You can Download Chapter 6 Molecular Basis of Inheritance Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance

2nd PUC Biology Molecular Basis of Inheritance Ncert Text Book Questions and Answers

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

Question 2.
If a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 1
Question 3.
If the sequence of one strand of DNA is written as follows:
5ATGCATGCATGC ATGCATGCATGC ATGC-3′
Write down the sequence of complementary strand in 5’—>3’ direction.
Answer:
5′ – ATGCATGC ATGC ATGC ATGC ATGC ATGC – 3′ (In question)
complement 3′ – TACG TACG TACG TACG TACG TACG – 5′
complement in 5′ – 3′ direction 5′ – GCAT GCAT GCAT GCAT GCAT GCAT GC AT – 3′.

Question 4.
If coding strand in a transcription unit is written as follows:
5′-ATGCATGCATGCATGCATGCATGCATGC – 3′ Write down the sequence of m-RNA.
Answer:
m RNA carries same code as coding strand except T is replaced by U.
(Question) transcription strand 5′ – ATGC ATGC ATGC ATGC ATGC ATGC ATGC – 3′ m RNA → 5′ – AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 3′.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer:
2 strands of DNA has complementary base pairs which run in opposite direction. Property was that during DNA replication one strand was conserved while the complementary is formed new.

KSEEB Solutions

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

  • DNA polymerases for DNA replication.
  • RNA polymerases for RNA synthesis or transcription

RNA Template

  • RNA dependent RNA polymerases for synthesis of RNA is some RNA viruses.
  • Reverse transcriptase to synthesize DNA over RNA template.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
They raised 2 types of bacteriophages

  • On radioactive phosphorous (32P)
  • On radioactive sulphur (35S).

35S gets into protein and 32P into DNA When both bacteriophages infected bacteria differently and by shaking them, viral protein coat was separated

After raising these bacteria it was found that those infected with 32P bacteriophage → radioactivity was found. But with 35S → no radioactivity was found.

Question 8.
Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
Answer:

Repetitive DNA Satellite DNA
(i) sequence of Nbases present in more than 1 copy in a genome. (i) part of DNA  having repeated short sequences of N2 bases
(ii) Repeated DNA sequence may or may not be present in text. (ii) Repeated sequence occur in tandem
(iii) Variability may or may not be present (ii) Variability occurs

(b) mRNA and tRNA
Answer:

m-RNA t – RNA
(i) Large sized RNA with cap and tail.
(ii) Carries codon information
(i)   Small sized with 3-4 loops and a limb
(ii) Carries infor­mation for association of AA with anti Condon for incor­poration

(c) Template strand and Coding strand
Answer:

Template strand Coding strand
(i) The strand of DNA which takes part in transcription. (i) It does not takes part in transcription.

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

  • Provides sites for attachment of m-RNA and charged t – RNA for polypeptide synthesis.
  • One of its r – RNA functions as peptidyl transferase for peptide bond formation.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after the addition of lactose in the medium?
Answer:
Due to decrease in lactose substrate concentration, lac operon shuts down as it is inducible operon.

Question 11.
Explain (in one or two lines) the function of the followings:

(a) Promoter
Answer:
It is a gene that lies near operator which functions as binding site structural genes if operator allows.

(b) tRNA
Answer:
Functions as adapter molecule that picks  up a particular amino – acid from cellular pool and takes same over A site, m – RNA for incorporation into polypeptide

(c) Exons
Answer:
Coding segments present in primary transcript which on splicing by snRNP’s get joined to form functional m-RNA.

KSEEB Solutions

Question 12.
Why is the Human Genome project called a mega project?
Answer:
Because it involves

  • Sequencing of over 3 × 109 bp
  • Identification of all genes and alleles in human genome and their function.
  • Storage of data requires storage space equal to 3300 books 1000 pages, each page with 1000 letters.

Question 13.
What is DNA fingerprinting? Mention its applications.
Answer:
DNA fingerprinting is technique of determing similarity and dissimilarity of VNTR’s between 2 samples of DNA so as to bring out relationship if any VNTR’s are specific for each individual. They are derived from parents in 50: 50 ratio.
Applications

  • Identification – of criminals
  • Paternity and Maternity disputes.
  • Migrations
  • Human Lineage.

Question 14.
Briefly describe the following:
(a) Transcription
Answer:
Transcription – Formation of-RNA over template of DNA.
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 2
The snRNA formed has codon information similar to sense or coding strand of DNA with just U replacing T.

  • The DNA strand which function as a template is template or anti-sense strand.
  • Only one of the DNA strands is transcribed
  • The main enzyme taking part in transcription is RNA polymerase.

2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 3

(b) Polymorphism
Answer:
Polymorphism It is the occurrence of more than one form of genetic material
Types →

  • Allelic polymorphism – multiple alleles in a gene. So it alters structure and function of protein formed.
  • SNP or single nucleotide Polymorphism – Unique in every human being. Useful in locating specific alleles, disease associated sequences.
  • RFLP or Restriction Fragment Length polymorphism – Different sized fragments are formed by cleavage with same enzyme most RFPL have no effect on phenotypes.

(c) Translation
Answer:
Translation (Biosynthesis of proteins) Coded genetic message brought by m – RNA from DNA is charged into polypeptide chain (proteins).
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 4
DNA transcription mRNA translation Proteins Materials required for translation are ribosomes, AA, t RNA’s, aninoacylt – RNA synthetase, m – RNA and same factors Step in Translation . Activation of Amino Acids → charging of t RNA→ Initiation → Elongation → Termination → Modification.

(d) Bioinformatics
Answer:
Science of handling, storing as databases, analysing, modelling and providing access to various aspects of biological information, especially molecules connected with genomies and proteomics
Applications

  • Organisation of Biological Data
  • Functional Genomics
  • Proteomics, Pharmacogenomics
  • Medical informatics, chemoinformatics
  • Faster drey research.

KSEEB Solutions

2nd PUC Biology Molecular Basis of Inheritance Additional Questions and Answers

2nd PUC Biology Molecular Basis of Inheritance One Mark Questions

Question 1.
Expand the term DNA, RNA and tRNA.
Answer:
DNA – Deoxy ribonucleic acid RNA – Ribo nucleic acid tRNA – transfer – RNA

Question 2.
Define genetic code.
Answer:
Relationship between the sequence of nucleotides or bases on mRNA and the sequence of amino acids in the polypeptide.

Question 3.
What are Okazaki fragments?
Answer:
The small stretches of DNA formed due to the opposite running of DNA template at the time of replication is called Okazaki fragments.

Question 4.
What is codon?
Answer:
It is the triplet nitrogenous base sequence which code for one amino acid. It lies on the m- RNA.

Question 5.
What is anti codon?
Answer:
It in the triplet nitrogenous bases in the t-RNA complementary to an m-RNA codon. It identifies a particular codon on mRNA.

Question 6.
What is a translation?
Answer:
It is the process of the formation of protein directed by a m-RNA molecule.

Question 7.
Define Transcription?
Answer:
It is the process of the formation of m-RNA on a DNA template.

Question 8.
Name the enzyme that catalyses
(a) Replication of DNA and
(b) Formation of RNA. (CBSE 1995)
Answer:
(a) Topoisomerase.
(b) RNA polymerase.

Question 9.
What is the genetic material of tobacco mosaic virus (TMV)
Answer:
RNA.

Question 10.
Name the bond present between two adjacent nucleotides
Answer:
Phosphodiester bond.

Question 11.
Due to mistake during transcription, ATG forms UAG in m-RNA. What change would occur in polypeptide chain translated by this m-RNA? (Cbse 1996)
Answer:
UAG is a termination codon and so at point protein synthesis will get stopped.

KSEEB Solutions

Question 12.
Name three different non – sense codons.
Answer:
UAA, UAG and UGA.

Question 13.
If the base sequence of one strand of DNA is CAT TAG TAC GAC, what will be the base sequence
(a) Of complementary DNA strand, and
(b) Of its complementary RNA strand? (CBSE 1991)
Answer:
(a) GTA ATC ATG CTG
(b) GUA AUCAUGCUG.

Question 14.
Name the scientist who proposed one gene one enzyme hypothesis. (CBSE 1997)
Answer:
Beadle and Tatum.

Question 15.
Name the enzyme that joints the short pieces in the lagging strand during synthesis of DNA.
Answer:
DNA legase. (CBSE 1998)

Question 16.
In which direction 5′ – 3′ or 3′ -5′ are the new strands of DNA formed during replication ? (CBSE 1992, 2K)
Answer:
5′ – 3′ direction

Question 17.
Give the present-day representation of central dogma.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 5

Question 18.
State Chargaff s base complementary rule.
Answer:
The total molar amount of adenine in any specimen of DNA is always equal to that of thymine. In a given DNA A = T and G = C.

Question 19.
Define mutation
Answer:
It is an abrupt and distinct change in the structure of base pair. It is a discontinuous inheritable and sudden change in an organism.

Question 20.
What do you call the kind of mutation in which a single base is added to a base strand : (CBSE 2K)
Answer:
Frame shift mutation.

Question 21.
Sickle cell anaemia is caused due to abnormal haemoglobin. Which chain of haemoglobin is responsible for this disease?
Answer:
β – chain of haemoglobin

Question 22.
What do the triplets AUG and UGA respectively code for during proteins synthesis?
Answer:
AUG – Methionine UGA – Termination codon (Nonsense codon)

Question 23.
Name the technique used by Watson and Crick to propose the double-helical structure of DNA molecule?
Answer:

  • X-ray crystallography
  • X-ray diffraction method.

Question 24.
Who discovered nucleic acid DNA? What was it called them?
Answer:
Fredrich Meischer. It was called as nuclein.

Question 25.
(a) What is the length of the pitch of helix?
(b) What is the distance between 2 base pairs is a stand of DNA?
Answer:
(a) 3.4 nm
(b) 34 nm.

Question 26.
Why is the distance between the 2 nucleotide chains in a DNA maintained almost constant?
Answer:
The complementary base pairing between the 2 strands i.e. adenine and thymine (double bond) and Guanine and cytosine (triple bond) is responsible for the uniform distance between the 2 strands.

Question 27.
Name the process which occurs in virus where the formation of DNA occurs from RNA.
Answer:
Reverse transcription.

Question 28.
Name the components ‘a’ and ‘b’ in the nucleotides with a parent given below. (CBSE, Delhi 2008)
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 6

Question 29.
What are histones?
Answer:
Histones are positively charges proteins found in association with DNA in eukaryotic cell.

Question 30.
Name 2 organisms where RNA as the genetic material.
Answer:
QB Bacteriophage, Tobacco mosaic virus.

Question 31.
Name the main enzyme involved in the replication of DNA.
Answer:
DNA dependent DNA polymerase enzyme.

KSEEB Solutions

Question 32.
Name the types of synthesis ‘a’ and ‘b’ occurring in the replication of DNA as shown below (CBSE 2008)
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 7

  • a – continuous synthesis
  • b – discontinuous synthesis.

Question 33.
What is replication fork in DNA?
Answer:
During DNA replication the unwinding of DNA leads to the formation of ‘Y’ shaped structure to the 2 strands of DNA duplex. This is known as the replication fork.

Question 34.
Define a cistron
Answer:
A cistron is defined as the length of mRNA, that codes for a polypeptide.

Question 35.
What is meant by hnRNA?
Answer:
The hnRNA is the precursor of mRNA transcribed by RNA polymerase II in eukaryotic cells hn represents heteronuclear RNA.

Question 36.
When and at what end does the ‘tailing’ of hnRNA take place? (AI 2009)
Answer:
After splicing, tailing occurs at the 3′ end of hn RNA.

Question 37.
Why is hn RNA required to undergo splicing?
Answer:
Since hn RNA contain both the coding sequences (exon) and the non-coding sequences (introns), hn RNA has to undergo splicing for the removal of introns.

Question 38.
Who proposed the operon concept?
Answer:
Francois Jacob and Jacque Monod proposed the operon concept.

Question 39.
Name the induce of lac operon in E. Coli?
Answer:
Lactose.

KSEEB Solutions

Question 40.
Given below is a schematic representation of a lac operon in the absence of an inducer. Identify ‘a’ and ‘b’ in it
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 8
a – Repressor
b – operator.

Question 41.
What term is given to a single base DNA differences?
Answer:
When the repressor binds to the operator, the operon is switched off and transcription is stopped.

Question 42.
Expand VNTR.
Answer:
Variable Number of Tandem Repeats.

Question 43.
Regulation of lac operon by repressor is referred to as negative why is it so?
Answer:
When the repressor binds to the operator, the operon is switched off and transcription is stopped.

Question 44.
Name 2 plants whose genome have been sequenced.
Answer:
Rice and Arabidopsis.

Question 45.
Define DNA polymorphism
Answer:
Inheritable mutations at high frequency in a population. It refers to the variation at genetic level.

Question 46.
Who discovered the techniques of DNA fingerprinting
Answer:
Alec Jeffreys.

Question 47.
What is a probe in DNA-finger printing?
Answer:
A probe is a short stretch of DNA, with the nucleotide sequence that is complementary to that of VNTR sequence.

Question 48.
Name the branch of science that HGP is closely associated with.
Answer:
Bioinformatics.

KSEEB Solutions

Question 49.
Given below is the sequence of steps of transcription in eukaryotic cell. Fill up the blank 1, 2, 3, 4 left in the sequence.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 9
Single nucleotide polymorphism (SNPS).
1 – RNA polymerase
2 – hn
3 – processed RNA
4 – tail.

Question 50.
Write any 3 unusual bases present in Yeast’s alanine tRNA with their sources.
(CBSE 2004)
Answer:
All tRNA have 2 unusual bases – dihydro uridine (derived from uracil) and pseudo uridine (from uraicl. The third common unusual base is hypoxanthine (from adenine).

2nd PUC Biology Molecular Basis of Inheritance Two Marks Questions

Question 1.
What are 5′ end, 3′ end of a polynucleotide chain?
Answer:

  • The poly nucleotide chain has at one end a free phosphate at the 5′ end of the pentose sugar. Which is referred to as 5′ end.
  • The other end of polynucleotide chain has a free. 3′ – OH group at the 3rd end of the pentose sugar which is referred to as 3′ end of polynucleotide chain.

Question 2.
Label the diagram 1, 2, 3, 4, 5, 6. (CBSE 2004)
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 10

  • 5′ end
  • Ribosome binding site
  • Start
  • Stop signal
  • Open reading frames (ORFs)
  • 3′ end.

Question 3.
Name the components (parts) A and B of the transcription unit ‘given below.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 11
A – Promoter
B – Coding strand.

Question 4.
(a) Who first proposed semi-conservative mode of replication of DNA.
(b) Which organisms is used in this experiment?
(c) Name the techniques used
(d) What is the result of first, second and 3rd generation?
Answer:
(a) Watson and Crick (1953)

  • Eschirisia Coli

(i) Use heavy isotope 15N instead of normal 14
(ii) Use of Cesium chloride based density gradient configuration with ethidium bromide as flarochrome.
(iii) First generation DNA is hybrid as with intermediate heaviness due to presence of both 14N and 15N strands. The parent generation is heaviest.
Second generation 50% light (14N/14N) and 50% intermediate heavy (14N/15N). Third generation75% light (14N/14N) and 25% of intermediate heaviness (14N/15N).          This is possible only if DNA is double stranded with semi conservative replication. (one parent strand and other new strand) UACGAG AGAUUUi

Question 5.
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 12
Study the messenger RNA segment given above which is complete to be translated into a polypeptide chain

  • Write the codons and ‘a’ and ‘b’
  • What do they code for?
  • How is peptide bond formed between 2 amino acids in the ribosome (CBSE 2008)

Answer:
(i) a – AUG
b – UAA | UAG | UGA

(ii) AUG codes for methionine.
UAA/ UAG/ UGA is stop / nonsense codon.

(iii) Peptide bond is formed between -COOH group of P- site amino acid and NH2 – group of A – site amino acid with the help of ribozine peptidyl transferase provided by ribosome.

Question 6.
State any one reason to explain why RNA virus mutate and evolve faster than other viruses.
Answer:
RNA is an unstable highly reactive molecule due to its single-stranded structure and exposure of its nitrogen bases. But DNA is stable, the molecule, as its, nitrogen bases are not exposed, became its double-helical nature.

Repressor binds to the operator region
(o)  and prevent RNA polymerase from transcribing the given.
Look at the figure above depicting lac operon in E-Coii.
(a) What could be a series of events when an inducer is present in the medium in which E. Coli is growing.
(b) Name the inducer.
Answer:
(a) When the inducer is present in the medium it is absorbed at first slowly into the bacterium. The inducer binds with the repressor attached to the operator gene the repressor leaves the operator gene and allows the RNA polymerase to pass from promotor to the structural genes for transcription or formation of a polycistronic mRNA.
(b) Lactose or galactoside.

KSEEB Solutions

Question 7.
Draw schematically a single polynucleotide strand (with at least three nucleotides)
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 13

Question 8.
Differentiate between euchromatin and heterochromatin.
Answer:
Euchromation

  • These are the regions where chromatin is loosely packed.
  • Euchromation stains lighter
  • This is transcriptionally more active.

Hetero chromatin

  • These are the regions where chromatin is tightly packed
  • Hetero chromatin stains darker
  • This is transcriptionally less active or inert.

Question 9.
Write the functions of DNA polymerase.
Answer:

  • DNA polymerase catalyses the polymerisation of nucleotides into a polynucleotide strand of DNA.
  • It carries out proof reacting in prokaryotes; it moves back and removes any wrong bases added before polymerisation continues.

Question 10.
Compare the roles of the enzymes DNA polymerase and DNA ligase in the replication fork of DNA.
Answer:
DNA polymerases the nucleotides in the 5’ —> 3′ direction, as a continuous stretch on the template strand with 3′ —> 5′ polarity and short stretches on the template strand with 5′ —> 3′ polarity. —> DNA ligase joins the short stretches of DNA formed on the template strand with 5′-3′ polarity.

Question 11.
Write the differences between mono-cistronic and polycistronic mRNAs.
Answer:

Monocistromic mRNA Polycistromic mRNA
(1) It is the mRNA that can code for only one polypeptide i.e. it has one cistron.
(2) it is normally found in eukaryotic cell.
(1) It is the mRNA that can code for more than one polypeptide i.e. it has more than one cistron.
(2) It is found in prokaryotic cell.

Question 12.
Differentiate between Exons and Introns.
Answer:
Exons (HOTS)

  • They are the coding sequence of DNA/RNA transcript, that form parts of mRNA and code for different regions of the polypeptide.
  • They are joined together during splicing to make the information continuous.

Introns

  • Introns are non coding sequences of DNA/ RNA transcript that do not become part of mRNA.
  • They are removed during splicing.

Question 13.
Genetic code is specific and nearly universal justify. (Delhi 2008)
Answer:
Since a codon codes only for a particular amino acid, genetic code is specific or unambiguous. A codon codes for one particular amino acid in all living, be it a bacterium or a human, hence it is universal.

KSEEB Solutions

Question 14.
Why is that transcription and translation can be coupled in prokaryotic cell but not in eukaryotic cells?
Answer:
(a) In prokaryotes, mRNA does not require any processing to become active

(b) Transcription and translation occur in the same compartment cytosol, as there is no well defined nuclear membrane. There for it can be coupled.

In eukaryotic m-RNA has to be processed (splicing) before it become active. Since RNA is synthesised inside the nucleus and translation occurs in the cytoplasm, coupling of transcription and translation is not possible.

Question 15.
Why is it essential that RNA binds to both an amino acid and an mRNA codon, during protein synthesis?
Answer:
(a) By binding an amino acid at its 3′ end the tRNA transports the amino acid to the site of protein synthesis.

(b) It binds to the codon on mRNA through hydrogen bonds with its anticodon as it is recognised by the codon of the amino acid it carries, it is to release the amino acid at the correct site for polypeptide elongation.

Question 16.
What is an operon? How does lactose act as an inducer in the lac operon? (Al 2008)
Answer:
All the genes controlling a metabolic pathway, collectively constitute an operon. Lactose binds to the repressor and inactivates it, and prevents it from binding to the operator. As a result, the RNA polymerase gets access to the promoter and transcription proceeds, ie., the operon is induced to function.

Question 17.
Explain VNTR as the basis of DNA finger printing.
Answer:

  • Variable number of tandem repeats belong to the class of satellite DNA referred to as mini satellite.
  • The number of repeats shows very high degree of polymorphism as a result the size of VNTR varies.
  • After hybridisation of DNA sample with VNTR probe, an autoradiogram developed which gives many bands give the characteristic pattern of an individual DNA.

Question 18.
Give the applications of DNA fingerprinting?
Answer:

  • To identify criminals in the forensic labs.
  • To determine the biological parent in case of dispute.
  • To verify whether an immigrant is really a close relative of the mentioned resident.
  • To identify racial groups to rewrite the biological evolution.

Question 19.
How is nucleosome formed? Draw a diagram of the nucleosome.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 14
In eukaryotes, histones which are positively charged proteins, become organised as a unit of 8 molecules, called histone octamer. The negatively charged DNA is wraped around the positively charged histone octamer, to form the structure, called nucleosome.

KSEEB Solutions

Question 20.
What is meant by R – cells and S – cell with which Frederick Griffith carried out his experiments on Diplococcus pneumonial? What did he prove from these experiments?
Answer:
R – cells are those bacterial cells that form rough colonies, without a capsule and are non-virulent. S – cells are those cells, that form smooth colonies, with a capsule and are virulent. He proved that virulence of S – cells had some how been transferred into R – cells, which became transformed into S- cells, it is the genetic material that had effected transformation.

Question 21.
Draw the schematic representation to show continuous or discontinuous synthesis of DNA (replication of DNA) and label it.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 15

Question 22.
Explain frames shift mutation.
Answer:
(a) It is the type of mutation where insertion or deletion of one or two bases changes the reacting frame from the point of insertion or deletion.

(b) When three or multiples of three bases are added there is addition of one or more amino acids and the reacting of the frame remains unaltered after that. This provides the genetic proof that codons are triplets.

Question 23.
A t-RNA is charged with the amino acid phenylalanine.
(a) At what end of the tRNA is the amino acid attached ?
(b) Give the mRNA codon that codes for phenylalanine.
(c) Which enzyme is responsible for this attachment ?
Answer:
(a) Amino acid is attached to the 3′ end of t-RNA.
(b) UUC or UUU are the codes for phenylalanine.
(c) Amino Acyl tRNA synthetase enzyme is responsible for this attachment.

Question 24.
(a) Draw the schematic diagram of tRNA showing the following.
(i) Methionine attached to the amino acid accept site.
(ii) Correct base sequence at the anticodon loop
(b) Write the role of “untranslated regions” on mRNA segment play in protein synthesis?
Answer:
(a)
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 16
(b) They are needed for efficient translation

Question 25.
Draw a labelled diagram depiciting schematically the process of elongation in translation
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 17
Question 26.
Describe the goals of Human genoine project.
Answer:
The major goals of Human genome project are

  • Determine the sequence of the 3 billion base pairs present in Human DNA.
  • Identify the genes in the human DNA.
  • Store the information in data bases.
  • Improve the tool for data analyses.
  • Address ELSI (ethical, leagal, social issues) that arise from project.
  • Transfer the technologies to other sectors.

2nd PUC Biology Molecular Basis of Inheritance Five Marks Questions

Question 1.
Describe the features of double helical model of DNA.
Answer:
The features of double helical DNA are

  • It is made up of 2 polynucleotide of sugar phosphate and the nitrogen bases inside.
  • The two chains have antiparallel polarity ie., one has 5′ —> 3’ polarity and the other with 3′ —> 5′ polarity.
  • The bases of 2 strands are joined by double hydrogen bond between adenine and thymine and triple hydrogen bond between guanine cytosine.
  • The distance between 2 base pairs is 0.34nm and the distance between each turn is 3.4nm. Each turn consists of 10 base pairs.
  • The plane of one base pair stacks over the other in double helen this gives the stability of double helical structure.

KSEEB Solutions

Question 2.
(1) Represent diagramatically the Watson Crick model for semi conservation replication of DNA.
(2) Differentiate between continuous and discontinuous synthesis of DNA.
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 18
(2)

Continuous Discontinuous
(a) One stand of DNA is synthesised as a continuous strech in the
5′ —>3′ direction.
(a) Short stretches  are synthesised in the 5’—>3′ direction from replication fork
(b) The template strand of DNA strand is with 3’—>5′ polarity. (b) The DNA strand with 5′ —> 3′ polarity is the template strand  for this.
(c) No need of enzyme ligase (for joining) (c) DNA ligase  enzyme is required  for joining short stretches.
(d) There is no need for primers. (d) There is need for primers.

Question 3.
Write short notes on different types of RNAs.
Answer:
The RNAs are of 3 types

  • Messenger RNA (m – RNA)
  • Transfer RNA (t – RNA)
  • Ribosomal RNA (r – RNA)

m – RNA It provides the template for polypeptide synthesis, ie., it decides the sequence of amino acids in the polypeptide through the sequence of bases on it. t – RNA It has the shape of clover leaf in two dimensional structure It transports the amino acids to the site of protein synthesis it recognises the codon on m RNA. r – RNA It plays the structural and catalytic role during translation.

Question 4.
Define genetic code and write its salient features.
Answer:
Genetic code is the relationship between the sequence of nucleotides on mRNA and sequence of amino acids in the polypeptide.
Its features are

  • The genetic code is universal ie., the codons code for one amino acid is same in all organism.
  • Codons are triplet codons and there are 64 codons. 61 codons code for twenty different amino acids while the other three (UAA, UAG, UGA) are termination codons which do not code for any amino acids.
  • Each codon codes for only one particular amino acid. Therefore it is unambiguous.
  • Some amino acids are coded by more than one codon there for it is said to be degenerate.
  • The codons are read in a continuous manner, without any punctuation ie., codons are comma less.
  • AUG has dual functions of coding for metheonine as well as functioning as initiation codon.

KSEEB Solutions

Question 5.
(1) Represent schematically the process of transcription in eukaryotic cells.
(2) How does DNA polymerase function in the replication fork of DNA?
Answer:
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 19
(2) The DNA polymerase can catalase the polymerisation of nucleotides only in one direction is 5′ —> 3′ direction.
On the template stand with 3’—>5′ polarity the polymerisation occurs continuously and on the template strand with 5′ —> 3′ polarity, polymerisation occurs in short stretches (discontinuous synthesis).

Question 6.
(a) Explain with the help of schematic representation, the lac operon in E-Coli.
(b) Mention the role of lactose in this operon.
Answer:
(a) Jacob and Monod explained that lactose inducer the expression of genes leading to its catabolism.
2nd PUC Biology Question Bank Chapter 6 Molecular Basis of Inheritance 20
(b) Role of lactose

  • Lactose is a substrate for the enzyme (3- galactosidase.
  • It function as the inducer and regulates the switching on and off of the operon.
  • When lactose is present, it combines with the repressor protein which otherwise has a high affinity for the operator.
  • This inactivates the repressor from binding to the operator and hence transcription continuous ie., the operon is switched on.

Leave a Reply

Your email address will not be published. Required fields are marked *