the fewest (231).

PDF Summary - Class 12 Biology Molecular Basis of Inheritance Notes (Chapter 6)

Notes of chapter 6 biology class 12 .

Students can download and read revision notes Class 12 biology Chapter 6 for completely free. Following the latest CBSE rules and guidelines, experienced tutors have written important class 12 biology chapter 6 revision notes to make your learning easier for you. These handy revision notes will help you to achieve the desired scores. It will also improve your learning experience to an extent.

Class 12 Biology Molecular Basis of Inheritance Revision Notes

In the following sections, students will get a brief of all the topics provided in the revision notes Class 12 biology Chapter 6 PDF. Molecular basis of inheritance involves the study of genes. Further, the study includes DNA, DNA as genetic material, DNA replication, transcription, translation, genetic code, regulation of gene expression and many more. 

As you have read it, DNA has a double-helical structure, discovered by James Watson & Francis Crick in the 1950s with their theory, model and experiment. Each strand of DNA helix is composed of repeating units of nucleotides.

Nucleotides are Consisting of 3 Major Components

Ribose or deoxyribose sugar

Nitrogenous base (purines or pyrimidines)

After mentioning all components of Nucleotides in Chapter 6, Purines and Pyrimidines has been discussed. 

There are 2 categories of Purines, 

There are three types of pyrimidines.

Discussing Nucleotides and the components in the very first section, Polynucleotide chain has been explained, which also gives an idea about the bonding between sugar & phosphate. At the end of the first section, students will know about the features of a double helix structure.

Structure of Polynucleotide Chain

A nucleotide has three components, which are:

 a nitrogenous base

a pentose sugar

a phosphate group

Nitrogenous bases are of two types — Pyrimidines and Purines. Cytosine, Uracil, and Thymine are Purines, in which Cytosine is present in DNA and RNA both while Thymine is only present in DNA. Similarly, pentose sugar is classified into two types — ribose and deoxyribose, of which ribose is present in RNA and deoxyribose is present in DNA. The phosphate group is formed by the nucleotide and nucleoside. 

A polynucleotide is formed by the combination of more than two nucleotide groups. Nucleotide groups connected with each other through 3’-5’phosphodiester linkage. 

Class 12 Biology Chapter 6 Revision Notes: Packaging of DNA Helix  

Positively charged basic protein which surrounds DNA is called Histones. The second section talks about Histones. This section also describes the amino acids which form Histones. These are called lysine and arginine. Along with this, you will also know about histone octamer formation. You will learn about Nucleosomes with functions. Packaging of DNA helix section ends with a description of (NHC) proteins, Non – Histone Chromosomal proteins with a discussion of Euchromatin and Heterochromatin.

Class 12 Biology Chapter 6 Revision Notes: DNA as Genetic Material

Frederick Griffith performed a scientific experiment known as the Transforming experiment. You will learn about this experiment in detail. The experiment successfully proved that bacteria are capable of transferring genetic information through this process known as transformation.

Class 12 Notes, Molecular Basis of Inheritance: Central Dogma of Molecular Biology

In this section of Class 12 Biology chapter 6 revision notes, students will get an explanation of how the flow of genetic information occurs in a biological system. This explains how DNA replicates and gets converted into messenger RNA (mRNA) through the process of transcription. This section also mentioned a process of translation where mRNA forms protein. 

Class 12 Notes, Molecular Basis of Inheritance: DNA Replication

The process of producing two identical copies of DNA from a single DNA molecule is called DNA replication. This section describes this whole process in details, as it is a process of biological inheritance. Students will be familiar with many different terminologies which are associated with the process of DNA replication. You will know about semi-conservative replication, replication fork. This section further explains about leading strand and lagging strand, mentioning the origin of replication at the end of this section.

Class 12 Notes, Molecular Basis of Inheritance: Transcription

After explaining the process of DNA replication, Chapter 6 explains the process, transcription, where the formation of RNA takes place before the occurrence of gene expression and protein synthesis. You will get an idea about what happens during the transcription process as you read this section. A promoter, structural gene and a terminator are all three different regions in a transcription unit, which are mentioned in this section. The process of transcription consists of three steps – initiation, elongation and termination. This section describes each of these steps precisely to understand the process and the occurrence with diagrams and examples. 

Class 12 Notes, Molecular Basis of Inheritance: Translation

Here in the section of Class 12 Biology chapter 6 revision notes, in translation, you will be getting a brief explanation regarding the mentioned topic. The process of gene expression that occurs in the cytosol is called translation. Students will get to understand about codon and its structure. This section also mentions all three essential components of protein synthesis, which are ribosomes, messenger RNA (mRNA) and transfer RNA (tRNA). The section explains the activity of (tRNA) that is, linking the process of mRNA and amino acid sequence of proteins. You will get to learn about all four significant steps which are involved.

Biology Class 12 Chapter 6 Revision Notes: Genetic Code

This section discusses the genetic code and describes all salient features. In general, genetic code is a definitive set of rules used by living cells, to translate information encoded within the genetic material into proteins.

Biology Class 12 Chapter 6 Revision Notes: Regulation of Gene Expression

Regulatory genes control gene expression. This section describes different levels at which regulation occurs.

Class 12 Revision Notes Chapter 6: Lac Operon or Lactose Operon

Lactose operon is required for the transport and metabolism of lactose in enteric bacteria. This section teaches about Lac operon and the structural components.

Class 12 Revision Notes Chapter 6: Human Genome Project

Chapter-6 ends by explaining the Human genome project and the salient features in their final section.

By downloading revision notes of class 12 biology chapter 6, you will understand and grasp the knowledge about Molecular basis of inheritance in a more simplistic manner. The study material is prepared in a way which is easy to comprehend with diagrams.

So, without wasting any time, download Class 12 Biology chapter 6 revision notes PDF file drafted by Vedantu.

Students who are appearing for the Board exams must try to write the answers to the following questions in their notebooks.

Question 1: Differentiate between mRNA and tRNA.

Question 2: Differentiate between the Template strand and the Coding strand.

Question 3: Write the function of Exons and Promoter.

Question 4: Human Genome project is called a mega project. Explain why?

Question 5: Explain DNA fingerprinting. What are its applications?

Question 6: If a double-stranded DNA has 20 per cent of cytosine, calculate the per

cent of adenine in the DNA.

FAQs on Molecular Basis of Inheritance Class 12 Notes CBSE Biology Chapter 6 (Free PDF Download)

1. What is the process of translation?

The process of protein synthesis in which mRNA is used to synthesize protein is known as translation. It is the process of gene expression or protein synthesis that occurs in the cytosol. The mRNA sequence is decoded to specify the amino acid of a polypeptide. Transcription consists of 3 steps.

Termination 

2. What does the process of DNA replication involve?

DNA replication is the course of creating dual copies of DNA from a single DNA molecule. The original strand is known as parent strands, and the new strand is known as the daughter strands. It is a process of biological inheritance. The initial stage in DNA replication is to unzip the double helix assembly of the molecule.

3. What are the differences between the leading and lagging strand?

Strands which are synthesized continuously are known as leading strands, whereas the strands which are synthesized discontinuously are known as lagging strands.

Leading strand does not require DNA ligase for its growth.

In the case of lagging strand DNA ligase is vital for constructing the Okazaki fragment.

The direction of growth of a leading strand is 5’ – 3’.

The direction of growth of a lagging strand is 3’ – 5’.

Formation of a leading strand starts instantly at the start of replication.

Development of a lagging strand starts after that of the leading strand.

4. What is DNA?

DNA which stands for Deoxyribonucleic acid, is the genetic material of humans and other animals. The DNA was first found by Friedrich Meischer in 1869. James Watson and Francis Crick made the discovery of the double helix structure of DNA in 1953. DNA is a continuous polymer made from simple units called nucleotides that are kept together firm by sugar and phosphate groups. This backbone contains four types of molecules known as bases, which are Adenine, Thymine, Guanine and Cytosine.

5. How is DNA packaged in Eukaryotes?

Eukaryotes contain a well-defined nucleus made up of DNA, a negatively charged polymer. This negatively charged polymer is enfolded with a positively charged histone octamer to form a nucleosome. Histone octamer is basically formed by organizing eight molecules of histones (which are positively charged basic proteins). The histones attain positive charge as they are rich in amino acid residues such as lysine and arginine. The nucleosomes formed are then coiled even more, resulting in the development of chromatin fibers.

6. Where can I find the NCERT solution for Class 12 Biology Chapter 6?

You can obtain NCERT solutions for all the questions provided at the end of the Class 12 Biology Chapter 6 on Vedantu. You can avail the entire solutions in PDF format for free of cost. These solutions are written by the professional subject matter experts at Vedantu keeping the CBSE curriculum in mind. The solutions provided to you explains the concept in a brief and simple manner. The language used is quite simple to make it understandable for the students. 

7. Why is Class 12 biology Chapter 6 important for students?

Class 12 biology chapter 12 ‘molecular basis of inheritance’ is an essential chapter for the students. This chapter holds the majority of the marks weightage in the board exams. Therefore, students must make sure to analyze the entire chapter thoroughly. This chapter tells the basic concepts of human genetics including DNA, RNA, process of replication, transcription, translation etc. Learn all the important definitions and practice all the important diagrams carefully. It is important to practice all the questions provided in the NCERT textbook.

8. Why should I prefer Revision notes of Vedantu for Class 12 biology Chapter 6 ?

Vedantu, today has become one of the renowned online learning platforms that offers the students with exclusive study materials and study guide for their learning process. Vedantu also provides well-written revision notes that encloses all the significant topics of any chapter, including for Class 12 Biology Chapter 6 Revision Notes. These revision notes are curated by experienced tutors and are available in PDF format on Vedantu. You can refer to these notes to enhance your concepts and revise the whole chapter in brief at free of cost on the Vedantu app as well.

Study Materials for Class 12

Molecular Basis of Inheritance MCQ | Class 12 | Biology | Chapter 6

Molecular variation of inheritance mcq chapter 6.

Below are some of the very important NCERT Molecular Basis Of Inheritance MCQ Class 12 Biology Chapter 6 with Answers. These Molecular Basis of Inheritance MCQ have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of term 1 and term 2. We have given these Molecular Variation of Inheritance MCQ Class 12 Biology Questions with Answers to help students understand the concept.

MCQ Questions for Class 12 Biology Chapter 6 are very important for the latest CBSE term 1 and term 2 pattern. These MCQs are very important for students who want to score high in CBSE Board.

We have put together these NCERT Questions Molecular Basis of Inheritance MCQ for Class 12 Biology Chapter 6 with Answers for the practice on a regular basis to score high in exams. Refer to these MCQs Questions with answers here along with a detailed explanation.

1. DNA is a

(a) long polymer of deoxyribonucleotides (b) short polymer of deoxyribonucleotides (c) monomer polymer of deoxyribonucleotides (d) long polymer of ribonucleotides

2. If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6×10 9 bp then the length of the DNA is approximately

(a) 2.5 m (b) 2.2 m (c) 2.7 m (d) 2.0 m

3. Which of the following are nucleotides?

(a) adenosine, cytidilic acid, cytosine (b) adenylic acid, cytidilic acid, gunaylic acid (c) cytidine, adenine, adenylic acid (d) uracil, thymidine, thymidylic acid

4. A nucleoside differs from in nucleotide. It lacks the

(a) base (b) sugar (c) phosphate group (d) hydroxyl group

5. In a DNA strand the nucleotides are linked together by

(a) glycosidic bonds (b) phosphodiester bonds (c) peptide bonds (d) hydrogen bonds

6. In DNA 20% bases are adenine what percent of bases are pyrimidine?

(a) 30% (b) 60% (c) 50% (d) 20%

7. Nucleosome is the repeating unit of _______ in a nucleus.

(a) chromosome (b) genes (c) chromatin (d) chromosome

8. Nucleosome consists of

(a) nucleolus (b) genes (c) microfilaments (d) histones

9. The packaging of chromatin at higher level requires additional set of proteins that are collectively referred to as

(a) histone protein (b) non-histone protein (c) basic protein (d) histone octamer

10. Lightly stained part of chromatin which remains loosely packed is

(a) euchromatin (b) heterochromatin (c) chromatosome (d) chromonemata

11. Densely packed and stain transcription inactive part of chromatin is

(a) euchromatin (b) chromatosome (c) heterochromatin (d) chromosome

12. Who introduced the transforming principle?

(a) Federal Griffith (b) Oswald Avery (c) Collin McLeod (d) Maclyn McCarty

13. S-type strain of Streptococcus pneumoniae is

(a) capsulated, virulent, smooth (b) non-capsulated, avirulent, rough c (c) capsulated, avirulent, rough (d) non-capsulated, virulent, smooth 

14. What happened when heat killed S cells along with live R cells were injected into mice?

(a) mice survived and showed live S cells (b) mice died and showed life S cells (c) mice survived and showed live R cells (d) mice died and showed live R cells

15. Transformation experiment of Griffith was approved by

(a) Griffith himself (b) Avery, MacLead, McCarty (c) Meselson (d) Breadle and Tatum

16. The result of which of the following reaction experiments carried out by Avery et at., on Streptococcus pneumoniae has proved conclusively that DNA is the genetic material?

(a) Live R-strain + DNA from S-strain + RNase (b) Live R-strain + DNA from S-strain + DNase (c) Live R-strain + Denatured DNA of S-strain + Protease (d) Heat killed R-strain + DNA from S-strain + DNase

17. Hershey and Chase used 35 S and 32 P to prove that DNA is a genetic material. Their experiments prove that DNA as genetic material because

(a) progeny viruses retained 32 P, but not 35 S (b) retention of 32 P in progeny viruses indicated that DNA was passed on (c) loss of 35 S in progeny viruses indicated that proteins were not passed on (d) All of the above

18. Match the following.

(a) (A) – (ii), (B) – (iv) , (C) – (iii), (D) – (i) (b) (A) – (i), (B) – (iv) , (C) – (iii), (D) – (ii) (c) (A) – (i), (B) – (ii) , (C) – (iii), (D) – (iv) (d) (A) – (i), (B) – (iii) , (C) – (iv), (D) – (ii)

19. Which group present in RNA nucleotides is very reactive and makes RNA liable and easily degradable than DNA?

(a) 3-OH’ group at every nucleotide (b) 2-OH’ group on ribose sugar (c) 3-OH’ group on ribose sugar (d) 4-OH’ group on ribose sugar

20. A molecule to act as a genetic material has the following properties

(I) should be able to replicate (II) should be structurally more stable (III) should be more reactive liable (IV) should provide scope for slow changes

Choose the correct option.

(a) I, II and III are correct (b) III alone is correct (c) III and IV are correct (d) I, II and IV are correct

21. The first genetic material could be

(a) protein (b) carbohydrate (c) DNA (d) RNA

Click Below To Learn Biology Term-1 Chapter Wise MCQs

• Chapter-2: Sexual Reproduction in Flowering Plants
• Chapter-3: Human Reproduction
• Chapter-4: Reproductive Health
• Chapter-5: Principles of Inheritance and Variation
• Chapter-6: Molecular Basis of Inheritance

22. Which one of the following option is correct?

(I) DNA has evolved from RNA with chemical modification (II) DNA being complementary double-stranded resists strangers by a process of repair (III) RNA being a catalyst is reactive and unstable

(a) I & II (b) II & III (c) I & III (d) I, II & III

23. Who experimentally proved the semi conservative mode of DNA replication?

(a) Matthew Meselson (b) Franklin Stahl (c) both (a) and (b) (d) Watson and Crick

24. If Meselson and Stahl’s experiment is continued for four generation in bacteria, the ratio of

15 N/ 15 N : 15 N/ 14 N : 14 N/ 14 N

containing DNA in the fourth generation would be

(a) 1:1:0 (b) 1:4:0 (c) 0:1:3 (d) 0:1:7

25. Given diagram depicts the experiment of Meselson and Stahl. Identify the type of isotopic DNA formed after 40 minutes (A, B, C & D)

(a) A- 14 N-DNA, B- 15 N-DNA, C- 14 N-DNA, D- 15 N-DNA (b) A- 14 N-DNA, B- 15 N-DNA, C- 14 N-DNA, D- 14 N-DNA (c) A- 14 N-DNA, B- 14 N-DNA, C- 14 N-DNA, D- 15 N-DNA (d) A- 14 N-DNA, B- 15 N-DNA, C- 15 N-DNA, D- 15 N-DNA

26. During DNA replication, okazaki fragments are used to elongate

(a) the leading strand towards replication fork (b) the lagging strand towards replication fork (c) the leading strand away from replication fork (d) the lagging strand away from the replication fork

27. During DNA replication the term leading strand is applied to the one which replicates in direction continuously.

(a) true (b) false (c) cannot say (d) partially true or false

28. Discontinuous synthesis of DNA occurs in one strand, because

(a) DNA molecule been synthesized is very long (b) DNA dependent DNA polymerase cattle lysis catalyzes polymerization only in one direction (5′ → 3′) (c) it is more efficient process (d) DNA ligase has to have a rule

29. Deoxyribonucleoside triphosphate serves dual purposes of

(I) acting as a substrate (II) acting as an enzyme (III) providing energy for polymerization (IV) increasing the rate of reaction

(a) I & II (b) II & III (c) III & IV (d) I & III

30. The process of copying genetic information from one strand of the DNA into are in is termed as

(a) translation (b) transamination (c) replication (d) transcription

31. Both the strands of DNA are not copied during transcription because RNA molecules with different sequences will be formed?

(a) true (b) false (c) cannot say (d) partly true or false

32. In the given figure find out A-E

(A) promoter site (B) structural gene (C) terminator site (D) template strand (E) coding strand

(a) (A) – 5, (B) – 1, (C) – 4, (D) – 2, (E) – 3 (b) (A) – 5, (B) – 1, (C) – 4, (D) – 3, (E) – 2 (c) (A) – 5, (B) – 4, (C) – 1, (D) – 2, (E) – 3 (d) (A) – 1, (B) – 4, (C) – 5, (D) – 2, (E) – 3

33. If the coding strand has the sequence 5′-ATCGATCH-3′ then find out the sequence of non coding strand.

(a) 3′-TAGCTAGC-5′ (b) 5′-TACGTACG-3′ (c) 5′-UAGCUAGC-3′ (d) 5′-UACFUACG-3′

34. The segment of DNA coding for a polypeptide is called

(a) muton (b) recon (c) cistron (d) exon

35. With regard to mature mRNA in eukaryotes

(a) exons and introns do not appear in the mature RNA (b) exons appear, but introns do not appear in the mature RNA (c) introns appear, but exons do not appear in the mature RNA (d) both exons and introns appear in the mature RNA

36. Match the following.

(a) (A) – (I), (B) – (III), (C) – (II) (b) (A) – (I), (B) – (II), (C) – (III) (c) (A) – (II), (B) – (III), (C) – (I) (d) (A) – (III), (B) – (II), (C) – (I)

37. Match the following.

(a) (A) – (IV), (B) – (I), (C) – (II), (D) – (III) (b) (A) – (II), (B) – (III), (C) – (IV), (D) – (I) (c) (A) – (I), (B) – (III), (C) – (IV), (D) – (II) (d) (A) – (III), (B) – (II), (C) – (I), (D) – (IV)

38. Genetic code

(a) is a relationship between sequence of DNA or mRNA to polypeptide (b) triplet based on mRNA (c) determines the sequence of amino acid in polypeptide (d) all of the above

39. From the following, identify the correct combination of salient features of genetic code.

(a) universal, non-ambigous, overlapping (b) degenerate, overlapping, commaless (c) universal, ambiguous, degenerate (d) degenerate, universal, non-ambigous

40. Because most of the amino acid are represented by more than one codon, the genetic code is

(a) overlapping (b) wobbling (c) degenerate (d) generate

41. Codons are non-ambiguous, which means that one codon codes for

(a) more than one amino acid (b) to amino acid (c) only one amino acid (d) nonsense amino acids

42. The terminator codons are

(a) UAA, UAG, UGA (b) AUG, UAG, UGA (c) UAC, AUG, UAG (d) DCC, UAA, CAC

43. Match the following.

(a) (A) – (IV), (B) – (V), (C) – (II), (D) – (III) (b) (A) – (V), (B) – (IV), (C) – (III), (D) – (II) (c) (A) – (I), (B) – (III), (C) – (IV), (D) – (V) (d) (A) – (II), (B) – (III), (C) – (IV), (D) – (V)

44. Point mutation may occur due to

(a) alteration in DNA sequence (b) change in a single base pair of DNA (c) deletion of segment of DNA (d) gain of segment in DNA

45. Which mutation of the genetic basis give the proof that codon is triplet and reads in a contagious manner?

(a) frameshift mutation (b) point mutation (c) both (a) and (b) (d) inversion mutation

46. Which are true about tRNA?

(I) It binds with an amino acid at its 3′ end (II) It has five double-stranded regions (III) It has a codon at one end which recognises the anticodon on messenger RNA (IV) it looks like clover leaf in the three-dimensional structures

(a) I only (b) II & III (c) III & IV (d) I & IV

47. The process of polymerisation of amino acids to form a polypeptide is

(a) transcription (b) replication (c) translation (d) polymerization

48. The first phase of translation is

(a) recognition of DNA molecule (b) amino acylation of tRNA (c) recognition of an anticodon (d) binding of mRNA to ribosome

49. Charging (aminoacylation) of tRNA involves the attachment of

(a) amino acid to mRNA (b) amino acid to tRNA (c) amino acid to rRNA (d) acidic amino acid to ribosome

50. Aminoacylation of tRNA helps in binding to ribosome?

(a) D-loop (b) T-loop (c) Variable loop (d) none of these

MCQ Answers

The distance between two consecutive base pair is 0.34 nm. The length of DNA double helix in a typical mammalian cell can be calculated by multiplying the total number of bp with distance between the two consecutive bp, i.e. 6.6 x 10 9 bp x 0.34 x 10 -9 m/bp = 2.2 m

Adenylic acid, cytidilic acid and guanylic acid are nucleotides. A nucleotide is composed of three components which are nitrogen base (adenosine, cytosine, guanine), ribose sugar and a phosphate group. These are monomer unit of nuclear nucleic acid RNA and DNA.

Adenine, cytidine, thymidine are nucleosides. Uracil, cytosine and adenosine are nitrogenous base.

Nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a nucleoside,i.e.

Nucleoside = Nitrogen base + Pentose Sugar

When a phosphate group is attached to the 5′-OH of a nucleoside through phosphodiester linkage, a nucleotide is formed, i.e. Nucleotides = Nitrogen base + Pentose sugar + phosphate (PO 4 )

Therefore, a nucleosides differs from a nucleotides as it lacks phosphate group.

The percentage of bases which are pyrimidines is 50 %. According to Chargaff’s rule

+G (Purines) = C+ (Pyrimidines) = 50%

i.e., If = 20%, then T = 20%

C + T = 50%

C = 50% – 20% = 30%

Total percentage of pyrimidine bases (T + C) are 20 + 30 = 50%

Chromatin are thread likes stained bodies seen in nucleus. The nucleosome is repeating unit in chromatin and seen as ‘beads on string’ when viewed under electron microscope.

Histones are main structural proteins found in Eukaryotic cells. The nucleosome core is made of four types of histone proteins, i.e. H 2 A, H 2 B, H 3 and H 4 occurring in pairs. 200 bp of DNA helix wraps around the nucleosome by turns, plugged by H1 histone proteins. So, nucleosome consists of histones.

The packaging of chromatin and higher level requires an additional set of basic proteins called non-histone protein. These are heterogeneous group of proteins that play important role in nucleosome remodeling, DNA processing etc.

The loosely packed form of DNA in the chromosome is called euchromatin.

At some places chromatin is densely packed to form darkly stained heterochromatin. It is transcriptionally inactive.

Transforming principle (Griffith’s experiment) was introduced by Frederick Griffith. He conducted a series of experiments with Streptococcus pneumoniae and found that living organism had changed in physical form.

As a part of his experiment, Griffith tried to inject the mice with heat killed S bacteria. Unsurprisingly, the heat kill as bacteria did not cause disease in mice.

However, when harmless R Bacteria were combined with harmless heat kill S bacteria and injected into another mice, it developed pneumonia and died. When Griffith took a blood sample from the dead mouse, he found that it contains a living S bacteria.

Oswald Avery, Colin MacLeod and Maclyn McCarty work to determine the biochemical nature of transforming principle in Griffith experiment.

R-strain is rough and harmless, while the S-strain is smooth and virulent form of Streptococcus pneumoniae . In their experiment, Avery et al ., found out that only when DNA from S-type bacteria was added to a culture of R-type bacteria are time got converted to S-type strain.

This transformation of our into S type did not occur in addition of carbohydrate and protein from S type bacteria. Also, when DNase enzyme was added, i.e., live R-strain + DNA (S-strain) + DNase, the transformation did not occur. It proved conclusively that DNA, indeed is the genetic material.

A molecule to act as a genetic material must fulfill the following criteria

• It should be able to replicate a form with carbon copies
• genetic material should be able to express itself to formation of specific biochemicals
• there are occasionally changes or mutation in the structure and functioning of its genes which are of permanent nature and inheritable. Mutations are essential for evolution and adaptability.
• it should be able table both chemically and physically

The first genetic material could be RNA we know that RNA is present as a genetic material in some viruses and it also works as a catalyst. But RNA being a catalyst is reactive and hence unstable. Hence, it is considered that DNA has evolved from RNA thereby making RNA the first genetic material.

Meselson and Stahl found that DNA of the first generation was hybrid or intermediate ( 15 N and 14 N). It settled in caesium chloride at a level higher than the fully labelled DNA of parents bacteria ( 15 N 15 N).

Second generation of bacteria after 40 minutes contain two types of DNA, 50% light (N 14 N 14 ) and 50% intermediate (N 15 N 14 ).

The third generation of bacteria after 60 minute contained two types of DNA, 25% intermediate (N 15 N 14 ) and 75% light (N 14 N 14 ) in ratio 1:3.

It can be assumed that 4th generation after 80 minutes could would contain 12.5% N 15 N 14 and 87.5% N 14 N 14 DNA in 1:7 ratio.

Thus, if Meselson and Stahl’s experiment is continued for 4th generation in bacteria, the ratio of 15 N/ 15 N : 15 N/ 14 N : 14 N/ 14 N containing DNA in the fourth generation would be 0:1:2.

Okazaki fragments are short segments of replicating DNA. These have 1000-2000 bp in prokaryotes and hundred 100-200 bp and eukaryotes. These fragments are formed discontinuously and are used to elongate the lagging strand away from the replication fork.

The phosphodiester nucleotide are deATP, deCTP, deTTP.

These triphosphates of base is serve dual purposes. They act as a substrate as well as provide energy for polymerization of nucleotides by releasing energy after dissociating the phosphate group.

The process in living cell in which genetic information of DNA is transferred to a molecule of messenger RNA is the first step in protein synthesis. This is known as transcription. It takes place in the cell nucleus of nuclear region and regulated by transcription factors .

(A) Promoter site – (V) (binding of RNA polymerase) (B) Structural gene – (IV) (formation of functional protein) (C) Terminal site – (I) Stopping of transcription) (D) Template strand – (II) (Part of DNA to which RNA transcribed) (E) Coding strand – (III) (complementary strand of DNA to RNA)

In mRNA of eukaryotes exons appear, but introns do not. This is because introns are intervening and non-coding sequences exams are coding and expressed sequences. Through slicing introns are removed and exons are joined to form mRNA.

All amino acids are specified by more than one codon (except tryptophan and methionine). Hence, they are degenerate. Since, there are 64 possible combination of the four different nucleotides in set of three, there are redundancy in the system which means that most amino acids can be coded by more than triplet.

UAA (ochre), UAG (amber) and UGA (opal) are the three codons, which bring about termination of polypeptide chain and thus, called terminator codons.

Point mutation or gene mutation involves only the replacement of one nucleotide with another or change in a single base pair of DNA.

Frameshift mutation of the genetic basis gives the proof that codon is triplet and reads in continuous manner. Deletion or addition of a base pay disturb the reading frame of DNA or mRNA.

Charging or Aminoacylation of tRNA is essential for protein synthesis, i.e. polypeptide formation through formation of peptide bonds between amino acids.

Assertion-Reason Based MCQ

• Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of the Assertion (A).
• Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• Assertion (A) is true, but Reason (R) is false.
• Assertion (A) is false, but Reason (R) is true.

1. Assertion DNA acts as a genetic material in all organisms

Reason It is a double-stranded by molecule in most organisms

2. Assertion S. pneumoniae produced two types of colonies, therefore, smooth and rough

Reason S-type bacteria from smooth colony due to the absence of polysaccharide coat

3. Assertion DNA has two chains having antiparallel polarity

Reason In one chain of DNA at one and has a free phosphate moiety 5′ end of ribose sugar and at other and the ribose has a free 3′ OH group

4. Assertion Adenine cannot pair with cytosine

Reason Adenine and cytosine do not have complementary between their respective hydrogen donor and hydrogen acceptor sites

5. Assertion Histones are basic in nature

Reason These are rich in the amino acids lysine and arginine

6. Assertion Heterochromatin is transcriptionally inactive

Reason It is densely packed

7. Assertion Viruses having RNA genome and shorter lifespan, mutate and evolve faster

Reason RNA is unstable and thus mutates faster

8. Assertion Replication on one strand of DNA is continuous and on another it is discontinuous

Reason The DNA polymerases works on direction 3’→5′ direction

9. Assertion Replication and transcription occurs in the nucleus, but translation takes place in the cytoplasm

Reason mRNA is transferred from the nucleus into cytoplasm where ribosomes and amino acids are available for protein synthesis

10. Assertion hnRNA is larger than mRNA

Reason hnRNA has non coding in terms which are not required for translation

Assertion-Reason Based MCQ Answers

DNA serves as the genetic material in must organisms, but in some viruses like TMV, RNA acts as the genetic material. DNA is a double-stranded biomolecule, but can also exist as a single-stranded biomolecule.

S-type bacteria form a smooth colony because they possess polysaccharide (mucus) coat around themselves, whereas R-type bacteria do not form any covering around themselves.

The two chains of DNA have antiparallel polarity. This is because one chain has a free phosphate moiety at the 5′ end of the ribose sugar and another chain has a free phosphate moiety at the 3′ end.

Adenine can’t pair with cytosine. It pairs up with thymine. It is because adenine pairs with thymine with two hydrogen bonds , having two hydrogen donor / hydrogen acceptor sites whereas cytosine has three hydrogen donor. Thus, due to lack of complementary between the hydrogen donor and hydrogen acceptor sites between adenine and cytosine, these can’t pair.

Histones are basic in nature because these are rich in amino acids lysine and arginine which are basic in nature.

Heterochromatin is densely packed and inaccessible to transcription factors. hence, it is rendered transcriptionally silent or inactive.

RNA is an unstable catalytic molecule. It mutates at a faster rate than DNA. Thus, ciruses having RNA genome and shorter lifespan, mutate and evolve faster due to this unstability.

In eukaryotes, replication ans transcription take place in the nucleus.

The fully processed hnRNA now called mRNA, is transferred from the nucleus into the cytoplasm where translation occurs.

The primary transcript in eukaryotes, i.e. the hnRNA is much longer as it contains both introns and exons. It is precursor of mRNA. During post-transcriptional modification. Introns, which don’t code for proteins are removed and all exons are joined to form fully prossessed mRNA.

Case-Study Based MCQ

1. Read the following passage and answer accordingly.

The Meselson and Stahl experiment was an experiment to prove that DNA replication was semiconservative and i twas first shown in Escherichia coli and subsequently in higher organisms, such as plants and human cells.

Semiconservative replication means that when the double stranded DNA helix was replicated, each of the two double-stranded helices consisted of one-stranded helices consisted of one strand coming from the parental helix and one is newly synthesized.

(i) The heavy isotope used by Meselson and Stahl for proving the semiconservative mode of DNA is/are

(a) 15 NH 2 Cl (b) 14 NH 2 Cl 2 (c) 13 NH 2 Cl 3 (d) All of the above

(ii) Heavy DNA can be differentiated from normal DNA by which centrifugation technique?

(a) AgCl density gradient (b) CsCl density gradient (c) CaSO4 density gradient (d) KCl density gradient

(iii) Similar experiments like Meselson and Stahl was performed by Taylor in 1958. The experimental organism of Taylor was

(a) Vicia faba (b) Fungi (c) E. coli (d) Protista

(iv) Radioisotope used by Taylor in his experiment was

(a) iron (b) titanium (c) thymidine (d) copper

2. Observe the given figure and answer accordingly.

(i) The number of nucleosomes present in human cells is

(a) 3.3 x 10 7 (b) 1.1 x 10 7 (c) 6.6 x 10 7 (d) Indefinite

(ii) Which amino acids are present in histones?

(a) Lysine and histadine (b) Valine and histadine (c) Arginine and lysine (d) Arginine and histidine

(iii) Linker DNA is

(a) a part of nucleosome (b) a part that joins two octamer cores (c) ssDNA (d) Both (a) and (b)

(iv) The association of histone H1 with a nucleosome indicates

(a) transcription is occuring (b) DNA replication is occuring (c) the DNA is condensed into a chromatin fibre (d) the DNA double helix is exposed

3. Observe the given figure and answer accordingly.

(i) In Hershey and Chase experiment, bacteriophage nucleic acids were labelled as

(a) 32 P labelled phosphate (b) 3 H labelled H 2 O (c) 35 S labelled sulphate (d) 14 C labelled CO 2

(ii) Bacteriophage protein coat was labelled by _______ in Hershey and Chase experiment.

(a) 35S labelled sulphur (b) 32S labelled sulphate (c) 30S labelled sulphur (d) 32P labelled sulphate

(iii) In Hershey and Chase experiment, radioactive 32P was used to culture bacteriophage which resulted in radioactive

(a) viral DNA (b) bacterial capsule (c) viral protein (d) plasma membrane of bacteria

(iv) DNA with labelled thymidine is added to a medium where E. coli is growing. After 5 minutes of growth

(a) all the DNA strands of parents and daughters will shoe DNA with labelled thymidine (b) only parental strands will show thymidine labelled DNA (c) all the strands of daughters will be thymidine labelled (d) half the daughter strands will have labelled and half strands without labelled thymidine

Case-Study Based MCQ Answers

1. (i)(a) (ii)(b) (iii)(a) (iv)(c) 2. (i)(a) (ii)(c) (iii)(d) (iv)(c) 3. (i)(a) (ii)(a) (iii)(a) (iv)(c)

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Final Words

From the above article, you have practiced Molecular Basis of Inheritance MCQ of Class 12 Biology Chapter 6. We hope that the above-mentioned MCQs for term 1 of chapter 6 Molecular Basis of Inheritance MCQ would will surely help you in your exam. 

If you have any doubts or queries regarding Molecular Basis of Inheritance MCQ (Multiple Choice Questions) with answers, feel free to reach us and we will get back to you as early as possible.

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CBSE Important Questions Class 12 Biology Chapter 6

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Important Questions Class 12 Biology Chapter 6

Important questions for cbse class 12 biology chapter 6 – molecular basis of inheritance.

The study of genes in the body, including DNA and its various processes such as replication, transcription, translation, genetic code, regulation, and others, is known as molecular inheritance. This concept explains why offspring resemble their parents.

The concept of molecular inheritance is covered in Chapter 6 of Class 12 Biology . This section is included in the Extramarks Important Questions Class 12 Chapter 6 which is compiled by subject matter experts from NCERT books. These questions are written in the most straightforward manner possible so that students can easily understand and learn during exams.

CBSE Class 12 Biology Chapter-6 Important Questions

Study important questions for class 12 biology chapter 6 – molecular basis of inheritance.

A sample of the important questions for Class 12 Biology Chapter 6 is given below. For the set of questionnaires, access the link given below. 

Very Short Answer Questions(One Mark)

Q1. RNA viruses mutate and evolve more quickly than other types of viruses. Why?

• Because the -OH group on RNA is reactive, it is unstable and mutates more quickly.

Q2. What exactly does aminoacyl tRNA synthetase do?

• Aminoacyl tRNA synthetase catalyses the activation of amino acids and their attachment to the 3-end of specific tRNA molecules.

Q3. What exactly is point mutation?

• Ans: A point mutation occurs when a single base pair in a DNA sequence changes.
• What exactly is a codon?
• A codon is a triplet sequence of bases that codes for a single amino acid.

Short Answer Questions (2 Marks)

Q1. Why is it so important for tRNA to bind to both amino acids and mRNA codons during protein synthesis?

• It is critical that tRNA binds to both amino acids and mRNA codons because tRNA functions as an adapter molecule that picks up a specifically activated amino acid from the cytoplasm and transports it to the ribosomal within the cytoplasm where proteins are synthesised. It binds to the ribosome in the sequence specified by mRNA and then transfers its amino acid to a new polypeptide chain.

Q2. Mention any four significant features of the genetic code.

• A genetic codon has the following important characteristics:
• Each codon is a triplet with three bases.
• Each codon encodes only one amino acid, making it unambiguous.
• Some amino acids are coded, and it is said that more than one codon is degenerative.
• Codons are read in a continuous direction with no punctuation.

Q3. Why can transcription and translation be coupled in prokaryotic cells but not eukaryotic cells?

• In prokaryotes, the mRNA synthesised does not require any processing to become active; both transcription and translation occur in the same cytosol; however, in eukaryotes, the primary transcript contains both exons and introns and is subjected to a process known as splicing, in which introns are removed or exons are joined in a specific order to form mRNA.

Q1. Two claimant fathers sued a woman who claimed to be the father of her only daughter. How could identifying the true biological father resolve this case?

• The DNA – fingerprinting technique could have resolved this case to identify the true biological father. In this method:-
• First and foremost, the DNA of the two claimants who must be tested is isolated.
• Isolated DNA is then digested with a restriction enzyme and the digest is run through gel electrophoresis.
• Alkali treatment denatures ds DNA fragments, resulting in ss DNA.
• The electrophoresed DNA is then transferred into a nitrocellulose filter paper and fixed.
• A known sequence of DNA is prepared and labelled with the radioactive isotope 32p before being added to nitrocellulose paper.
• Autoradiography is used to photograph nitrocellulose paper on X-ray film. The film is examined for the presence of hybrid nucleic acid.
• The DNA fingerprints of the two claimants are then compared with the DNA fingerprints of the lady and her daughter, and whoever matches is declared to be the biological father of her daughter.

Q2. What are the three types of RNA and what role do they play in protein synthesis?

• There are three kinds of RNA:
• Messenger RNA (mRNA): It is a single-stranded RNA that transports the genetic information from DNA transcribed on it to the cell for protein synthesis.
• Transfer RNA (tRNA): A molecule with a “anticodon loop” on one end that reads the code and an amino acid acceptor end that binds to the specific amino acid on the other.
• Ribosomal RNA (rRNA): Ribosomes serve as a site for protein synthesis and catalyse the formation of peptide bonds.

Long Answer Questions (3 Marks)

Q1. What do you mean by DNA replication’s semi-conservative nature? Who and how did they prove it?

• The semiconservative nature of DNA replication suggested that during replication, two strands separate and each acts as a template for the synthesis of a new complementary strand, so that after complete replication, each DNA molecule would have one parental strand and one newly synthesised strand, preserving half of the information over a generation. Mathew Messselson and Franklin Stahl used Escherichia coli to demonstrate that DNA replication is semiconservative. They grew E.coli in a medium containing 15NH4Cl until 15N was incorporated into the two strands of newly synthesised DNA. Heavy DNA can be separated from normal DNA by centrifugation in a CsCl density gradient. 

The cells were then placed in a medium containing normal X14NH4Cl, samples were taken at various time intervals, DNA was extracted, and the cells were centrifuged to determine their densities. After one generation, the DNA is extracted from the cells in order to transfer from X15N medium to X14N medium.

Molecular Basis of Inheritance Important Questions

Extramarks Class 12 Chapter 6 Important Questions are highly recommended and preferred for making preparation easy for the board examinations and competitive exams. 

Subject matter experts have compiled all the extra questions of Chapter 6 Biology Class 12 so that no important topic is skipped while studying. These important questions come with solutions, so students can improve their answers and aim for the highest possible score.

Molecular Basis of Inheritance Class 12 Questions – Summary

DNA (Deoxyribonucleic Acid) (Deoxyribonucleic Acid)

Except for certain viruses, which have an RNA genome, most species’ genetic material is DNA. TMV is a good example of this (Tobacco mosaic virus).

RNA is primarily a messenger, an adaptor, and a catalytic agent. The nucleotide or base pair count of DNA is specified by its length (bp).

Structure of a Polynucleotide Chain

The polynucleotide chain structure is made up of three fundamental components:

Nitrogenous Substance:

• Purines – Adenine (A) and Guanine (G) can be found in both DNA and RNA.
• Pyrimidines are RNA’s cytosine and uracil, and DNA’s cytosine (C) and thymine (T). Thymine, also known as 5-methyl uracil, is responsible for increased DNA molecule stability.
• Pentose sugar- As the name implies, ribose is found in RNA (ribonucleic acid) and deoxyribose is found in DNA.

Phosphate Group:

• Nucleoside:  A nitrogen base linked to the hydroxyl group of 1 ‘C pentose sugar via an N-glycosidic bond.
• Nucleotide: A phospho-ester bond connects the phosphate group to the hydroxyl group at 5 ‘C of the nucleoside.

In this way, each strand of DNA has a “backbone” of phosphate-sugar-phosphate-sugar-phosphate.

Double Helix Model Given for the DNA Structure

Watson and Crick proposed the DNA double-helix structure in 1953.

• The adenine and thymine to guanine and cytosine ratio is one and tends to remain constant. Ervin Chargaff stated as much.
• DNA is made up of two polynucleotide chains with sugar-phosphate backbones and bases within.
• There is the polarity between the two chains, namely one with 3′-5′ polarity and the other with 5′-3′ polarity.
• The base pair (bp) is formed by hydrogen bonding between the nitrogen bases found on two of the polypeptide chains.
• A purine base from one nucleotide chain is frequently connected to a pyrimidine base from another nucleotide chain to form a base pair.
• Adenine and Thymine (or Uracil in RNA) are linked by two hydrogen bonds (A=T), while Guanine and Cytosine are linked by three hydrogen bonds (A=T) (G).

Replication

According to Watson and Crick, DNA replication is semi-conservative in nature. Meselson and Stahl demonstrated it experimentally in 1958. The replication of DNA begins at the origin of replication and ends with the formation of a replication fork. It results in the formation of two strands, leading and lagging. Furthermore, the formation of some fragments known as the Okazaki fragments occurs.

Transcription

Transcription is the first step in gene expression. To make an RNA molecule, the DNA sequence of a gene must be copied. This is accomplished by the enzyme RNA polymerase, which attaches nucleotides to form an RNA chain, which is then transcribed (using a DNA strand as a template). Transcription occurs at three stages: initiation, elongation, and termination.

Genetic Code

• Amino acids are the sequences of bases found in mRNA that are responsible for coding for a specific amino acid.
• Each code is made up of three nucleotides and is known as a triplet. Codons are almost universal, with the exception of some protozoan and mitochondrial codons.
• Because more than one triplet can code for the same amino acid, the code is said to have degenerated.
• There are 64 codons in total, 61 of which code for amino acids.
• There are three different stop codons that do not code for any amino acids: UAA, UAG, and UGA.
• AUG, in addition to being a starting codon, also codes for methionine.

Point Mutation: A single base pair shift, for example, causes a point mutation. Sickle cell anaemia is caused by a single point mutation in the gene that codes for the -globin chain. As a result, glutamate in the regular protein is converted to valine in the sickle cell.

Frameshift mutation occurs when one or two base pairs are lost or gained, causing the reading frame to shift at the point of insertion or deletion.

Translation

The translation is the process of converting a messenger RNA (mRNA) molecule sequence during protein synthesis into an amino acid chain. The translation process is completed in four steps: activation, initiation, elongation, and termination. These words describe the progression of the amino acid chain (polypeptide). Amino acids are transferred to ribosomes and combined to form proteins.

Human Genome Project

In 1990, the Human Genome Project (HGP) was launched in order to decode the entire DNA sequence of the human genome.

Using genetic engineering techniques, the DNA section was separated and cloned in order to determine the DNA sequence.

The project was completed in 2003, and the sequence of chromosome 1 was completed in May 2006.

DNA Fingerprinting

DNA fingerprinting is a type of test that represents an individual’s or any other living thing’s genetic makeup. It is the technique used to establish a link between a suspect and biological evidence in a criminal investigation. A DNA sample from a crime scene is compared to a DNA sample from a suspect. If the two DNA profiles match, then the evidence comes from that perpetrator.

The Molecular Basis of Inheritance Class 12 Important Questions are crucial in terms of exam preparation. Students will learn about DNA (Deoxyribonucleic acid), the structure of a polynucleotide chain, the Double Helix Model for DNA Structure, replication, transcription, genetic code, mutation, translation, the human genome project, and DNA fingerprinting in this chapter. Students can access the set of important questions available on the website at their convenience.

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FAQs (Frequently Asked Questions)

1. what is dna polymorphism.

DNA polymorphism is a process of variation in DNA caused by mutations occurring at non-coding sequences, i.e., DNA Polymorphism refers to the various DNA sequences found in living organisms. There are numerous variations occurring at the DNA level, such as base pair changes, repeated sequences, and so on.

2. Why both strands of DNA are not copied during DNA transcription? Explain.

Both strands of DNA are not copied during transcription because of the following reasons: 

I)If both strands code for RNA, two different RNA molecules and two different proteins are formed, making genetic machinery more complicated.

• II) Because the two RNA molecules are complementary, they will wind together to form dsRNA without undergoing translation, implying that transcription is pointless.

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Important Questions for CBSE Class 12 Biology Chapter 6 - Molecular Basis of Inheritance

Cbse class 12 biology chapter-6 important questions – free pdf download.

Free PDF download of Important Questions with Answers for CBSE Class 12 Biology Chapter 6 – Molecular Basis of Inheritance prepared by expert Biology teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination. You can also Download  Biology Revision Notes Class 12  to help you to revise complete Syllabus and score more marks in your examinations. Jump to 2 Marks Questions Jump to 3 Marks Questions Jump to 5 Marks Questions

CBSE Class 12 Biology Important Questions Chapter 6 – Molecular Basis of Inheritance

1 mark questions.

Chapter 6 Molecular Basis of Inheritance 

1 Marks Questions 1. Name the factors for RNA polymerase enzyme which recognises the start and termination signals on DNA for transcription process in Bacteria. Ans. Sigma (s) factor and Rho(p) factor)

2. Mention the function of non-histone protein. Ans. Packaging of chromatin

3. During translation what role is performed by tRNA Ans.   (i)  Structural role (ii)  Transfer of amino acid.

4. RNA viruses mutate and evolve faster than other viruses. Why? Ans.  -OH group is present on RNA, which is a reactive group so it is unstable and mutate faster.

6. Mention the dual functions of AUG. Ans. (i)  Acts as initiation codon for protein synthesis (ii)  It codes for methionine.

7. Write the segment of RNA transcribed from the given DNA – 3´ -A T G C A G T A C G T C G T A ‘5´- Template Strand 5´ – T A C G T C A T G C A G C A T ‘3´ – Coding Strand. Ans.  5’- U A C G U C A U G C A G C A U – 3’ (In RNA ‘T’ is replaced by‘U’)

8.Name the process in which unwanted mRNA regions are removed & wanted regions are joined. Ans. RNA splicing.

9.Give the initiation codon for protein synthesis. Name the amino acid it codes for? Ans. Initiation codon – AUG & it code for methionine.

10.In which direction, the new strand of DNA synthesised during DNA replication. Ans. 5’ → →  3

11.What is the function of amino acyl tRNAsynthetase. Ans. Amino acyl tRNAsynthetasecatalyses activation of amino and attachment of activated amino acids to the 3-end of specific tRNA molecule.

12.What is point mutation? Ans. Mutation due to change in a single base pair in a DNA sequence is called point mutation.

13.Name the enzyme that joins the short pieces in the lagging strand during synthesis of DNA? Ans. Ligase.

14.Name the enzyme which helps in formation of peptide bond? Ans. Peptidyltransferase

15.Who experimentally prove that DNA replication is semi conservative. Ans. Messelson&stahl.

16.What is a codon? Ans. Triplet sequence of bases which codes for a single amino is called a codon.

17.Name the three non-sense codons? Ans. UAA, UAG, UGA

18.What is the base pairing pattern of DNA? Ans. In DNA, adenine always binds with thymine & cytosine always binds with Guanine.

19.Mention the dual functions of AUG? Ans. AUG codes for amino acid methionine & also acts as an initiator codon.

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2 Mark Questions

2. Complete the blanks a, b, c and d on the basis of Frederick Griffith Experiment. S Strain  → →  inject into mice  → →  (a) R strain  → →  inject into mice  → →  (b) S strain (heat killed)  → →  inject into mice  → →  (c) S strain (heat killed) + R strain (live)  → →  inject into mice  → →  (d) Ans.(a)  Mice die (b)  mice live (c)  mice live (d)  mice die

3. Give two reasons why both the strands of DNA are not copied during transcription. Ans. (a)  If both the strands of DNA are copied, two different RNAs(complementary to each other) and hence two different polypeptideswill produce; If a segment of DNA produces two polypeptides, thegenetic information machinery becomes complicated. (b)  The two complementary RNA molecules (produced simultaneously)would form a doublestranded RNA rather than getting translated intopolypeptides. (c)  RNA polymerase carries out polymerisation in 53’direction andhence the DNA strand with 35’ polarity acts as the template strand.(Any two)

4. Mention any two applications of DNA fingerprinting. Ans.(i)  To identify criminals in the forensic laboratory. (ii)  To determine the real or biological parents in case of disputes. (iii)  To identify racial groups to rewrite the biological evolution. (Any two)

5. State the 4 criteria which a molecule must fulfill to act as a genetic material. Ans.(i)  It should be able to generate its replica. (ii)  Should be chemically and structurally stable. (iii)  Should be able to express itself in the form of Mendelian characters. (iv)  Should provide the scope for slow changes (mutations) that are necessary for evolution.

6.“DNA polymerase plays a dual function during DNA replication” comment on statement? Ans. DNA polymerase plays a dual function –it helps in synthesis of new strand & also helps in proof reading i.e replacement of RNA strands lay DNA fragments.

7.Three codons on mRNA are not recognised by tRNA what are they? What is the general term used for them what is their significance in protein synthesis? Ans. UAG UAA & UGA are the three codons that are not recognised by tRNA these are known as stop codon or non-sense codon. Since these three codons are not recognised by any tRNA they help in termination of protein chain during translation.

8.Give two reasons why both the strands of DNA are not copied during DNA transcription? Ans. I)If both the strands code for RNA two different RNA molecules & two different proteins wouldbe formed hence genetic machinery would become complicated II) Since the two RNA molecules would be complementary to each other, they would wind togetherto form dsRNA without carrying out translation which means process of transcription would befutile

9.Why is it essential that tRNA binds to both amino acids & mRNA codon during protein synthesis? Ans. It is essential that tRNA binds to both amino acids & mRNA codon because tRNA acts as an adapter molecule with picks up a specific activated aminoacid from the cytoplasm & transferred it to the ribosomal in the cytoplasm where proteins are synthesized. It attracts itself to ribosome with the sequence specified by mRNA & finally it transmits its amino acid to new polypeptide chain.

11.Explain what happens in frameshift mutation? Name one disease caused by the disorder? Ans. Frameshift mutation is a type of mutation where addition or deletion of one or two bases changes the reading from the site of mutation, resulting in protein with different set of amino acid.

12.What do you mean by “Central Dogma of Molecular genetics?” Ans. The central dogma of molecular genetics is the flow of genetic information from DNA to DNA through replication, DNA to mRNA through transcription & mRNA to proteins through translation. ReplicationDNA  → →  mRNA  → →  proteins. transcription translation

13.Give two reasons why both the strands are not copied during transcription? Ans. i) If both the strands codes for RNA, two different RNA molecules & two different proteins areformed hence genetic machinery would be complicated. ii)Since two RNA molecules produced would be complementary to each other, they would wind together to form ds-RNA.

14.Why is human Genome project considered as mega project? Ans. Human Genome project was called mega project for the following facts.

• The human genome has approximately 3.3 x 109bp, if the cost of sequencing is US  3 p e r b p , t h e a p p r o x i m a t e c o s t i s a b o u t U S 3perbp,theapproximatecostisaboutUS  g billion.
• If the sequence obtained were to be stored in a typed form in books & if each page contained1000 letters & each book contained 1000 page than 3300 such books would be needed to store complete in formation
• The enormous quantity of data expected to be generated also necessitates the use of high speedcomputational devices for data storage, retrieval & analysis.

15.Why is DNA & not RNA is the genetic material in majority of organisms? Ans. The -OH group in the nucleotides of RNA is much more reactive & makes RNA labile & easily degradable thus, DNA and not RNA acts as genetic material in majority of organisms.

16.Mention any four important characteristics of genetic code. Ans. Genetic codon has following imp-features :-

• Each codon is a triplet consisting of three bases.
• Each codon codes for only one amino acid i.e. – unambiguous.
• Some amino acids are coded lay more than one codon  ∴ ∴  said to be degenerative.
• Codons are read in a continuous manner in direction & have no punctuation.

17.Why it is that transcription & translation could be coupled in prokaryotic cell but not in eukaryotic cell? Ans. In prokaryotes the mRNA synthesised does not require any processing to become active &both transcription & translation occurs in the same cytosol but In Eukaryotes, primary transcriptcontains both exon & intron & is subjected to a process called splicing where introns are removed &exons are joined in a definite order to form mRNA.

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3 Mark Questions

3 Marks Questions 1. Give six points of difference between DNA and RNA in their structure/chemistry and function. Ans.

2. Explain how does the hnRNA becomes the mRNA. OR Explain the process of splicing, capping and tailing which occur during transcription in Eukaryotes. Ans. hnRNA is precursor of mRNA. It undergoes (i) Splicing : Introns are removed and exons are joined together. (ii) Capping : an unusual nucleotide (methyl guanosine triphosphate isadded to the 5´ end of hnRNA. (iii) Adenylate residues (200-300) are added at 3´ end of hnRNA.

3. Name the three major types of RNAs, specifying the function of each inthe synthesis of polypeptide. Ans. (i)  mRNA-(Messenger RNA) : decides the sequence of amino acids. (ii)  tRNA-(Transfer RNA) : (a) Recognises the codon on mRNA (b) transport the aminoacid to the site of protein synthesis. (iii)  rRNA (Ribosomal RNA) : Plays the structural and catalytic role during translation.

4. Enlist the goals of Human genome project. Ans.  The Human Genome Project (HGP) is an international scientific research project with the goal of determining the sequence of chemical base pairs which make up human DNA, and of identifying and mapping all of the genes of the human genome from both a physical and functional standpoint

5. A tRNA is charged with the amino acid methionine. (i) Give the anti-codon of this tRNA. (ii) Write the Codon for methionine. (iii) Name the enzyme responsible for binding of amino acid to tRNA. Ans.   (a)  UAC  (b)  AUG  (c)  Amino-acyltRNAsynthetase.

6. Illustrate schematically the process of initiation, elongation and termination during transcription of a gene in a bacterium. Ans. In bacteria, the mRNA provides the template, tRNA brings aminoacids and reads the genetic code, and rRNAs play structural and catalytic role during translation. There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria. RNA polymerase binds to promoter and initiates transcription (Initiation) It somehow also facilitates opening of the helix and continues elongation Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination

7.What is transformation? Describe Grifith’s experiment to show transformation? What did he prove from his experiment? Ans. Transformation means change in genetic makeup of an individual. Fredrick Grifith conducted aseries of experiments on streptococcus pneumoniae. He observed two strains of this bacterium –one forming smooth colonies with capsule (s-type) & other forming rough colonies without capsule (R-type) (i) when live s-type cells are infected into mice, they produced pneumonia & mice dies. (ii) When live R-type cells are infected into mice, disease was not produced did not appear. (iii) When heat – killed S-type cells were infected into mice, the disease did not appear. (iv) When heat killed S-type cells were mixed with live R-cells & infected into mice, the mice died. He concluded that R-strain bacteria had somehow been transformed by heat –killed S-strain bacteria which must be due to transfer of genetic material

8.The base sequence on one strand of DNA is ATGTCTATA (i) Give the base sequence of its complementary strand. (ii) If an RNA strand is transcribed from this strand what would be the base sequence of RNA? (iii) What holds these base pairs together? Ans. (i)  TACAGATAT. (ii)  UACAGAUAU (iii)  Hydrogen bonds hold these base pairs together. Adenine & thymine are bounded by twohydrogen bonds & cytosine & Guanine are bonded by three hydrogen bonds.

9.Two claimant fathers filed a case against a lady claiming to be the father of her only daughter. How could this case be settled identifying the real biological father? Ans. This case to identify the real biological father could lee settled lay DNA – finger printingtechnique. In this technique :-

• first of all, DNA of the two claimants who has to be tested is isolated.
• Isolated DNA is then digested with suitable restriction enzyme & digest is subjected to gelelectrophoresis.
• The fragments of ds DNA are denatured to produce ss DNA by alkali treatment.
• The electrophoresed DNA is then transferred from get into a nitrocellulose filter paper where itis fixed.
• A known sequence of DNA is prepared called probe – DNA & is labelled with radioactive esotope32p & then probe is added to nitrocellulose paper.
• The nitrocellulose paper is photographed on X –ray film through auto radiography. The film isanalysed to determine the presence of hybrid nucleic acid.

Then, the DNA fingerprints of the two claimants is compared with the DNA fingerprint of the lady &her daughter, whosoever matches with each other would be declared as biological father of herdaughter.

11.A tRNA is charged with amino acid methionine. i) At what site in the ribosome will the tRNA bind? ii) Give the anticodon of this tRNA? iii) What is the mRNA codon for methionine? iv) Name the enzyme responsible for this binding? Ans. (i)  P- site (ii)  UAC (iii)  AUG (iv)  Amino acyl tRNASynthetase

13.What are the three types of RNA & Mention their role in protein Synthesis? Ans. There are three types of RNA :

• Messenger RNA (mRNA) :- It is a single – stranded RNA which brings the genetic information ofDNA transcribed on it for protein synthesis.
• Transfer RNA (tRNA) :- It has a clover leaf like structure which acts as an adapter moleculewhich contains an “anticodon loop” on one end that reads the code on one hand &” an amino acid acceptor end which binds to the specific amino acid on other hand.
• Ribosomal RNA (rRNA) :- Ribosomes provides the site for synthesis of protein &catalyse theformation of peptide bond.

14. Define bacterial transformation? Who proved it experimentally & how? Ans.  The transformation is a mode of exchange or transfer of genetic information betweenorganism or from one organism to another. Fredrick Griffith tested the virulence of two strains of Diplococci to show transformation in thefollowing steps :-

• When S-III strains of bacteria are injected into mice. It developed pneumonia & died.
• When R-II strains are infected into mice, they did not develop pneumonia & survive.
• When heat – killed S-III strains of bacteria are injected into mice, No symptoms of pneumoniadevelops& mice remain healthy.
• When a mixture of heat – killed S-III strain & lives R-II strain is injected into mice, theydeveloped pneumonia & died.

From these results, Griffith concluded that the presence of heat – killed S-III bacteria must convertliving R-II type bacteria to type S-III so as to restore them the capacity for capsule formation. Thiswas called “BACTERIAL TRANSFORM ATION” S strain  → →  Inject into mice  → →  Mice die R strain  → → Injct into mice  → →  Mice live S strain (heat-killed) → → Inject into mice  → →  Mice live S strain (heat-killed) + R strain (live)  → →  Inject into mice  → → Mice die

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5 Marks Questions

5 Marks Questions 1. What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally? Ans. Meselson and Stahl, performed an experiment using E.coli to prove that DNA replication is semi conservative.

• They grew E.coli in a medium containing  15 N H 4 C l 15NH4Cl .
• Then separated heavy DNA from normal (14N) by centrifugation in CsCl density gradient.
• The DNA extracted, after one generation of transfer from 15N medium to 14N medium, had an intermediate density.

-The DNA extracted after two generations consisted of equal amounts of light and hybrid DNA. -They proved that DNA replicates in a semiconservative manner.

2. What does the lac operon consist of? How is the operator switch turned on and off in the expression of genes in this operon? Explain. Ans. Lac Operon consists of the following :

• Structural genes : z, y, a which transcribe a polycistronic mRNA.
• gene ‘z’ codes for b-galactosidase
• gene‘y’ codes for permease.
• gene‘a’ codes for transacetylase.
• Promotor : The site where RNA polymerase binds for transcription.
• Operator : acts as a switch for the operon
• Repressor : It binds to the operator and prevents the RNAPolymerase from transcribing.
• Inducer : Lactose is the inducer that inactivates the repressor by binding to it.
• Allows an access for the RNA polymerase to the structural gene andtranscription.

NCERT Solutions for class 12th Biology Chapter 6 Molecular Basis of Inheritance

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

Sol. Nitrogenous bases are Adenine, Uracil, Cytosine, Thymine. Nucleosides (Nitrogenous base+ sugar) are Cytidine and Guanosine.

Question 2. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.

Hint: Apply Chargaff s rule.

Sol. According to Chargafl’s rule, in a double-stranded DNA, the ratios between cytosine and guanine & adenine and thymine are constant and equals one, i . e . , number of cytosine is equal to the number of guanine and the number of adenine is equal to number of thymine.

As per question, double-stranded DNA has 20% cytosine then according to Chargafl’s rule, it will have 20% guanine.

Total percentage of G + C content= 20% + 20% = 40% Total percentage of A+ T content= 100% -40% = 60%

Since adenine and thymine are present in equal number, content of adenine willbe30%.

Question 3. If the sequence of one strand of DNA is written as follows:

5′-AT GCATGCATGCATGCATGCATGCATGC-3′

Write down the sequence of the complementary strands in 5′  3′ direction.

Sol. According to the complementary base pairing property of DNA, adenine present in one DNA strand always base pairs with thymine of the complementary strand and guanine in a DNA strand always base pairs with cytosine of the complementary strand and vice versa.

If the sequence ofone DNA strand is:

5′-ATGCATGCATGCATGCATGCATGCATGC-3′

Since two strands in a double-stranded DNA are antiparallel, the sequence of complementary strand in 3′ to 5′ direction will be:

3′-TACGTACGTACGTACGTACGTACGTACG-5′

The sequence of complementary strand in 5′ to 3′ direction will be: 5′-GCATGCATGCATGCATGCATGCATGCAT-3′

Question 4. If the sequence of the coding strand in a transcription unit is written as follows:

5′-AT G CAT GCA T GCA T GC AT GCATGCATGC-3′

Write down the sequence of mRNA.

Sol. Coding strand in a transcription unit is the DNA strand with polarity 5′  3′ and does not code for any protein. It has the base sequence same as RNA except thymine at the place of uracil.

If the sequence of coding strand is:

The sequence of mRNA will be:

5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

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

Sol. Watson and Crick observed that the nitrogenous bases are in complementary pairing in two strands of double helix of DNA molecule. Such an arrangement of DNA molecule led them to hypothesize the semiconservative mode of replication of DNA.

During DNA replication, the two DNA strands separate and act as a template for the synthesis of new complementary strands. The sequence of newly synthesized DNA is determined by the sequence of bases on the template strand based on the complementary base pairing rule. After the completion of replication, each daughter DNA molecule has one parental and one newly synthesized strand. Since only one parental strand is present from one generation to the next, this mode of replication is called semiconservative.

Question 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.

Sol. Different types of nucleic acid polymerases are as follows:

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

Sol. Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages. In 1952, they chose a bacteriophage known as T2 as their experimental material to prove that DNA is the genetic material. To differentiate between DNA and protein, they grew some viruses on a medium that contained radioactive phosphorus (32p) and some others on a medium that contained radioactive sulfur (35s). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus-containing nucleotides but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulfur and proteins contain sulphur-containing amino acids.

Radioactive phages were allowed to attach to E. coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.

Bacteria which was infected with viruses 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 indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.

Question 8. Differentiate between the following:

• Repetitive DNA and Satellite DNA
• mRNA and tRNA
• Template strand and Coding strand

Sol. (a)  Differences between repetitive DNA and satellite DNA

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

Sol. A ribosome is a compact ribonucleoprotein particle which has two subunits, one large subunit and one small subunit. They associate with different mRNAs for the synthesis of different polypeptides.

Ribosome perform the following essential roles during protein synthesis or translation:

• The ribosome provides the binding site for mRNA in small subunit and two binding sites for subsequent amino acids in the large subunit and thus, be close enough to each other for the formation of a peptide bond. It also moves along mRNA by three nucleotides each time to add amino acids sequentially to the growing polypeptide. In this way, it provides an environment in which recognition between a codon of mRNA and anticodon of tRNA occurs and aminoacyl-tRNAs add amino acids to the growing chain in response to the corresponding triplet codons.
• The ribosome also acts as a catalyst. The large subunit of ribosome has peptidyl transferase activity for the formation of peptide bond between two amino acids. This activity is associated with 23S rRNA in large subunit of ribosome of bacteria so this enzyme is a ribozyme.

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 some time after addition of lactose in the medium?

Sol. Lac operon is an operon in bacteria that contains structural genes responsible for metabolism of lactose. It consists of one regulatory gene, gene lac i and three structural genes, lac z, lacy and lac a (which codes for three enzymes -galactosidase, permease and transacetylase respectively). Lactose is the substrate for the enzyme -galactosidase and it regulates switching on and off of the operon. Hence, it is called inducer. In the absence of a preferred carbon source such as glucose, if lactose is provided in the growth medium of the bacteria, the lactose is transported into the cells through the action of permease. Lactose binds to the repressor and prevents its binding to operator. This allows RNA polymerase access to the promoter and transcription of structural genes proceeds leading to the formation of ­ galactosidase. This enzyme breaks down the lactose into glucose and galactose.

After some time, lac operon shuts down due to the following reasons:

• As inducer is completely metabolized, lactose is not available to inactivate the repressor. This results in binding of repressor to the operator gene and in turn, preventing RNA polymerase from transcribing the operon. Hence, the transcription is stopped and the operon is switched off.
• Lac operon cannot be switched on by lactose due to the accumulation of glucose which is a preferred energy source over lactose.

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

Sol. (a) Promoter: Promoter is one of the three regions of transcription unit located towards 5′ end of the structural gene. It is a DNA sequence that provides binding site for RNA polymerase to initiate transcription. Also, the position of a promoter in a transcription unit defines the template and coding strands.

tRNA: tRNA or transfer RNA is a 75 to 85 long ribonucleic acid folded to form a secondary structure shaped like a cloverleaf Each tRNA molecule has an anticodon loop that has bases complementary to the code, and an amino acid acceptor end to which it binds to amino acids. Each tRNA has only one type of anticodon and attaches to a specific amino acid. Its function is to bind to an activated amino acid, transfer it from cellular pool to ribosomes where tRNA binds to correct codon in mRNA and helps assemble amino acids into polypeptides.

Exons: Exons are coding sequences interrupted by non-coding sequences (called introns) in the split genes of eukaryotes. Introns are removed and exons are joined to produce functional RNA by splicing. The exons in functional mRNA are transcribed into functional polypeptide.

Question 12. Why is the Human genome project called a mega project?

Sol. Human genome project aimed to sequence every base in human genome. It was a 13-year project coordinated by the U.S. Department of Energy and the National Institute of Health. The project was completed in 2003.

It was a mega project due to the following reasons:

• Large size of human genome and high invested cost of sequencing: Human genome is said to have approximately 3 x 109 bp and the total estimated cost of the project would be approximately 9 billion US dollars, and if the cost of sequencing required is US$ 3 per bp (the estimated cost in the beginning).
• Requirement of high-speed computational devices for data storage: The enormous amount of data would be expected to generate from DNA sequencing from a single human cell. This necessitated the use of high speed computational devices for data storage and retrieval, and analysis. HGP was closely associated with the rapid development of a new area in biology called bioinformatics.

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

Sol. •  DNA fingerprinting is a technique used to identify individuals based on the differences in some specific regions in DNA sequence called as repetitive DNA. It works on the principle of polymorphism in DNA sequences. This technique was initially developed by Sir Alec Jeffreys who discovered that certain regions of DNA showed variations in the number of tandem repeats known as variable number of tandem repeats (VNTRs). He used a satellite DNA as probe that shows very high degree of polymorphism.

• It can be applied in forensic science to identify an individual in criminal and civil cases. DNA fingerprinting is the basis of paternity testing, in case of disputes. It is used to identify genes connected with hereditary diseases. It is used to identify and protect the commercial varieties of crops and livestock.
• It finds its application in evolutionary biology where it is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

Question 14. Briefly describe the following:

• Transcription
• Polymorphism
• Translation
• Bioinformatics

Sol. (a) Transcription

• It is the process in which genetic information from DNA is copied into RNA.
• In transcription, only a segment of DNA is copied into RNA. The segment of DNA that codes for RNA is called transcription unit. A transcription unit in DNA has three regions namely promoter, structural gene and a terminator. The promoter and terminator flank the structural gene in a transcription unit. The promoter is a DNA sequence that provides binding site for RNA polymerase. The terminator defines the end of the process of transcription. Also during transcription, one of the strands of DNA acts a template to direct the synthesis of complementary RNA.
• RNA is transcribed from DNA strand with polarity 3′ 5′ by the enzyme DNA-dependent RNA polymerase in 5′ 3′ direction using ribonucleotides.
• The process of transcription completes in three steps: initiation, elongation and termination. During initiation, RNA polymerase binds to promoter and initiates transcription (initiation). It uses nucleoside triphosphates as substrate.

(b) Polymorphism

• Polymorphism refers to the variation at genetic level arising due to mutations.
• DNA polymorphism is an inheritable mutation observed in a population at high frequency. The probability of such variation is higher in non-coding DNA sequence as mutations in these sequences may not have any immediate effect or impact in an individual’s reproductive ability.
• These mutations keep on accumulating generation after generation, and form one of the bases of variability or polymorphism. There is a variety of different types of polymorphisms ranging from single nucleotide change to very large scale changes. For evolution and speciation, such polymorphisms play very important role.
• Polymorphism in DNA sequence also forms the basis of genetic mapping of human genome as well as of DNA fingerprinting.
• Translation is the process of protein synthesis from mRNA. The mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNAs play structural and catalytic role during translation.
• During translation, amino acids are polymerised to form a polypeptide. The order and sequence of amino acids in a polypeptide are defined by the sequence of bases in the mRNA.
• The ribosome binds to mRNA at a specific area.
• The ribosome starts matching tRNA anticodon sequences to the mRNA codon sequence.
• Each time a new tRNA comes into the ribosome, the amino acid that it was carrying gets added to the elongating polypeptide chain. The ribosome continues until it hits a stop sequence, then it releases the polypeptide and the mRNA.
• The polypeptide forms into its native shape and starts acting as a functional protein in the cell.

Note: In bacteria, transcription and translation take place in the same compartment and many times the translation can begin much before the mRNA is fully transcribed because there is no separation of cytosol from nucleus in bacteria. In contrast, in eukaryotes, transcription takes place in nucleus and translation occurs in cytoplasm.

• Bio informaticsis an inter disciplinary field mainly involving molecular biology and genetics, computer science, mathematics, and statistics. This field develops methods and software tools for understanding biological data. It helps to analyse and compare protein sequences, search for genes, assemble genomes, compare species based on their genetic material, forecast protein structures and find active parts of proteins, analyse gene expression, and inverse engineer genetic networks.
• It also helps to relate the genome to its phenotype and can trace evolution back to its roots to find adaptations and changes in the organisms.

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• NCERT Solutions for Class 12 (All Subjects)
• NCERT Solutions for Class 12 Biology
• Molecular Basis of Inheritance Class 12 Notes Biology Chapter 6

• CBSE Notes For Class 12
• Biology Notes Class 12
• Chapter 6: Molecular Basis Of Inheritance

Molecular Basis of Inheritance Class 12 Notes

According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 5.

What Is DNA?

DNA- Deoxyribonucleic Acid is considered the molecule of inheritance as it carries genetic information in all living organisms. It is a long polymer chain of deoxyribonucleotides. Its length depends on the number of nucleotide base pairs present in it.

Watson and Crick were the first scientists who proposed a double-helical model for DNA based on X-ray crystallography of the molecule. Each strand of DNA is a polymer of nucleotides, every nucleotide consists of a deoxyribose sugar, a nitrogen base and a phosphate.

According to the central dogma of molecular biology, genetic information flows from DNA to RNA to protein.

The complete DNA structure looks like a twisted ladder. The two strands of DNA are held together by weak hydrogen bonds between the nitrogen bases. A purine base always pairs with a pyrimidine base, i.e., adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C).

Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:

• Molecular Basis of Inheritance Study Notes
• Molecular Basis of Inheritance MCQ Practice Questions
• Molecular Basis of Inheritance MCQ Practice Solutions

Structure of Polynucleotide

A nucleotide consists of three elements – a nitrogenous base, sugar and a phosphate group. Nitrogenous bases are in the form of purines(Adenine, Guanine) and Pyrimidines(Cytosine and Thymine). The sugar part is constituted by the pentose sugar(ribose in RNA and deoxyribose in DNA), while the phosphate group is constituted by the nucleoside and nucleotide.

See Also:  Important Questions for Class 12 Chapter 6: Molecular Basis of Inheritance

What Is a Gene?

A gene is the functional unit of inheritance. In all eukaryotic organisms, DNA consists of both coding and non-coding sequences of nucleotides. The coding sequences are defined as exons, and non-coding sequences are defined as introns. These exons appear in the matured RNA, but the introns do not appear.

RNA- Ribonucleic Acid. It is a single strand of nucleic acid present in all living cells and acts as a messenger carrying instructions from DNA for controlling protein synthesis. The three primary types of RNA molecules are,

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

Packaging DNA Helix

• DNA in prokaryotes is arranged as a large loop in the nucleoid region wherein the negatively charged DNA is firmly held by positively charged proteins.
• DNA in prokaryotes is observed as a complex organisation of DNA in chromosomes. DNA is wound around the core of histone octamer (a unit with 8 histone molecules) to form a  nucleosome.
• Positively charged proteins in the form of histones are observed that are rich in basic amino acids, lysine and arginine. They are of 5 types – H1, H2A, H2B, H3 and H4. The histone octamer has 2 molecules of 4 histone proteins vital in gene regulation.
• Nucleosome is a repeating unit in the chromatin preventing DNA from tangling containing nearly 200 bp of DNA.
• The further packaging of chromatin is facilitated by NHC (Non-histone chromosomal proteins).
• Euchromatin:  transcriptionally active areas, wherein chromatin is loosely packed and they take up the light stain.
• Heterochromatin:  transcriptionally inactive areas, wherein chromatin is densely packed and takes up the dark stain.

RNA was the first genetic material, and supporting this is enough evidence suggesting vital life processes evolved around RNA. It is used to act as a genetic material and a catalyst as well. But as a catalyst, RNA was very reactive and, therefore, unstable. Hence, DNA evolved from RNA with chemical alterations making it more stable.

Replication

• Watson and Crick proposed that DNA replication is semiconservative.
• Meselson and Stahl, in 1958, proved experimentally that DNA replicates semi conservatively.
• Taylor et al., in another experiment on faba beans (Vicia faba) using radioactive thymidine, proved that the replication of DNA is semiconservative.
• Enzyme DNA polymerase catalyses DNA replication. It can polymerise only in 5’→3’ direction.
• Replication is continuous in a strand with 5’→3’ direction, called the leading strand, where the template strand has 3’→5’ polarity, called the leading strand template.
• Replication is discontinuous in a lagging strand template where the template strand has 5’→3’ polarity.

Transcription

The genetic information present in DNA (one segment only) is copied into RNA. Adenine pairs with Uracil instead of Thymine in RNA. Transcription of DNA involves three regions –  the structural gene, promoter, and terminator. The RNA polymerase catalyses transcription, while the direction of transcription is the same as that of replication by DNA polymerase, i.e. 5’→3’ direction. The template strand has a 3’→5’ polarity acting as a template for RNA formation, known as an antisense strand. The coding strand has a 5’→3’ polarity and is also known as a sense strand. In addition, it consists of the structural gene, promoter, terminator, exons and introns.

Genetic Code

These are the sequences of bases in mRNA coding for a specific amino acid in the synthesis of proteins. Here every code is composed of three nucleotides known as triplets. Totally, there are 64 codons, where 61 code for amino acids. The rest 3 are known as stop codons, as they do not code for any amino acid. AUG is the start codon and codes for the amino acid methionine as well.

Mutations and Genetic Code

A change in the single base pair causes point mutation. Example – sickle cell anaemia in the gene coding for the 𝛽-globin chain. As a result, Glutamate in the normal protein gets converted to Valine in the sickle cell. The reading frame at the point of deletion or insertion is changed when there is a loss or gain of one or two base pairs. This is known as frameshift mutation.

Translation

It is the process of amino acid polymerisation. Amino acids are joined by peptide bonds. All three RNAs(mRNA, tRNA and rRNA) have a different role in the process of translation. The first stage in this process of translation is the aminoacylation of tRNA. Ribosomes are a protein manufacturing factory, acting as a catalyst in the formation of a peptide bond. The translation process is in the 5’→3’ direction always.  There are two sites in the large subunit of a ribosome accommodating two tRNAs with amino acids close enough to form a peptide bond.

Regulation of Gene Expression

• At the time of formation of a primary transcript, i.e. transcription,
• At the time of processing or splicing,
• At the time of transportation of mRNA from the nucleus to the cytosol, and
• At the time of protein synthesis, i.e., translation.
• Gene expression is regulated by environmental, physiological and metabolic conditions.
• The development and differentiation of an embryo is a result of coordinated regulation and expression of several sets of genes.
• Control of gene expression in prokaryotes is mainly at the initiation of transcription.
• The activity of RNA polymerase at the start site is regulated by regulatory proteins, which can be a repressor or activator.
• The accessibility of the promoter region is regulated by an operator sequence adjacent to it that binds with the specific protein, mostly a repressor. There is a specific operator and repressor protein in a specific operator.

Read More: DNA Fingerprinting

For more information on Heredity and Recombinant DNA Technology, watch the below videos

A Few Important Questions

• What is Polymorphism?
• What is DNA fingerprinting?
• List out the functions of the Promoter.
• Differentiate between mRNA and tRNA.
• List out the goals of the Human Genome Project.

Learn more about heredity, the Law of Inheritance, chromosomes and genes from the topics given below:

Frequently Asked Questions on CBSE Class 12 Biology Notes Chapter 6 Molecular Basis of Inheritance

What are the uses of polynucleotide.

Polynucleotides are used in biochemical experiments such as polymerase chain reaction (PCR) or DNA sequencing.

What is a DNA Helix?

A double helix is the description of the structure of a DNA molecule. A DNA molecule consists of two strands that wind around each other like a twisted ladder.

What is gene expression regulation?

Regulation of gene expression, or gene regulation, includes a wide range of mechanisms that are used by cells to increase or decrease the production of specific gene products (protein or RNA).

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Molecular Basis of Inheritance Class 12 MCQ Online Test

Free mcq test, table of content, molecular basis of inheritance test - 101.

Duration: 10 Mins

Maximum Marks: 10

Read the following instructions carefully.

1. The test contains 10 total questions.

2. Each question has 4 options out of which only one is correct .

3. You have to finish the test in 10 minutes.

4. You will be awarded 1 mark for each correct answer.

5. You can view your Score & Rank after submitting the test.

6. Check detailed Solution with explanation after submitting the test.

7. Rank is calculated on the basis of Marks Scored & Time

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Duration: 15 Mins

Maximum Marks: 15

1. The test contains 15 total questions.

3. You have to finish the test in 15 minutes.

4. There is No Negative marking .

5. You will be awarded 1 mark for each correct answer.

6. You can view your Score & Rank after submitting the test.

7. Check detailed Solution with explanation after submitting the test.

8. Rank is calculated on the basis of Marks Scored & Time.

Molecular Basis of Inheritance Test - 22

Duration: 10 Mins

1. The test contains 10 total questions.

3. You have to finish the test in 10 minutes.

Molecular Basis of Inheritance Test - 21

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Duration:  10 Mins

1.  The test contains  10 total questions.

2.  Each question has  4 options  out of which  only one is correct .

3.  You have to finish the test in  10 minutes.

4.  There is  No Negative marking .

5.  You will be awarded  1 mark  for each correct answer.

6.  You can view your  Score & Rank  after submitting the test.

7.  Check  detailed Solution  with explanation after submitting the test.

8.  Rank is calculated on the basis of  Marks Scored & Time.

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Duration:  30 Mins

Maximum Marks:  20

1.  The test contains  20 total questions.

3.  You have to finish the test in  30 minutes.

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Duration:  20 Mins

Maximum Marks:  19

1.  The test contains  19 total questions.

3.  You have to finish the test in  20 minutes.

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In Class 12 Biology, Molecular Basis of Inheritance is one of the important chapters. To help you practise questions on this topic, we have Molecular Basis of Inheritance Class 12 MCQ in online format. Molecular Basis of Inheritance Class 12 MCQ is created by subject matter experts of selfstudys.com, It can help you measure your significant educational outcomes which includes knowledge, understanding, judgement and problem solving skills.

Class 12 MCQ on Molecular Basis of Inheritance is created as per the latest CBSE Syllabus. The students who practise the MCQ test of Molecular Basis of Inheritance will be able to secure good marks. 

As you are aware that Molecular Basis of Inheritance is a very significant topic, it requires a lot of practice and objective knowledge. 

It is highly advisable for all the students to solve Molecular Basis of Inheritance Class 12 MCQ Test as this will build their confidence and will also increase their problem solving skills. 

Molecular Basis of Inheritance Class 12 Biology MCQ: Format 

Practising Multiple Choice Questions allows students to explore knowledge and assess high order thinking. We have Molecular Basis of Inheritance Class 12 MCQ. It will help students to understand the concepts well. These MCQs can be a great way to do revision.

Class 12 MCQ on Molecular Basis of Inheritance are created as per the format of last year's question papers. 

Students often get worried whether they are prepared for the exam or not. Molecular Basis of Inheritance Class 12 MCQs help them to know their preparation well and also assess knowledge and understanding of complex concepts.

How to attempt Molecular Basis of Inheritance Class 12 MCQ?

Follow are the steps on how you can attempt Class 12 MCQ on Molecular Basis of Inheritance 

• Visit the website i.e. ( www.selfstudys.com ) 
• Click on the three lines on the upper left and tap on the CBSE option, scroll down and you will find MCQ tests. Click on that.

• Scroll down and select the Class 12

• Select Biology and click on the chapter Molecular Basis of Inheritance.

• After clicking, you can attempt Molecular Basis of Inheritance Class 12 MCQ. 

Instructions for Molecular Basis of Inheritance Class 12 MCQ

In our online MCQ test of Molecular Basis of Inheritance Class 12 Biology we have mentioned some instructions so that you can better give your MCQ test.

• In Molecular Basis of Inheritance Class 12 MCQ, the total number of questions will be 10. 
• Each question will have 4 options out of which only 1 will be correct. 
• Duration of Class 12 MCQ on Molecular Basis of Inheritance will be 10 minutes to ensure that a student manages their time effectively. 
• For each correct answer, the student will be awarded 1 mark. 
• After submitting the test, you can see your score. 
• After submitting Class 12 MCQ on Molecular Basis of Inheritance, you can check the solutions of the test with a detailed explanation. 
• Rank of Molecular Basis of Inheritance Class 12 MCQ will be calculated on the basis of marks obtained in the test and time taken to complete it. 
• You can also reattempt Class 12 MCQ on Molecular Basis of Inheritance test. 

How to prepare for Molecular Basis of Inheritance Class 12 MCQ? 

Before solving the MCQ of Class 12 Molecular Basis of Inheritance, try to prepare for the test in this manner.

• Memorising Important Notes: Begin by reading your notes of Molecular Basis of Inheritance again and again. Read it slowly as rushing will not help and only make you stressed. 
• Create Acronyms: Another great way to prepare for Molecular Basis of Inheritance Class 12 MCQ is that you can make acronyms to help you learn the names and topics. For that, you have to take the first letter of the important word and place it into a word so that you can remember it.
• Use the word associations: Another great way to prepare for Class 12 MCQ on Molecular Basis of Inheritance is to use the word associations to understand meanings and remember things. 

Molecular Basis of Inheritance Biology Class 12 MCQ for Improvement in Biology 

After completing the syllabus, the first thing a student wants to know is the status of their preparation. Molecular Basis of Inheritance Class 12 MCQ can help students know their strengths and weaknesses in Biology and to know where is the need for them to improve in the subject.

Understand The Format of Molecular Basis of Inheritance Biology Class 12 MCQ

Molecular Basis of Inheritance Biology Class 12 is important for all the Class 12 Students as they will have objective questions to solve. Molecular Basis of Inheritance Biology Class 12 MCQ test will have questions and four options. The 3 options are just to confuse the student to test his knowledge. So, while preparing for board exams, students must prepare for Molecular Basis of Inheritance Class 12 MCQ test to avoid getting confused or distracted. 

Also, it is advised to do a revision of the chapter Molecular Basis of Inheritance before attempting Molecular Basis of Inheritance Biology MCQ for Class 12 Test. Molecular Basis of Inheritance Class 12 MCQ is just perfect for students who have completed their preparation and now want to test their learning. 

Benefits of Molecular Basis of Inheritance Class 12 MCQ Tests

Let’s have a look at the benefits of the Class 12 MCQ on Molecular Basis of Inheritance

• Flexible Questioning Technique: Multiple Choice Questions are a flexible questioning technique which can be helpful in the learning process of a student. Molecular Basis of Inheritance Class 12 MCQ are versatile which helps a student to learn lessons with HIgh Order Thinking. 
• Time Management: Class 12 MCQ on Molecular Basis of Inheritance are delivered quickly as compared to lectures which means students can have more time for topic discussion. Class 12 MCQ on Molecular Basis of Inheritance can be answered quickly and the students will have more time for revision. This will help students in managing the time. 
• Created by Highly Qualified Subject Matter Experts: Molecular Basis of Inheritance Class 12 MCQ are created by the Highly Qualified Subject Matter Experts of Selfstudys which have years of experience in the education field and are aware of the format of the questions asked in exams. 
• Fast and Easy to Score: Unlike pen and paper exams , Molecular Basis of Inheritance Class 12 MCQ are fast and easy to score which can give a quick update to the students of their performance and identify the areas where they are lacking. 
• Gives Idea to the Students: Class 12 MCQ on Molecular Basis of Inheritance are very Helpful to give you a basic idea of the questions which could be asked in exams. 
• Improve the skills of the Students: Class 12 MCQ on Molecular Basis of Inheritance improves the thinking and reasoning skills of the students along with the problem solving skills. 

Tips and Strategies for Multiple Choice Questions of Living World Class 12 Biology

Let’s discuss the tips and strategies which you should always keep in mind while attempting Molecular Basis of Inheritance Class 12 MCQ- 

• Read the entire question: It is advisable for all the students to read the entire question of Class 12 MCQ on Molecular Basis of Inheritance to understand it better. Students often get excited by looking at the question and choose the most logical answer and do not read the entire question due to which they can make mistakes. 
• Answer it in your mind first: After looking at the question in Molecular Basis of Inheritance Class 12 MCQ, try answering it without looking at the options. This will help you to be 100% sure whether the answer which you have in your mind is correct or not. 
• Answer the questions you know first: If you don’t know the answer of a particular question in Molecular Basis of Inheritance Class 12 MCQ, skip it for the time being and move on to the question for which you are sure that you know the answer. This will help in time management as you will have time for the questions which requires time. 
• Make an educated guess: If it will not cost your marks, make an educated guess in Class 12 MCQ on Molecular Basis of Inheritance. (In a lot of exams, incorrect answers result in negative marking. For example, a correct answer gives 2 marks, a skipped question 0 marks and an incorrect answer -1 marks). In Class 12 MCQ on Molecular Basis of Inheritance, you can make an educated guess. 
• It is generally best to stick with your first-choice but not every time: It is best to stick with the option you decided after reading Molecular Basis of Inheritance Class 12 MCQ. It is opposite to second guess yourself and change the option which you chose in the first place. However, this does not mean the first option you chose was the correct one. The options provided in the Multiple choice questions will include the most common wrong answer among the options that look correct but are not.

How to choose the Right Answers to Molecular Basis of Inheritance Class 12 MCQ? 

• Eliminate wrong answers: After reading the questions and looking at the answer options in Molecular Basis of Inheritance Class 12 MCQ, eliminate the answers which you think are not right. Even if you think you are aware of the right option, eliminating the wrong options will give you surety that your selected option is right. 
• Options like “All of the above” and “None of the above”: If you come across options like ‘All of the above” and “None of the above” in the Class 12 MCQ on Molecular Basis of Inheritance, do not select them if you are not completely sure as students feel that this might be the correct answer. This is a very common mistake which students make.
• Read every answer option: Read every option of the question before selecting the final answer. This is a common mistake which most of the students make in Molecular Basis of Inheritance Class 12 MCQ. There is usually a best answer to every multiple-choice question, you may end up not selecting the best answer. 
• Look for the answers hidden in questions: Before selecting the option, read Class 12 MCQ on Molecular Basis of Inheritance 2-3 times, this is because a lot of times, the answers are found in the questions themselves. 
• True or false test: Read Class 12 MCQ on Molecular Basis of Inheritance and if you get confused, give each option a true or false test, Eliminate the false options. This is a great way to find out the correct answer. 
• Pay attention to these words: Be aware of these words in Molecular Basis of Inheritance Class 12 MCQ: not, sometimes, always and never . The option which includes always must be incorrect. The same thing goes for the word never. 
• When there are apparently two correct answers: When two answers seem correct in Class 12 MCQ on Molecular Basis of Inheritance with an “All of the above option”, then probably it is the correct option. 

How Class 12 MCQ on Molecular Basis of Inheritance helps in Time Management? 

Time Management is very important when one is solving Molecular Basis of Inheritance Class 12 MCQ. The first step is to look at the total number of questions in Class 12 MCQ on Molecular Basis of Inheritance and then divide your time into segments to avoid last minute stress. Do not check time after every question as it is a waste of time. You need to have awareness of the clock to make sure that you answer all the questions. 

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