NCERT Solutions for the molecular basis of inheritance class 12 notes

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NCERT Solutions molecular basis of inheritance class 12 notes


Class12th 
Chapter No06
ProvidingQuestions And Answers, Notes & Numericals PDF
Chapter Namemolecular basis of inheritance
BoardCBSE
Book NCERT
SubjectBiology
Medium English
Study MaterialsFree VVI Study Materials are Available
Download PDF molecular basis of inheritance handwritten notes pdf download

molecular basis of inheritance class 12 notes


1. Nucleic acids are long polymers of nucleotides. 

Nucleotides

(i) Nitrogenous base

(ii) Pentose sugar

(iii) Phosphate group

2. Two types of nucleic acids are found in living organisms, e.g., Deoxyribonucleic Acid (DNA) and Ribonucleic Acid (RNA).

3. DNA acts as the genetic material in most organisms. It is the information molecule that codes for all the metabolic processes of life. Sups

4. RNA helps in the transfer and expression of information. It functions as a messenger RNA (mRNA) for effecting the translation of proteins. emios AMIL am had to countimo at best.  It also acts as a genetic material in some viruses.

(I) It may also function as an adapter, structurally and as a catalytic molecule. 

(ii) Although both RNA and DNA can act as genetic material, DNA being chemically and structurally more stable is a better choice.

(iii) RNA is the first to evolve and DNA was derived from RNA. 

(iv) The hallmark of the double-stranded helical structure of DNA is the hydrogen bonding between the bases from opposite strands.

5. Adenine pairs with thymine through two H-bonds and guanine with cytosine through three H-bonds. This makes one strand complementary to the other.

6. The DNA replicates semiconservative, which means the replicated molecule will have one strand new and one of the original.

The process is guided by the complementary H-bonding. An enzyme DNA-dependent, DNA polymerase, uses each DNA strand as a template to catalyse the polymerisation of deoxynucleotides, on both the strands. These enzymes are highly efficient as they have to catalyse the polymerisation of a large number of nucleotides in a very short time. 

NCERT Solutions for the molecular basis of inheritance class 12 notes 1024x615 1

7. A segment of DNA that codes for RNA or a polypeptide, can be referred to as a gene.

(i) During transcription also, one of the strands of DNA acts as a template to direct the synthesis of complementary RNA. 

(ii) In bacteria, the transcribed mRNA is functional, hence can directly be translated.

(iii) In eukaryotes, the genes are split. The coding sequences, i.e., exons, are interrupted by non-coding sequences, i.e., introns. 

(iv) Introns are removed and exons are spliced back together to produce functional RNA.

(v) The messenger RNA contains the base sequences that are read in a combination of three (to make triplet genetic code) to code for an amino acid. These are called codons.

(vi) The genetic code is read again on the principle of complementarity by tRNA which acts as an adapter molecule. There are specific tRNAs for every amino acid.

(vii) The RNA binds to the specific amino acid at one end and pairs through H-bonding with codons on mRNA through its anticodons,

(viii) The site of translation (protein synthesis) is ribosomes, which bind to mRNA and provide a platform for the joining of amino acids.



(ix) One of the RNA acts as a catalyst for peptide bond formation, which is an example of an RNA enzyme (ribozyme). Since transcription and translation are energetically very expensive processes, these have to be tightly regulated.

8. Regulation of transcription is the primary step for the regulation of gene expression. In bacteria, more than one gene is arranged together and regulated in units called operons.

9. Lac operon is the prototype operon in bacteria, which codes for genes responsible for the metabolism of lactose. The operon is regulated by the amount of lactose in the medium, where the bacteria are grown. Therefore, this regulation can also be viewed as the regulation of enzyme synthesis by its substrate.

10. Human Genome Project was a mega project that aimed to sequence every base in the human genome. This project has yielded much new information. Many new areas and avenues have opened up as a consequence of the project.

11. DNA fingerprinting is a technique to find out variations in individuals of a population at the DNA level. It works on the principle of polymorphism in DNA sequences. It has immense applications in the field of forensic science, genetic biodiversity and evolutionary biology, and Kinship relationships.


Class 12 molecular basis of Inheritance neet questions


Question 1. Group the following as nitrogenous bases and Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.

Answer Nitrogenous bases are Adenine, Thymine and Uracil. Nucleosides are Cytidine, Guanosine and Cytosine.

Question 2. If a double-stranded DNA has 20% of cytosine, calculate the per cent of adenine in the DNA.

Answer As in DNA molecule, AT and C= G. Therefore, if C is 20% then G is also 20% And C+G is 40%. Now, A+T will be 60% (100-40 = 60). As A = T so, both will be 30% each. 

Question 3. If the sequence of one strand of DNA is written as follows 5-ATGCATGCATGCATGCATGCATGCATGC-3 Write down the sequence of the complementary strands in the 5′ →→ 3′ direction. 

Answer It will be 3-TACGTACGTACGTACGTACGTACGTACG-5′

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

Answer If coding strand sequence is given, we’ll reduce the template strand sequence from this, which will be as follows:

3-TACGTACGTACGTACGTACGTACGTACG-5′ Now, the RNA strand will be complementary to this, with U replacing T as follows

5-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3 

Question 5. Which property of DNA double helix led Watson and Crick to hypothesize a semi-conservative mode of DNA replication? Explain.

Answer The property that DNA has

(i) two strands running opposite to each other, wherein bases will always pair with their counterpart-A with T and G with C (specific pairing). 

(ii) if H bonds break and bases of one strand lie exposed, or unpaired, they will easily pair up with free nucleotides as well. This led them to hypothesise that H-bonds will break and expose the bases of two strands. Free nucleotides present in the cytoplasm will pair up with the bases of each strand to make a copy, with the help of enzymes. As each new DNA molecule will have one old strand (that served as a template) and one newly constructed, the replication was called semi-conservative.

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

(0) DNA-dependent RNA polymerases I, II and III (for RNA synthesis from DNA template). (1) DNA polymerases I and II (for DNA replication using DNA as a template). (i) RNA-dependent DNA polymerase (for DNA synthesis using RNA as a template).

Question 7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?

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Answer 

(1) They grew some bacteriophages in a medium that contained radioactive phosphorus and some in another medium that contained radioactive sulphur.

(ii) Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein as phosphorus is present only in DNA. 

(iii) Viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

(iv) It was found that bacteria that were infected with bacteriophages that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria.

(v) Bacteria that were infected with viruses that had radioactive proteins were not radioactive. This indicated that proteins did not enter the bacteria from the viruses.

(vi) This was clear-cut proof that DNA is the genetic material that is passed from virus to bacteria.

Question 8. Differentiate between the following

(a) Repetitive DNA and satellite DNA

(b) mRNA and t RNA

(c) Template strand and coding strand

Answer 

S.NORepetitive DNASatellite DNA
11 It is the non-coding DNA with multiple copies of identical sequences which may lie in tandem or interspersed.Refers to non-coding tandem repeat sequences.
22. These can be a few base pairs These are generally short sequences of hundreds or thousands of base pairs. These are generally Short sequence repeats (up to 60 base pair long).
33. In CSCI density gradient analysis, it appears as small dark bands. appears as light bands.This appears as small dark bands.
S.NOmRNA RNA
11. It is called messenger RNA and carries the codes for amino acid sequences.It is called transfer RNA as it carries amino acids to the site of protein synthesis.
22. It is a long chain of bases with large molecular weight (25-200000 dalton). It is a clover leaf-like structure, with a small molecular weight (25000 daltons).
33. It is synthesised by RNA polymerase IISynthesised by RNA polymerase III.
S.NO Template strand Coding strand
1It is the strand that is transcribed into RNAIt has the same sequence as the mRNA
2It is called the anti-sense strand.It is called sense or non-template strand,
3It has 3-5′ polarity.It has a 3-5′ polarity.

Question 9. List two essential roles of ribosomes during translation.

Answer

(1) To serve as a platform for protein synthesis by binding mRNA in the smaller sub-unit. 

(ii) Larger subunit has an enzyme peptidyl transferase on its P-site that facilitates the formation of peptide bonds between amino acids to help build the polypeptide chain.

Question 10. In the medium, where E. coli was growing, lactose was added, which induced the lac operon. Then, why does the lac operon shut down sometime after the addition of lactose in the medium?

NCERT BIO XII C06 E01 010 S01

The lactose then induces the operon in the following manner: 

(i) The repressor of the operon is synthesized all the time from the i’s gene.

(ii) It binds the repressor protein which binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon 

(iii) In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer.

(iv) This allows RNA polymerase access to the promoter and transcription proceeds.

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

(a) Promoter (b) tRNA

(c) Exons

Answer

(a) Promoter is an essential component of the transcription unit. It is located at the beginning of 5-end. It provides a site for the attachment of transcription factors and RNA polymerase.

(b) tRNA is a small-sized RNA molecule that takes part in transcription. It physically picks up activated amino acids from the cytoplasm and carries (transfers) them to ribosomes, where they join together through peptide bonds and leave the RNA to fetch more amino acids. 

(c) Exons are the coding sequences of DNA that are transcribed and translated. 

Question 12. Why is the human genome project called a mega project?

Answer Human genome project is called a mega project because 

(i) it aimed to determine the nucleotide sequence of the complete human genome which was a task of enormous magnitude. 

(ii) a total of 3 x 10° base pairs were to be sequenced and the cost was about 9 billion US dollars.

(iii) it required bioinformatics database techniques and other contemporary devices for the analysis, storage and retrieval of information. 

(iv) Many countries worked jointly to complete this timed project.

Question 13. What is DNA fingerprinting? Mention its application

Answer DNA fingerprinting is a technique based on the fact that the DNA of any two people is alike in some regions but is significantly different in others.

This difference is the result of enormous variations in some non-coding sequences due to mutations. This is called polymorphism. Being sexually reproducing species, these sequences are inherited from the parents.

Although every individual’s DNA sequences are unique just like fingerprints, still there is some sharp resemblance with parent DNA, that can be observed in electrophoresis patterns. bet.

This is the basis of its application as a test to 

(i) Sort out maternity /paternity issues Child’s DNA fingerprints can be matched with those of the parent.

(ii) Solve forensic medicine cases A sample from the crime scene (a few drops of blood, semen or even hair follicle cells) can be used to identify the perpetrator.

(ii) Lineage studies It is used to study the human lineage by making comparative studies with other problems. northwind sends nodes 

(iv) Population and genetic diversities

(v) It is also being used to study the cause of disease at the molecular/genetic level.

Question 14. Briefly describe the following:

(a) Transcription

(b) Polymorphism

(c) Translation

(d) Bioinformatics

Answer

(a) Transcription is the process of copying information from one of the strands of DNA to assemble RNA that is required for the synthesis of proteins.

RNA is assembled simply based on the complementarity of the DNA strand, only uracil is substituted in place of thymine. Only a small segment of DNA that codes for a polypeptide is copied,

(b) Polymorphism refers to the variation in DNA arising through mutation at non-coding sequences. Such variations are unique to specific sites of DNA and can occur due to deletion, insertion or substitution of bases.

Polymorphism can be observed by making fragments of DNA samples and separating them through electrophoresis. It is the basis for DNA fingerprinting.

(c) Translation is the process of polymerization of polypeptide chain using mRNA as a template, with the help of ribosomal units. The sequence of amino acids is dictated by the sequence of triplet codons on the mRNA transcript.

(d) Bioinformatics is the application of computer science and Information technology in biology and medicine to process information, analyse data and create new knowledge.


molecular basis of inheritance exercise


Question 1. What is the function of histones in DNA packaging? 

Answer Histones are proteins rich in basic amino acids, that are intimately associated with DNA molecules and help in their extensive coiling around to facilitate organisation into tightly packed chromosomes.

Question 2. Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active? 

Answer Differences between heterochromatin and euchromatin

S.NOHeterochromatin Euchromatin
11. It is a darkly stained region of the chromatin (chromosome).It is a lightly stained region.
22 It is the compactly coiled region and thus, has more DNAIt is a loosely coiled region and thus has less DNA
3It is transcriptionally inert and can not be transcribed into mRNA due to the very tight coilingit is transcriptionally active and is transcribed Into mRNA

Question 3. The enzyme DNA polymerase in E. coil is a DNA-dependent polymerase and also can proofread the DNA strand being synthesised. Explain. Discuss the dual polymerase. 

Ans:- In bacteria, three types of DNA polymerases are there. All of them can add nucleotides in the 5′-3′ direction. They possess exonuclease activity as well.

DNA polymerase III can proofread the newly synthesised strand and senses the wrong base insertions. It deletes wrong bases and helps correct the mistake by putting in the right one.

The only mistake it can not correct is the substitution of uracil in place of thymine. DNA polymerase I can repair any damages done to DNA by UV exposure, etc., or the leftover proofreading mistakes. It detects mutations caused by UV, removes mismatched pairs and puts back the right ones.

Question 4. What is the cause of the discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?

Answer One of the DNA strands opens in 3→ 5′ direction and it can be replicated easily by the addition of nucleotides in the 5′-3′ direction as this is the preferred direction of nucleotide addition by the polymerase enzyme and hence, this new strand builds on continuously.

The other strand with a polarity of 5’3′ replicates discontinuously as only short segments can be made in this direction. These small stretches called Okazaki fragments are joined together by DNA ligase enzyme that closes the nicks.


molecular basis of inheritance handwritten notes


Question 1. Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as the genetic material. 

Answer In Griffith’s experiment transformation can be defined as a change in the genetic constitution of an organism by picking up DNA from the environment (from other dead organisms).

It helped in the identification of DNA as a genetic material. When heat was used to kill the virulent bacteria, they die but not their DNA. 

This DNA when picked up by non-virulent bacteria made them capable of causing infection. Since the ability to cause infection could be passed on by these organisms to their progeny, it was deciphered that DNA was the material that was inherited. 

Question 2. Discuss the significance of the heavy isotope of nitrogen in Meselson and Stahl’s experiment.

Answer The Meselson and Stahl’s experiment used a heavy isotope of 15N in the nutrient medium to grow Escherichia coli (E. coli), for several generations.

They wanted to test the three alternative models of DNA replication, i.e., conservative, semi-conservative and dispersive. After growing the bacteria with a heavy isotope, they then transferred them into a new medium with a lighter isotope of nitrogen, 14N.  newly synthesised DNA

It was seen that each DNA molecule after replication contained both heavy nitrogen and light nitrogen, which proved Watson and Crick’s

The bacteria grown in heavy nitrogen got this isotope incorporated in their Now, they wore transferred to a growth medium containing lighter isotope N for one round of replication. This lighter isotope would incorporate into any semiconservative model of DNA replication. 

Question 3. Define a cistron. Giving examples differentiate between monocistronic and polycistronic transcription units. I to zero

Answer A cistron is a stretch of base sequences that codes for one polypeptide chain including adjacent control regions. It may also code for RNA.

RNA molecule or may perform other specific functions including regulating functions of other cistrons. This term has replaced the definition of a gene.

Monocistronic transcription unit will have all the regulatory and coding sequences for a single polypeptide, whereas polycistronic may have coding sequences for more than one polypeptide. In eukaryotic cells almost all the messenger RNAs are monocistronic. In prokaryotes, the lac operon coding sequence would be an example of a polycistronic DNA region.

Question 4. During DNA replication, why is it that the entire molecule does not open in one go? Explain the replication fork. What are the two functions that the monomers (dNTPs) play?

Answer: While replicating, the entire DNA molecule to keep the whole molecule stabilised does not open in one go because it would be highly expensive energetically. 

Unwinding creates tension in the molecule as uncoiled parts start forming supercoils due to the interaction of exposed nucleotides. Instead, the helicase enzyme acts on the double strand at our site and a small stretch is unzipped. Immediately, it is held and stabilised by single-strand

binding proteins. Slowly with the help of enzymes, exposed strands are copied as a point of unwinding moves ahead in both directions. It gives the appearance of a Y-shaped structure which is called a replication fork. The two functions that the monomer units of NTPs play are: 

(i) They pair up with the exposed nucleotides of the template strand and make phosphodiester linkages and release a pyrophosphate. 

(ii) hydrolysis of this pyrophosphate by the enzyme pyrophosphatase releases energy that will facilitate making hydrogen bonds between free nucleotides and bases of the template strand.

Question 5. Retroviruses do not follow central dogma. Comment

Answer: Retroviruses do not follow the central dogma of biology (DNA RNA protein) because their genetic material is not DNA. Instead, they have RNA that is converted to DNA by the enzyme reverse transcriptase,

Question 6. What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?

Answer If histones were mutated and made rich in acidic amino acids, they will not be able to serve the purpose of keeping the DNA coiled around them. This is because DNA is a negatively charged molecule and histones are positively charged because of basic amino acids. So, they are attracted to each other. If histones become negatively charged, instead of binding. they will rather repel DNA

Question 7. There is only one possible sequence of amino acids when deduced from given nucleotides. But multiple nucleotides sequence can be deduced from a single amino acid sequence. Explain this phenomenon.

Answer This is because a given nucleotide sequence would be read in triplet codons and each codon would specify a particular amino acid. Whereas one amino acid can be coded as more than one codon.

For example, if the nucleotide sequence is UUUUCAUCG it would be read in triplets as UUU-phenylalanine, UCA-serine and UCG-serine. But, if we have this amino acid sequence and try to decipher the nucleotide sequence-Phenylalanine, this single amino acid could mean UUU or UUC. Similarly, serine would mean nucleotide sequence could be anything from UCU, UCC, UCA, UCG, AGU

Question 8. 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 answer.

Answer: A single base mutation in a gene will not result in loss/gain of function as long as it does not change the codon specifying a particular amino acid. 

For example, if the sequence changes from UUU to UUC, the codon would still specify phenylalanine and so, the resulting polypeptide/ protein would remain unaltered. On the other hand, if the codon is changed in a way that now specifies

another amino acid, it may alter the protein’s function as it happens in the case of beta globulin of haemoglobin protein. 

A substitution of valine instead of glutamate causes a change in its structure and function, resulting in sickle cell anaemia.

Question 9. A low level of expression of the lac operon occurs all the time. Can you explain the logic behind this phenomenon? 

Answer A low-level expression of the lac operon is required all the time because the permease enzyme has to be present to let the lactose enter the cell. Permease will be present in the cell only if the lac operon is on. Permease is one of the three enzyme proteins that the lac operon codes for.


molecular basis of inheritance important questions


Question 1. During evolution why DNA was chosen over RNA as genetic material? Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the relighting of biochemical differences between DNA and RNA.

Answer A molecule that can act as a genetic material must fulfil the following:

(1) 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.

Biochemical differences between DNA and RNA 

(i) Both nucleic acids (DNA and RNA) can direct their duplication proteins fail for the first criteria.

(ii) RNA is reactive, it also acts as a catalyst, Hence, DNA is less reactive and structurally more stable than RNA. 

(iii) The presence of thymine at the place of uracil also confers additional stability to DNA. 

Question 2. Give an account of the methods used in sequencing the human genome.

Answer Sequencing of the human genome has made it possible to understand the link between various genes and their functions. If any gene defects express as disorders or that increase the susceptibility of an individual to a disease then specific gene therapies can be worked out.


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