Please refer to Molecular Basis of Inheritance Class 12 Biology Important Questions given below. These solved questions for Molecular Basis of Inheritance have been prepared based on the latest CBSE, NCERT and KVS syllabus and books issued for the current academic year. We have provided important examination questions for Class 12 Biology all chapters.
Class 12 Biology Molecular Basis of Inheritance Important Questions
Very Short Answer Questions
Question. How many base pairs would a DNA segment of length 1.36 nm have?
0.34 ×10-6 ×1.36
4 ×106 bp
Answer. Distance between two base pairs = 0.34 nm or 0.34×10–6 nm Number of base pairs in 1.36 nm DNA segment
Question. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments?
Answer. By density gradient centrifugation.
Question. Why hnRNA is required to undergo splicing?
Answer. hnRNA undergoes splicing in order to remove introns which are intervening or non-coding sequences and exons are joined to form functional mRNA.
Question. 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–Repressor bound to the operator and prevents transcription of structural genes.
Question. Mention the contribution of genetic maps in human genome project.
Answer. Genetic maps have played an important role in sequencing of genes, DNA fingerprinting, tracing human history, chromosomal location for disease associated sequences (Any one).
Question. Why are proteins either positively or negatively charged?
Answer. If the proteins are rich in basic amino acids, they are positively charged, and if the proteins are rich in acidic amino acids, they are negatively charged.
Question. Group 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. If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer. Cytosine = 20%, therefore Guanine = 20%
According to Chargaff’s rule,
A + T = 100 – (G + C)
A + T = 100 – 40. Since both adenine and thymine are in equal amounts,
∴ Thymine = Adenine = 60/2=30%
Question. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer.
Question. If the sequence of one strand of DNA is written as follows :
5′—ATGCATGCATGCATGCATGCATGCATGC—3′
Write down the sequence of complementary strand in 5′→3′ direction.
Answer. In 3′→5′ direction, 3′—TACGTACGTACGTACGTACGTACGTACG—5′
In 5′→3′ direction, 5′—GCATGCATGCATGCATGCATGCATGCAT—3′
Question. At which ends do ‘capping’ and ‘tailing’ of hnRNA occur, respectively?
Answer. Capping occurs at 5′-end and tailing occurs at 3′-end.
Short Answer Questions
Question. Draw a schematic representation of dinucleotide. Label the following:
(i) The components of a nucleotide
(ii) 5′ end
(iii) N-glycosidic linkage
(iv) Phosphodiester linkage.
Answer. Nucleotide = Ribose sugar + Base + phosphate group.
Question. Describe the structure of a RNA polynucleotide chain having four different types of nucleotides.
Answer.
Question. Differentiate between codon and an anticodon.
Answer.
| Codon | Anticodon |
| The sequence of 3 nitrogen bases on mRNA that codes for a particular amino acid during translation is called codon. | The sequence of 3 nitrogenous bases on tRNA that are complementary to the codon on mRNA for a particular amino acid during translation is called anticodon. |
Question. A DNA segment has a total of 1500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
Answer. According to Chargaff’s rule A/G = T/G = 1
G = C, G = 410, hence C = 410
G + C = 410 + 410
= 820
So, A + T = 1500 – 820
= 680
A = T, so T = 680/2 = 340
So, Pyrimidines = C + T
= 410 + 340
= 750
Question. A DNA segment has a total of 2,000 nucleotides, out of which 520 are adenine containing nucleotides. How many purine bases this DNA segment possesses?
Answer. [A] = [T]
A + T = 520 + 520
= 1040
Total number of nucleotides = 2000
∴ G + C = 2000 – 1040
= 960
G = 960/2 = 480
∴ Total number of purines (A + G) = 520 + 480
= 1000.
Question. Describe the structure of a nucleosome.
Answer. Roger Kornberg (1974) reported that chromosome is made up of DNA and protein.
• Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins.
• The proteins associated with DNA are of two types— basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins.
• The negatively charged DNA molecule wraps around the positively charged histone proteins to form a structure called nucleosome.
• The nucleosome core is made up of four types of histone proteins—H2A, H2B, H3 and H4 occurring in pairs.
Question. Explain the role of 35S and 32P in the experiments conducted by Hershey and Chase.
Answer. Viruses grown in the medium containing 32P contained radioactive DNA but not radioactive protein because DNA contains phosphorus but proteins do not contain phosphorus. Similarly,viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
Question. Recall the experiment done by Frederick Griffith, Avery, Macleod annd McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat killed strain of Streptococcus have tranformed the R-strain into virulent strain? Explain your Answerwer.
Answer. RNA is more labile and prone to degradation (owing to the presence of 2′–OH group in its ribose).
Hence heat-killed S-strain may not have retained its ability to transform the R-strain.
Question. How do histones acquire positive charge?
Answer. Histones are rich in the basic amino acid residues lysines and arginines, which carry positive charges in their side chains. Therefore, histones are positively charged.
Question. 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 Answerwer.
Answer. The statement is correct because of degeneracy of codons, mutations at third base of codon,usually doe not result into any change in phenotype. This is called silent mutations but at other times it can lead to loss or formation of malformed protein changing the phenotype.
Question. Explain the dual function of AUG codon. Give the sequence of bases it is transcribed from and its anticodon.
Answer. The dual function of AUG codon:
(a) It codes for amino acid methionine.
(b) It is an initiation codon.
The sequence of bases from which it is transcribed is TAC. Its anticodon is UAC.
Question. Protein synthesis machinery revolves around RNA but in the course of evolution it was replaced by DNA. Justify.
Answer. Since RNA was unstable and prone to mutations, DNA evolved from RNA with chemical modifications that makes it more stable.
DNA has double stranded nature and has complementary strands. These further resist changes by evolving a process of repair.
Question.
Why do you see two different types of replicating strands in the given DNA replication fork? Explain. Name these strands.
Answer. The DNA-dependent DNA polymerase catalyses polymerisation only in one direction i.e., 5′→3′.
Therefore, in one strand with polarity 3′→5′ continuous replication takes place whereas the other strand with polarity 5′→3′ carries out discontinuous replication.
The strand with polarity 3′→5′ is called leading strand and the strand with polarity 5′→3′ is called lagging strand.
Long Answer Questions
Question. How is the translation of mRNA terminated? Explain.
Ans. • When the A-site of ribosome reaches a termination codon, which does not code for any amino acid, no charged tRNA binds to the A-site.
• Dissociation of polypeptide from ribosome takes place, which is catalysed by a ‘release factor’.
• There are three termination codons namely UGA, UAG and UAA.
Question. (a) A DNA segment has a total of 1000 nucleotides, out of which 240 of them are adenine containing nucleotides. How many pyrimidine bases this DNA segment possesses?
(b) Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Ans. (a) A = T, A = 240, hence T = 240
A + T = 240 + 240 = 480
So, G + C = 1000 – 480 = 520
G = C, so C 520/2 =260
So, pyrimidines = C + T
= 260 + 240
= 500
(b)
Question. (a) Explain DNA polymorphism as the basis of genetic mapping of human genome.
(b) State the role of VNTR in DNA fingerprinting.
Ans. (a) Genetic polymorphism means occurrence of genetic material in more than one form. It is of three major types, i.e., allelic, SNP and RFLP.
Allelic polymorphism: Allelic polymorphism occurs due to multiple alleles of a gene. Allele possess different mutations which alter the structure and function of a protein formed by them as a result, change in phenotype may occur.
SNP or single nucleotide polymorphism: SNP is very useful for locating alleles, identifying disease-associated sequence and tracing human history.
(b) Variable Number Tandem Repeats (VNTRs) are used in DNA fingerprinting as markers.
VNTRs vary from person to person and are inherited from one generation to the next.
Therefore, only closely related individuals have similar VNTRs.
Question. (a) Why did Hershey and Chase use radioactive sulphur and radioactive phosphorus in their experiment?
(b) Write the conclusion they arrived at and how.
Ans. • Procedure:
(i) Some bacteriophage virus were grown on a medium that contained radioactive phosphorus (32P) and some in another medium with radioactive sulphur (35S).
(ii) Viruses grown in the presence of radioactive phosphorus (32P) contained radioactive DNA.
(iii) Similar viruses grown in presence of radioactive sulphur (35S) contained radioactive protein.
(iv) Both the radioactive virus types were allowed to infect E. coli separately.
(v) Soon after infection, the bacterial cells were gently agitated in blender to remove viral coats from the bacteria.
(vi) The culture was also centrifuged to separate the viral particle from the bacterial cell.
Question. “A very small sample of tissue or even a drop of blood can help determine paternity”. Provide a scientific explanation to substantiate the statement.
Ans. (i) DNA from all cells of an individual shows the same degree of polymorphism and therefore becomes a useful identification tool.
(ii) Polymorphs are heritable and the child inherits 50% of the chromosome from each parent.
(iii) With the help of PCR the small amount of DNA from blood can be amplified and be used in DNA finger printing to identitfhye paternity.
Question. (Img 244)
(a) Identify strands ‘A’ and ‘B’ in the diagram of transcription unit given above and write the basis on which you identified them.
(b) Write the functions of RNA polymerase-I and RNA polymerase-III in eukaryotes.
Ans. (a) A—Template strand
B—Coding strand
The templates are identified on the basis of polarity with respect to promoter. Template strand has polarity 3′ → 5′ and coding strand has polarity 5′ → 3′.
(b) RNA polymerase-I transcribes rRNAs.
RNA polymerase-III transcribes tRNA, 5srRNA and snRNA.
Question. State any two structural differences and one functional difference between DNA and rRNA.
Ans.
Question. DNA polymerase and RNA polymerase differ in their requirement while functioning Explain.
Ans.
| S.No. | RNA polymerase | DNA polymerase |
| (i) | It cannot carry out proofreading. | It carries out proofreading for DNA repair mechanism. |
| (ii) | RNA polymerase does not require RNA primer for synthesis of RNA. | DNA polymerase requires RNA primer for synthesis of DNA. |
| (iii) | It uses ribonucleotides for RNA synthesis. | It uses deoxyribonucleotides for DNA synthesis. |
Question. The average length of a DNA double helix in a typical mammalian cell is approximately 2.2 metres and the dimension of the nucleus is about 10–6 m.
(a) How is it possible that long DNA polymers are packed within a very small nucleus?
(b) Differentiate between euchromatin and heterochromatin.
(c) Mention the role of non-histone chromosomal protein.
Ans. (a) Packaging of DNA in eukaryotes
• Roger Kornberg (1974) reported that chromosome is made up of DNA and protein.
• Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins.
• The proteins associated with DNA are of two types—basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins.
• The negatively charged DNA molecule wraps around the positively charged histone proteins to form a structure called nucleosome.
• The nucleosome core is made up of four types of histone proteins—H2A, H2B, H3 and H4 occurring in pairs.
(c) The packaging of chromatin at higher level requires the presence of non-histone chromosomal protein.
Question. List the criteria a molecule that can act as genetic material must fulfill. Which one of the criteria are best fulfilled by DNA or by RNA thus making one of them a better genetic material than the other? Explain.
Ans. A molecule that can act as a genetic material must fulfill the following criteria:
(i) 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 Characters’.
In DNA the two strands being complementary if separated by heating come together, when appropriate conditions are provided. Further, 2’-OH group present at every nucleotide in RNA is also now known to be catalytic, hence reactive. Therefore DNA chemically is less reactive and structurally more stable when compared to RNA. Therefore, among the two nucleic acids, the DNA is a better genetic material. The presence of thymine at the place of uracil also confers additional stability to DNA.
Both DNA and RNA are able to mutate. In fact, RNA being unstable, mutate at a faster rate. RNA can directly code for the synthesis of proteins, hence can easily express the characters. DNA, however, is dependent on RNA for synthesis of proteins. The protein synthesising machinery has evolved around RNA.
Question. (a) How are the following formed and involved in DNA packaging in a nucleus of a cell?
(i) Histone octomer
(ii) Nucleosome
(iii) Chromatin
(b) Differentiate between Euchromatin and Heterochromatin.
Ans. (a) (i) Eight molecules of positively charged basic proteins called histones are organised to form histone octomer.
(ii) Negatively charged DNA is wrapped around positively charged histone octamer to give rise to nucleosome.
(iii) Nucleosome constitute the repeating unit of a structure called chromatin.
(b)
| S.No. | Euchromatin | Heterochromatin |
| (i) | Regions of chromatin, which are loosely packed during interphase are called euchromatin. | Regions of chromatin, which are densely packed during cell division are called heterochromatin. |
| (ii) | When chromosomes are stained with Feulgen stain (specific for DNA), these appear as lightly stained chromatin. | When chromosomes are stained with Feulgen stain, these appear as intensely stained chromatin. |
| (iii) | Euchromatin contains active genes. | Heterochromatin contains inactive genes. |
| (iv) | They do not contain repetitive DNA sequences. | They are enriched with highly repetitive tandemly arranged DNA sequences. |
| (v) | It is transcriptionally active. | It is transcriptionally inactive. |
Question. What background information did Watson and Crick had available with them for developing a model of DNA? What was their own contribution?
Ans. Watson and Crick had the following informations which helped them to develop a model of DNA:
(i) Chargaff’s Law suggesting A=T and C G.
(ii) Wilkins and Franklin’s X-ray diffraction studies on DNA’s physical structure.
Based on these information, Watson and crick proposed
(i) complementary base-pairing of nitrogenous bases
(ii) semi-conservative mode of replication
(iii) occurrence of mutation through tautomerism.
Q. 5. Describe the packaging of DNA helix in a prokaryotic cell and an eukaryotic nucleus.
Ans. (i) Packaging of DNA in prokaryotes
• In prokaryotes, well-defined nucleus is absent so DNA is present in a region called nucleoid.
The negatively charged DNA is coiled with some positively charged non-histone basic proteins.
• DNA in nucleoid is organised in large loops held by proteins.
(ii) Packaging of DNA in eukaryotes
• Roger Kornberg (1974) reported that chromosome is made up of DNA and protein.
• Later, Beadle and Tatum reported that chromatin fibres look like beads on the string, where beads are repeated units of proteins.
• The proteins associated with DNA are of two types—basic proteins (histones) and acidic non-histone chromosomal (NHC) proteins.
• The negatively charged DNA molecule wraps around the positively charged histone proteins to form a structure called nucleosome.
Question. (a) How did Griffith explain the transformation of R-strain (non-virulent) bacteria into S-strain (virulent)?
(b) Explain how MacLeod, McCarty and Avery determined the biochemical nature of the molecule responsible for transforming R-strain bacteria into S-strain bacteria.
OR
(a) Describe the various steps of Griffith’s experiment that led to the conclusion of the ‘Transforming Principle’.
(b) How did the chemical nature of the ‘Transforming Principle’ get established?
Ans. (a) • Frederick Griffith (1928) conducted experiments with Streptococcus pneumoniae (bacterium causing pneumonia).
• He observed two strains of this bacterium—one forming smooth shiny colonies (S-type) with capsule,while other forming rough colonies (R-type) without capsule.
(b) • Oswald Avery, Colin MacLeod and Maclyn McCarty repeated Griffith’s experiment in an in vitro system in order to determine biochemical nature of transforming principle.
• They purified biochemicals (proteins, DNA, RNA) from heat-killed S-type cells, and checked which of these could transform live R-type cells into S-type cell. They observed that DNA alone from S-type cells caused transformation of R-type cells into virulent S-type cells.
• They also discovered that proteases (protein digesting enzymes) and RNases (RNA digesting enzymes) did not affect transformation while DNases inhibited the process.
• They concluded that DNA is the hereditary material.
Question. (a) Write the scientific name of the bacterium used by Frederick Griffith in his experiment.
(b) How did he prove that some ‘transforming principle’ is responsible for transformation of the non-virulent strains of bacteria into the virulent form?
(c) State the biochemical nature of ‘transforming principle’.
(d) Name the scientists who proved it.
Ans. (a) Streptococcus pneumoniae
(b) Refer to Basic Concepts Point 6.
(c) ‘The transforming principle’ was nucleic acid, i.e., DNA.
(d) It was proved by O. Avery, C. MacLeod and M. McCarty.
Question. (a) Explain the experiment performed by Griffith on Streptococcus pneumoniae. What did he conclude from this experiment?
(b) Name the three scientists who followed up Griffith’s experiments.
(c) What did they conclude and how?
Ans. (a) Refer to Basic Concepts Point 5.
(b) Oswald Avery, Colin MacLeod and Maclyn McCarty.
(c) • Oswald Avery, Colin MacLeod and Maclyn McCarty repeated Griffith’s experiment in an in vitro system in order to determine biochemical nature of transforming principle.
• They purified biochemicals (proteins, DNA, RNA) from heat-killed S-type cells, and checked which of these could transform live R-type cells into S-type cell. They observed that DNA alone from S-type cells caused transformation of R-type cells into virulent S-type cells.
• They also discovered that proteases (protein digesting enzymes) and RNases (RNA digesting enzymes) did not affect transformation while DNases inhibited the process.
• They concluded that DNA is the hereditary material.
Question. (a) What did Meselson and Stahl observe when
(i) they cultured E. coli in a medium containing 15NH4Cl for a few generations and centrifuged the content?
(ii) they transferred one such bacterium to the normal medium of NH4Cl and cultured for 2 generations?
(b) What did Meselson and Stahl conclude from this experiment? Explain with the help of diagrams.
(c) Which is the first genetic material? Give reasons in support of your answer.
Ans. (a) (i) Meselson and Stahl observed that in the E. coli bacterium the DNA becomes completely labelled with 15N medium after few generations.
(ii) After two generations, they observed that density changed and showed equal amount of light DNA (14N) and dark hybrid DNA (15N–14N).
(b) They concluded that DNA replicates semi-conservatively.
(c) RNA is the first genetic material.
Reasons:
(i) RNA is highly reactive and acts as a catalyst as well as a genetic material.
(ii) Essential life processes such as metabolism, translation and splicing evolved around RNA.
(iii) It expresses itself through proteins.
Question. Describe the Hershey−Chase experiment. Write the conclusion they arrived at after the experiment.
OR
How did Hershey and Chase established that DNA is transferred from virus to bacteria?
Ans. In some viruses, RNA is the genetic material (e.g., Tobacco Mosaic virus). RNA also performs functions of messenger and adapter.
(i) DNA and RNA have the ability to direct their duplications because of rule of base pairing and complementarity but proteins fail to fulfill first criteria itself.
(ii) Genetic material should be stable so as not to change with different stages of life cycle, age or change in physiology of organism.
(iii) RNA being unstable mutates at a faster rate. Thus, viruses having RNA genome and having shorter life span mutate and evolve faster.
(iv) RNA can code directly for protein synthesis and hence can easily express characters. But DNA is dependent on RNA for protein synthesis. Protein synthesizing machinery has evolved around RNA.
Question. You are repeating the Hershey–Chase experiment and are provided with two isotopes: 32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different?
Ans. Use of 15N will be inappropriate because method of detection of 35P and 15N is different (32Pbeing a radioactive isotope while 15N is not radioactive but is the heavier isotope of nitrogen).
Even if 15N was radioactive then its presence would have been detected both inside the cell (l5N incorporated as nitrogenous base in DNA) as well as in the supernatant because 15N would also get incorporated in amino group of amino acids in proteins). Hence, the use of 15N would not give any conclusive results.
Question. A criminal blew himself up in a local market when was chased by cops. His face was beyond recognition. Suggest and describe a modern technique that can help establish his identity.
Ans. The identity can be established by the technique of DNA fingerprinting.
For method: • Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases.
• Human genome has 3 × 109 bp. 99.9% of base sequences among humans are the same, which makes every individual unique in phenotype.
• Polymorphism: The genome consists of small stretches of DNA which are repeated many times.
These are called repetitive DNA and comprise of satellite DNA. Satellite DNA does not code for any proteins but form large portion of human genome. These sequences show high degree of polymorphism. As polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing. Polymorphism arises due to mutations. New mutations may arise in somatic cells or in germ cells. If mutation occurs in germ cells; it is passed on to offsprings. If an inheritable mutation is observed in a population at high frequency, it is
called as DNA polymorphism. Polymorphism ranges from single nucleotide change to very large scale changes.
Question. Answer the following questions based on Hershey and Chases’s experiments:
(a) Name the kind of virus they worked with and why.
(b) Why did they use two types of culture media to grow viruses in? Explain.
(c) What was the need for using a blender and later a centrifuge during their experiments?
(d) State the conclusion drawn by them after the experiments.
Ans. (a) They worked with bacteriophage because when it attacks a bacteria it only inserts its genetic material in its body.
(b) They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive sulphur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.
(c) Blender was used to agitate the bacteria to remove the viral coats from them. Centrifuge was used to separate virus particle from the bacteria.
(d) Bacteria which was infected with viruses that had radioactive DNA were radioactive, indicating that DNA was the material that passed from the virus to the bacteria. Bacteria that
were infected with viruses that had radioactive proteins were not radioactive. This indicates that proteins did not enter the bacteria from the viruses. DNA is therefore the genetic material that is passed from virus to bacteria.
