In This Post we are providing Chapter-MOLECULAR BASIS INHERITANCE NCERT MOST IMPORTANT QUESTIONS for Class 12 BIOLOGY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.
NCERT MOST IMPORTANT QUESTIONS ON MOLECULAR BASIS INHERITANCE
1.Two claimant fathers filed a case against a lady claiming to be the father of her only daughter. How could this case be settled identifying the real biological father?
Ans.This case to identify the real biological father could lee settled lay DNA – finger printingtechnique. In this technique :-
- first of all, DNA of the two claimants who has to be tested is isolated.
- Isolated DNA is then digested with suitable restriction enzyme & digest is subjected to gelelectrophoresis.
- The fragments of ds DNA are denatured to produce ss DNA by alkali treatment.
- The electrophoresed DNA is then transferred from get into a nitrocellulose filter paper where itis fixed.
- A known sequence of DNA is prepared called probe – DNA & is labelled with radioactive esotope32p & then probe is added to nitrocellulose paper.
- The nitrocellulose paper is photographed on X –ray film through auto radiography. The film isanalysed to determine the presence of hybrid nucleic acid.
Then, the DNA fingerprints of the two claimants is compared with the DNA fingerprint of the lady &her daughter, whosoever matches with each other would be declared as biological father of herdaughter.
2.The length of DNA in an 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 positivelycharged histone octamer into a structure called nucleosome. Atypical nucleosome consists of 200bp of DNA helix. Thenucleosomes are the repeating units that form chromatin fibres.
These chromatin fibres condense at metaphase stage of celldivision to form chromosomes. The packaging of chromatin at higher level requires additional set ofproteins called non-histone chromosomal proteins thus in nucleus, certain regions of the chromatinare loosely packed & they Stain lighter than the other region, these are called euchromatin. Theother region are lightly packed & they stain darker & are called heterochromatin
3.Describe the continuous & discontinuous Synthesis of DNA?
Ans.Synthesis of new strand of DNA takes place lay additionof fresh nucleotides to the 3 – OH group of the last nucleotideof the primer. This synthesis takes place in 5 direction&enzyme that catalyses this is DNA – polymerase
∴∴ synthesis of strand called leading strand iscontinuous.
The replication of second strand of the DNA molecule is
DISCONTINOUS on strand called lagging strand.
Primase initiates primer synthesis on strand near the fork. The RNA – primer thus formedprovides free for replication of single stranded region on lagging strand the newcomplementary strand is formed in small fragments of DNA called Okazaki fragments. It is calleddiscontinuous because it has to be initiated several times & every time an Okazaki fragment isproduced.
4.What are the three types of RNA & Mention their role in protein Synthesis?
Ans. There are three types of RNA :
- Messenger RNA (mRNA) :- It is a single – stranded RNA which brings the genetic information ofDNA transcribed on it for protein synthesis.
- Transfer RNA (tRNA) :- It has a clover leaf like structure which acts as an adapter moleculewhich contains an “anticodon loop” on one end that reads the code on one hand &” an amino acid acceptor end which binds to the specific amino acid on other hand.
- Ribosomal RNA (rRNA) :- Ribosomes provides the site for synthesis of protein &catalyse theformation of peptide bond.
5. Define bacterial transformation? Who proved it experimentally & how?
Ans. The transformation is a mode of exchange or transfer of genetic information betweenorganism or from one organism to another.
Fredrick Griffith tested the virulence of two strains of Diplococci to show transformation in thefollowing steps :-
- When S-III strains of bacteria are injected into mice. It developed pneumonia & died.
- When R-II strains are infected into mice, they did not develop pneumonia & survive.
- When heat – killed S-III strains of bacteria are injected into mice, No symptoms of pneumoniadevelops& mice remain healthy.
- When a mixture of heat – killed S-III strain & lives R-II strain is injected into mice, theydeveloped pneumonia & died.
From these results, Griffith concluded that the presence of heat – killed S-III bacteria must convertliving R-II type bacteria to type S-III so as to restore them the capacity for capsule formation. Thiswas called “BACTERIAL TRANSFORM ATION”
S strain →→ Inject into mice →→ Mice die
R strain →→Injct into mice →→ Mice live
S strain (heat-killed)→→Inject into mice →→ Mice live
S strain (heat-killed) + R strain (live) →→ Inject into mice →→Mice die
6. What is meant by semi conservative replication? How did Meselson and Stahl prove it experimentally?
Ans .Meselson and Stahl, performed an experiment using E.coli to prove that DNA replication is semi conservative.
- They grew E.coli in a medium containing 15NH4Cl15NH4Cl.
- Then separated heavy DNA from normal (14N) by centrifugation in CsCl density gradient.
- The DNA extracted, after one generation of transfer from 15N medium to 14N medium, had an intermediate density.
-The DNA extracted after two generations consisted of equal amounts of light and hybrid DNA.
-They proved that DNA replicates in a semiconservative manner.
7. What does the lac operon consist of? How is the operator switch turned on and off in the expression of genes in this operon? Explain.
Ans.Lac Operon consists of the following :
- Structural genes : z, y, a which transcribe a polycistronic mRNA.
- gene ‘z’ codes for b-galactosidase
- gene‘y’ codes for permease.
- gene‘a’ codes for transacetylase.
- Promotor : The site where RNA polymerase binds for transcription.
- Operator : acts as a switch for the operon
- Repressor : It binds to the operator and prevents the RNAPolymerase from transcribing.
- Inducer : Lactose is the inducer that inactivates the repressor by binding to it.
- Allows an access for the RNA polymerase to the structural gene andtranscription.
8. What is an operon? Describe the major steps involved in an operon?
Ans.Operon is a group of controller & structural genes which controls the catabolism of the cell geneticallyeg lactose operon / lac operon.
(i) When inducer or lactose is absent :-
The lac regulator gene synthesize a repressor protein by transcription & translation. This repressor protein binds with operator site of lac operon & blocks RNA polymerase. Thus, RNA polymerase unable to transcribe mRNA & structural gene unable to translate enzyme B-galactosidase.
(ii) When inducer or lactose is present :_
The lac regulator gene transcribe mRNA &synthesise active lac repressor protein & at the same time lactose is converted into isomer allolactose. Allolactose binds to active lac repressor due to which it is converted to inactive repressor. This inactive repressor is released from operator site of lac operon & RNA polymerase binds to promoter & starts to transcribe mRNA & forms β-galactosidase are which converts lactose into glucose &galactose.
Thus, presence of lactose determines whether or not lac. Repressor is bound to operator & genes are expressed on not.
9.Where do transcription & translation takes place in a prokaryotic cell? Describe the three stepsinvolved in translation?
Ans.In a prokaryotic cell both transcription & translation occurs in cytoplasm. It consist offollowing steps :-
(i)ACTIVATION OF AMINO ACIDS :- amino acids are activated in the presence of ATP lay enzaminoacyltRNASynthetase.
(ii)BINDING OF ACTIVATED AMINOACID WITH tRNA :- Activated amino acids binds with specific tRNA to form charged tRNA .
(iii)INITIATION OF POLYPEPTIDE CHAIN :- Initiation codon is AUG which codes for methionine. Initiation codon of mRNA binds to p-site of ribosome with the help of initiation factors.
(iv)ELONGATION OF POLYPEPTIDE CHAIN :-
(a)Second activated amino acid along itstRNA reaches the ‘A’ site & binds to mRNA codon next to AUG.
(b)A peptide bond is formed betweentwo amino acid by peptidyl transferase.
(c) Ribosomes translocation mRNA in -direction due to which free tRNA slips away &peptidyltRNA reaches at P – site. Now third amino acid reaches at A – site & process continues.
(d)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.
(i)
10.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 Harshey& chase whoworked with bacteriophage.
The bacteriophage on infection injects only the DNA into the bacterial cell & not the protein coat.
Bacterial cell treats the viral DNA as its own & subsequently manufactures more virus particles.
They grew some viruses on a medium that ‘contained radioactive Phosphorus & some other on medium that contained radioactive sulphur. Virus grown in the presence of radioactive phosphorus contained radioactive DNA but not proteins because DNA contains phosphorus. Similarly virus grown on radioactive sulfur contained radioactive protein because DNA does not contain sulfur.
Radioactive phages are allowed to infect E. coli bacteria & soon after infection the cultures weregently agitated in a blender to separate the adhering protein coat of virus from bacterial cell.It was found that when phage containing radioactive DNA was used to infect the bacteria itsradioactivity was found in bacterial cells indicating that DNA has been injected into bacterial cell so,the DNA is the genetic material & not proteins
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