CBSE Class 12 Biology Molecular Basis Of Inheritance Worksheet Set E

Very Short Answer Questions

Question: In a nucleus, the number of RNA nucleoside triphosphates is 10 times more than the number of DNA nucleoside triphosphates, still only DNA nucleotides are added during the DNA replication, and not the RNA nucleotides. Why? 
Answer. DNA polymerase is highly specific to recognise only deoxyribonucleoside triphosphates.
Therefore it cannot hold RNA nucleotides.

Question: Name the negatively charged and positively charged components of a nucleosome.
Answer. In a nucleosome, the negatively charged component is DNA and positively charged component is histone octamer.

Question: Suggest a technique to a researcher who needs to separate fragments of DNA.
Answer. Gel electrophoresis is used to separate DNA fragments.

Question: When and at what end does the ‘tailing’ of hnRNA take place?
Answer. ‘Tailing’ of hnRNA takes place during conversion of hnRNA into functional mRNA after transcription. It takes place at the 3′-end.

Question: Mention one difference to distinguish an exon from an intron. 
Answer. Exon is the coded or expressed sequence of nucleotides in mRNA.
Intron is the intervening sequence of nucleotides not appearing in processed mRNA.

Short Answer Questions

Question: Retrovirus do not follow central dogma. Comment.
Answer. Genetic material of retrovirus is RNA. At the time of synthesis of protein, RNA is reverse transcribed to its complementary DNA first, then transcriped to RNA and proteins. Hence, retrovirus are not known to follow central dogma.

Question: State the difference between the structural genes in a transcription unit of prokaryotes and eukaryotes.
Answer.

Question: Discuss the role of enzyme DNA ligase plays during DNA replication.
Answer. DNA ligase joins or seals the discontinuous DNA fragments.

Question: Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer. The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest semi-conservative mechanism of DNA replication in which one strand of parent is conserved while the other complementary strand formed is new.

Question: A template strand is given below. Write down the corresponding coding strand and the mRNA strand that can be formed, along with their polarity.
3′ ATGCATGCATGCATGCATGCATGC 5′
Answer. Coding strand: 5′ TACGTACGTACGTACGTACGTACG 3′mRNA strand: 5′ UACGUACGUACGUACGUACGUACG 3′

Question: Differentiate between the following:
(a) Repetitive DNA and satellite DNA
(b) mRNA and tRNA
(c) Template strand and coding strand

Question: State the dual role of deoxyribonucleoside triphosphates during DNA replication.
OR
Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerisation.
Answer. (i) Deoxyribonucleoside triphosphates act as substrates for polymerisation.
(ii) These provide energy from its two terminal phosphates for polymerisation reaction.
Answer. (a) M is the repressor.
(b) When repressor binds with the operator, transcription stops.

Question: If the sequence of the coding strand in a transcription unit is written as follows:
5′—ATGCATGCATGCATGCATGCATGCATG—3′
Write down the sequence of mRNA.
Answer. 5′—AUGCAUGCAUGCAUGCAUGCAUGCAUG—3′

Answer. (a) Differences between repetitive DNA and satellite DNA

(b) Differences between mRNA and tRNA

(c) Differences between template strand and coding strand

Question: Differentiate between the genetic codes given below:
(a) Unambiguous and Universal.
(b) Degenerate and Initiator
Answer. 

Question: Comment on the utility of variability in number of tandem repeats during DNA fingerprinting.
Answer. Tandemness in repeats provides many copies of the sequence for fingerprinting and variability in nitrogen base sequence in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

Question: (i) Name the scientist who suggested that the genetic code should be made of a combination of three nucleotides.
(ii) Explain the basis on which he arrived at this conclusion. 
Answer. (i) George Gamow.
(ii) He proposed that there are four bases and 20 amino acids So, there should be atleast 20 different genetic codes for these 20 amino acids.
The only possible combinations that would meet the requirement is combinations of 3 bases that will give 64 codons.

Question: Name the category of codons UGA belongs to. Mention another codon of the same category.Explain their role in protein synthesis.
Answer. UGA is a stop or termination codon.
UAA, UAG are the other stop codons of the category.
They prevent the elongation of the polypeptide chain by terminating translation.

Question: Explain the process of transcription in a bacterium.
Answer.  - In prokaryotes, the structural gene is polycistronic and continuous.
- In bacteria, the transcription of all the three types of RNA (mRNA, tRNA and rRNA) is catalysed by single DNA-dependent enzyme, called the RNA polymerase.
- All three RNA’s are needed to synthesize a protein in cell. mRNA provides the template, tRNA brings amino acids and reads the genetic code, and rRNA plays structural and catalytic role durinng translation.
- The transcription is completed in three steps: initiation, elongation and termination.
- Initiation: s (sigma) factor recognises the start signal and promotor region on DNA which then along with RNA polymerase binds to the promoter to initiate transcription. It uses nucleoside triphosphates as substrate and polymerises in a template-dependent fashion following the rule of complementarity.

Long Answer Questions

Question: State the conditions when ‘genetic code’ is said to be
(i) degenerate,
(ii) unambiguous and specific,
(iii) universal. 
Answer. (i) Degenerate—When some amino acids are coded by more than one amino acids.
(ii) Unambiguous and speci—ficWhen one codon codes for only one specific amino acid.
(iii) A particular codon codes for same amino acid in all organisms except in mitochondria and few protozoa.

Question: In a maternity clinic, for some reasons the authorities are not able to hand over the two newborns to their respective real parents. Name and describe the technique that you would suggest to sort out the matter. 
Answer. The technique is DNA fingerprinting. It includes the following steps:
Methodology and Technique
(i) DNA is isolated and extracted from the cell or tissue by centrifugation.
(ii) By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification. 
(iii) DNA is cut into small fragments by treating with restriction endonucleases.
(iv) DNA fragments are separated by agarose gel electrophoresis.
(v) The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye.
(vi) The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting.
(vii) VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation.

Question: Explain the process of translation in a bacterium.
Answer.  - Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
- A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide,a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.

Question: Unambiguous, universal and degenerate are some of the terms used for the genetic code.
Explain the salient features of each one of them. 
Answer. Unambiguous code means that one codon codes for only one amino acid, e.g., AUG codes for only methionine.
Universal code means that codon and its corresponding amino acid are the same in all organisms, e.g., from bacteria to human, UUU codes for phenylalanine.
Degenerate code means that same amino acids are coded by more than one codon, e.g., UUU and UUC code for phenylalanine.

Question: Draw a schematic diagram of lac operon in its ‘switched off’ position. Label the following:
(i) The structural genes (ii) Repressor bound to its correct position
(iii) Promoter gene (iv) Regulatory gene. 
Answer. 
(i) z, y and a are structural genes.
(iii) p is the promoter sequence.
(iv) i is the regulatory gene.

Question: How is hnRNA processed to form mRNA? 
Answer. The hnRNA undergoes the following processes to form mRNA:
(i) Capping: Addition of methyl guanosine triphosphate at 5’-end.
(ii) Tailing: Addition of 200-300 adenylate residues at 3’-end.
(iii) Splicing: Removal of introns and rejoining of exons.

Question: “The codon is a triplet and is read in a contiguous manner without punctuations.” Provide the genetic basis for the statement.
Answer. Since there are only four bases which code for twenty amino acids, the code should be made up of three bases, i.e., (4 × 4 × 4) = 64 codons; a number more than the required.
If the codon consists of four letters, only (4 × 4), only sixteen codons are possible, which is less than the required. Hence the codon is a triplet.
As the ribosome moves on mRNA, continuously without break, the codons are read in a contiguous manner.

Question: Write any six salient features of the human genome as drawn from the human genome project. 
Answer. (i) The human genome contains 3164.7 million nucleotide bases.
(ii) The average gene consists of 3000 bases; the largest known human gene being dystrophin at 2.4 million bases.
(iii) The total number of genes is estimated to be 30,000 and 99.9 per cent nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50 per cent of the discovered genes.
(v) Less than 2 per cent of the genome codes for proteins.
(vi) The human genome contains large repeated sequences, repeated 100 to 1000 times.
(vii) The repeated sequence is thought to have no direct coding functions but they throw light on chromosome structures, dynamics and evolution.
(viii) Chromosome 1 has most genes (2968) and the Y has the fewest genes (231).
(ix) Scientists have identified about 1.4 million locations where single base DNA sequence differences called SNPs or single nucleotide polymorphism (pronounced as ‘snips’) occur in humans.This information promises to revolutionise the processes of finding chromosomal locations for disease—associated sequences and tracing human history.

Question: Identify giving reasons, the salient features of genetic code by studying the following nucleotide sequence of mRNA strand and the polypeptide translated from it.
AUG UUU UCU UUU UUU UCU UAG
Met – Phe – Ser – Phe – Phe – Ser
Answer. 

Question: (a) Explain the observations of Meselson and Stahl 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 does the above experiment prove?
(c) Which is the first genetic material identified?
Answer. (a) (i) Meselson and Stahl observed that in the E. coli bacterium the DNA becomes completely labelled with 15N medium by centrifugation for few generations.
(ii) After two generations, 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) Ribonucleic acid (RNA) was the first genetic material.

Question: Observe the representation of genes involved in the lac operon given below: (Img 254)

(a) Identify the region where the repressor protein will attach normally.
(b) Under certain conditions repressor is unable to attach at this site. Explain.
(c) If repressor fails to attach to the said site what products will be formed by z, y and a?
(d) Analyse why this kind of regulation is called negative regulation.
Answer. (a) The repressor protein will attach to operator region, o.
(b) In presence of an inducer, lactose, repressor is unable to attach.
(c) z—b galactosidase.
y—Permease
a—Transacetylase
(d) It is called negative regulation as it involves constitutive (all the time) repressor. The operon is always in off position due to presence of repressor and is switched on only in presence of an inducer. Inducer Lactose or allolactose interacts with repressor making it inactive.

Question: (a) Explain the process of aminoacylation of tRNA. Mention its role in translation.
(b) How do ribosomes in the cells act as factories for protein synthesis?
(c) Describe ‘initiation’ and ‘termination’ phases of protein synthesis.
Answer. (a) Aminoacylation is the process by which amino acids become activated by binding with its aminoacyl tRNA synthetase in the presence of ATP.
If two charged tRNAs come close during translation process, the formation of peptide bond between them in energetically favourable.
(b) The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists
as two subunits: a large subunit and a small subunit. When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins. There are two sites in
the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst (23S rRNA in
bacteria is the enzyme ribozyme) for the formation of peptide bond.
(c) - Translation is the process of synthesis of protein from amino acids, sequence and order of amino acids being defined by sequence of bases inm RNA. Amino acids are joined by peptide bonds.
- A translational unit in mRNA from 5′ → 3′ comprises of a start codon, region coding for a polypeptide, a stop codon and untranslated regions (UTRs). UTRs are additional sequences of mRNA that are not translated. They are present at both 5′ end (before start codon) and at 3′ end (after stop codon) for efficient translation process.

Question: Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and write the procedure that would help in the identification of kinship. 
OR
A number of passengers were severely burnt beyond recognition during a train accident.
Name and describe a modern technique that can help to hand over the dead to their relatives.
Answer. DNA fingerprinting can help in identification of kinship.
For procedure, - 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.

Question: What is hnRNA? Explain the changes hnRNA undergoes during its processing to form mRNA.
Answer. hnRNA is the precursor of mRNA that is transcribed by RNA ploymerase II and is called heterogenous nuclear RNA.
Changes:
- The hnRNA undergoes two additional processes called capping and tailing.
- In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5’-end of hnRNA.
- In tailing, adenylate residues (about 200–300) are added at 3’-end in a template independent manner.
- Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing.
- Fragments on one of the template strands.