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Get here all the Important questions for Class 12 Biology chapter wise as free PDF download. Here you will get Extra Important Questions with answers, assertion reasoning and Multiple Choice Questions (MCQ's) chapter wise in Printable format. Class 12 Biology has 16 important chapters covering various important topics related to human physiology evolution, diseases, genetics, organisms, populations, etc.Solving Chapter wise questions is one of the best ways to prepare for the examination. Students are advised to understand the concepts and theories of Biology properly before the exam. You can easily find 1 Mark, 2 marks, 3 marks, and 5 marks questions from each chapter of Class 12 Biology and prepare for exam more effectively. These preparation material for Class 12 Biology , shared by teachers, parents and students, are as per latest NCERT and CBSE Pattern syllabus and assure great success in achieving high score in Final CBSE Board Examinations.

Latest MCQ's and Important Questions for CBSE Class 12 Biology

Reproduction in Organisms class 12 important questions pdf Sexual Reproduction in Flowering Plants class 12 important questions pdf Human Reproduction class 12 important questions pdf Reproductive Health class 12 important questions pdf Principles of Inheritance and Variation class 12 important questions pdf Molecular Basis of Inheritance class 12 important questions pdf Evolution class 12 important questions pdf Human Health and Diseases class 12 important questions pdf Strategies for Enhancement in Food Production class 12 important questions pdf Microbes in Human Welfare class 12 important questions pdf Biotechnology Principles and Processes class 12 important questions pdf Organisms and Populations class 12 important questions pdf Ecosystem class 12 important questions pdf Biodiversity and Conservation class 12 important questions pdf Environmental Issues class 12 important questions pdf important questions of chapter 1 biology class 12 important questions of chapter 2 biology class 12 important questions of chapter 3 biology class 12 important questions of chapter 4 biology class 12 important questions of chapter 5 biology class 12 important questions of chapter 6 biology class 12 important questions of chapter 7 biology class 12 important questions of chapter 8 biology class 12 important questions of chapter 9 biology class 12 important questions of chapter 10 biology class 12 important questions of chapter 11 biology class 12 important questions of chapter 12 biology class 12 important questions of chapter 13 biology class 12 important questions of chapter 14 biology class 12 important questions of chapter 15 biology class 12 important questions of chapter 16 biology class 12 Reproduction in Organisms class 12 mcq Sexual Reproduction in Flowering Plants class 12 mcq Human Reproduction class 12 mcq Reproductive Health class 12 mcq Principles of Inheritance and Variation class 12 mcq Molecular Basis of Inheritance class 12 mcq Evolution class 12 mcq Human Health and Diseases class 12 mcq Strategies for Enhancement in Food Production class 12 mcq Microbes in Human Welfare class 12 mcq Biotechnology Principles and Processes class 12 mcq Biotechnology: and its Application class 12 mcq Organisms and Populations class 12 mcq Ecosystem class 12 mcq Biodiversity and Conservation class 12 mcq Environmental Issues class 12 mcq case based questions class 12 biology chapter 1 case based questions class 12 biology chapter 2 case based questions class 12 biology chapter 3 case based questions class 12 biology chapter 4 case based questions class 12 biology chapter 5 case based questions class 12 biology chapter 6 case based questions class 12 biology chapter 7 case based questions class 12 biology chapter 8 case based questions class 12 biology chapter 9 case based questions class 12 biology chapter 10 case based questions class 12 biology chapter 11 case based questions class 12 biology chapter 12 case based questions class 12 biology chapter 13 case based questions class 12 biology chapter 14 case based questions class 12 biology chapter 15 case based questions class 12 biology chapter 16 Assertion Reason questions Biology Class 12 Chapter 1 Assertion Reason questions Biology Class 12 Chapter 2 Assertion Reason questions Biology Class 12 Chapter 3 Assertion Reason questions Biology Class 12 Chapter 4 Assertion Reason questions Biology Class 12 Chapter 5 Assertion Reason questions Biology Class 12 Chapter 6 Assertion Reason questions Biology Class 12 Chapter 7 Assertion Reason questions Biology Class 12 Chapter 8 Assertion Reason questions Biology Class 12 Chapter 9 Assertion Reason questions Biology Class 12 Chapter 10 Assertion Reason questions Biology Class 12 Chapter 11 Assertion Reason questions Biology Class 12 Chapter 12 Assertion Reason questions Biology Class 12 Chapter 13 Assertion Reason questions Biology Class 12 Chapter 14 Assertion Reason questions Biology Class 12 Chapter 15 Assertion Reason questions Biology Class 12 Chapter 16

Total Papers :

Class 12 Biology Marks Distribution
Units	Marks
Reproduction	14
Genetics and Evolution	18
Biology and Human Welfare	14
Biotechnology and its Applications	12
Ecology and Environment	12
Total	70

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.

Structure of CBSE Biology Sample Paper for Class 12 Science is

Type of Question	Marks per Question	Total No. of Questions	Total Marks
Very Short Answer Type Questions	1	5	5
Short Answer Type Questions - 1	2	7	14
Short Answer Type Questions - 2	3	12	36
Long Answer Type Questions	3	5	15
Total		27	70

For Preparation of exams students can also check out other resource material

CBSE Class 12 Biology Sample Papers

CBSE Class 12 Biology Worksheets

CBSE Class 12 Biology Question Papers

CBSE Class 12 Biology Test Papers

CBSE Class 12 Biology Revision Notes

Question Bank of Other Subjects of Class 12

Importance of Question Bank for Exam Preparation?

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

• Since Important questions included in question bank are collections of questions that were asked in previous exams and tests thus when a student tries to attempt them they get a complete idea about what type of questions are usually asked and whether they have learned the topics well enough. This gives them an edge to prepare well for the exam.Students get the clear idea whether the questions framed from any particular chapter are mostly either short or long answer type questions or multiple choice based and also marks weightage of any particular chapter in final exams.
• 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.
• Solving a lot of different types of important questions gives students a clear idea of what are the main important topics of any particular chapter that needs to focussed on from examination perspective and should be emphasised on for revision before attempting the final paper. So attempting most frequently asked questions and important questions helps students to prepare well for almost everything in that subject.
• Although students cover up all the chapters included in the course syllabus by the end of the session, sometimes revision becomes a time consuming and difficult process. Thus, practicing important questions from Question Bank allows students to check the preparation status of each and every small topic in a chapter. Doing that ensures quick and easy insight into all the important questions and topics in each and every individual. Solving the important questions also acts as the revision process.

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Class 12th Biology - Molecular Basic of Inheritance Case Study Questions and Answers 2022 - 2023

Class 12th Biology - Molecular Basic of Inheritance Case Study Questions and Answers 2022 - 2023 Study Materials Sep-08 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12 Biology Subject - Molecular Basic of Inheritance, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Molecular basic of inheritance case study questions with answer key.

Final Semester - June 2015

 Case Study 

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Unit 6: The Molecular Basis Of Inheritance

About this unit.

Learn about the history, structure and replication of DNA and RNA, transcription and translation. This unit is aligned to the Class 12 NCERT curriculum.

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)
• Molecular mechanism of DNA replication (Opens a modal)
• DNA structure and replication review (Opens a modal)
• DNA structure and replication 4 questions Practice
• Replication 4 questions Practice

Transcription and RNA processing

• DNA replication and RNA transcription and translation (Opens a modal)
• Central dogma of molecular biology (Opens a modal)
• Intro to gene expression (central dogma) (Opens a modal)
• Overview of transcription (Opens a modal)
• Transcription and mRNA processing (Opens a modal)
• Post-transcriptional regulation (Opens a modal)
• Eukaryotic pre-mRNA processing (Opens a modal)
• Central dogma 4 questions Practice
• Transcription 4 questions Practice
• Transcription and RNA processing 4 questions Practice

Translation

• The genetic code (Opens a modal)
• Translation (mRNA to protein) (Opens a modal)
• Overview of translation (Opens a modal)
• Impact of mutations on translation into amino acids (Opens a modal)
• Translation 4 questions Practice
• Codons and mutations 4 questions Practice

Regulation of Gene Expression

• DNA and chromatin regulation (Opens a modal)
• Operons and gene regulation in bacteria (Opens a modal)
• Overview: Gene regulation in bacteria (Opens a modal)
• Lac operon (Opens a modal)
• The lac operon (Opens a modal)
• Gene regulation in bacteria 4 questions Practice

Human genome project and DNA fingerprinting

• What is the human genome project? (Opens a modal)
• How was the human genome sequenced? (Opens a modal)
• DNA polymorphism (Opens a modal)
• Satellite DNA (Opens a modal)
• Gel electrophoresis (Opens a modal)
• Southern blot (Opens a modal)
• DNA fingerprinting (Opens a modal)
• Human genome project 4 questions Practice
• DNA fingerprinting 4 questions Practice

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Important Questions for CBSE Class 12 Biology Chapter 6 - Molecular Basis of Inheritance 2024-25

• Class 12 Important Question
• Chapter 6: Molecular Basis Of Inheritance

CBSE Class 12 Biology Chapter-6 Important Questions - Free PDF Download

Important questions of Chapter 6 Biology Class 12 PDF would be really helpful for the students in preparing themselves for the examination. This PDF is well organized, according to the ease of learning. These important questions of Chapter 6 Biology Class 12 are presented in the simplest and easiest way such that the students enjoy it while learning. All the major part s o f Chapter 6 Biology Class 12 important questions have been covered in this PDF to make all the concepts clear and cover the major part of the chapter. 

Download CBSE Class 12 Biology Important Questions 2024-25 PDF

Also, check CBSE Class 12 Biology Important Questions for other chapters:

CBSE Class 12 Biology Important Questions
Sl.No	Chapter No	Chapter Name
1	Chapter 1
2	Chapter 2
3	Chapter 3
4	Chapter 4
5	Chapter 5
6	Chapter 6	Molecular Basis of Inheritance
7	Chapter 7
8	Chapter 8
9	Chapter 9
10	Chapter 10
11	Chapter 11
12	Chapter 12
13	Chapter 13
14	Chapter 14
15	Chapter 15
16	Chapter 16

Related Chapters

Study Important Questions for Class 12 Biology Chapter 6 - Molecular basis of Inheritance

Very short answer questions                                                                             1 mark.

1. Name the factors for RNA polymerase enzymes which recognize the start and termination signals on DNA for the transcription process in Bacteria.

Ans: Sigma (s) factor and Rho(p)  factor)

2. Mention the function of the 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 mutates faster.

5. Name the parts ‘X’ and ‘Y’ of the transcription unit given below.

Ans: X - Template strand, Y - Terminator.

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’- UAACAAGGAGCAUCC – 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 & its code for methionine.

10. In which direction, the new strand of DNA synthesized during DNA replication. 

Ans: 5’ to 3’

11. What is the function of aminoacyl tRNA synthetase?

Ans: Aminoacyl tRNAsynthetase catalyses activation of amino and attachment of activated amino acids to the 3-end of specific tRNA molecules.

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 the synthesis of DNA?

Ans: Ligase.

14. Name the enzyme which helps in the formation of peptide bonds? 

Ans: Peptidyl Transferase

15. Who experimentally proved that DNA replication is semi-conservative. 

Ans: Meselson & stahl.

16. What is a codon?

Ans: Triplet sequence of bases that codes for a single amino is called a codon.

17. Name the three nonsense 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.

Short Answer Questions                                                                                    2 Marks

1. The process of termination during transcription in a prokaryotic cell is being represented here. Name the label a, b, c, and d.

Ans: (a) DNA molecule

(b) mRNA transcript

(c) RNA polymers

(d) Rho factor

2. Complete the blanks a, b, c and d on the basis of the 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 RNA which are complementary  to each other and hence two different polypeptides will be produced; If  a segment of DNA produces two polypeptides, the genetic information machinery becomes complicated.

To identify criminals in the forensic laboratory.

To determine the real or biological parents in case of disputes.

To identify racial groups to rewrite biological evolution. (Any two)

(b) The two complementary RNA molecules (produced simultaneously)would form a double-stranded RNA rather than getting translated into polypeptides.

(c)  RNA polymerase carries out the polymerization in 53 direction and hence 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 identify the real or biological parents in case of disputes.

(iii) To identify racial groups to rewrite the biological evolution.

5. State the 4 criteria which a molecule must fulfill to act as genetic material.

Ans: (i) Replica should be generated.

(ii) It should be chemically and structurally stable.

(iii) It should be able to express itself in the form of Mendelian characters.

(iv) It 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 the statement?

Ans: DNA polymerase plays a dual function –it helps in the synthesis of new strands & also helps in proofreading i.e replacement of RNA strands lay DNA fragments.

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

Ans: UAG UAA & UGA are the three codons that are not recognized by tRNA; these are known as stop codons or nonsense codons. Since these three codons are not recognized by any tRNA they help in the termination of the 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 would be formed hence genetic machinery would become complicated

II) 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.

9. Why is it essential that tRNA binds to both amino acids & mRNA codons 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 specifically activated amino acid from the cytoplasm & further it is transferred to the ribosomal within the cytoplasm where proteins are synthesized. It attracts itself to the ribosome with the sequence specified by mRNA & finally, it transmits its amino acid to a new polypeptide chain.

10. What is a peptide bond? How is it formed?

Ans: A peptide bond is formed between the carboxylic group (COOH) of the first amino acid & the amino group (-NH 2 ) of the second amino acid. This reaction is catalyzed by peptidyltransferase.

11. Explain what happens in frameshift mutation? Name one disease caused by the disorder?

Ans: A frameshift mutation is a type of mutation where the addition or deletion of one or two bases changes the reading from the site of mutation, resulting in a protein with different sets of amino acids.

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.

Central Dogma of Biology

13. Give two reasons why both the strands are not copied during transcription?

Ans: i) If both the strands code for RNA, two different RNA molecules & two different proteins are formed hence genetic machinery would be complicated.

ii) The two RNA molecules which are complementary to each other would bind together to form ds-RNA.

14. Why is the human Genome project considered a megaproject? 

Ans: The Human Genome project was called a mega project for the following facts.

The human genome has approximately 3.3 x 109bp if the cost of sequencing is the US g billion.

If the sequence which was obtained to be stored in a typed form in books & if each page contained 1000 letters & each book contained 1000 pages then 3300 such books would be needed to store complete information.

The enormous quantity of data expected to be generated also necessitates the use of high-speed computational devices for data storage, retrieval & analysis.

15. Why is DNA & not RNA the genetic material in the 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 the majority of organisms.

16. Mention any four important characteristics of the genetic code. 

Ans: Genetic codon has the following important 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 and more than one codon is said to be degenerative.

Codons are read in a continuous manner in direction & have no punctuation.

17. Why is it that transcription & translation could be coupled in the prokaryotic cells but not in eukaryotic cells?

Ans: In prokaryotes the mRNA synthesized does not require any processing to become active both transcription & translation occurs in the same cytosol but In Eukaryotes, the primary transcript contains both exon & intron & is subjected to a process called splicing where introns are removed or exons are joined in a definite order to form mRNA.

Short Answer Questions                                                                                    3 Marks

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

Ans: Below given are the six points of difference between DNA and RNA in their structure/chemistry and function:

DNA	RNA
Double-stranded molecules	Single-stranded molecules
Thymine as pyrimidine base	Uracil is pyrimidine base
Pentose sugar is Deoxyribose	Sugar is ribosome.
Quite stable and not very reactive	2´-OH makes it reactive
Dictates the synthesis of Polypeptides	Perform their functions in protein synthesis.
Found in the nucleus.	They are transported into the cytoplasm

2. Explain how the hnRNA becomes the mRNA. OR Explain the process of splicing, capping, and tailing which occur during transcription in Eukaryotes.

Ans : hnRNA is the precursor of mRNA. It undergoes:

Splicing: Introns are removed and exons are joined together.

Capping: an unusual nucleotide (methyl guanosine triphosphate is added to the 5´ end of hnRNA.

Adenylate residues (200-300) are added at the 3´ end of hnRNA.

3.  Name the three major types of RNAs, specifying the function of each in the synthesis of the 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 amino acid to the site of protein synthesis.

(iii) rRNA (Ribosomal RNA): Plays the structural and catalytic role during translation.

4. Enlist the goals of the Human genome project.

Ans: The Human Genome Project (HGP) is an international scientific research project whose goal is to determine the sequence of chemical base pairs that make up human DNA, as well as to discover and map all of the human genome's genes, both physically and functionally.

5. A tRNA is charged with the amino acid methionine.

i. Give the anticodon of this tRNA.

ii. Write the Codon for methionine.

iii.  Name the enzyme responsible for the binding amino acid to tRNA. 

Ans: Amino-acyltRNAsynthetase

6. Illustrate schematically the process of initiation, elongation, and termination during transcription of a gene in a bacterium.

Ans: During translation in bacteria, mRNA serves as a template, tRNA transports amino acids and reads the genetic information, and rRNAs serve as structural and catalytic components.

There is a single DNA-dependent RNA polymerase that catalyzes the transcription of all types of RNA in bacteria.

RNA polymerase connects to the promoter and starts transcription (Initiation). It also helps the helix open and continues elongation in some way.

The nascent RNA, as well as the RNA polymerase, falls off once the polymerases approach the terminator region. This results in termination.

Transcription of a Bacterial Gene

7. What is transformation? Describe Griffith’s experiment to show transformation? What did he prove from his experiment?

Ans: Transformation means a change in the genetic makeup of an individual. Fredrick Griffith conducted a series of experiments on streptococcus pneumoniae. He observed two strains of this bacterium –one forming smooth colonies with capsule (s-type) & the other forming rough colonies without capsule (R-type).

When live s-type cells are injected into mice, they produce pneumonia & mice die.

When liveR-type cells are injected into mice, the disease was not produced and did not appear.

When heat-killed S-type cells were injected into mice, the disease did not appear.

When heat-killed S-type cells were mixed with live R-cells & injected 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 the transfer of genetic material.

8. The base sequence on one strand of DNA is ATGTCTATA

(i) Give the base sequence of its complementary strand.

Ans: TAC AGATAT

(ii) If an RNA strand is transcribed from this strand what would be the base sequence of RNA?

Ans: UACA GUAU

(iii) What holds these base pairs together?

Ans: Hydrogen bonds hold these base pairs together. Adenine & thymine are bounded by two hydrogen 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 by identifying the real biological father?

Ans: This case to identify the real biological father could have settled lay DNA – fingerprinting technique. In this technique:-

First of all, the DNA of the two claimants who have to be tested is isolated.

Isolated DNA is then digested with a suitable restriction enzyme & digest is subjected to gel electrophoresis.

The fragments of ds DNA are denatured to produce ss DNA by alkali treatment.

The electrophoresed DNA is then transferred from getting into a nitrocellulose filter paper where it is fixed.

A known sequence of DNA is prepared called probe – DNA & is labeled with radioactive isotope 32p & then the probe is added to nitrocellulose paper.

The nitrocellulose paper is photographed on X-ray film through autoradiography. The film is analyzed to determine the presence of hybrid nucleic acid.

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

10. The length of DNA in a eukaryotic cell is N 2.2 m How can such a huge DNA be packaged in a nucleus of micrometer in diameter.

Ans: In eukaryotes, the DNA is wrapped around a positively charged histone octamer into a structure called a nucleosome. An atypical nucleosome consists of 200bp of DNA helix.

The repeating units of nucleosomes form chromatin fibers.

These chromatin fibers condense at the metaphase stage of cell division to form chromosomes. The packaging of chromatin at a higher level requires an additional set of proteins called non-histone chromosomal proteins thus in the nucleus, certain regions of the chromatin are loosely packed & they Stain lighter than the other region, these are called euchromatin. The other regions are lightly packed & they stain darker & is called heterochromatin.

11. A tRNA is charged with amino acid methionine.

i. At what site in the ribosome will the tRNA bind?

Ans: P- site

ii. Give the anticodon of this tRNA?

iii. What is the mRNA codon for methionine?

iv. Name the enzyme responsible for this binding?

Ans: Aminoacyl tRNA synthetase

12. Describe the continuous & discontinuous Synthesis of DNA?

Ans: Synthesis of a new strand of DNA takes place to lay the addition of fresh nucleotides to the 3 – OH group of the last nucleotide of the primer. This synthesis takes place in 5 direction enzymes that catalyze this DNA – polymerase synthesis of a strand called leading strands is continuous. The replication of the second strand of the DNA molecule is discontinuous on a strand called a lagging strand.

Primase initiates primer synthesis on the strand near the fork. The RNA – primer thus formed provides free for replication of single-stranded regions on the lagging strand; the new complementary strand is formed in small fragments of DNA called Okazaki fragments. It is called discontinuous because it has to be initiated several times & every time an Okazaki fragment is produced.

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 that brings the genetic information of DNA transcribed on it for protein synthesis.

Transfer RNA (tRNA):- It has a clover leaf-like structure that acts as an adapter a molecule that contains an “anticodon loop” on one end that reads the code on one hand &” an amino acid acceptor end that binds to the specific amino acid on the other hand.

Ribosomal RNA (rRNA):- Ribosomes provide the site for the synthesis of protein & catalyze the formation of the peptide bond.

14. Define bacterial transformation? Who proved it experimentally & how?

Ans: The transformation is a mode of exchange or transfer of genetic information between organisms or from one organism to another.

Fredrick Griffith tested the virulence of two strains of Diplococci to show transformation in the following steps:-

When S-III strains of bacteria are injected into mice. It developed pneumonia & died.

When R-II strains are injected into mice, they do not develop pneumonia & survive.

When heat-killed S-III strains of bacteria are injected into mice, No symptoms of pneumonia develop & mice remain healthy.

When a mixture of heat-killed S-III strain & live R-II strain is injected into mice, they develop pneumonia & die.

From these results, Griffith concluded that the presence of heat-killed S-III bacteria must convert living R-II type bacteria to type S-III so as to restore the capacity for capsule formation. This was called “BACTERIAL TRANSFORMATION”

S strain →Inject into mice R strain→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

Long Answer Questions                                                                                    5 Marks

1. What is meant by semi-conservative replication? How did Meselson and Stahl prove it experimentally?

Ans: An experiment was performed by  Meselson and Stahl using E.coli to prove that DNA replication is semi-conservative in nature. They grew E.coli in a medium. Then heavy DNA was separated 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 extracted DNA, after two generations, consisted of equal amounts of light and hybrid DNA. Thus, it was 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' 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 RNA polymerase from transcribing.

Inducer: Lactose is the inducer that inactivates the repressor by binding to it. Allows access for the RNA polymerase to the structural gene and transcription.

3. What is an operon? Describe the major steps involved in an operon?

Ans: An operon is a group of controllers & structural genes which control the catabolism of the cell genetically lactose operon/lac operon.

When inducer or lactose is absent:-

The lac regulator gene synthesizes a repressor protein by transcription & translation. This repressor protein binds with the operator site of lac operon & blocks RNA polymerase. Thus, RNA polymerase is unable to transcribe mRNA & structural genes are unable to translate enzyme B- galactosidase.

When inducer or lactose is present:

The lac regulator gene transcribes mRNA & synthesizes active lac repressor protein & at the same time lactose is converted into isomer allolactose. Allolactose binds to an active lac repressor due to which it is converted to an inactive repressor. This inactive repressor is released from the operator site of lac operon & RNA polymerase binds to promoter & starts to transcribe mRNA & forms β-galactosidase which converts lactose into glucose vs galactose.

Thus, the presence of lactose determines whether or not lac. The repressor is bound to the operator & genes are expressed on not.

4. What do you mean semi-conservative nature of DNA replication? Who proved it & how?

Ans: The semiconservative nature of DNA replication suggested that during replication two strands would separate & 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 & one newly synthesized strand thus, half the information is conserved over a generation. Mathew Messselson& Franklin Stahl have performed an experiment using Escherichia coli to prove that DNA replication is semiconservative. They grew E.coli in a medium containing 15NH 4 Cl until 15N was incorporated in the two strands of newly synthesized DNA this heavy DNA can be separated from normal DNA lay centrifugation in CsCl density gradient. Then they transferred the cells into a medium with normal X 14 NH 4 Cl & took samples at various time intervals & extracted DNA & centrifuged them to measure their densities. The DNA is extracted from the cells after one generation to transfer from X 15 N medium to X 14 N

5. Where do transcription & translation take place in a prokaryotic cell? Describe the three steps involved in translation?

Ans: In a prokaryotic cell, both transcription & translation occurs in the cytoplasm. It consists of the following steps:-

Activation of Amino Acids:- amino acids are activated in the presence of ATP lay enzaminoacy tRNA Synthetase.

Binding of Activated Amino Acid With tRNA: - Activated amino acids bind with specific tRNA to form charged tRNA.

Initiation of Polypeptide Chain:- Initiation codon is AUG which codes for methionine. The initiation codon of mRNA binds to the p-site of the ribosome with the help of initiation factors.

Elongation of Polypeptide Chain:-

 Second, activated amino acid along its tRNA reaches the ‘A’ site & binds to mRNA codon next to AUG.

 A peptide bond is formed between two amino acids by a peptidyl transferase.

Ribosomes translate mRNA in -direction due to which free tRNA slips away &peptidyltRNA reaches P – site. Now the third amino acid reaches the A – site & the process continues.

 Termination of Polypeptide Chain: - When a termination codon (UAA, UAG, UGA) reaches at A- site translation terminates Since there is no specific tRNA for these codons.

6. Who performed the blender experiment? What does this experiment prove? Describe the steps followed in this experiment?

Ans: The proof for DNA as the genetic material came from the experiments of Hershey & chase who worked with bacteriophages.

The bacteriophage on infection injects only the DNA into the bacterial cell & not the protein coat.

Bacterial cells treat the viral DNA as their own & subsequently manufacture more virus particles.

They grew some viruses on a medium that contained radioactive Phosphorus & some others on a medium that contained radioactive sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not proteins because DNA contains phosphorus. Similarly, viruses grown on radioactive sulfur contain radioactive protein because DNA does not contain sulfur.

Radioactive phages are allowed to infect E. coli bacteria & soon after infection the cultures were gently agitated in a blender to separate the adhering protein coat of the virus from the bacterial cells. It was found that when a phage containing radioactive DNA was used to infect the bacteria its radioactivity was found in bacterial cells indicating that DNA has been injected into the bacterial cells. So, the DNA is the genetic material & not proteins.

Important Questions Class 12 Biology Chapter 6

Molecular basis of inheritance important questions.

Vedantu has compiled crucial issues on the molecular basis of heredity in the form of a PDF. This PDF is highly recommended and desired for making final test and entrance exam preparation easier. Vedantu has worked hard to compile all of the supplementary questions from Chapter 6 Biology Class 12 so that no information is missed when learning. Important class 12 bio ch 6 questions give a conceptual comprehension of this chapter. The produced PDF is convenient and can thus be simply accessible in any system. Also, printouts can be taken so that they can be conveniently analysed. Important Class 12 Biology Chapter 6 questions may also be learned through Vedantu's live tuition courses.

Molecular Basis of Inheritance Class 12 Questions - Summary

DNA (Deoxyribonucleic Acid)

In most species, DNA is the genetic material, except for certain viruses that have an RNA genome. An example of the same may include the TMV (Tobacco mosaic virus).

RNA functions primarily as a messenger, an adaptor and have a catalytic role. The length of DNA specifies the nucleotides or number of base pairs (bp).

Structure of a Polynucleotide Chain

The polynucleotide chain structure is composed of three basic elements:

Nitrogenous Base:

Purines - The presence of Adenine (A) and Guanine (G) is observed in both DNA and RNA. 

Pyrimidine - Cytosine and Uracil in RNA and Cytosine (C) and Thymine (T) in DNA. Thymine can also be termed as 5-methyl uracil and is responsible for greater DNA molecule stability.

Sugar: 

Pentose sugar- As the name suggests, Ribose in RNA (ribonucleic acid), deoxyribose in DNA 

Phosphate Group:

Nucleoside: a nitrogen base connected by the N-glycosidic bond to the hydroxyl group of 1 'C pentose sugar

Nucleotide: The phosphate group is bound by a phospho-ester bond to the hydroxyl group which is present at 5 'C of the nucleoside.

Double Helix Model Given for the DNA Structure

In 1953, Watson and Crick suggested the DNA double-helix structure.

The ratio of adenine and thymine to that of guanine and cytosine is one and tends to remain constant. This was stated by Ervin Chargaff.

DNA is formed by two polynucleotide chains, where the backbone is sugar-phosphate and bases are present within.

There exists opposite polarity between the two chains, i.e. and the one with 3'-5' polarity other with 5'-3' polarity.

The base pair (bp) is created by the formation of hydrogen bonding amongst the nitrogen bases that present on two of the polypeptide chains.

To make a base pair, a purine base of one nucleotide chain is often connected to a pyrimidine base of another nucleotide chain or vice versa.

Two hydrogen bonds (A=T) pair Adenine with Thymine (or Uracil in RNA) while three hydrogen bonds pair Guanine with Cytosine (G).

Replication

Watson and Crick have stated that DNA replication is semi-conservative in nature. It was experimentally proven by Meselson and Stahl, in 1958. The replication of DNA begins from the region called the origin of replication and results in the formation of a replication fork. It leads to the formation of two strands, namely leading and lagging strands. Moreover, there occurs the formation of some fragments called as the Okazaki fragments. 

Transcription

The first step in gene expression is transcription. It requires copying the DNA sequence of a gene to create an RNA molecule. This process is carried out by the enzyme termed as RNA polymerase, which attaches nucleotides to form an RNA chain, and are transcribed (using a DNA strand as a template). There are three levels of transcription: initiation, elongation, and termination.

Genetic Code

Amino acids can be defined as the sequence of bases present within the mRNA and are responsible for coding for a specific amino acid.

Each of the code is composed of three nucleotides and are termed as a triplet. Apart from the codons of some protozoans and mitochondrial, codons are almost universal.

More than one triplet can code for one particular type of amino acid and thus the code is said to have degenerated.

In total, there are 64 codons, 61 of which code for amino acids.

There are three different stop codons, named - UAA, UAG, UGA that would not code for any amino acids.

Along with being a starting codon, AUG also codes for methionine.

Point Mutation : The shift in the single base pair leads in the points mutation, e.g. The consequence of a point mutation in the gene coding for the β-globin chain is sickle cell anaemia. As a consequence, in the sickle cell, glutamate in the regular protein is transformed to Valine.

Frameshift Mutation: If one or two base pairs are lost or gained, the reading frame shifts at the point of insertion or deletion, leading to the frameshift mutation.

Translation

The translation is the method of translating a messenger RNA (mRNA) molecule sequence during protein synthesis into a chain of amino acids. The process of translation is completed in four steps, namely: activation, initiation, elongation and termination. These words characterize the development of the chain of amino acids (polypeptide). Amino acids are transferred to ribosomes and placed together into proteins.

Human Genome Project

To decode the full DNA sequence of the human genome, a project was launched, named the Human Genome Project (HGP), in 1990.

The DNA section was separated and cloned for the determination of the DNA sequence using genetic engineering techniques.

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

DNA Fingerprinting

DNA fingerprinting is a kind of test which represents the genetic makeup of an individual or any other living thing. It can be defined as the technique used to create a correlation in a criminal investigation of a suspect and biological evidence. A sample of DNA obtained from a crime scene is matched with a suspect's DNA sample. If there's a match between the two DNA profiles, then the evidence comes from that perpetrator.

Molecular basis of inheritance Class 12 questions are very important for the exam point of view. In this chapter, we studied about DNA (Deoxyribonucleic acid), Structure of a polynucleotide chain, Double Helix Model given for the DNA Structure, replication, transcription, genetic code, mutation, translation, human genome project and DNA fingerprinting. Download the Extra questions of Chapter 6 Biology Class 12 from our site for better result.

Important Related Links for CBSE Class 12 Biology

CBSE Class 12 Biology Study Materials

FAQs on Important Questions for CBSE Class 12 Biology Chapter 6 - Molecular Basis of Inheritance 2024-25

1. What are the important questions in Chapter 6 of Class 12 Biology?

Being familiar with important questions is greatly beneficial to students because these questions cover topics that are of importance for the exams. For Chapter 6 of Biology in Class 12, there are a lot of websites that offer PDFs of these important questions, and of them, one of the best can be downloaded from Vedantu’s official website. These topics cover the important questions in Chapter 6 of Class 12 Biology:

Structure of Polynucleotide Chain

2. What is DNA polymorphism?

DNA Polymorphism is a process of variation in DNA that arises because of mutations happening at non-coding sequences, that is, DNA Polymorphism is the different sequences of DNA in living organisms. There are a whole lot of variations happening at the DNA level like base pair changes, repeated sequences, etc. The topic of DNA Polymorphism constitutes one of the important questions usually asked in the Biology Board exam from chapter 6. The detailed answer is available on Vedantu’s website. These solutions are available at free of cost on Vedantu(vedantu.com) and mobile app.

3. What is DNA fingerprinting according to Chapter 6 of Class 12 Biology?

In a procedure known as DNA fingerprinting, DNA is often compared from many sources to determine an identity. This is essentially a chemical test used to examine a person's or other living organism's genetic code. The test detects variations in certain sections of the DNA sequence. This is what is known as repetitive DNA. DNA fingerprinting is often used to identify bodies, provide evidence, find remedies for ailments, and so on.

4. What is molecular inheritance?

The notion of molecular inheritance is covered in Chapter 6 of Class 12 Biology. This portion is included in the Important Questions that students are given. These questions are written in the most straightforward manner possible so that students may easily follow them. The study of genes in the body, including DNA and its numerous activities such as replication, transcription, translation, genetic coding, control, and others, is known as molecular inheritance. This notion explains why offspring resemble their parents.

5. How many questions of NEET are from ‘Molecular Basis of Inheritance’?

Biology is one of the most important and concentrated courses on the NEET test. As a result, preparation for this topic must be faultless, and students may do so by referring to crucial questions produced particularly for biology, in this case, for chapter 6 titled "Molecular Basis of Inheritance." The subject comprises 45 questions in total, with this chapter having a weightage of roughly 8% to 9% and being included as a sub-topic of Genetics and Evolution.

CBSE Class 12 Biology Important Questions

Cbse study materials.

• Biology Important Questions
• Class 12 - Biology
• Chapter 6: Molecular Basis Inheritance

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

Genes are the basic unit of heredity. Most of the genes comprises strands of genetic material called DNA. DNA comprises all the hereditary information of an individual. This information is passed on from one generation to the other in the form of homologous chromosomes. The DNA is transcribed into mRNA and translated into proteins. This is known as central dogma.

Explore important questions on the Molecular Basis of Inheritance for a better understanding of the concept.

Very Short Answer Type Questions

Q.1. State the function of histones in DNA packaging .

• They play a role in gene regulation.
• They help the DNA to wind around it.
• The histones are positively charged proteins, which can easily bind to the negatively charged DNA.

Also Read   DNA packaging .

Q.2. What is the difference between heterochromatin and euchromatin ?

Heterochromatin is a tightly packed DNA, which can be identified when stained in an extreme nuclear stain.

Euchromatin is a lightly packed DNA, which can be identified when stained in a less nuclear stain.

More Details:   Difference between heterochromatin and euchromatin

Q.3. Name any three viruses with RNA as the genetic material.

A.3. The viruses in which the genetic material is RNA is called the RNA virus. The three examples of the RNA virus.

• Influenza Virus.
• Hepatitis C Virus.
• Human Immunodeficiency Virus.

Explore More: Virus

Q.4. Give a reason for the discontinuous synthesis of DNA on one of the parental strands?

A.4. The biological process of DNA synthesis naturally occurs in 5′ to 3′ direction. In the double-stranded DNA, the strands are parallel and antiparallel to each other. During the synthesis of DNA, both the strands act as templates and only one (3′ to 5′ direction) can synthesize the parallel strand in 5’→3′ direction. The other strand 5′ to 3′ is synthesized in the opposite direction producing small stretches of DNA known as Okazaki fragments. This is the reason for the discontinuous synthesis of DNA on one of the parental strands.

Q.5. The sequence of the coding strand of DNA in a transcription unit is mentioned below.

3′ AATGCAGCTATTAGG 5′

Write the sequence for:

• Its complementary strand
• The complementary strand is  5′ TTACGTCGATAATCC 3′
• The mRNA is  5′ AAU GCAGCUAUUAGG 3′

Q.6. What is DNA polymorphism?

A.6. DNA’s polymorphism is the variation in the DNA sequence arising due to mutation at non-coding sequences.

Q.7. Retroviruses do not follow central dogma.  Comment on this statement

A.7. Retroviruses do not follow central dogma, because, they possess RNA as genetic material instead of the DNA, which is later converted into DNA by the enzyme reverse transcriptase.

Explore more: Retroviruses

Q.8. Sometimes, the young ones born have an extremely different set of eyes or limbs. Give a relevant explanation for the abnormality.

A.8. This abnormality is caused by many factors, including alcohol abuse by the mother during her pregnancy, medicine side effects or reactions caused to the womb,  environmental factors, such as maternal exposure to the chemicals, radiations, virus, and it can also be due to the genes and non-coordination in the regulation of expression in the set of genes associated with the development of organs.

Q.9. Explain about the dual polymerase present in E.coli.

A.9. The DNA polymerase present in E.coli is a DNA dependent polymerase. This DNA polymerase helps in the:

• Replication process.
• Performs the 5′ to 3′ polymerase activity as well as 3′ to 5′ exonuclease activity.
• The DNA polymerase III also has the ability to proofread the wrong nucleotides and substitutes it with the correct one.

Q.10. What are the functions of the :

• Methylated guanosine cap
• Poly-A tail

Methylated guanosine cap plays a primary role in the attachment of the mRNA to the smaller sub-units of the ribosome during translation initiation.

The Poly-A tail functions by increasing the length of the mRNA and also provides longevity to the mRNA.

Q.11. Mention any two functions of AUG codon.

A.11. The AUG codon is also called the start codon. The two important functions of AUG codon include:

• It codes for methionine.
• It acts as an initiation codon for protein synthesis.

Q.12. What is the function of amino acyl t-RNA synthase?

A.12.  Amino acyl t-RNA synthase plays a major role in the biosynthesis of proteins by attaching an appropriate amino acid on to the tRNA molecules.

Long Answer Type Questions

Q.1. Enumerate the post-transcriptional modifications in a eukaryotic mRNA.

A.1. Transcription is the process of conversion of DNA to mRNA. The post-transcriptional modifications include:

• Capping at 5’-end
• Poly-A tail at 3’-end
• mRNA splicing

The 5′ cap protects the RNA from ribonuclease. The poly-A tail protects the mRNA from enzymatic degradation. The introns are spliced during mRNA splicing and the exons are joined together to form a continuous sequence that codes for a functional protein.

Refer more:   Transcription

Q.2. Explain the process of translation.

A.2. The translation is the process of protein synthesis in which the mRNA is used to synthesize proteins. The mRNA sequence is decoded to specify the amino acid of a polypeptide. The process of translation is carried out in the following steps:

• Initiation.
• Elongation.
• Termination.

Also Read :  Translation

Q.3. Explain the process of DNA fingerprinting .

A.3. DNA fingerprinting is a technique that is used to analyze the genetic makeup of living beings. It is widely used for DNA analysis in forensic tests and paternity tests to identify the biological parents of the child, and also to identify the criminal during forensic investigations.

Extended Reading: DNA fingerprinting

Q.4. What is an operon? Explain an inducible operon.

A.4. An operon is the functional unit of DNA that contains a cluster of genes controlled by a single promoter. It consists of the following components:

• The DNA fragment that transcribes the mRNA.
• Regulator gene that codes for a repressor protein.
• Inducer that prevents the repressor from binding to the operator.
• A promoter where the RNA polymerase binds and initiates the transcription.
• An operator that is a DNA sequence adjacent to the promoter where the repressor protein binds.

The lac operon of E.coli is an inducible operon.

Q.5. Explain the process of DNA replication .

A.5. DNA replication is a biological process of producing two identical strands of DNA from the original strand. The original strand is known as the parent strand and the new strands are known as the daughter strands. This is achieved by a number of enzymes such as DNA polymerase, helicase, primase, topoisomerase, and ligase.

Read More:  DNA replication

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Molecular Basis Of Inheritance Class 12 Important Questions with Solutions

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Assertion Reason Questions for Biology Chapter 28 Molecular Basis of Inheritance

• Last modified on: 3 years ago
• Reading Time: 14 Minutes

Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as: (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true but Reason is false. (d) If both Assertion and Reason are false.

Q.1. Assertion : In a DNA molecule, A–T rich parts melt before G–C rich parts. Reason : In between A and T there are three H–bond, whereas in between G and C there are two H-bonds. [AIIMS 2010, 2015]

Q.2. Assertion: The two chains of DNA have anti-polarity. Reason: In one chain of DNA, ribose sugar at 5′ end consists of a free phosphate moiety while at the other end the ribose has a free 3′ OH group.

Q.3. Assertion: Adenine cannot pair with cytosine. Reason: Adenine and cytosine do not have a perfect match between hydrogen donor and hydrogen acceptor sites. Hence, they cannot pair.

Q.4. Assertion: The sugar phosphate backbone of two chains in DNA double helix show anti-parallel polarity. Reason: The phosphodiester bonds in one strand go from a 3′ carbon of one nucleotide to a 5′ carbon of adjacent nucleotide, whereas those in complementary strand go vice versa.

Q.5. Assertion: DNA is considered to be better genetic material than RNA for most organisms. Reason: 2′-OH group present in DNA makes it labile and less reactive.

Q.6. Assertion : Histones are basic proteins of major importance in packaging of eukaryotic DNA. DNA and histones comprise chromatin forming the bulk of eukaryotic chromosome. Reason : Histones are of five major types H1, H2A, H2B, H3 and H4. [AIIMS 2000]

Q.7. Assertion : DNA is associated with proteins. Reason : DNA binds around histone proteins that form a pool and the entire structure is called a nucleosome. [AIIMS 2013]

Q.8. Assertion: Histones are basic in nature. Reason: Histones are rich in the amino acids lysine and arginine.

Q.9. Assertion: DNA acts as a genetic material in all organisms. Reason: It is a single-stranded biomolecule.

Q.10. Assertion: In Griffith’s experiment, a mixture of heat-killed virulent bacteria R and live non-virulent bacteria S, lead to the death of mice. Reason: ‘Transforming principle’ got transferred from heat killed R strain to S strain and made it virulent.

Q.11. Assertion: Template or antisense strand, having 3′ → 5′ polarity takes part in transcription. Reason: Non-template or sense strand, having 5′ → 3′ polarity, does not take part in transcription.

Q.12. Assertion: The uptake of DNA during transformation is an active, energy requiring process. Reason: Transformation occurs only in those bacteria, which possess the enzymatic machinery involved in the active uptake and recombination.

Q.13. Assertion: Killer strain of Paramecium aurelia can kill sensitive strain. Reason: If sensitive strain is provided kappa particle, it becomes killer.

Q.14. Assertion: Scaffold proteins are nonhistone chromosomal proteins. Reason: They are rich in lysine and arginine.

Q.15. Assertion: Viruses having RNA genome have shorter life span and mutate faster. Reason: RNA is unstable and thus mutates faster.

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Case Based Questions Test: Molecular Basis of Inheritance - 2 - NEET MCQ

15 questions mcq test - case based questions test: molecular basis of inheritance - 2.

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Read the following and answer any four questions from 3(1) to 3(v) given below: DNA replication is a complex multistep process that requires enzymes, protein factors and metal ions. DNA replication in eukaryotes occurs in the nucleus during the S-phase of the cell cycle. It is semi discontinuous in eukaryotes. In prokaryotes, replication takes place in the cytoplasm. DNA replication in bacteria occurs prior to fission. Nucleoid or viral chromosome is a single molecule of nucleic acid, it may be linear or circular. Nucleic acid in a virus is either DNA or RNA but never both. In viral DNA, how many origins of replication are present?

• C. Multiple

Read the following and answer any four questions from 2(i) to 2(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. Name the linkage present between the nitrogen base and pentose sugar in DNA.

• A. Phosphodiester bond
• B. Glycosidic bond
• C. Hydrogen bond
• D. None of these

Read the following and answer any four questions from 3(1) to 3(v) given below: DNA replication is a complex multistep process that requires enzymes, protein factors and metal ions. DNA replication in eukaryotes occurs in the nucleus during the S-phase of the cell cycle. It is semi discontinuous in eukaryotes. In prokaryotes, replication takes place in the cytoplasm. DNA replication in bacteria occurs prior to fission. Nucleoid or viral chromosome is a single molecule of nucleic acid, it may be linear or circular. Nucleic acid in a virus is either DNA or RNA but never both. Select the main enzyme involved in DNA replication.

• A. DNA ligase
• B. DNA dependent DNA polymerase
• C. Topoisomerase
• D. Helicase

Read the following and answer any four questions from 2(i) to 2(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 he double helix structure of DNA was proposed by

James Watson and Francis Crick

Earwin Chargaff

Federick Griffith

Hershey and Chase.

The correct structure of D N A 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 pairing and study of X-ray diffraction patterns of DNA produced by Maurice Wilkins and Rosalind Franklin which helped Watson and Crick to design the 3-dimensional structure of DNA.

Read the following and answer any four questions from 3(1) to 3(v) given below: DNA replication is a complex multistep process that requires enzymes, protein factors and metal ions. DNA replication in eukaryotes occurs in the nucleus during the S-phase of the cell cycle. It is semi discontinuous in eukaryotes. In prokaryotes, replication takes place in the cytoplasm. DNA replication in bacteria occurs prior to fission. Nucleoid or viral chromosome is a single molecule of nucleic acid, it may be linear or circular. Nucleic acid in a virus is either DNA or RNA but never both.

DNA strand, built up of Okazaki fragments is called

• A. lagging strand
• B. leading strand
• C. complementary strand
• D. parental strand.

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

• A. right and left part
• B. major and minor grooves
• C. upper and lower sides
• D. linear and circular part.

Select the incorrect statement about DNA polymerase in eukaryotes.

• A. Polymerase a is required for initiation of replication.
• B. When the RNA primer gets removed the gap is filled by DNA polymerase β in eukaryotes.
• C. Polymerase ɛ helps in elongation of lagging strand
• D. Polymerase is the largest and main enzyme for DNA replication in eukaryotes.

Read the following and answer any four questions from 4(i) to 4(v) given below: The process o f copying genetic information from a template strand of DNA into RNA is called transcription. It is mediated by RNA polymerase. Transcription takes place in the nucleus of eukaryotic cells. In transcription, only a segment of DNA and only one of the strands is copied into RNA.

Monocistronic structural genes are found in which organisms?

• A. Prokaryotes
• B. Bacteria
• D. Eukaryotes

Read the following and answer any four questions from 5(i) to 5(v) given below: Translation is the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence bases in the mRNA. The amino acids are joined by a bond called peptide bond. Ribosome is the site of protein synthesis.

Which ion is essential for association of both units of ribosome at the time of protein formation?

Which enzyme helps in tailing or polyadenylation?

• A. Poly-A polymerase
• B. Exonucleases
• C. RNA polymerase I
• D. RNA polymerase II

During translation, how many initiation factors are required in eukaryotes for initiation reactions?

Read the given list of materials.

1. RNA polymerase enzyme

2. DNA template

3. RNA primers

4. Okazaki segments

5. Four types of ribonucleotides triphosphates

6. Divalent metal ions Mg 2+ as a cofactor.

Which of the above given materials are required for transcription?

• A. (1), (2), (3) and (4)
• B. (1), (2), (3), (5) and (6)
• C. (1),(2), (5) and (6)
• D. All of these

Name the enzyme that helps in combining amino acid to its particular tRNA.

• A. Activating enzyme
• B. Amino-acyl /TRNA-synthetase
• C. Peptidyl transferase
• D. Both (a) and (b)

Read the following and answer any four questions from 6(i) to 6(v) given below: I he process of translation requires transfer of genetic information from a polymer of nucleotides to synthesise a polymer of amino acids. Ihe 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.

What is a codon?

• A. A length of DNA which codes for a particular protein.
• B. A part of the fRNA molecule to which a specific amino acid is attached.
• C. A part of the fRNA molecule which recognises the triplet code on the messenger RNA.
• D. A part of the messenger RNA molecule that has a sequence of bases coding for an amino acid.

From the given list, select the translation machinery.

2. Ribosomes

3. Amino acids

5. Peptidyl transferase

6. Aminoacyl tRNA synthetase

7. Pyrophosphatase

• A. (1), (2), (3), (4) and (6)
• B. (1), (2), (3), (4) and (5)
• C. (1), (2), (3), (4), (5) and (6)
• D. (1), (2), (3), (4), (5), (6) and (7)

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Biology MCQs for Class 12 with Answers Chapter 6 Molecular Basis of Inheritance

November 15, 2019 by Veerendra

Free PDF Download of CBSE Biology Multiple Choice Questions for Class 12 with Answers Chapter 6 Molecular Basis of Inheritance. Biology MCQs for Class 12 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Biology Molecular Basis of Inheritance MCQs Pdf with Answers to know their preparation level.

Molecular Basis of Inheritance Class 12 Biology MCQs Pdf

Question 1. In a DNA strand the nucleotides are linked together by (a) glycosidic bonds (b) phosphodiester bonds (c) peptide bonds (d) hydorgen bonds. Answer: (b) phosphodiester bonds

Question 2. The net electric charge on DNA and histones is (a) both positive (b) both negative (c) negative and positive, respectively (d) zero. Answer: (c) negative and positive, respectively

Question 3. Which of the following statements is the most appropriate for sickle cell anaemia ? (a) It cannot be treated with iron supplements. (b) It is a molecular disease. (c) It conferes resistance to acquiring malaria. (d) All of the above. Answer: (d) All of the above.

Question 4. The first genetic material could be (a) protein (b) cabohydrates (c) DNA (d) RNA. Answer: (d) RNA.

Question 5. The human chromosome with the highest and least number of genes in them are respectively (a) chromosome 21 and Y (b) chromosome 1 and X (c) chromosome 1 and Y (d) chromosome X and Y. Answer: (c) chromosome 1 and Y

Question 6. Who amongst the following scientist had no contribution in the development of the double helix model for the structure of DN A ? (a) Rosalind Franklin (b) Maurice Wilkins (c) Erwin Chargaff (d) Meselson and Stahl Answer: (b) Maurice Wilkins

Question 7. Which of the following steps in transcription is catalysed by RNA polymerase ? (a) Initiation (b) Elongation (c) Termination (d) All of the above Answer: (d) All of the above

Question 8. Control of gene expression takes place at the level of (a) DNA-replication (b) transcription (c) translation (d) none of the above. Answer: (b) transcription

Question 9. Which was the last human chromosome to be completely sequenced ? (a) Chromosome 1 (b) Chromosome 11 (c) Chromosome 21 (d) Chromosome X Answer: (a) Chromosome 1

Question 10. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called (a) A – DNA (b) B – DNA (c) cDNA (d) rDNA. Answer: (c) cDNA

Question 11. If the sequence of initrogen bases of the coding strand of DNA in a transcription unit is: 5’ – ATGAATG – 3’, the sequence of bases in its RNA transcript would be (a) 5’ – AUG A AUG – 3’ (b) 5’ – UACUU AC – 3’ (c) 5’ – CAUUCAU – 3’ (d) 5’ – GUAAGUA – 3’. Answer: (d) 5’ – GUAAGUA – 3’.

Question 12. The RNA polymerase holocnzyme transcribes (a) the promoter, structural gene and the terminator region. (b) the promoter and the terminator region (c) the structural gene and the terminator region (d) the structural gene only. Answer: (b) the promoter and the terminator region

Question 13. If the base sequence of a codon in mRNA is 5’ – AUG – 3′ the sequence of tRNA pairing with it must be (a) 5’ – UAC – 3’ (b) 5’ – CAU – 3’ (c) 5’-AUG – 3’ (d) 5’ – GUA – 3’ Answer: (b) 5’ – CAU – 3’

Question 14. The amino acid attaches to the tRNA at its (a) 5’- end (b) 3’ – end (c) anticodon site (d) DHUloop. Answer: (b) 3’ – end

Question 15. To initiate translation, the wiRNA first bind to (a) the smaller ribosomal sub-unit (b) the larger ribosomal sub-unit (c) the whole ribosome (d) no such specificity exists. Answer: (a) the smaller ribosomal sub-unit

Question 16. In E. colt, the lac operon gets switched on when (a) lactose is present and it binds to the repressor (b) repressor binds to operator (c) RNA polymerase binds to the operator (d) lactose is present and it binds to RNA polymerase. Answer: (a) lactose is present and it binds to the repressor

Question 17. In DNA strand, the nucleotides are linked together by (a) glycosidic bonds (b) phosphodiester bonds (c) peptide bonds (d) hydrogen bonds. Answer: (b) phosphodiester bonds

Question 18. If a double stranded DNA has 20% of cytosine, what will be the percentage of adenine in it ? (a) 20% (b) 40% (c) 30% (d) 60% Answer: (c) 30%

Question 19. If the sequence of bases in one strand of DNA is ATGCATGCA, what would be the sequence of bases on complementary strand ? (a) ATGCATGCA (b) AUGCAUGCA (c) TACTACGT (d) UACGUACGU Answer: (c) TACTACGT

Question 20. How far is each base pair from the next one in DNA double helix model ? (a) 2 nm (b) 3.4 nm (c) 34 nm (d) 0.34 nm Answer: (d) 0.34 nm

Question 21. Synthesis of DNA from RNA is explained by (a) central dogma reverse (b) reverse transcription (c) feminism (d) all of these. Answer: (d) all of these.

Question 22. Histone proteins are (a) basic, negatively charged (b) basic, positively charged (c) acidic, positively charged (d) acidic, negatively charged Answer: (b) basic, positively charged

Question 23. The structure in chromatin seen as ‘beads-on string’ when viewed under electron microscope are called (a) nucleotides (b) nucleosides (c) histone octamer (d) nucleosomes. Answer: (d) nucleosomes.

Question 24. Find out the wrong statement about heterochromatin, (a) It is densely packed (b) It stains dark. (c) It is transcriptionally active. (d) It is late replicating. Answer: (c) It is transcriptionally active.

Question 25. They year 2003 was celebrated as the 50th anniversary of discovery of (a) transposons by Barbare Me Clintock (b) structure of DNA by Watson and Crick (c) Mendel’s laws of inheritance (d) biotechnology by Kary Muliis. Answer: (b) structure of DNA by Watson and Crick

Question 26. The process of transofrmation is not affected by which of the following enzymes ? A. DNase B. RNase C. Peptidase D. Lipase (a) A, B (b) A, B, C, D (c) B, C, D (d) A, B, C Answer: (c) B, C, D

Question 27. The three codons which result in the termination of polypeptide chain synthesis are (a) UAA, UAG, GUA (b) UAA, UAG, UGA (c) UAA, UGA, UUA (d) UGU,UAG,UGA Answer: (b) UAA, UAG, UGA

Question 28. Amino acids which are specified by single codons are (a) phenylalanine and arginine (b) tryptophan and methionine (c) valine and proline (d) methionine and aroinine. Answer: (b) tryptophan and methionine

Question 29. Which out of the following statements is incorrect ? (a) Genetic code is ambiguous. (b) Genetic code is deqenerate. (c) Genetic code is universal. (d) Genetic code is non-overlanning. Answer: (a) Genetic code is ambiguous.

Question 30. Some amino acids are coded by more than one codon, hence the genetic code is (a) overlapping (b) degenerate (c) wobbled (d) unambiguous. Answer: (d) unambiguous.

Question 31. The mutations that involve addition, deletion or substitution of a single pair in a gene are referred to as (a) point mutations (b) lethal mutations (c) silent mutations (d) retrogressive mutations. Answer: (a) point mutations

Question 32. Sickle cell anemia results from a single base substitution in a gene, thus it is an example of (a) point mutation (b) frame-shift muttion (c) silent mutation (d) both (a) and (b). Answer: (a) point mutation

Question 33. Select the incorreclty matched pair. (a) Initation codons – AUG,GUG (b) Stop codons – UAA, UAG, UGA (c) Methionine – AUG (d) Anticodons – mRNA Answer: (d) Anticodons – mRNA

Question 34. Amino acid acceptor end of tRNA lies at (a) 5’ end (b) 3’ end (c) T VC loop (d) DHUloop. Answer: (b) 3’ end

Question 35. Which RNA carries the amino acids from the amino acid pool to mRnA during protein synthesis ? (a) rRNA (b) mRNA (c) /RNA (d) hnRNA Answer: (c) /RNA

Question 36. During translation, activated amino acids get linked to tRNA. This process is commonly called as (a) charging of tRNA (b) discharging of tRNA (c) aminoacylation of tRNA (d) both (a) and (c) Answer: (b) discharging of/RNA

Question 37. To prove that DNA is the genetic material, which radioactive isotopes were used by Hershey and Chase (1952) in experiments ? (a) 33S and 15N (b) 32P and 35S (c) 32P and 15N (d) 14N and 15N Answer: (d) 14N and 15N

Question 38. RNA is the genetic material in (a) prokaryotes (b) eukaryotes (c) Tabacco Mosaic Virus (TMV) (d) E.coli. Answer: (c) Tabacco Mosaic Virus (TMV)

Question 39. Which one among the following was the first genetic material ? (a) DNA (b) RNA (c) Protein (d) Nuclein Answer: (b) RNA

Question 40. Which of the following life processes is evolved around RNA ? (a) Metabolism (b) Translation (c) Splicing (d) All of these Answer: (b) Translation

Question 41. Chemically, RNA is (i) reactive and (ii) stable as compared to DNA. (a) (i) equally, (ii) equally (b) (i) less, (ii) more (c) (i) more, (ii) less (d) (i) more, (ii) equally Answer: (c) (i) more, (ii) less

Question 42. Which of the following phenomena was experimentally proved by Meselson and Stahl ? (a) Transformation (b) Transduction (c) Semi-conservative DNA replication (d) Central dogma Answer: (c) Semi-conservative DNA replication

Question 43. First experimental proof for semi-conservative DNA replication was shown in (a) Streptococcus pneumoniae (b) Escherichia coli (c) Neurospora crassa (d) Rattus rattus. Answer: (b) Escherichia coli

Question 44. Select the correct match of enzyme with its related function. (a) DNA polymerase – Synthesis of DNA strands (b) Helicase – Unwinding of DNA helix (c) Ligase – Joins together short DNA segments (d) All of these Answer: (d) All of these

Question 45. Other than DNA polymerase, which are the enzymes involved in DNA synthesis ? (a) Topoisomerase (b) Helicase (c) RNA primase (d) All of these Answer: (d) All of these

Question 46. DNA replication takes place at _________ phase of the cell cyle. (a) G 1 (b) S (c) G 2 (d) M Answer: (b) S

Question 47. The process of copying genetic information from one strand of DNA to RNA is termed as _________ . (a) replication (b) transcription (c) translation (d) reverse transcription Answer: (b) transcription

Question 48. The enzyme DNA dependent RNA polymerase catalyses the polymerisation reaction in ___________direction. (a) only 5’ → 3’ (b) only 3’ → 5’ (c) both (a) and (b) (d) none of these Answer: (a) only 5’ → 3’

Question 49. If the sequence of bases in coding strand of DNA is ATTCGATG, then the sequence of bases in mRNA will be (a) TAAGCTAC (b) UAAGCUAC (c) ATTCGATG (d) AUUCGAUG. Answer: (d) AUUCGAUG.

Question 50. If the sequence of bases in DNA is GCTTAGGCAA then the sequence of bases in its transcript will be (a) GCTTAGGCAA (b) CGAATCCGTT (c) CGAAUCCGUU (d) AACGGAUUCG. Answer: (c) CGAAUCCGUU

Question 51. Transcription unit (a) starts with TATA box (b) starts with pallendrous regions and ends with rho factor. (c) starts with promoter region and ends in terminator region (d) starts with CAAT region. Answer: (c) starts with promoter region and ends in terminator region

Question 52. During transcription, the site of DNA molecule at which RNA polymerase binds is called (a) promoter (b) regulator (c) receptor (d) enhancer. Answer: (a) promoter

Question 53. Polycistronic messenger RNA (mRNA) usually occurs in (a) bacteria (b) prokaryotes (c) eukaryotes (d) both (a) and (b) Answer: (d) both (a) and (b)

Question 54. In transcription in eukaryotes, heterogenous nuclear RNA (hnRNA) is tmascribed by (a) RNA polymerase I (b) RNA polymerase II (c) RNA poly merase II (d) all of these. Answer: (b) RNA polymerase II

Question 55. Methyl guanosine triphosphate is added to the 5’ end of hnRNA in a process of (a) splicing (b) capping (c) tailing (d) none of these Answer: (b) capping

Question 56. In eukaryotes, the process of processing of primary transcript involves (a) removal of introns (b) capping at 5’end (c) tailing (polyadenlation) at 3’ end (d) all of these. Answer: (b) capping at 5’end

Question 57. In a n/RNA molecule, untranslated regions (UTRs) are present at (a) 5’ – end (before start codon) (b) 3’ – end (after stop codon) (c) both (a) and (b) (d) 3’- end only. Answer: (c) both (a) and (b)

Question 58. UTRs are the untranslated regions present on (a) rRNA (b) hnRNA (c) mRNA (d) hnRNA. Answer: (c) mRNA

Question 59. Which of the following statements is correct regarding ribosomes ? (a) Most of a cell’s DNA molecule are stored there. (b) Complete polypeptide is released from there. (c) mRNAs are produced there. (d) DNA replication takes place there. Answer: (b) Complete polypeptide is released from there.

Question 60. Regulation of gene expression occurs at the level of (a) transcription (b) processing/splicing (c) translation (d) all of these. Answer: (d) all of these.

Question 61. During expression of an operon, RNA polymerase binds to (a) structural gene (b) regulator gene (c) operator (d) promoter. Answer: (d) promoter.

Question 62. The sequence of structural genes in lac operon is (a) Lac A, Lac Y, Lac Z (b) Lac A, Lac Z, Lac Y (c) Lac Y, Lac A, Lac A (d) Lac Z, Lac Y, Lac A Answer: (d) Lac Z, Lac Y, Lac A

Question 63. Which of the following cannot act as inducer ? (a) Glucose (b) Lactose (c) Galactos (d) Both (a) and (c) Answer: (d) Both (a) and (c)

Question 64. Human genome consists of approximately (a) 3 × 10 9 bp (b) 6 × 10 9 bp (c) 20,000 – 25,000 bp (d) 2.2 × 10 4 bp. Answer: (a) 3 × 10 9 bp

Question 65. Estimated number of genes in human beings is (a) 3,000 (b) 80,000 (c) 20,500 (d) 3 × 10 9 Target Series Objective Guide Science (English Medium) Answer: (c) 20,500

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Extra Questions of Class 12 Biology Molecular Basis of Inheritance. myCBSEguide has just released Chapter Wise Question Answers for class 12 Biology. There chapter wise Test papers with complete solutions are available for download in myCBSEguide   website and mobile app. These Extra Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Biology Extra questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in  CBSE Class 12 Biology   syllabus  and questions are framed as per the latest marking scheme and blue print issued by CBSE for Class 12.

CBSE Class 12 Biology Ch – 6 Practice Questions

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Class 12 Biology Chapter 6 Test Papers

• Used to understand several diseases like Alzheimer’s Parkinson’s diseases, etc.
• Used as gene therapy
• Improves diagnosis of diseases
• All of these.
• Matching the fingerprint of different persons
• Analyzing VNTR samples of DNA obtained from body fluid or cells
• Finding similarities of finger shapes of different individuals
• Comparing the DNA of two individuals
• Highly sensitive electron microscope
• X-ray scanners
• DNA segments having radioactive isotopes
• Protein analysis b.  gene counting c.  chromosome counting d.  DNA fingerprinting
• Structural gene b.  Regulator gene c.  Operator gene d.  Promoter gene
• What is the distance between two base pairs in DNA?
• Khorana and his colleagues synthesized an RNA molecule with repeating sequence of AG nitrogenous bases (AG AG AG AG AG AG). It produced a tetrapeptide with alternating sequence of valine and tyrosine. It proves that codon for valine and tyrosine are?
• What is teminism?
• Mention two functions of the codon AUG.
• Differentiate between the process of transcription in prokaryotes and eukaryotes.

A Couple quarrelled with the hospital authority on suspicion that their child had been exchanged after birth. The couple based their argument on the fact that their child is O blood group whereas they are A and B blood groups respectively. The doctor smiled and explained.

• What values of the doctor is reflected here?
• How can the child be O blood group as explained by the doctor?
• Which test method can be considered authentic to identify the biological parents of the child?
• Name the other blood group(s) which the child could have inherited.
• Differentiate between leading strand and lagging strand.
• How did Hershey and Chase prove that DNA is the hereditary material? Explain their experiment with suitable diagrams.

Ch-6 Molecular Basis of Inheritance

• All of these,  Explanation:  The application of the power of molecular genetics to the problems of human disease plays an important role in many of the research programs in the Department of Biology. Several complementary approaches are used by our research groups. The power of genomic analysis is used to identify, isolate and characterize genes which cause and contribute to the etiology of human diseases. Human disease is also studied through the functional analysis of key genes: low density lipoprotein receptors in atherosclerosis and stroke; a broad spectrum of tumor suppressors and oncogenes in cancer, genes directly leading to the disease etiology in neuromuscular disorders such as Alzheimer’s, Huntington’s disease, and muscular dystrophies.
• Analyzing VNTR samples of DNA obtained from body fluid or cells Explanation:  VNTR analysis (sometimes called DNA fingerprinting) can be used to identify and match cell samples. In order to use VNTRs, we must have a source of DNA. The DNA can come from any nucleated cell, e.g., white blood cells, skin samples, semen samples, or hair folicles. Red blood cells cannot be used as they do not have a nucleus. For a VNTR analysis, the DNA is extracted and cut using restriction endonucleases. A Southern Blot is then performed using various probes.
• DNA segments having radioactive isotopes,  Explanation:  In finger printing the nylon membrane was incubated with radioactive probes. DNA Probes are small fragments of minisatellite DNA tagged with radioactive phosphorous. The probes only attach to the pieces of DNA that they are complementary to – in this case they attach to the minisatellites in the genome. Radioactive isotopes can be easily identified using scanner.
• DNA fingerprinting,  Explanation:  DNA paternity testing is the most accurate form of paternity testing possible. If DNA patterns between the child and the alleged father do not match on two or more DNA probes, then the alleged father can be totally ruled out. If the DNA patterns between mother, child, and the alleged father match on every DNA probe, the likelihood of paternity is 99.9 percent.
• Regulator gene,  Explanation:  Regulator gene is a gene that regulates the expression of one or more structural genes by controlling the production of a protein (such as a genetic repressor) which regulates their rate of transcription. Regulation of lac operon by repressor is called negative regulation.
• The diameter of the B-DNA is ~20 Angstroms, and the distance between base pairs is ~3.4 Angstroms. The base pairing of opposite strands is stereochemically selective, Adenine always pairing with Thymine, and Guanine with Cytosine.
• Codon for valine – AGA Codon for tyrosine – GAG
• ‘a’ – synthesis of leading / continuous strand. ‘b’ – synthesis of lagging / discontinuous strand.
• Reverse transcription is known as teminism, which was first reported by Temin and Baltimore. RNA of some viruses (Retroviruses) first synthesizes DNA through reverse transcription. The DNA then transfers information to RNA, which takes part in translation of coded information to form polypeptide.
• (i) It codes for methionine amino acid. (ii) It acts as initiator codon for protein synthesis

(i) Products of transcription become effective in situ.	(i) Products of transcription come out of the nucleus for functioning in cytoplasm.
(ii) There is only one RNA-polymerase.	(ii) Three RNA polymerases take part in it.
(iii) mRNA is polycistronic.	(iii) mRNA is monocistronic.
(iv) Splicing is not required.	(iv) Splicing is required for removing introns.

• The doctor was assertive, patient and pragmatic.

A	A	A, O
A	B	A, AB, B, O
A	AB	A, AB, B
AB	AB	A, AB, B
B	B	B, O
B	AB	A, AB, B
O	O	O
O	A	A, O
O	AB	A, B
Rh+	Rh-	RH+, Rh-
Rh-	Rh-	Rh-
Rh+	Rh+	Rh+, RH-

• DNA fingerprinting

1. It is a replicated strand of DNA which grows continuously without any gap.	1. The lagging strand is a replicated strand of DNA which is formed in short segments called discontinuous.
2. It does not require DNA ligase for its growth	2. DNA ligase is required for joining okazaki fragments.
3. The direction of growth of a leading strand is 5′ →→ 3′	3. The direction of the lagging strand is 3′ →→ 5′
4. Only a single RNA primer is required	4. Starting of each okazaki fragment requires a new RNA.
5. Its template opens in 3′ →→ 5′ direction	5. Its template opens in 5′ →→ 3′ direction
6. Formation of leading strand begins immediately at the beginning of replication.	6. Formation of lagging strand begins a bit later than that of leading strand.

Class 12 Biology Chapter Wise Test Paper

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

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• Practice Questions of Class 12 Biology Human Health and Disease
• Class 12 Biology Evolution Chapter 7 Important Questions
• Class 12 Biology Human Reproduction Extra Questions

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Class 12 Biology Case Study of Chapter 5 Principles of Inheritance and Variation

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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 5 Principles of Inheritance and Variation 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 Principles of Inheritance and Variation  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.

Principles of Inheritance and Variation Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 12 Biology  Chapter 5 Principles of Inheritance and Variation

Case Study/Passage-Based Questions

(a) A and B only

(b) B only

(c) A and C only

(d) A, Band C

Answer: (c) A and C only

(ii) Teacher asked to conduct an experiment on Drosophila because

(a) the male and female flies are easily distinguishable	(b) it completes its life cycle in about two weeks
(c) a single mating could produce a large number of progeny flies	(d) all of these.

Answer: (d) all of these.

(iii) Genes white-eyed and yellow-bodied located very close to one another on the same chromosome tend to be transmitted together and are called

(a) allelomorphs

(b) identical genes

(c) linked genes

(d) recessive genes

Answer: (c) linked genes

(iv) Select the correct statement regarding the given experiment.

(a) The physical distance between two genes determines the strength of linkage

(b) The physical distance between two genes determines the frequency of crossing over

(c) The two linked genes always segregate independently of each other

(d) Both (a) and (b)

Answer: (a) The physical distance between two genes determines the strength of linkage

(v)  Assertion : When yellow-bodied, white-eyed Drosophila females were hybridized with brown-bodied, red-eyed males; and FI progeny was intercrossed, the F2 ratio deviated from 9: 3: 3: 1. Reason:  When two genes in a dihybrid are on the same chromosome, the proportion of parental gene combinations is much higher than in the non-parental type.

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

(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion

(c) Assertion is true but the reason is false

(d) Both assertion and reason are false

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

Case Study 2: Turner’s syndrome is an example of mono somy. It is formed by the union of an allosome free egg and a normal ‘X’ containing sperm or a normal egg and an allosome free sperm. The individual has 2n = 45 chromosomes (44 + X0) instead of 46. Such individuals are sterile females who have rudimentary ovaries, under developed breasts, small uterus, short stature, webbed neck and abnormal intelligence. They may not menstruate or ovulate. This disorder can be treated by giving female sex hormone to the women from the age of puberty to make them develop breasts and have menstruation. This makes them feel more normal.

What is Turner’s syndrome an example of? A) Trisomy B) Monosomy C) Polysomy D) Karyotype alteration

How many chromosomes are typically present in an individual with Turner’s syndrome? A) 44 B) 46 C) 45 D) 47

What is the chromosomal configuration in Turner’s syndrome? A) 44 + XX B) 44 + XY C) 44 + X0 D) 44 + XXY

Which of the following is a characteristic of Turner’s syndrome? A) Overdeveloped Breasts B) Tall Stature C) Normal Ovaries D) Webbed Neck

What is the primary treatment approach for women with Turner’s syndrome from the age of puberty? A) Giving Male Sex Hormone B) Administering Insulin C) Providing Female Sex Hormone D) Performing Surgery

Individuals with Turner’s syndrome often experience which of the following reproductive issues? A) Overactive Ovaries B) Frequent Ovulation C) Sterility and Rudimentary Ovaries D) Multiple Pregnancies

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 12 Biology Principles of Inheritance and Variation Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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Molecular Basis of Inheritance Class 12 Biology Important Questions

Please refer to Molecular Basis of Inheritance Class 12 Biology Important Questions given below. These solved questions for Molecular Basis of Inheritance have been prepared based on the latest CBSE, NCERT and KVS syllabus and books issued for the current academic year. We have provided  important examination questions for Class 12 Biology  all chapters.

Class 12 Biology Molecular Basis of Inheritance Important Questions

Very Short Answer Questions

Question. How many base pairs would a DNA segment of length 1.36 nm have? 0.34 ×10 -6 ×1.36 4 ×10 6 bp Answer . Distance between two base pairs = 0.34 nm or 0.34×10 –6 nm Number of base pairs in 1.36 nm DNA segment

Question. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments? Answer.  By density gradient centrifugation.

Question. Why hnRNA is required to undergo splicing?  Answer.  hnRNA undergoes splicing in order to remove introns which are intervening or non-coding sequences and exons are joined to form functional mRNA.

Question. Given below is a schematic representation of a lac operon in the absence of an inducer. Identify ‘a’ and ‘b’ in it.

Answer.  a–Repressor b–Repressor bound to the operator and prevents transcription of structural genes.

Question. Mention the contribution of genetic maps in human genome project.  Answer.  Genetic maps have played an important role in sequencing of genes, DNA fingerprinting, tracing human history, chromosomal location for disease associated sequences (Any one).

Question. Why are proteins either positively or negatively charged? Answer.  If the proteins are rich in basic amino acids, they are positively charged, and if the proteins are rich in acidic amino acids, they are negatively charged.

Question. Group the following as nitrogenous bases and nucleosides:  Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine. Answer . Nitrogenous bases—Adenine, Thymine, Uracil and Cytosine. Nucleosides—Cytidine and Guanosine.

Question. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA. Answer.  Cytosine = 20%, therefore Guanine = 20% According to Chargaff’s rule, A + T = 100 – (G + C) A + T = 100 – 40. Since both adenine and thymine are in equal amounts, ∴ Thymine = Adenine = 60/2=30%

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

Question. If the sequence of one strand of DNA is written as follows : 5′—ATGCATGCATGCATGCATGCATGCATGC—3′ Write down the sequence of complementary strand in 5′→3′ direction. Answer.  In 3′→5′ direction, 3′—TACGTACGTACGTACGTACGTACGTACG—5′ In 5′→3′ direction, 5′—GCATGCATGCATGCATGCATGCATGCAT—3′

Question. At which ends do ‘capping’ and ‘tailing’ of hnRNA occur, respectively? Answer.  Capping occurs at 5′-end and tailing occurs at 3′-end.

Short Answer Questions

Question. Draw a schematic representation of dinucleotide. Label the following: (i) The components of a nucleotide (ii) 5′ end (iii) N-glycosidic linkage (iv) Phosphodiester linkage.  Answer.  Nucleotide = Ribose sugar + Base + phosphate group.

Question. Describe the structure of a RNA polynucleotide chain having four different types of nucleotides.  Answer.  

Question. Differentiate between codon and an anticodon. Answer.  

The sequence of 3 nitrogen bases on mRNA that codes for a particular amino acid during translation is called codon.	The sequence of 3 nitrogenous bases on tRNA that are complementary to the codon on mRNA for a particular amino acid during translation is called anticodon.

Question. A DNA segment has a total of 1500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this DNA segment possesses? Answer.  According to Chargaff’s rule A/G = T/G  = 1 G = C, G = 410, hence C = 410 G + C = 410 + 410          = 820 So, A + T = 1500 – 820                = 680 A = T, so T = 680/2 = 340 So, Pyrimidines = C + T                        = 410 + 340                        = 750

Question. A DNA segment has a total of 2,000 nucleotides, out of which 520 are adenine containing nucleotides. How many purine bases this DNA segment possesses?   Answer . [A] = [T] A + T = 520 + 520          = 1040 Total number of nucleotides = 2000 ∴ G + C = 2000 – 1040              = 960    G = 960/2 = 480 ∴ Total number of purines (A + G) = 520 + 480                                                   = 1000.

Question. Describe the structure of a nucleosome. Answer.  Roger Kornberg (1974) reported that chromosome is made up of DNA and protein. • Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins. • The proteins associated with DNA are of two types— basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins. • The negatively charged DNA molecule wraps around the positively charged histone proteins to form a structure called nucleosome. • The nucleosome core is made up of four types of histone proteins—H2A, H2B, H3 and H4 occurring in pairs.

Question. Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase. Answer.  Viruses grown in the medium containing 32P contained radioactive DNA but not radioactive protein because DNA contains phosphorus but proteins do not contain phosphorus. Similarly,viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

Question. Recall the experiment done by Frederick Griffith, Avery, Macleod annd McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Streptococcus have tranformed the R-strain into virulent strain? Explain your Answerwer. Answer.  RNA is more labile and prone to degradation (owing to the presence of 2′–OH group in its ribose). Hence heat-killed S-strain may not have retained its ability to transform the R-strain.

Question. How do histones acquire positive charge? Answer . Histones are rich in the basic amino acid residues lysines and arginines, which carry positive charges in their side chains. Therefore, histones are positively charged.

Question. A single base mutation in a gene may not ‘always’ result in loss or gain of function. Do you think the statement is correct? Defend your Answerwer. Answer.  The statement is correct because of degeneracy of codons, mutations at third base of codon,usually doe not result into any change in phenotype. This is called silent mutations but at other times it can lead to loss or formation of malformed protein changing the phenotype.

Question. Explain the dual function of AUG codon. Give the sequence of bases it is transcribed from and its anticodon. Answer.  The dual function of AUG codon: (a) It codes for amino acid methionine. (b) It is an initiation codon. The sequence of bases from which it is transcribed is TAC. Its anticodon is UAC.

Question. Protein synthesis machinery revolves around RNA but in the course of evolution it was replaced by DNA. Justify.  Answer.  Since RNA was unstable and prone to mutations, DNA evolved from RNA with chemical modifications that makes it more stable. DNA has double stranded nature and has complementary strands. These further resist changes by evolving a process of repair.

Why do you see two different types of replicating strands in the given DNA replication fork?  Explain. Name these strands. Answer.  The DNA-dependent DNA polymerase catalyses polymerisation only in one direction i.e., 5′→3′. Therefore, in one strand with polarity 3′→5′ continuous replication takes place whereas the other strand with polarity 5′→3′ carries out discontinuous replication. The strand with polarity 3′→5′ is called leading strand and the strand with polarity 5′→3′ is called lagging strand.

Long Answer Questions

Question. How is the translation of mRNA terminated? Explain. Ans .  •  When the A-site of ribosome reaches a termination codon, which does not code for any amino acid, no charged tRNA binds to the A-site. •  Dissociation of polypeptide from ribosome takes place, which is catalysed by a ‘release factor’. •  There are three termination codons namely UGA, UAG and UAA.

Question. (a) A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides. How many pyrimidine bases this DNA segment possesses? (b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.

Ans.  (a) A = T, A = 240, hence T = 240 A + T = 240 + 240 = 480 So, G + C = 1000 – 480 = 520 G = C, so C 520/2 =260 So, pyrimidines = C + T = 260 + 240 = 500

Question. (a) Explain DNA polymorphism as the basis of genetic mapping of human genome. (b) State the role of VNTR in DNA fingerprinting.  Ans . (a) Genetic polymorphism means occurrence of genetic material in more than one form. It is of  three major types, i.e., allelic, SNP and RFLP. Allelic polymorphism: Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur. SNP or single nucleotide polymorphism: SNP is very useful for locating alleles, identifying disease-associated sequence and tracing human history. (b) Variable Number Tandem Repeats (VNTRs) are used in DNA fingerprinting as markers. VNTRs vary from person to person and are inherited from one generation to the next. Therefore, only closely related individuals have similar VNTRs.

Question. (a) Why did Hershey and Chase use radioactive sulphur and radioactive phosphorus in their experiment? (b) Write the conclusion they arrived at and how. Ans. •   Procedure: (i) Some bacteriophage virus were grown on a medium that contained radioactive phosphorus (32P) and some in another medium with radioactive sulphur (35S). (ii) Viruses grown in the presence of radioactive phosphorus (32P) contained radioactive DNA. (iii) Similar viruses grown in presence of radioactive sulphur (35S) contained radioactive protein. (iv) Both the radioactive virus types were allowed to infect E. coli separately. (v) Soon after infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria. (vi) The culture was also centrifuged to separate the viral particle from the bacterial cell.

Question. “A very small sample of tissue or even a drop of blood can help determine paternity”. Provide a scientific explanation to substantiate the statement.  Ans.  (i) DNA from all cells of an individual shows the same degree of polymorphism and therefore becomes a useful identification tool. (ii) Polymorphs are heritable and the child inherits 50% of the chromosome from each parent. (iii) With the help of PCR the small amount of DNA from blood can be amplified and be used in DNA finger printing to identitfhye paternity.

Question. (Img 244)

(a) Identify strands ‘A’ and ‘B’ in the diagram of transcription unit given above and write the basis on which you identified them. (b) Write the functions of RNA polymerase-I and RNA polymerase-III in eukaryotes. Ans.  (a) A—Template strand B—Coding strand The templates are identified on the basis of polarity with respect to promoter. Template strand has polarity 3′ → 5′ and coding strand has polarity 5′ → 3′. (b) RNA polymerase-I transcribes rRNAs. RNA polymerase-III transcribes tRNA, 5srRNA and snRNA.

Question. State any two structural differences and one functional difference between DNA and rRNA. Ans.

Question. DNA polymerase and RNA polymerase differ in their requirement while functioning Explain. Ans . 

(i)	It cannot carry out proofreading.	It carries out proofreading for DNA repair mechanism.
(ii)	RNA polymerase does not require RNA primer for synthesis of RNA.	DNA polymerase requires RNA primer for synthesis of DNA.
(iii)	It uses ribonucleotides for RNA synthesis.	It uses deoxyribonucleotides for DNA synthesis.

Question. The average length of a DNA double helix in a typical mammalian cell is approximately 2.2 metres and the dimension of the nucleus is about 10 –6  m. (a) How is it possible that long DNA polymers are packed within a very small nucleus? (b) Differentiate between euchromatin and heterochromatin. (c) Mention the role of non-histone chromosomal protein. Ans . (a) Packaging of DNA in eukaryotes •  Roger Kornberg (1974) reported that chromosome is made up of DNA and protein. •  Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins. •  The proteins associated with DNA are of two types—basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins. •  The negatively charged DNA molecule wraps around the positively charged histone proteins to form a structure called nucleosome. •  The nucleosome core is made up of four types of histone proteins—H2A, H2B, H3 and H4 occurring in pairs.

(c) The packaging of chromatin at higher level requires the presence of non-histone chromosomal protein.

Question. List the criteria a molecule that can act as genetic material must fulfill. Which one of the criteria are best fulfilled by DNA or by RNA thus making one of them a better genetic material than the other? Explain.  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 chemically and structurally be stable. (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’. In DNA the two strands being complementary if separated by heating come together, when appropriate conditions are provided. Further, 2’-OH group present at every nucleotide in RNA is also now known to be catalytic, hence reactive. Therefore DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material. The presence of thymine at the place of uracil also confers additional stability to DNA. Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA.

Question. (a) How are the following formed and involved in DNA packaging in a nucleus of a cell? (i) Histone octomer (ii) Nucleosome (iii) Chromatin (b) Differentiate between Euchromatin and Heterochromatin.  Ans.  (a) (i) Eight molecules of positively charged basic proteins called histones are organised to form histone octomer. (ii) Negatively charged DNA is wrapped around positively charged histone octamer to give rise to nucleosome. (iii) Nucleosome constitute the repeating unit of a structure called chromatin. (b)

(i)	Regions of chromatin, which are loosely packed during interphase are called euchromatin.	Regions of chromatin, which are densely packed during cell division are called heterochromatin.
(ii)	When chromosomes are stained with Feulgen stain (specific for DNA), these appear as lightly stained chromatin.	When chromosomes are stained with Feulgen stain, these appear as intensely stained chromatin.
(iii)	Euchromatin contains active genes.	Heterochromatin contains inactive genes.
(iv)	They do not contain repetitive DNA sequences.	They are enriched with highly repetitive tandemly arranged DNA sequences.
(v)	It is transcriptionally active.	It is transcriptionally inactive.

Question. What background information did Watson and Crick had available with them for developing a model of DNA? What was their own contribution? Ans . Watson and Crick had the following informations which helped them to develop a model of DNA: (i) Chargaff’s Law suggesting A=T and C G. (ii) Wilkins and Franklin’s X-ray diffraction studies on DNA’s physical structure.

Based on these information, Watson and crick proposed (i) complementary base-pairing of nitrogenous bases (ii) semi-conservative mode of replication (iii) occurrence of mutation through tautomerism.

Q. 5. Describe the packaging of DNA helix in a prokaryotic cell and an eukaryotic nucleus. Ans . (i) Packaging of DNA in prokaryotes •  In prokaryotes, well-defined nucleus is absent so DNA is present in a region called nucleoid. The negatively charged DNA is coiled with some positively charged non-histone basic proteins. •  DNA in nucleoid is organised in large loops held by proteins.

(ii) Packaging of DNA in eukaryotes •  Roger Kornberg (1974) reported that chromosome is made up of DNA and protein. •  Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins. •  The proteins associated with DNA are of two types—basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins. •  The negatively charged DNA molecule wraps around the positively charged histone proteins to form a structure called nucleosome.

Question. (a) How did Griffith explain the transformation of R-strain (non-virulent) bacteria into S-strain (virulent)? (b) Explain how MacLeod, McCarty and Avery determined the biochemical nature of the molecule responsible for transforming R-strain bacteria into S-strain bacteria. OR (a) Describe the various steps of Griffith’s experiment that led to the conclusion of the ‘Transforming Principle’. (b) How did the chemical nature of the ‘Transforming Principle’ get established? Ans.  (a)  •  Frederick Griffith (1928) conducted experiments with Streptococcus pneumoniae (bacterium causing pneumonia). •  He observed two strains of this bacterium—one forming smooth shiny colonies (S-type) with capsule,while other forming rough colonies (R-type) without capsule. (b)  •  Oswald Avery, Colin MacLeod and Maclyn McCarty repeated Griffith’s experiment in an in vitro system in order to determine biochemical nature of transforming principle. •  They purified biochemicals (proteins, DNA, RNA) from heat-killed S-type cells, and checked which of these could transform live R-type cells into S-type cell. They observed that DNA alone from S-type cells caused transformation of R-type cells into virulent S-type cells. •  They also discovered that proteases (protein digesting enzymes) and RNases (RNA digesting enzymes) did not affect transformation while DNases inhibited the process. •  They concluded that DNA is the hereditary material.

Question. (a) Write the scientific name of the bacterium used by Frederick Griffith in his experiment. (b) How did he prove that some ‘transforming principle’ is responsible for transformation of the non-virulent strains of bacteria into the virulent form? (c) State the biochemical nature of ‘transforming principle’. (d) Name the scientists who proved it.  Ans.  (a) Streptococcus pneumoniae (b) Refer to Basic Concepts Point 6. (c) ‘The transforming principle’ was nucleic acid, i.e., DNA. (d) It was proved by O. Avery, C. MacLeod and M. McCarty.

Question. (a) Explain the experiment performed by Griffith on Streptococcus pneumoniae. What did he conclude from this experiment?  (b) Name the three scientists who followed up Griffith’s experiments. (c) What did they conclude and how? Ans.  (a) Refer to Basic Concepts Point 5. (b) Oswald Avery, Colin MacLeod and Maclyn McCarty. (c)  •  Oswald Avery, Colin MacLeod and Maclyn McCarty repeated Griffith’s experiment in an in vitro system in order to determine biochemical nature of transforming principle. •  They purified biochemicals (proteins, DNA, RNA) from heat-killed S-type cells, and checked which of these could transform live R-type cells into S-type cell. They observed that DNA alone from S-type cells caused transformation of R-type cells into virulent S-type cells. •  They also discovered that proteases (protein digesting enzymes) and RNases (RNA digesting enzymes) did not affect transformation while DNases inhibited the process. •  They concluded that DNA is the hereditary material.

Question. (a) What did Meselson and Stahl observe when (i) they cultured E. coli in a medium containing 15NH4Cl for a few generations and centrifuged the content? (ii) they transferred one such bacterium to the normal medium of NH4Cl and cultured for 2 generations? (b) What did Meselson and Stahl conclude from this experiment? Explain with the help of diagrams. (c) Which is the first genetic material? Give reasons in support of your answer. Ans.  (a) (i) Meselson and Stahl observed that in the E. coli bacterium the DNA becomes completely labelled with 15N medium after few generations. (ii) After two generations, they observed that density changed and showed equal amount of light DNA (14N) and dark hybrid DNA (15N–14N). (b) They concluded that DNA replicates semi-conservatively.

(c) RNA is the first genetic material. Reasons: (i) RNA is highly reactive and acts as a catalyst as well as a genetic material. (ii) Essential life processes such as metabolism, translation and splicing evolved around RNA. (iii) It expresses itself through proteins.

Question. Describe the Hershey−Chase experiment. Write the conclusion they arrived at after the experiment. OR How did Hershey and Chase established that DNA is transferred from virus to bacteria? Ans.  In some viruses, RNA is the genetic material (e.g., Tobacco Mosaic virus). RNA also performs functions of messenger and adapter. (i) DNA and RNA have the ability to direct their duplications because of rule of base pairing and complementarity but proteins fail to fulfill first criteria itself. (ii) Genetic material should be stable so as not to change with different stages of life cycle, age or change in physiology of organism. (iii) RNA being unstable mutates at a faster rate. Thus, viruses having RNA genome and having shorter life span mutate and evolve faster. (iv) RNA can code directly for protein synthesis and hence can easily express characters. But DNA is dependent on RNA for protein synthesis. Protein synthesizing machinery has evolved around RNA.

Question. You are repeating the Hershey–Chase experiment and are provided with two isotopes: 32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different?  Ans.  Use of 15N will be inappropriate because method of detection of 35P and 15N is different (32Pbeing a radioactive isotope while 15N is not radioactive but is the heavier isotope of nitrogen). Even if 15N was radioactive then its presence would have been detected both inside the cell (l5N incorporated as nitrogenous base in DNA) as well as in the supernatant because 15N would also get incorporated in amino group of amino acids in proteins). Hence, the use of 15N would not give any conclusive results.

Question. A criminal blew himself up in a local market when was chased by cops. His face was beyond recognition. Suggest and describe a modern technique that can help establish his identity. Ans.  The identity can be established by the technique of DNA fingerprinting. For method:  •  Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases. •  Human genome has 3 × 109 bp. 99.9% of base sequences among humans are the same, which makes every individual unique in phenotype. •  Polymorphism: The genome consists of small stretches of DNA which are repeated many times. These are called repetitive DNA and comprise of satellite DNA. Satellite DNA does not code for any proteins but form large portion of human genome. These sequences show high degree of polymorphism. As polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing. Polymorphism arises due to mutations. New mutations may arise in somatic cells or in germ cells. If mutation occurs in germ cells; it is passed on to offsprings. If an inheritable mutation is observed in a population at high frequency, it is called as DNA polymorphism. Polymorphism ranges from single nucleotide change to very large scale changes.

Question. Answer the following questions based on Hershey and Chases’s experiments: (a) Name the kind of virus they worked with and why. (b) Why did they use two types of culture media to grow viruses in? Explain. (c) What was the need for using a blender and later a centrifuge during their experiments? (d) State the conclusion drawn by them after the experiments. Ans.  (a) They worked with bacteriophage because when it attacks a bacteria it only inserts its genetic material in its body. (b) They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulphur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur. (c) Blender was used to agitate the bacteria to remove the viral coats from them. Centrifuge was used to separate virus particle from the bacteria. (d) 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.

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8.1: Case Study: Genes and Inheritance

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• Page ID 16760

• Suzanne Wakim & Mandeep Grewal
• Butte College

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Case Study: Cancer in the Family

People tend to look similar to their biological parents, as illustrated by the family tree in Figure \(\PageIndex{1}\). But, you can also inherit traits from your parents that you can’t see. Rebecca becomes very aware of this fact when she visits her new doctor for a physical exam. Her doctor asks several questions about her family's medical history, including whether Rebecca has or had relatives with cancer. Rebecca tells her that her grandmother, aunt, and uncle, who have all passed away, all had cancer. They all had breast cancer, including her uncle, and her aunt additionally had ovarian cancer. Her doctor asks how old they were when they were diagnosed with cancer. Rebecca is not sure exactly, but she knows that her grandmother was fairly young at the time, probably in her forties.

Rebecca’s doctor explains that while the vast majority of cancers are not due to inherited factors, a cluster of cancers within a family may indicate that there are mutations in certain genes that increase the risk of getting certain types of cancer, particularly breast and ovarian cancer. Some signs that cancers may be due to these genetic factors are present in Rebecca’s family, such as cancer with an early age of onset (e.g. breast cancer before age 50), breast cancer in men, and breast cancer and ovarian cancer within the same person or family.

Based on her family medical history, Rebecca’s doctor recommends that she see a genetic counselor because these professionals can help determine whether the high incidence of cancers in her family could be due to inherited mutations in their genes. If so, they can test Rebecca to find out whether she has the particular variations of these genes that would increase her risk of getting cancer.

When Rebecca sees the genetic counselor, he asks how her grandmother, aunt, and uncle with cancer are related to her. She says that these relatives are all on her mother’s side — they are her mother’s mother and siblings. The genetic counselor records this information in the form of a specific type of family tree, called a pedigree, indicating which relatives had which type of cancer and how they are related to each other and to Rebecca. He also asks her ethnicity. Rebecca says that her family, on both sides, are Ashkenazi Jews, meaning Jews whose ancestors came from central and eastern Europe. “But what does that have to do with anything?” she asks. The counselor tells Rebecca that mutations in two tumor-suppressor genes called BRCA1 and BRCA2, located on chromosome 17 and 13, respectively, are particularly prevalent in people of Ashkenazi Jewish descent and greatly increase the risk of getting cancer. About 1 in 40 Ashkenazi Jewish people have one of these mutations, compared to about 1 in 800 in the general population. Her ethnicity, along with the types of cancer, age of onset, and the specific relationships between her family members who had cancer indicate to the counselor that she is a good candidate for genetic testing for the presence of these mutations.

Rebecca says that her 72-year-old mother never had cancer, and nor had many other relatives on that side of the family, so how could the cancers be genetic? The genetic counselor explains that the mutations in the BRCA1 and BRCA2 genes, although dominant, are not inherited by everyone in a family. Also, even people with mutations in these genes do not necessarily get cancer — the mutations simply increase their risk of getting cancer. For instance, 55 to 65% of women with a harmful mutation in the BRCA1 gene will get breast cancer before age 70, compared to 12% of women in the general population who will get breast cancer sometime over the course of their lives.

Rebecca is not sure she wants to know whether she has a higher risk of cancer. The genetic counselor understands her apprehension but explains that if she knows that she has harmful mutations in either of these genes, her doctor will screen her for cancer more often and at earlier ages. Therefore, any cancers she may develop are likely to be caught earlier when they are often much more treatable. Rebecca decides to go through with the testing, which involves taking a blood sample, and nervously waits for her results.

Chapter Overview: Genetics

At the end of this chapter, you will find out Rebecca ’s test results. By then, you will have learned how mutations in genes such as BRCA1 and BRCA2 can be passed down and cause disease. Especially, you will learn about:

• How Gregor Mendel discovered the laws of inheritance for certain types of traits.
• The science of heredity, known as genetics, and the relationship between genes and traits.
• Simple and more complex inheritance of some human traits.
• Genetic Disorders.

As you read this chapter, keep Rebecca’s situation in mind and think about the following questions:

• What do the BRCA1 and BRCA2 genes normally do? How can they cause cancer?
• Are BRCA1 and BRCA2 considered linked genes? And are they on autosomes or sex chromosomes?
• After learning more about pedigrees, draw the pedigree for cancer in Rebecca’s family. Use the pedigree to help you think about why it is possible that her mother does not have one of the BRCA gene mutations, even if her grandmother, aunt, and uncle did have it.
• Why do you think certain gene mutations are prevalent in certain ethnic groups?

Attributions

• Caelius and Valerius family tree by Ann Martin , licensed CC BY 2.0 via Flickr
• Text adapted from Human Biology by CK-12 licensed CC BY-NC 3.0