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

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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 of 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 2.
If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Answer:

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

Question 4.
If coding strand in a transcription unit is written as follows:
5′-ATGCATGCATGCATGCATGCATGCATGC – 3′ Write down the sequence of mRNA.
Answer:
UACGUACGUACGUACGUACGUACGUACG

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesize a semi-conservative model of DNA replication? Explain.
Answer:
The complementary base pairing property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication. Watson & Crick observed that the nitrogenous bases are in complementary pairing in two strands of the double helix of DNA molecule. Such an arrangement of DNA molecules led them to hypothesize the semi-conservative mode of replication of DNA.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
The types of nucleic acid polymerases required for the synthesis of DNA and RNA are :

1. DNA polymerase I, II, and III – They help in the replication of DNA.
2. RNA dependent DNA polymerase – It helps in the synthesis of DNA from RNA (reverse transcription).
3. DNA dependent RNA polymerase – It helps in the synthesis of RNA from DNA (transcription).

In eukaryotes, there are at least three RNA polymerases in addition to those found in cell organelles.

• RNA polymerase I transcribes rRNA (28S, 18S, and 5.8 S).
• RNA polymerase II transcribes the precursor of mRNA called heterogeneous nuclear RNA (hnRNA).
• RNA polymerase III transcribes tRNA, 5SrRNA, and snRNAs.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
A. D. Hershey and Martha Chase performed an experiment to prove that DNA is the genetic material not the protein. They used T₂ bacteriophages which infect the Escherichia coil. To differentiate between DNA and protein during the experiment, they used radioactive isotopes of phosphorus (³²P) and sulphur (³⁵S) in the culture medium of bacterial hosts. They could not use radioactive isotnpes of carbon (¹⁴C) etc., because they are present in both molecules and would not distinguish the two. They grew two cultures of Eschcrichia cou.

One culture was supplied with radioactive sulphur, ³⁵S. The other culture was provided with radioactive phosphorus, ³²P. Radioactive sulphur gets incorporated into sulphur containing amino acids and therefore, becomes part of bacterial proteins. Radioactive phosphorus gets incorporated into nucleotides which form nucleic acids, mostly DNA. Therefore, bacteria of both cultures became labelled.

They then introduced bacteriophage T₂ in both bacterial cultures. The virus entered the bacteria where it multiplied. The viral progeny was tested in both cases. It was labelled, one type with radioactive protein and the other types with radioactive DNA. These labelled phages were introduced in new bacterial cells. It was found that phages with labelled protein did not make the bacterial host labelled, while those with labelled DNA made the host labelled. It indicated that viral DNA is the genetic material (not proteins) which is transferred to the infected host.

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

Repetitive DNA	Satellite DNA
(i) the sequence of N2 bases 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 the text.	(ii) Repeated sequence occur in tandem
(iii) Variability may or may not be present	(ii) Variability occurs

(b) mRNA and tRNA
Answer:

mRNA	tRNA
(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 the association of AA with anti-Condon for incor­poration

(c) Template strand and Coding strand
Answer:

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

Question 9.
List two essential roles of ribosomes during translation.
Answer:
Two essential roles of the ribosome during translation are as follows:

• They provide a surface for binding of mRNA in the groove of the smaller subunit of the ribosome.
• As a larger subunit of the ribosome has peptidyl transferase on its ‘P’ site, therefore it helps in joining amino acids by forming peptide bonds.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down sometime after the addition of lactose in the medium?
Answer:
Lac operon is switched on adding lactose in the medium, as lactose acts as inducer and make repressor inactive. Due to this switch on of lac operon system, (3-galactosidase is formed which converts lactose into glucose and galactose. As soon as lactose is consumed, the repressor again becomes active and causes a switch off (shut down) of the system.

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

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

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

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

Question 12.
Why is the Human Genome project called a megaproject?
Answer:
HGP (Human Genome Project) is called a megaproject because:

• it involved many countries (USA, UK, Japan, France, Germany, China) for determining the nucleotide sequences of genes
• it involved sequencing 3 x 10⁹ base pairs costing 9 billion US dollars
• it required bioinformatics databasing and other high-speed computational devices for analysis, storage, and retrieval of information.

Question 13.
What is DNA fingerprinting? Mention its applications.
Answer:
DNA Fingerprinting: Every human individual is characterized by unique print at the fingertips. The study of fingers, palm, and sole print is called dermatologyphics’.

Like prints of the fingertips, each individual has a unique DNA fingerprint. Unlike the prints of a finger, the DNA fingerprints can not be altered by surgery. The latter is exactly similar in all the cells and tissues of an individual. It can not be changed by medical treatment. The distinction of individuals on the basis of DNA fingerprint is due to the sequence of nucleotides in whole genomic DNA.

The technique to identify a person on the basis of his/her DNA specificity is called DNA fingerprinting. This was invented by Sir Alec Jeffreys in 1984 at Leicester University U.K. In India, Dr. V. K. Kashyap and Dr. Lalji Singh started this technique at CCMB, Hyderabad.

Question 14.
Briefly describe the following:
(a) Transcription
Answer:
Transcription – Formation of-RNA over the template of DNA.

The snRNA formed has codon information similar to the sense or coding strand of DNA with just U replacing T.

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

(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 the structure and function of the 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 a polypeptide chain (proteins).

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, modeling 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.

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

2nd PUC Biology Molecular Basis of Inheritance One Mark Questions

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

Question 2.
Capping and tailing are seen during the transcription of RNA.
a. How is this process done?
b. What is the use of this process?
Answer:
a. Capping is done by adding methylated guanosine at the 5′ end. Tailing is done by adding 200-300 adenylate residues at the 3′ end.
b. Capping and tailing are used to process the hnRNA. The fully processed hnRNA is called mRNA.

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

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

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

Question 6.
In the lac operon, there is no apo-repressor and co-repressor. But it is present in tryptophan operon.
a. Who proposed tryptophan operon?
b. What is the significance of the two types of repressors given above?
Answer:
a. Francois Jacob and Jacques Monod
b. Apo-repressor: This is a protein produced by the regulator gene which is unable to bind the operator gene.
Co-repressor: When tryptophan binds with the apo-repressor, it is known as co-repressor.

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

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

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

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

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

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

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

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:

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:
Frameshift 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 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:

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

Question 30.
Name 2 organisms where RNA is 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.

Question 32.
Name the types of synthesis ‘a’ and ‘b’ occurring in the replication of DNA as shown below (CBSE 2008)
Answer:

• a – continuous synthesis
• b – discontinuous synthesis.

Question 33.
What is a replication fork in DNA?
Answer:
During DNA replication the unwinding of DNA leads to the formation of a ‘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 contains 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.

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:

a – Repressor
b – operator.

Question 41.
What term is given to a single base DNA difference?
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 a 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 has 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 the 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 the VNTR sequence.

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

Question 49.
Given below is the sequence of steps of transcription in eukaryotic cells.
Fill up the blank 1, 2, 3, 4 left in the sequence.
Answer:

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 pseudouridine (from uracil. The third common unusual base is hypoxanthine (from adenine).

2nd PUC Biology Molecular Basis of Inheritance Two Marks Questions

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

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

Question 2.
Label the diagram 1, 2, 3, 4, 5, 6. (CBSE 2004)
Answer:

• 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:

A – Promoter
B – Coding strand.

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

• Use heavy isotope ¹⁵N instead of normal ¹⁴
  
• Use of Cesium chloride-based density gradient configuration with ethidium bromide as a fluorochrome.
• First-generation DNA is hybrid as with intermediate heaviness due to the presence of both ¹⁴N and ¹⁵N strands. The parent generation is the 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 strands) UACGAG AGAUUUi

Question 5.

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 a 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 the promotor to the structural genes for transcription or formation of a polycistronic mRNA.
(b) Lactose or galactoside.

Question 7.
Draw schematically a single polynucleotide strand (with at least three nucleotides)
Answer:

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

• These are the regions where chromatin is loosely packed.
• Euchromatin 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 notes on the structural gene, regulator gene, and operator gene.
Answer:
Structural gene – It is a type of gene containing the code for the synthesis of enzymes necessary for lactose catabolism.
Regulator gene – The regulator gene is a type of control gene which is involved in lactose catabolism.
Operator gene – It is also a type of control gene involved in lactose metabolism.

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 cells.	(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 cells.

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.
Amino acids are the building blocks of protein. How is a protein synthesised from aminoacids?
Answer:
By linking the aminoacids with the help of peptide bonds.

Question 14.
Why is that transcription and translation can be coupled in the 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. Therefore it can be coupled.

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

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

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

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

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

Question 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 fingerprinting.
Answer:

• Variable number of tandem repeats belong to the class of satellite DNA referred to as minisatellite.
• The number of repeats shows very high degree of polymorphism as a result the size of VNTR varies.
• After hybridization 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 the nucleosome formed? Draw a diagram of the nucleosome.
Answer:

In eukaryotes, histones which are positively charged proteins, become organised as a unit of 8 molecules, called histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer, to form the structure, called a nucleosome.

Question 20.
What is meant by R – cells and S – a cell with which Frederick Griffith carried out his experiments on Diplococcus pneumonia? 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 somehow been transferred into R – cells, which became transformed into S- cells, it is the genetic material that had affected transformation.

Question 21.
Draw the schematic representation to show the continuous or discontinuous synthesis of DNA (replication of DNA) and label it.
Answer:

Question 22.
Explain frames shift mutation.
Answer:

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

Question 23.
A tRNA 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 tRNA.
(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 the 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)

(b) They are needed for efficient translation

Question 25.
Draw a labeled diagram depicting schematically the process of elongation in translation.
Answer:

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 databases.
• Improve the tool for data analyses.
• Address ELSI (ethical, legal, social issues) that arise from the project.
• Transfer the technologies to other sectors.

2nd PUC Biology Molecular Basis of Inheritance Five Marks Questions

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

• It is made up of 2 polynucleotides 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 a double hydrogen bond between adenine and thymine and a 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 the double-helical structure.

Question 2.
(1) Represent diagrammatically the Watson Crick model for semi conservation replication of DNA.
(2) Differentiate between continuous and discontinuous synthesis of DNA.
Answer:

(2)

Continuous	Discontinuous
(a) One strand of DNA is synthesised as a continuous stretch in the 5′ → 3′ direction.	(a) Short stretches  are synthesised in the 5’→ 3′ direction from the replication fork
(b) The template strand of the DNA strand is with 3’→ 5′ polarity.	(b) The DNA strand with 5′ → 3′ polarity is the template strand for this.
(c) No need for enzyme ligase (for joining)	(c) DNA ligase enzyme is required for joining short stretches.
(d) There is no need for primers.	(d) There is a 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 a cloverleaf in a two-dimensional structure It transports the amino acids to the site of protein synthesis it recognizes 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 the sequence of amino acids in the polypeptide.
Its features are

• The genetic code is universal ie., the codons code for one amino acid is the same in all organisms.
• 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 that 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 methionine as well as functioning as initiation codon.

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:

(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.

(b) Role of lactose

• Lactose is a substrate for the enzyme (3- galactosidase.
• Its functions 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.