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Samacheer Kalvi solutions for Biology (Zoology) [English] Class 12 TN Board chapter 5 - Molecular Genetics [Latest edition]

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Samacheer Kalvi solutions for Biology (Zoology) [English] Class 12 TN Board chapter 5 - Molecular Genetics - Shaalaa.com
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Solutions for Chapter 5: Molecular Genetics

Below listed, you can find solutions for Chapter 5 of Tamil Nadu Board of Secondary Education Samacheer Kalvi for Biology (Zoology) [English] Class 12 TN Board.


Evaluation
Evaluation [Pages 84 - 86]

Samacheer Kalvi solutions for Biology (Zoology) [English] Class 12 TN Board 5 Molecular Genetics Evaluation [Pages 84 - 86]

Evaluation | Q 1. | Page 84

Hershey and Chase experiment with bacteriophage showed that ____________.

  • Protein gets into the bacterial cells

  • DNA is the genetic material

  • DNA contains radioactive sulphur

  • Viruses undergo transformation

Evaluation | Q 2. | Page 84

DNA and RNA are similar with respect to ____________.

  • Thymine as a nitrogen base

  • A single-stranded helix shape

  • Nucleotide containing sugars, nitrogen bases and phosphates

  • The same sequence of nucleotides for the amino acid phenyl alanine

Evaluation | Q 3. | Page 84

A mRNA molecule is produced by ____________.

  • Replication

  • Transcription

  • Duplication

  • Translation

Evaluation | Q 4. | Page 84

The total number of nitrogenous bases in human genome is estimated to be about ____________.

  • 3.5 million

  • 35000

  • 35 million

  • 3.1 billion

Evaluation | Q 5. | Page 84

E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?

  • One high and one low density band.

  • One intermediate density band.

  • One high and one intermediate density band.

  • One low and one intermediate density band.

Evaluation | Q 6. | Page 84

What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?

  • Origin of replication occurs only at the 5' end of the molecules.

  • DNA ligase works only in the 3' → 5' direction

  • DNA polymerase can join new nucleotides only to the 3' end of the growing stand.

  • Helicases and single-strand binding proteins that work at the 5' end.

Evaluation | Q 7. | Page 84

Which of the following is the correct sequence of event with reference to the central dogma?

  • Transcription, Translation, Replication

  • Transcription, Replication, Translation

  • Duplication, Translation, Transcription

  • Replication, Transcription, Translation

Evaluation | Q 8. | Page 85

Which of the following statements about DNA replication is not correct?

  • Unwinding of DNA molecule occurs as hydrogen bonds break.

  • Replication occurs as each base is paired with another exactly like it.

  • Process is known as semi conservative replication because one old strand is conserved in the new molecule.

  • Complementary base pairs are held together with hydrogen bonds.

Evaluation | Q 9. | Page 85

Which of the following statements is not true about DNA replication in eukaryotes?

  • Replication begins at a single origin of replication.

  • Replication is bidirectional from the origins.

  • Replication occurs at about 1 million base pairs per minute.

  • There are numerous different bacterial chromosomes, with replication ocurring in each at the same time.

Evaluation | Q 10. | Page 85

The first codon to be deciphered was __________ which codes for ________.

  • AAA, proline

  • GGG, alanine

  • UUU, Phenylalanine

  • TTT, arginine

Evaluation | Q 11. | Page 85

Meselson and Stahl’s experiment proved ____________.

  • Transduction

  • Transformation

  • DNA is the genetic material

  • Semi-conservative nature of DNA replication

Evaluation | Q 12. | Page 85

Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for _________ and the larger subunit has two binding sites for two __________.

Evaluation | Q 13. | Page 85

An operon is a: ______

  • Protein that suppresses gene expression

  • Protein that accelerates gene expression

  • Cluster of structural genes with related function

  • Gene that switched other genes on or of

Evaluation | Q 14. | Page 85

When lactose is present in the culture medium:

  • Transcription of lac y, lac z, lac a genes occurs.

  • Repressor is unable to bind to the operator.

  • Repressor is able to bind to the operator.

  • Both Transcription of lac y, lac z, lac a genes occurs and Repressor is unable to bind to the operator are correct.

Evaluation | Q 15. | Page 85

Give reasons: Genetic code is ‘universal’.

Evaluation | Q 16. | Page 85

Name the parts marked ‘A’ and ‘B’ in the given transcription unit:

Evaluation | Q 17. | Page 85

Differentiate - Leading stand and lagging strand.

Evaluation | Q 18. | Page 85

Differentiate between the following:

Template strand and Coding strand

Evaluation | Q 19. | Page 85

Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.

Evaluation | Q 20. | Page 85

State any three goals of the human genome project.

Evaluation | Q 21. | Page 86

In E.coli, three enzymes β- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.

Evaluation | Q 22. | Page 86

Distinguish between structural gene, regulatory gene and operator gene.

Evaluation | Q 23. | Page 86

A low level of expression of lac operon occurs at all the time. Justify the statement.

Evaluation | Q 24. | Page 86

HGP is the window for the treatment of various genetic disorders. Justify the statement.

Evaluation | Q 25. | Page 86

Why is the Human Genome project called a mega project?

Evaluation | Q 26. | Page 86

From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability, and mutation?

Evaluation | Q 27. | Page 86

Why tRNA is called an adapter molecule?

Evaluation | Q 28. | Page 86

What are the three structural differences between RNA and DNA?

Evaluation | Q 29. | Page 86

Name the anticodon required to recognize the following codon: AAU.

Evaluation | Q 29. | Page 86

Name the anticodon required to recognize the following codon: CGA

Evaluation | Q 29. | Page 86

Name the anticodon required to recognize the following codon: UAU

Evaluation | Q 29. | Page 86

Name the anticodon required to recognize the following codon: GCA.

Evaluation | Q 30. | Page 86

a) Identify the figure given below

b) Redraw the structure as a replicating fork and label the parts


c) Write the source of energy for this replication and name the enzyme involved in this process.

d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.

Evaluation | Q 31. | Page 86

If the coding sequence in a transcription unit is written as follows:

5' TGCATGCATGCATGCATGCATGCATGC 3'

Write down the sequence of mRNA.

Evaluation | Q 32. | Page 86

How is the two stage process of protein synthesis advantageous?

Evaluation | Q 33. | Page 86

Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?

Evaluation | Q 34. | Page 86

Explain the formation of a nucleosome.

Evaluation | Q 35. | Page 86

It is established that RNA is the first genetic material. Explain giving three reasons.

Solutions for 5: Molecular Genetics

Evaluation
Samacheer Kalvi solutions for Biology (Zoology) [English] Class 12 TN Board chapter 5 - Molecular Genetics - Shaalaa.com

Samacheer Kalvi solutions for Biology (Zoology) [English] Class 12 TN Board chapter 5 - Molecular Genetics

Shaalaa.com has the Tamil Nadu Board of Secondary Education Mathematics Biology (Zoology) [English] Class 12 TN Board Tamil Nadu Board of Secondary Education solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Samacheer Kalvi solutions for Mathematics Biology (Zoology) [English] Class 12 TN Board Tamil Nadu Board of Secondary Education 5 (Molecular Genetics) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

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Concepts covered in Biology (Zoology) [English] Class 12 TN Board chapter 5 Molecular Genetics are Gene as the Functional Unit of Inheritance, Introduction of Search for Genetic Material, The Genetic Material is a DNA, Chemistry of Nucleic Acids, The RNA World, Properties of Genetic Material (DNA Versus RNA), Packaging of DNA Helix, DNA Replication, Introduction of Transcription, Genetic Code, tRNA – the Adapter Molecule, Translation, Human Genome Project, Regulation of Gene Expression, DNA Fingerprinting Technique.

Using Samacheer Kalvi Biology (Zoology) [English] Class 12 TN Board solutions Molecular Genetics exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Samacheer Kalvi Solutions are essential questions that can be asked in the final exam. Maximum Tamil Nadu Board of Secondary Education Biology (Zoology) [English] Class 12 TN Board students prefer Samacheer Kalvi Textbook Solutions to score more in exams.

Get the free view of Chapter 5, Molecular Genetics Biology (Zoology) [English] Class 12 TN Board additional questions for Mathematics Biology (Zoology) [English] Class 12 TN Board Tamil Nadu Board of Secondary Education, and you can use Shaalaa.com to keep it handy for your exam preparation.

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