Questions

3 Marks Question

Take a timed test

116 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
What is DNA fingerprinting? Mention its application.
Answer
DNA Fingerprinting is a technique to find out variations in individuals of a population at DNA level.
Its applications are as follows:
  1. Used in forensic science to identify suspects.
  2. Used to find out history of an organism.
  3. Used to find out paternity and family relations.
View full question & answer
Question 23 Marks
Differentiate between the following:mRNA and tRNA.
Answer
mRNA and tRNA.
mRNA tRNA
It is called messenger RNA and carries the codes for amino acid sequence. It is called transfer RNA as it carries amino acids to the site of protein synthesis.
It is a linear molecule. It has clover leaf shape.
It is synthesised by RNA polymerase II It is synthesised by RNA polymerase III.
View full question & answer
Question 33 Marks
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer
The property of DNA double helix led Watson and Crick are:
  • Two strands running opposite to each other, wherein bases will always pair with their counterpart-A with T and G with C (specific pairing).
  • If H bonds break and bases of one strand lie exposed, unpaired, they will easily pair up with free nucleotides as well. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. where the two strands separate and act as a template for the synthesis of a new complementary strand.
View full question & answer
Question 43 Marks
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer
Alfred Hershey and Martha Chase (1952) worked with viruses that infect bacteria called bacteriophages. In 1952, they chose a bacteriophage known as T2 for their experimental material. They grew some viruses on a medium that contained radioactive phosphorus (p32) and some others on medium that contained radioactive sulphur (s35). Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protem does not Similarly. viruses grown on radioactive sulphur contained radioactive protein but not radio'active DNA because DNA does not contain sulphur. Radioactive phages were allowed to attach to E.coli bacteria. Then, as the infection proceeded, the viral coats were removed from the bacteria by agitating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge., 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.
View full question & answer
Question 53 Marks
Differentiate between the following:Template strand and Coding strand.
Answer
Template strand and Coding strand.
Template strand Coding strand
It is the strand which is transcribed into RNA. It has the same sequence as mRNA.
It is called anti sense strand. It is called sense or non-template strand.
It has 3' → 5' polarity. It has 5' → 3' polarity
View full question & answer
Question 63 Marks
Differentiate between the following:Repetitive DNA and Satellite DNA.
Answer
Repetitive DNA and Satellite DNA.
Repetitive DNA Satelite DNA
It is the non-coding DNA with multiple copies of identical sequences which may lie in tandem or interspersed. It refers to non-coding tandem repeat sequences.
These can be few base pairs to hundreds or thousands of base pairs. These are generally short sequence repeats (up to 60 base pair long).
It appears as light bands. It appears as small dark bands.
View full question & answer
Question 73 Marks
  1. Why does DNA replication occur in small replication forks and not in its entire length?
  2. Why is DNA replication continuous and discontinuous in a replication fork?
  3. Explain the importance of origin of replication' in a replication fork.
Answer
  1. Replication of DNA occurs in small replication forks, because DNA is such a long molecule that the separation of the two strands along its entirelength requires a very high amount of energy.
  2. DNA polymerase can catalyse the polymerization of nucleotides only in 5' → 3' direction.
  1. So on the template strand with 3' → 5' polarity, DNA replication is continuous.
  2. On the template strand with 5' → 3' polarity, DNA synthesis occurs in short stretches as the opening of replication fork continues; later these short stretches are joined by the action of DNA ligases .
  1. Replication of DNA does not initiate randomly, and DNA polymerases on their own cannot initiate replication.
  1. So, there is a specific sequence on DNA, called origin of replication; DNA polymerase bindsto it and continues the process.
View full question & answer
Question 83 Marks
  1. Draw the structure of the initiator tRNA adaptor molecule.
  2. Why is tRNA called an adaptor molecule?
Answer
  1.  
  1. On one hand, it reads the code.
On the other hand, it binds to specific amino acid.
View full question & answer
Question 93 Marks
What are the functions of-
  1. Methylated guanosine cap?
  2. Poly-A 'tail' in a mature RNA?
Answer
Function of Methylated Guanosine Cap:
  1. Attachment of mRNA: it helps in the attachment of mRNA to smaller ribosomal subunit during initiation of translation.
  2. Function of poly A tail: It helps in the Protection of translational region.
View full question & answer
Question 103 Marks
"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.
View full question & answer
Question 113 Marks
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 -One codon codes for one amino acid, e.g. AUG (Methionine).Universal -Codon and its corresponding amino acid are the same in all organisms.
Example:- more-Bacteria to human UUU codes for phenylalanine (phe).
Degenerate - Some amino acids are by more than one codon.
Example:- UUU and UUC code for phenylalanine (phe).
View full question & answer
Question 123 Marks
It is established that RNA is the first genetic material. Explain giving three reasons.
Answer
Processes like metabolism, translation, splicing evolved around RNA, RNA is reactive and catalyses reaction. In some virus it is the hereditary material. It is so unstable and hence would have mutated to lead to evolution.
View full question & answer
Question 133 Marks
Write any three goals of human genome project.
Answer
Three goals of human genome project:
  1. To develop new and improved medicines.
  2. To predict and prevent diseases.
  3. To ensure that the diagnosis is accurate.
View full question & answer
Question 143 Marks
With respect to Messelson and Stahl’s Experiment, answer the following questions:
  1. Identify the method used to distinguish between heavy and light isotopes of nitrogen.
  2. With the help of diagrams, compare the results for the DNA isolated after 20 minutes of experiment with the DNA which was isolated after 40 minutes.
Answer
  1. Centrifugation in a CsCl density gradient.
  2.  
View full question & answer
Question 153 Marks
Discuss the significance of heavy isotope of nitrogen in the Meselson and Stahl’s experiment.
Answer
The heavy isotope of nitrogen was used in Meselson and Stahl’s experiment for various reasons. The heavy DNA molecule (containing $15_N$) could be easily distinguished from the normal DNA by centrifugation in a Cesium chloride (CsCl) density gradient. It could be easily separated from lighter nitrogen $(14_N)$ on the basis of density. Use of lighter and heavier nitrogen made the task of identifying transfer of DNAs through subsequent generation quite easier.
View full question & answer
Question 163 Marks
Given below is a part of the template strand of a structural gene:
TAC CAT TAG GAT
  1. Write its transcribed mRNA strand with its polarity.
  2. Explain the mechanism involved in initiation of transcription of this strand.
Answer
  1. 5’ AUG GUA AUC CUA 3’ (Polarity),
  2. RNA polymerase, blinds at 5 end, in the presence of (initiation factor) $\sigma$ factor, at the promoter site.
View full question & answer
Question 173 Marks
Expand 'BAC' and 'YAC'. Explain how they were used in sequencing of human genomes.
Answer
'BAC'- Bacterial Artificial Chromosome.
"YAC' - Yeast Artificial Chromosome.
  1. They are the commonly used vectors for cloning the DNA fragments in the hosts like bacteria and yeast.
  2. The cloning results into amplification of each fragment of DNA, so that they could be sequenced with ease.
View full question & answer
Question 183 Marks
‘‘A very small sample of tissue or even a drop of blood can help determine paternity’’. Provide a scientific explanation to substantiate the statement.
Answer
The technique help in solving a case of paternity dispute over the custody of a child by two different families is DNA – finger printing. The technique of DNA finger printing was initially developed. He used a satellite DNA as probe that shows very high degree of polymorphism.
Step II: Amplification: Many copies of the extracted DNA are made by Polymerase Chain Reaction (PCR).
  1. Separation of DNA sequence restriction fragments: The separated fragments can be visualized by Staining them with a dye that shows fluorescene under ultra violet radiation.
  2. Southern Blotling: The separated DNA sequence are transferred on to a nitro cellulose or nylon membrane of Sheet placed over the gel.
  3. Hybridization: The nylon membrane is immersed in a bath and radioactive proves are added: These probes target a specific nucleotide sequence that complementary to them.
  4. The Probs/Result: X-ray film and dark bands develop at the probs sites which resemble the barcda.

Scheme representation of DNA fingerprinting: Few representative chromosomes have been shown to contain different copy number of VNTR. For the sake of understanding different colour schems have been used to trace the origin of each band in the gel. The two alleles (paternal and maternal) of a chromosome also contan different copy nubers of VNTR. It is clear that the banding pattern of DNA from crime scene.
matches with individual B, and not with A.
View full question & answer
Question 193 Marks
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.
Answer
DNA fingerprinting (analysis)
  • Isolation and digestion of DNA by restriction endonuclease.
  • Separation of DNA fragments by electrophoresis and transferring them to synthetic membranes/nitrocellulose/nylon.
  • Hybridisation using labelled VNTR probe.
  • Detection of hybridised DNA fragments by autoradiography.
  • Matching banding pattern of DNA/DNA fingerprints/autoradiograms of the passengers killed and that of relatives.
View full question & answer
Question 203 Marks
AtRNA is charged with the amino acid phenylalanine.
  1. At what end of the tRNA is the amino acid attached?
  2. What is the mRNA codon that codes for phenylalanine?
  3. Name the enzyme responsible for this attachment.
Answer
  1. It is attached at the 3' end of RNA.
  2. UUU and UUC.
  3. Aminoacyl-tRNA synthetase.
View full question & answer
Question 213 Marks
Where is an 'operator' located in a prokaryote DNA? How does an operator regulate gene expression at transcriptional level in a prokaryote? Explain.
Answer
The operator region is located adjacent to promoter elements/prior to structural gene. In regulation of gene expression.
  • Switch off- The repressor binds to the operator region, & prevents transcription.
  • Switch on- In the presence of inducer the repressor is inactivated, (by the interaction with the inducer) and operator allows RNA polymerase access to the promoter, & transcription proceeds.
View full question & answer
Question 223 Marks
DNA polymerase and RNA polymerase differ in their requirement while functioning. Explain.
Answer
S. No RNA polymerase DNA polymerase
(i) It cannot carry out proofreading. It carries out proofreading for DNA repairmechanism.
(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.
View full question & answer
Question 233 Marks
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.
Answer
DNA finger printing:
Isolation of DNA and digestion of DNA by restriction endonucleases, separation of DNA fragments by (gel) electrophoresis and transferring (blotting) of separated DNA fragments to synthetic membrane or nitrocellulose or nylon, hybridization using VNTR probe and detection of hybridised DNA fragments by autoradiography, matching the banding pattern so obtained with that of relative.
View full question & answer
Question 243 Marks
Explain the process of charging of RNA. Why is it essential in translation?
Answer
Charging of tRNA:
  1. The amino acids are activated in the presence of ATP and linked to their cognate tRNA; this process is called charging of tRNA or amino acylation of tRNA.
  2. This process is necessary as the formation of peptide bond between the amino acids is favoured energetically, when they are brought together.
  3. The activation of amino acids by ATP provides the energy for the formation of peptide bond.
View full question & answer
Question 253 Marks
Give any six features of the human genome.
Answer
Six features of the human genome are as follows:
  • The human genome contains 3164.7 million nucleotides.
  • The average gene in the human genome contains 3000 bases.
  • The total number of genes is estimated to be 30,000.
  • Almost all (about 99.9%) of nucleotides are same in all human beings.
  • Less than 2 percent of the genome codes for protein.
  • Chromosome 1 has the most genes (2968) and chromosome Y has the least (231).
View full question & answer
Question 263 Marks
Now, sequencing of total genomes getting is getting less expensive day by the day. Soon it may be affordable for a common man to get his genome sequenced. What in your opinion could be the advantage and disadvantage of this development?
Answer
Advantages of Affordable Genome Sequencing: It can help in settling disputes which may arise in case of parentage of a child. This can also help in disputes of property inheritance by finding the bonafide beneficiary. Human genome can also help in preparing a database on people with criminal record. It can help in identifying the chances of genetic disorders in a family.
Disadvantages: Genome sequencing can have serious issues of privacy. Some employers may misuse the data to blackmail their employees. Many private matters may leak into public domain; creating embarrassment for the affected person.
View full question & answer
Question 273 Marks
  1. Expand VNTR and describe its role in DNA fingerprinting.
  2. List any two applications of DNA fingerprinting technique.
Answer
  1. VNTR stands for “Variable Number of Tandem Repeats”.
The VNTR belongs to a class of satellite DNA referred to as mini-satellite. A small DNA sequence is arranged tandemly in many copy numbers. The copy number varies from chromosome to chromosome in an individual. The numbers of repeat show very high degree of polymorphism. As a result th size of VNTR varies in size from 0.1 to 20kb. Consequently, after hybridization with VNTR probe, the autoradiogram gives many bands of differing sizes. These bands give characteristic pattern for an individual DNA which is used to identify individuals.
  1. Since DNA from every tissue (such as blood, hair-follicle, skin, bone, saliva, sperm etc.), from an individual show the same degree of polymorphism, they become very useful identification tool in forensic applications to identify criminals. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basic of paternity testing, in case of disputes.
View full question & answer
Question 283 Marks
Construct and label a transcription unit from which the RNA segment given below has been transcribed. Write the complete name of the enzyme that transcribed this RNA.
Answer
As per the question, the RNA strand given is having Thymine which is not possible. Hence, the question is wrong. Taking U (Uracil) instead of T (Thymine) in the given strand the possible solution shall be RNA Polymerase is the enzyme which is used during transcription.
View full question & answer
Question 293 Marks
Why both the strands of DNA are not copied during transcription? Explain.
Answer
Both the strands of DNA are not copied during transcription because if both strands act as a template, they would code for RNA with different sequence the two RNA molecules, if produced simultaneously would be complementary to each other, hence would form a double stranded RNA. Transcription is when RNA is made from DNA. During transcription, RNA polymerase makes a copy of a gene from the DNA to mRNA as needed. This process is similar in eukaryotes and prokaryotes. One difference, however, is that eukaryotic RNA polymerase associates with mRNA processing enzymes during transcription so that processing can proceed quickly after the start of transcription.
View full question & answer
Question 303 Marks

Study the mRNA Segment given above which is complete to be translated into a polypeptide chain.
  1. Write the codons ‘a’ and ‘b.'
  2. What do they code for?
  3. How is peptide bond formed between two amino acids in the ribosome?
Answer
  1.  
a - AUG.

b - UAA/UAG/UGA.
  1.  
AUG code for Methionine.

UAA/UAG/UGA - Stop codon/Nonsense codon/Does not code for any amino acid.
  1.  
Charged tRNA are brought closer together on mRNA in the ribosomes, and Ribosomes acts as a catalyst (ribozyme) forming peptide bond.
View full question & answer
Question 313 Marks
What are 5' end and 3' end of a polynucleotide chain?
Answer
The Polynucleotide Chain One end of the chain ends with a phosphate linked to the 5' carbon of the sugar and is called the 5' end. The other end of the chain ends with an hydroxyl group linked to the 3' carbon of the sugar and is called the 3' end.
View full question & answer
Question 323 Marks
  1. Why is RNA called an 'adapter'?
  2. Draw and label a secondary structure of RNA. How does the actual structure of RNA look like?
Answer
  1. Since 'RNA on one hand binds to a specific amino acid and on the other hand reads the codon of the amino acid bound to it through its anticodon, it is called an 'adapter'.
  2.  


It actually looks like an inverted L.
View full question & answer
Question 333 Marks
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
Properties of genetic code:
  1. AUG is the initiation codon and codes for methionine; methionine is the first amino acid in the given polypeptide.
  2. Genetic code is unambiguous and specific, i.e., each codon codes only for one particular amino acid.
e.g AUG - Methionine, GUG - Valine, etc.
  1. Genetic code is degenerate, i.e. some amino acids are coded by more than one codon.
​​​​​​​e.g. UUU and UUC code for phenylalanine
  1. Each codon is a triplet, i.e. made of three nucleotides e.g AUG, UUU, etc.
  2. UAG does not code for any amino acid; it is a termination codon.
View full question & answer
Question 343 Marks
The following is the flowchart highlighting the steps in DNA fingerprinting technique. Identify A, B, C, D, E and F.
Answer
A- Restriction endonuclease.
B- Agarose.
C- Nylon/ Nitrocellulose.
D- VNTR.
E- Hybridisation.
F- Autoradiography.
View full question & answer
Question 353 Marks
A number of passengers were severely burnt beyond recognition during a train accident. Name and describe a modern technique that can help hand over the dead to their relatives.
OR
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.
Answer
The technique that can help in the identification of victims is DNA fingerprinting which distinguishes between individuals of same species by using their DNA as sample. The chemical structure of DNA is same in everyone (99.9%) except the order of base pairs, i.e. only 0.1"of DNA makes every individual unique. DNA fingerprinting exploits the highly variable repeating sequences, i.e. VNTRs for profiling. These VNTRs are highly conserved among members of the same species.
The DNA polymorphism in simple-sequence DNA and VNTRs is revealed during DNA fingerprinting in establishing kinship since it is very remote possibility to have two individuals with same repeats of satellite DNA throughout the genome except they share a biological relationship. This is because a child gets half DNA from father and half from the mother. Restriction enzymes of genome produce the restriction fragments length polymorphisms, or RFLPs depending on the location of restriction sites which are then sorted by gel electrophoresis followed by southern blotting. The DNA sample from killed passengers and of the sample from possible relatives are compared for sequence homology using radioactive probes and kinship is established.
The presence of similarities between the victims and their relatives determines their association on the basis of which dead bodies can be identified and handed over to their families.
View full question & answer
Question 363 Marks
A tRNA is charged with amino acid phenylalanine:
  1. At what end of the tRNA is the amino acid attached?
  2. What is the mRNA codon that codes for phenylalanine?
  3. Name the enzyme responsible for this attachment.
Answer
  1. At the 3' end.
  2. UUU or UUC.
  3. Aminoacyl tRNA synthetase.
View full question & answer
Question 373 Marks
  1. Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical coding strand given below.
  1. Write the RNA strand transcribed from the above transcription unit along with its polarity.
Answer
  1.  
  1. The RNA transcribed is 5' AUGCAUGCAUAG 3'.
View full question & answer
Question 383 Marks
State any two structural differences and one functional difference between DNA and rRNA.
Answer
S. No
DNA
rRNA
Structural:
(i)
It is a double-stranded structure.
It is a single-stranded structure.
(ii)
It contains nitrogen bases, A, T, G, C.
It contains nitrogen base, A, U, G, C.
(iii)
It has deoxyribose sugar.
It has ribose sugar.
Functional:
 
It determines sequence of amino acid in a polypeptide by transcription and passes information from one generation to another.
It is the site of translation.
View full question & answer
Question 393 Marks
Monocistronic structural genes in eukaryotes have interrupted coding sequences. Explain. How are they different in prokaryotes?
Answer
  1. In eukaryotes, the hnRNA (primary transcript of mRNA) has coding sequences, called exons as well as non-coding sequences, called introns, i.e. the information is split.
  2. It undergoes a process, called splicing, in which the introns are removed and the exons are joined together in a particular manner, to form the functional mRNA.
  • In prokaryotes, the mRNA is polycistronic, i.e. codes for more than one polypeptide.
  1. The information is continuous and no splicing is required.
View full question & answer
Question 403 Marks
AtRNA is charged with the amino acid methionine.
  1. Name the process involved in attachment.
  2. Point out the mRNA codon and anticodon on tRNA for this amino acid.
  3. Name the enzyme responsible for this attachment.
Answer
  1. Aminoacylation of RNA.
  2. mRNA codon-AUG, RNA anticodon-UAC.
  3. Enzyme-aminoacyl RNA synthetase.
View full question & answer
Question 413 Marks
What background information did Watson and Crick have made available for developing a model of DNA? What was their contribution?
Answer
Wastson and Crick had the following informations which helped them to develop a model of DNA,
  1. Chargaffs’ Law suggesting A = T, and C = G.
  2. Wilkins and Rosalind Franklin’s work on DNA crystal’s X-ray diffraction studies about DNA’s physical structure.
  3. Watson and crick proposed:
  • How complementary bases may pair.
  • Semi conservative replication and.
  • Mutation through tautomerism.
View full question & answer
Question 423 Marks
The base sequence in one of the strands of DNA is TAGCATGAT
  1. Give the base sequence of its complementary strand.
  2. How are these base pairs held together in a DNA molecule?
  3. Explain the base complementarity rules. Name the scientist who framed this rule.
Answer
  1. ATCGTACTA.
  2. Through Hydrogen bonds, between A and T and C and G on the two strands.
  3. A= T and C$\equiv$ G, Watson and Crick/Chargaff.
View full question & answer
Question 433 Marks
Study the figure given below and answer the following questions.
  1. Name the molecule 'X' synthesised by i gene. How does this molecule get inactivated?
  2. Which one of the structural genes codes for B-galactosidase?
  3. When will the transcription of this gene stop?
Answer
  1. Molecule 'X' is a repressor protein. When an inducer combines with it, it is inactivated.
  2. 'z' gene.
  3. Transcription of gene stops.
  1. Substrate (lactose) is not available.
  2. Energy source (glucose) is available to the cells.
View full question & answer
Question 443 Marks
Define transformation in Griffith's experiment. Discuss how it helps in the identification of DNA as the genetic material.
Answer
During the course of Griffith’s experiment, bacteria changed its physical form. This was termed as transformation. In this experiment, the DNA of the S strain bacteria survived heating of bacteria. When a mice was injected with mixture of killed S strain and R strain, the mice died of pneumonia. This showed that DNA had the capability of surviving adverse circumstances and manifesting itself on return of favourable conditions. Stability and survival are key considerations for a material to be classified as genetic material. Thus, transformation in Griffith’s experiment helped in identification on DNA as genetic material.
View full question & answer
Question 453 Marks
Study the diagram given below.

Name the linkage X, Y, Z and the respective molecules formed by them.
Answer
X: N-glycosidic linkage.
Y : Phosphoester linkage.
Z: 3'-5' phosphodiester linkage.
X forms nucleoside.
Y forms nucleotide.
Z forms polynucleotide.
View full question & answer
Question 463 Marks
  1. 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?
  2. Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer
  1. Pyrimidine = 500,
  1. Calculation
A = T, A = 240 hence T = 240
A + T = 240 + 240 = 480
so G+C= 1000 - 480 = 520
G = C, so C $=\frac{520}{2}=260$
so pyrimidines = C + T = 260 + 240 = 500
  1. Purine A and G always pair with T and C respectively​​.
  2. $\frac{A}{G}=\frac{T}{C}=1$​​​​​ (Chargaff frule)​​​​​​.
  1.  
View full question & answer
Question 473 Marks
  1. What are the transcriptional products of RNApolymerase III?
  2. Differentiate between 'Capping' and 'Tailing'.
  3. Expand hnRNA.
Answer
  1. RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs, (small nuclear RNAs).
  2.  
S. No.
Capping
Tailing
1.
It is the process of addition of methyl guanosine triphosphate to the 5' end of hnRNA after splicing.
It is the process of addition of adenylate residues to the 3' end of hnRNA after splicing.
  1. Heterogenous nuclear ribonucleic acid (hnRNA).
View full question & answer
Question 483 Marks
During a fire in an auditorium, a large number of assembled guests got burnt beyond recognition. Suggest and describe a modern technique that can help hand over the dead to their relatives.
Answer
DNA finger printing
Isolation of DNA and digestion of DNA by restriction endonucleases, separation of DNA fragments by (gel) electrophoresis and transferring (blotting) of separated DNA fragments to synthetic membrane or nitrocellulose or nylon, hybridization using VNTR probe and detection of hybridised DNA fragments by autoradiography, matching the banding pattern so obtained with that of relative.
View full question & answer
Question 493 Marks
  1. List the two methodologies which were involved in human genome project. Mention how they were used.
  2. Expand ‘YAC’ and mention what was it used for.
Answer
  1. Expressed Sequence Tags, Identifying all the genes that are expressed as RNA.
Sequence Annotation, sequencing the whole set of genome coding or non-coding sequences and later assigning different region with functions.
  1. Yeast Artificial Chromosome used as cloning vectors. (cloning/amplification)
View full question & answer
Question 503 Marks
ATGGAGTACTTCGTGTGA is the coding strand of DNA in a transcription unit.
  1. Write the mRNA transcribed from this DNA segment.
  2. How many amino acids does it code for? Why?
Answer
  1. AUG GAG UAC UUC GUG UGA
  2. It will code for five amino acids.
The last codon, UGA is a termination codon. that does not code for any amino acid.
View full question & answer
Question 513 Marks
Identify by 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
S. No.
Salient features of genetic code
Reason
1. The condin is triplet. AUG, UUU, etc., are triplets.
2.
One codon codes for only one amino acid, so it is unambigus and specific.
UUU codes for serine, AUG, codes for methionine, etc.
3.
AUG has dual function as it codes for methionine and it also acts as initator codon.
AUG is seen at the beginning of the polypeptide chain.
4.
UAG act as a stop codon.
No amino acid is coded by UAG in the polypeptide chain given.
View full question & answer
Question 523 Marks
  1. A tRNA is charged with the amino acid methionine.
  1. Name the process involved in the attachment.
  2. Point out the mRNA codon and anticodon on RNA for this amino acid.
  1. What is heterochromatin?
Answer
  1.  
  1. Amino acylation of tRNA or charging of tRNA.
  2. mRNA codon-AUG.
tRNA anticodon-UAC.
  1. The part of the chromatin that is tightly packed and stains dark, is called heterochromatin.
View full question & answer
Question 533 Marks
With the help of a schematic diagram, explain the location and role of the following in a transcription unit.
Promoter, structural gene, terminator.
Answer
  1. Promoter- It is a DNA sequence that provides a binding site for RNA polymerase for the initiation of transcription.
  2. Structural gene- It codes for enzymes or proteins for structural functions.
  3. Terminator- It usually defines the end of the transcription process.
View full question & answer
Question 543 Marks
How has the sequencing of human genome opened new windows for treatment of various genetic disorders. Discuss amongst your classmates.
Answer
Sequencing of human genome has opened new windows for treatment of various genetic disorders. We know that genetic disorders are caused by some alteration in genes. At present, we do not have exact information about the base pair sequence where this alteration takes place. Hence, we are unable to devise any tool to prevent genetic disorders. By proper understanding of the particular sequence responsible for a particular genetic disorder, the scientist may be able to devise some tools to prevent genetic disorders. A future may come when nobody will be suffering from genetic disorders; especially those which create serious disability.
View full question & answer
Question 553 Marks
The base sequence on one of the strands of DNA is ATGTCTATA.
  1. Give the base sequence of its complementary strand.
  2. If an RNA strand is transcribed by this strand, what would be the base sequence of RNA?
  3. In what other respect, an RNA molecule differs from a DNA molecule?
Answer
  1. TACAGATAT
  2. UACAGAUAU
  3.  
S. No DNA RNA
(i) It is a double-stranded structure. It is a single-stranded structure.
(ii) It contains deoxyribose sugar. It contains ribose sugar.
(iii) Bases are A, T, G, C Bases are A, G, C, U
View full question & answer
Question 563 Marks
What are the three types of RNA? Mention their relation to protein synthesis.
Answer
Types of RNA:
S. No.
Types of RNA
Functions
(i)
Messenger RNA (mRNA)
1.
It stores the genetic information from DNA.
2.
It decides the sequence of amino acid in a polypeptide.
(ii)
Transfer RNA (tRNA)
1.
It acts as an adaptor molecule that at one end reads the code on mRNA and accordingly bind to amino acid on the other end.
2.
It recognises the codon on mRNA by its anticodon and leaves amino acid at the protein synthesis site.
(iii)
Ribosomal RNA (rRNA)
1.
It constitutes the ribosomal structure.
2.
It helps to form peptide bond.
View full question & answer
Question 573 Marks
What is an operon? Name any four operons in E.coli.
Answer
All the genes controlling a metabolic pathway, constitute an operon.Example:
Lac operon - trp operon. Val operon - his operon.
View full question & answer
Question 583 Marks
In a typical nucleus some regions of chromatin are stained light and others dark. Explain why it is so and what is its significance.
Answer
  1. The regions of a chromatin, which are stained light, are the regions where chromatin is loosely packed; they are called euchromatin.
  2. The regions of a chromatin, which are stained dark, are the regions where chromatin is tightly packed; they are called heterochromatin.
  3. Euchromatinis transcriptionally more active, while heterochromatin is inactive.
View full question & answer
Question 593 Marks
A burglar in a huff forgot to wipe off his blood-stains from the place of crime where he was involved in a theft and fight. Name the technique which can help in identifying the burglar from the blood-stains. Describe the technique.
Answer
DNA Fingerprinting:
  • Dr. Alec Jeffreys developed the technique of DNA fingerprinting in an attempt to identify DNA marker for inherited diseases.
  • DNA fingerprinting uses short nucleotide repeats called Variable Number Tandem Repeats (VNTRs) as markers. VNTRs vary from person to person and are inherited from one generation to the next. Only closely related individuals have similar VNTRs.
Methodology and Technique: DNA is isolated and extracted from the cell or tissue by centrifugation. By the process of polymerase chain reaction (PCR), many copies are produced. This step is called amplification. DNA is cut into small fragments by treating with restriction endonucleases.
  1. DNA fragments are separated by agarose gel electrophoresis.
  2. The separated DNA fragments are visualised under ultraviolet radiation after applying suitable dye.
  3. The DNA is transferred from electrophoresis plate to nitrocellulose or nylon membrane sheet. This is called Southern blotting.
  4. VNTR probes are now added which bind to specific nucleotide sequences that are complementary to them. This is called hybridisation.
  5. The hybridised DNA fragments are detected by autoradiography. They are observed as dark bands on X-ray film.

Schematic representation of DNA fingerprinting: Few chromosomes have been shown to contain different copy number of VNTR. The two alleles (paternal and maternal) of a chromosome also contain different copy numbers of VNTR. It is clear that the banding pattern of DNA from crime scene matches with individual B and not with A.Applications of DNA Fingerprinting:
  1. It is used as a tool in forensic tests to identify criminals and criminal investigations.
  2. It is used to settle paternity disputes and maternity disputes.
  3. It is used to determine population and genetic diversities to study evolution.
  4. It is used in the study of evolutionary biology.
View full question & answer
Question 603 Marks
Where is an operator' located in a prokaryotic DNA? How does an operator regulate gene expression at transcription level in a prokaryote? Explain.
Answer
  1. The operator' region is adjacent to the promoter region.
  2. The operator regulates the accessibility of promoter region to the RNA polymerase for transcription.
  3. The operator binds to the repressor protein coded by the regulatory gene and prevents the RNA polymerase from binding to the promoter; the operon is switched "off".
  4. When an inducer binds to the repressor and inactivates it, it cannot bind to the operator; this allows RNA polymerase access to the promoter and transcription continues, i.e., the operon is switched 'on'.
View full question & answer
Question 613 Marks
The following is the flow chart highlighting the steps in DNA fingerprinting technique. Identify a, b, c, d, e and f.
Answer
  1. Restriction endonuclease.
  2. Agarose gel.
  3. Nitrocellulose/ nylon/ synthetic membrane.
  4. Variable Number Tandem Repeats (VNTR).
  5. Hybridisation with VNTR probe.
  6. Autoradiography.
View full question & answer
Question 623 Marks
Describe the experiment that helped demonstrate the semi-conservative mode of DNA replication.
Answer
Grown E.coli in ${ }^{15} \mathrm{NH}_4 \mathrm{Cl}$ for many generations to get ${ }^{15} \mathrm{N}$ incorporated into DNA, Then the cells are transferred into ${ }^{14} \mathrm{NH}_4 \mathrm{Cl}$, The extracted DNA are centrifuged in CsCl and measured to get their densities, DNA extracted from the culture after one generation (20 minutes), showed intermediate hybrid density, DNA extracted after two generations (40 minutes) showed light DNA and hybridDNA. A correctly labelled diagrammatic representation in lieu of the  above explanation of experiment to be considered.
View full question & answer
Question 633 Marks
  1. Identify the polarity at A and B respectively in the figure given below.
  2. Explain the mechanism the figure represents.
Answer
  1. A-5', B-3
  2.  
  1. The figure represents the continuous and discontinuous synthesis of DNA strands at the replication fork, during replication of DNA.
  2. Replication fork is the Y-shaped structure formed with small opening of DNA double helix.
  3. DNA polymerase catalyses polymerisation of nucleotides only in 5' → 3' direction.
  4. Both the strands of parental) DNA act as templates for the synthesis of new strands.
  5. On the template strand with 3' → 5' polarity, the new strand is synthesised as a continuous stretch; it is called continuous synthesis.
  6. On the other strand with 5' → 3' polarity, DNA is synthesised as short stretches; it is called discontinuous synthesis.
  7. Later the short stretches of DNA are joined by DNA-ligases into a continuous strand.
View full question & answer
Question 643 Marks
  1. Identify the polarity from a to a', in the above diagram and mention how many more amino acids are expected to be added to this polypeptide chain.
  2. Mention the DNA sequence coding for serine and the anticodon of tRNA for the same amino acid.
  3. Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA?
Answer
  1. a to a' is 5' → 3'. No more amino acid will be added.
  2. TCA; anticodon is UCA.
  3.  
  1. The untranslated regions are required for efficient translation process.
  2. They are present before the initiation codon at the 5'-end and after the stop/ termination codon, at the 3'-end.
View full question & answer
Question 653 Marks
Name the three RNA polymerases found in eukaryotic cells and mention their functions.
Answer
In eukaryotes, there are three RNA-polymerases:
  1. RNA-polymerase I catalyses transcription of rRNAs (28S, 185 and 5.85).
  2. RNA-polymerase II catalyses transcription of the precursor of mRNA; it is called hnRNA.
  3. RNA- polymerase III catalyses transcription of RNA, 5SrRNA and shRNAs.
View full question & answer
Question 663 Marks
One of the codons on mRNA is AUG. Draw the structure of tRNA adaptor molecule for this codon. Explain the uniqueness of this tRNA.
Answer
This tRNA is specific for amino acid methionine and also acts as initiator tRNA.
View full question & answer
Question 673 Marks

Why do you see two different types of replicating strands in the given DNA replication fork? Explain. Name these strands.
Answer
  1. Both the strands of parent DNA function as template strands.
  2. On the template strand with 3' → 5' polarity, the new strand is synthesised as a continuous stretch as the DNA polymerase can carry out polymerisation of the nucleotides only in 5' → 3' direction; this is called continuous synthesis and the strand is called leading strand.
  3. On the other template strand with 5 → 3' polarity, the new strand is synthesised from the point of replication fork, also in 5' → 3' direction, but in short stretches; they are later joined by DNA-ligases to form a strand, called lagging strand.
View full question & answer
Question 683 Marks
Describe the discontinuous synthesis of DNA.
Answer
The discontinuous synthesis of DNA is as follows:
  1. At the replication site, unwinding of double stranded DNA takes place by DNA gyrase and helicase.
  2. ssBPs (single-stranded binding proteins) bind to the separated strands to avoid recoiling or to provide stability.
  3. Since DNA polymerase can synthesise DNA only in 5′ → 3′ direction, DNA synthesis occurs discontinuously on the lagging strand.
  4. These small fragments of DNA are called Okazaki fragments.
  5. The enzyme primase adds primers after every fragment is formed.
  6. These Okazaki fragments are then joined by DNA ligase.
View full question & answer
Question 693 Marks
A considerable amount of lactose is added to the growth medium of E.coli. How is the lac operon switched on in the bacteria? Mention the state of the operon when lactose is digested.
Answer
In lac operon, when lactose is added, it enters the cell wall with the help of permease, a small amount of which is already present in cell. Lactose binding to activates repressor and changes its structure. The repressor now fails to bind to the operator. Then, RNA polymerase starts transcription of operon by binding to promoter site-P. All the three enzymes for lactose metabolism are synthesised.
Finally, all the lactose molecules are used up. After sometime, when whole of lactose is consumed, there is no inducer present to bind to the repressor. Then the repressor becomes active again, attaches itself to the operator and finally switches off the operon.
View full question & answer
Question 703 Marks
Describe the initiation process of transcription in bacteria.
Answer

(DNA dependent) RNA polymerase, binds to the promoter, at 5’ end, associates transiently with initiation factor/sigma factor, using nucleoside triphosphates as substrate, and energy initiates transcription.
View full question & answer
Question 713 Marks
Identify A, B, C, D, E and F in the table given below:
S. No.
Component-I
Component-II
Chemical linkage bonding the two components
Product
(i)
A
B
C
Nucleotide
(ii)
Nucleotide
D
E
Nucleotide
(iii)
Nucleotide
Nucleotide
F
Dinucleotide
Answer
A- Nitrogenous base (Purine/ Pyrimidine).
B- Pentose sugar (Ribose/ Deoxyribose).
C- N-glycosidic linkage.
D- Phosphate E-Phosphoester linkage.
F- Phosphodiester linkage.
View full question & answer
Question 723 Marks
Explain the post-transcriptional modifications, the hnRNA undergoes in eukaryotic cells.
Answer
  1. Since eukaryotes have split gene arrangement, the hRNA has both coding sequences (exons) and non-coding sequences (introns) and is non-functional; so it has to undergo splicing, the process, in which introns are removed and exons are joined.
  2. It has to undergo two other processes, namely capping and tailing to become functional.
  3. In capping, an unusual nucleotide, called methyl guanosine triposphate residues are added at the 5' end of hnRNA.
  4. In-tailing, about 200-300 adenylate residues are added at the 3' end, in a template-independent manner.
  5. These changes take place in the nucleus of the cell.
View full question & answer
Question 733 Marks
  1. A DNA segment has a total of 1,500 nucleotides, out of which 410 are Guanine containing nucleotides. How many pyrimidine bases this segment possesses?
  2. Draw a diagrammatic sketch of a portion of DNA segment to support your answer.
Answer
  1. 750
  1. Calculation.
$\text{G} =\text{C},\ \text{G} = 410\ \ \ \text{hence}\text{ C} = 410$
$\text{G + C} = 410 + 410 = 820$
$\text{so A + T }= 1500 - 820 = 680 $
$\text{A} =\text{T},\text{so}\ \ \text{T}=\frac{680}{2}=340$
$\text{so pyrimidines} =\text{ C + T}$
$= 410 + 340 = 750$
  1. Purine A and G always pair with T and C respectively.
  2. $\frac{\text{A}}{\text{G}}=\frac{\text{T}}{\text{C}}=1\text{(Chargaff rule)}$
  • Diagram showing polarity
  • A-T
  • G-C
  • H-bond.
View full question & answer
Question 743 Marks
  1. Draw a labelled schematic diagram of a replication fork showing continuous and discontinuous replication of DNA strands.
  2. State a reason why is the replication continuous and discontinuous in the diagram drawn.
Answer
  1.  
  1. 2 strands are antiparallel , DNA polymerase acts only in one direction i.e. 5' $\rightarrow $ 3'
View full question & answer
Question 753 Marks
A number of passengers were severely burnt beyond recognition during a train accident. Name and describe a modern technique that can help hand over the dead to their relatives.
Answer
DNA fingerprinting
Isolation of DNA and digestion of DNA by restriction endonucleases, separation of DNA fragments by (gel) electrophoresis and transferring (blotting) of separated DNA fragments to synthetic membrane or nitrocellulose or nylon, hybridization using VNTR probe and detection of hybridised DNA fragments by autoradiography, matching the banding pattern so obtained with that of relative.
View full question & answer
Question 763 Marks
Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer
  1. They do not code for any proteins,
  2. They form large part of the human genome,
  3. They show high degree of polymorphism/Specific to each individual.
View full question & answer
Question 773 Marks
  1. Differentiate between unambiguous and degenerate codons.
  2. Write two functions of the codon AUG.
Answer
  1. Differences between unambiguous and degenerate codons are-
S. No.
Unambiguous codon
Degenerate codon
1.
No ambiguity for a particular codon.
Code is degenerate for a particular amino acid.
2.
A particular codon will always code for the same amino acid, where it is found.
One amino acid often has more than one code triplet.
3.
eg. GGA is an unambiguous codon, it codes only for glycine and no other amino acid.
e.g. Phenylalanine has two codons, i.e. UUU and UUC.
  1. Functions of codon AUG is as follows-
  1. Codes for methionine.
  2. Serves as a signal to initiate protein synthesis (initiator codon).
View full question & answer
Question 783 Marks
Explain the mechanism of DNA replication with the help of a replication fork. What role does the enzyme DNA-ligase play in a DNA replication fork?
Answer
  1. DNA replication occurs in small replication forks. It does not occur in its entire length in one time as DNA is a very large molecule and only that part of DNA opens up which is being replicated. The opening of the whole DNA molecule would be an energetically more expensive process.
  2. The main enzyme involved in DNA replication is the DNA-dependent DNA polymerase. This enzyme catalyzes the polymerization of deoxynucleotides along the 5′ → 3′ direction, and hence, replication is continuous along the 3′ → 5′ strand (leading strand) and discontinuous along the template, i.e., the 5′ → 3′ direction (lagging strand).
  3. Okazaki fragments are short DNA segments on the lagging strand, formed in the 5’ – 3’ direction, starting from RNA primers. A separate RNA primer is needed for the synthesis of each Okazaki fragment. These discontinuously synthesized fragments are later joined by the enzyme DNA ligase.
  4. Ori stands for Origin of replication. This site has the highly conserved sequence of DNA among various species. The replication of DNA starts here because this site attracts some proteins which help in the opening and unwinding of DNA and this leads to the initiation of replication.
The function of DNA Ligase is to join the two nucleotides. During the DNA replication process, it joins the Okazaki fragments of the daughter DNA to form the complete DNA molecule on the lagging strand.
View full question & answer
Question 793 Marks
The charge present on proteins plays an important role in gene regulation. State the biochemistry if histones were to be mutated and made rich in acidic amino acids in place of basic amino acids.
Answer
In eukaryotes, gene regulation takes place by different mechanisms. Here proteins (histones) play major role in gene regulation. Histones are gene repressors and these mark the gene activity in non-specific manner.
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 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. The packaging of DNA in eukaryotes would not happen. Consequently, the chromatin fibre would not be formed.
View full question & answer
Question 803 Marks
Describe the structure of RNA polynucleotide chain having four different types of nucleotides.
Answer
A RNA nucleotide has three main components-a nitrogenous base, a ribose sugar and a phosphate group.
  1. The ribose sugar and the phosphates form the backbone of a polynucleotide chain with nitrogenous base linked to sugar moiety and projecting from the backbone.
  2. Nitrogenous Bases- (i) parines (Adenine & Guanine) (ii) Pyrimidines- (Cytosine and Uracil).
  3. Nitrogenous base is linked to the ribose sugar through N-glycosidic linkages.
  4. Phosphate group is linked to 5’ OH of a nucleoside through phosphodiester linkage to form a corresponding nucleotide.
  5. Every nucleotide residue has an additional-OH group present at 2’-position in the ribose.
  6. Many nucleotides are linked through 3’-5’ phosphodiester linkage to each other to form the polynucleotide chain.
  7. The end of the chain which has free phosphate moiety at 5 'end of ribose sugar is referred to as 5' end the other end of the chain having a free 3'-OH group at the ribose sugar is referred to as 3' - end of the polynucleotide chain.
View full question & answer
Question 813 Marks
Draw a labelled schematic sketch of replication fork of DNA. Explain the role of the enzymes involved in DNA replication.
Answer
A.
B. DNA polymerase catalyses the polymerisation of nucleotides, Ligase joins the fragments of discontinuous synthesis.
View full question & answer
Question 823 Marks
One of the codons on mRNA is AUG. Draw the structure of tRNA adapter molecule for this codon. Explain the uniqueness of this tRNA?
Answer

Clover leaf shaped.
This tRNA is specific for methionine/can act as initiator tRNA.
View full question & answer
Question 833 Marks
You are repeating the Hershey-Chase experiment and are provided with two isotopes: $^{32}P$ and $^{15}N$ (in place of $^{35}S$ in the original experiment). How do you expect your results to be different?
Answer
Use of ${ }^{15} \mathrm{N}$ will be inappropriate because method of detection of ${ }^{35} \mathrm{P}$ and ${ }^{15} \mathrm{N}$ is different $\left({ }^{32} \mathrm{P}\right.$ being a radioactive isotope while ${ }^{15} \mathrm{N}$ is not radioactive but is the heavier isotope of nitrogen). Even if ${ }^{15} \mathrm{~N}$ was radioactive then its presence would have been detected both inside the cell ( ${ }^{15} \mathrm{N}$ incorporated as nitrogenous base in DNA) as well as in the supernatant because ${ }^{15} \mathrm{N}$ would also get incorporated in amino group of amino acids in proteins). Hence, the use of ${ }^{15} \mathrm{N}$ would not give any conclusive results.
View full question & answer
Question 843 Marks

  1. What is this diagram representing?
  2. Name the parts a, band c.
  3. In the eukaryotes the DNA molecules are organised within the nucleus.How is the DNA molecule organised in a bacterial cell in absence of a nucleus?
Answer
  1. Nucleosomes.
  2. a - Histone octamer
b - DNA.

c- H1 histone.
  1. In bacteria DNA in nucleoid, is organised in large loops held by proteins.
View full question & answer
Question 853 Marks
A typical mammalian cell has 2.2m long DNA molecule, whereas the nucleus in which it is packed measures about 10m. Explain how such a long DNA molecule is packed within a tiny nucleus in the cell.
Answer
  1. In the mammalian cells (or eukaryotes) there is a set of positively-charged basic proteins, called histones.
  2. Histones are organised to form a unit of eight molecules, called histone octamer.
  3. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure, called nucleosome.
  4. A typical nucleosome contains 200 bp of DNA helix.
  5. The nucleosomes constitute the repeating units of chromatin, which appear as beads-on-string structure under an electron microscope.
  6. These are further packaged to form the chromatin fibres, which condense to form chromosomes.
  7. The packaging of chromatin at higher levels requires additional set of proteins called non-histone chromosomal (NHC) proteins.
View full question & answer
Question 863 Marks
  1. Why did Hershey and Chase use radioactive sulphur and radioactive phosphorus in their experiment?
  2. Write the conclusion they arrived at and how.
Answer
  1. In order to label protein coat of virus with radioactive sulphur, label DNA with radioactive phosphorus.
  2. Bacteria which were infected with viruses having radioactive DNA were found to contain radioactive DNA later on-
Bacteria which were infected with viruses having radioactive protein coat were not found to contain radioactivity.

Conclusion- DNA is the genetic material.
View full question & answer
Question 873 Marks
Why did Hershey and Chase work with bacteriophages in their experiments to prove that DNA is the genetic material? Explain.
Answer
Hershey and Chase selected bacteriophage for their work because of the following reasons:
  1. A bacteriophage has only two components-protein and DNA.
  2. They worked to find whether it was protein or DNA from the virus that entered the bacteria.
  3. The bacteriophage attaches to the bacteria.
  4. Its genetic material enters the bacterial cell.
  5. The bacterial cell treats the viral genetic material as if it was its own and subsequently manufactures more virus particles. vi. Proving DNA to be the genetic material/by tracing the radioactivity on DNA.
View full question & answer
Question 883 Marks
Explain the following salient features of a DNA molecule:
  1. Complimentarity of base pairs of its two chains.
  2. Antiparallel polarity of the two chains.
Answer
  1. The two strands of a DNA are said to be complementary to each other, i.e., a purine of one strand always pair with a pyrimidine.
  1. Adenine (a purine) pairs with thymine (a pyrimidine) by forming two hydrogen bonds.
  2. Guanine (a pyrimidine) by forming three hydrogen bonds.
  1. The two chains of a DNA shows antiparallel polarity, which means that if one strand has 5' 3' polarity, the other strand has 3' → 5' polarity.
View full question & answer
Question 893 Marks
  1. In a human genome, which one of the chromosomes has the most genes and which one has the fewest?
  2. Scientists have identified about 1.4 million single nucleotide polymorphs in human genome. How is the information of their existence going to help the scientists?
Answer
  1. Chromosome 1 has the most genes and Y-chromosome has the fewest.
  2. It is expected to help in:
  1. Locating the disease-associated sequences of DNA on the chromosomes.
  2. Tracing human history.
View full question & answer
Question 903 Marks
  1. One of the codons on mRNA IS AUG. Draw the structure of RNA adapter molecule for this codon.
  2. Name the RNA polymerase that transcribes RNA in eukaryotes.
  3. What is unique about the amino acid this RNA binds with?
Answer
  1.  
  1. RNA polymerase III.
  2. The amino acid methionine is the initiator amino acid.
View full question & answer
Question 913 Marks
The enzyme DNA polymerase in E.coli is a DNA dependent polymerase and also has the ability to proof-read the DNA strand being synthesised. Explain. Discuss the dual polymerase.
Answer
DNA polymerase uses DNA template to catalyse the polymerization of deoxynucleotides and hence it is called DNA–dependent. When a new strand of DNA is being processed, this enzyme moves along to hasten the speed of polymerization. While doing so, it “proof reads” the strand being formed. By doing so, it helps in speeding up the process. Hence, its nature can be said as dual, i.e. of reading the template and then proof reading the new strand.
View full question & answer
Question 923 Marks
DNA separated from one cell, when introduced into another cell is able to bestow some of the properties of former to the latter. What is this change called in technical terms? Describe the experimental evidences which led to the discovery of the above phenomenon.
Answer
This change is called transformation.Transforming Principle:
  • 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.
  • When live S-type cells were injected into mice, they died due to pneumonia.
  • When live R-type cells were injected into mice, they survived.
  • When heat-killed S-type cells were injected into mice, they survived and there were no symptoms of pnuemonia.
  • When heat-killed S-type cells were mixed with live R-type cells and injected into mice, they died due to unexpected symptoms of pneumonia and live S-type cells were obtained from mice.
  • He concluded that heat-killed S-type bacteria caused a transformation of the R-type bacteria into S-type bacteria but he was not able to understand the cause of this bacterial transformation.
View full question & answer
Question 933 Marks
Why is a DNA molecule considered as a better hereditary material than RNA molecule?
OR
Why is DNA a better genetic material when compared to RNA?
Answer
DNA is a better genetic material for the following reasons:
  1. The genetic material should be stable and should not change with age or change in physiology; this stability is given to DNA by its.
  1. Double-stranded nature.
  2. Presence of thymine.
  3. The plane of base pair stacking over the other.
  1. Because the 2-OH group of RNA nucleotides is a reactive group, it makes RNA labile and easily degradable; it is absent in DNA.
  2. RNA (23S RNA) is also catalytic, i.e. it is reactive; but DNA is less reactive and more stable.
View full question & answer
Question 943 Marks
Describe the termination process of transcription in bacterium.
Answer
  1. Once the RNA polymerase reaches the terminator sequence of the DNA, it binds transiently to the termination factor (rho/ p).
  2. The RNA strand synthesised falls off, followed by the RNA polymerase, i.e., termination has occurred.
​​​​​​​
View full question & answer
Question 953 Marks
  1. How may codons code for amino acids and how many are unable to do so?
  2. Why are codes said to be.
  1. Degenerate.
  2. Unambiguous.
OR
  1. How many codons code for amino acids and how many do not?
  2. Explain the following giving one example of each.
  1. Unambiguous and specific codon.
  2. Degeneration codon.
  3. Universal codon.
  4. Initiator codon.
Answer
  1. Out of 64 codons 61 code for amino acids and rest 3 codons do not code for any amino acids. These function as stop codons.
  2.  
  1. As codon is a triplet. Out of 64 codons, 6l code for 20 amino acids and 3 codons (UAA, UGA and UAG) do not code for any amino acids. Thus, they function as terminating or stop codons.
  2. Genetic code is uaambiguous and specific. Thus, one codon codes for only one amino acid, e.g. CCU codes only for proline and not for any other amino acid.
  3. Since, rhe number of codons is greater, all of thc amino acids, with exceptions of methionine (AUG) and tryptophan (UGG), are coded by more than one codon, a fearure referred to as the degenerecy of genetic code, e.g. GGU, GGC, GGA and GGG all specifr the same amino acid glycine.
  4. The codon is read in mRNA in a contiguous fuhion, i.e. without puncoration. Thus, the code is commaless.
  5. The genetic code is nearly universal, i.e. particular codon codes for the same amino acid in all organisms except in mitochondrial codons and few Protozoans.
  6. AUG is a codon with dual fi.rncdons. It codes for the amino acid methionine (met) and also acts as an intiator codon.
View full question & answer
Question 963 Marks
Describe the experiments that established the identity of 'transforming principles' of Griffith.
Answer
  • Purification of biochemicals like Proteins, RNA & DNA from S cells. (heat killed)
  • Presence of Protein & RNA in medium did not affect transformation.
  • DNA alone from S Bacteria caused R Bacteria to transform.
  • Digestion with DNAase did inhibit transformation,
Conclusion: DNA is the transforming chemical/biochemical.
View full question & answer
Question 973 Marks
The length of a DNA molecule in a typical mammalian cell is calculated to be approximately 2.2m. How is the packaging of this long molecule done to accommodate it within the nucleus of the cell?
Answer
The length of DNA in humans can be calculated as
Distance between two consecutive base pairs is
$0.34 \mathrm{~nm}=0.34 \times 10^{-9} \mathrm{~m}$
Total number of base pairs in a DNA helix in a typical mammalian cell $=6.6 \times 10^9$
$\therefore$ Length of this DNA double helix $=6.6 \times 10^9 \times 0.34 \times 10^{-9} \mathrm{~m}$
$=2.2 \mathrm{~m}$ (approx)
Thus, we see that, the length of DNA is hr greater than the dimension of a typical nucleus, which is approx 10{m in diameter. Theerefore, there must bJ some special arrangement to accommodate such a long polymer in a cell of both prokaryotes and eukaryotes.
View full question & answer
Question 983 Marks
Study the diagram given below:
Name the linkages X, Y, Z and the respective molecules formed by them.
Answer
  1. X is N-glycosidic linkage; it forms a nucleoside.
  2. Y is phosphoester linkage; it forms a nucleotide.
  3. Z is a phosphodiester linkage; it forms a dinucleotide.
View full question & answer
Question 993 Marks
How are the structural genes activated in the lac operon in E. coli?
Answer
Lactose acts as the inducer, binds with repressor protein, frees operator gene, RNA polymerase freelymoves over the structural genes, transcribing lac mRNA, which in turn produces the enzymes responsible for the digestion of lactose A complete labelled diagram depicting the concept can be evaluated in lieu of explanation.
View full question & answer
Question 1003 Marks
Given below are the sequences of nucleotides in a particular mRNA and amino acids coded by it.
UUU AUG UUC GAG UUA GUG UAA
Phe - Met - Phe - Glu - Leu – Val
Write the properties of genetic code that can be and that cannot be correlated from the above given data.
Answer
  1. The properties that can be correlated:
  1. UAA does not code for any amino acid; it is a termination codon.
  2. Genetic code is specific and unambiguous, i.e. one codon codes for a particular amino acid only.
  3. Each codon is a triplet.
  4. Genetic code is degenerate, as one amino acid is coded by more than codon, e.g. UUU and UUC code for phenylalanine.
  5. Genetic code is read in a contiguous manner without any punctuation.
  1. The property that cannot be correlated:
AUG has a dual function; it is initiation codon as well as codes for methionine.
View full question & answer
Question 1013 Marks
What is DNA polymorphism? Why is it important to study it?
Answer
Polymorphism (variation at genetic level) arises due to mutations. Allelic sequence variation has traditionally been described as a DNA polymorphism if more than one variant (allele) at a locus occurs in human population with a frequency greater than 0.01. In simple terms, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism. Polymorphism become very useful identification tool in forensic applications. Further, as the polymorphisms are inheritable from parents to children, DNA fingerprinting is the basis of paternity testing, in case of disputes.
View full question & answer
Question 1023 Marks
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
Aspartic and glutamic acid are acidic amino acids, while lysine and arginine are basic amino acids. Lysine and arginine carry positive charge on their side chains which is not the case with aspartic acid and glutamic acid. DNA is negatively charged and hence is wrapped around the positively charge histone octamer. If acidic amino acids are present in histone because of mutation, DNA won’t be able to wrap around itself. Thus, long strand of DNA will not be able to fit inside the small space in the nucleus. This will mean an end to nuclear organization which is possible because of efficient packaging.
View full question & answer
Question 1033 Marks
How do initiation and termination of translation processes occur in bacteria? Where are untranslated regions located in an mRNA? Mention their role.
Answer
Initiation of translation:
  1. When the small subunit of ribosome binds to the mRNA the process of translation starts; in bacteria the ribosome also acts as a catalyst (23S rRNA is ribozyme) for peptide bond formation.
  2. The ribosome binds to mRNA at the start codon (AUG), that is recognised by the initiator RNA; it involves certain initiation factors.
Termination of translation:
  1. When the ribosome falls on a termination codon, a release factor binds to it, and translation is terminated and the polypeptide is released from the ribosome.
  2. Untranslated regions are present at the 5' end before the start codon and also at the 3' end after the termination codon.
  3. They are required for efficient translation.
View full question & answer
Question 1043 Marks
In a series of experiments with Streptococcus and mice F. Griffith concluded that R-strain bacteria had been transformed. Explain.
Answer
S strain bacteria when injected - mice die, R - mice live, heat killed S - mice live, heat killed S + live.
R - mice die, recovered living S from dead mice, R strain bacteria had been transformed to S strain by the genetic material of heat killed S strain.
View full question & answer
Question 1053 Marks
What is mutation? Explain with the help of an example how does a point mutation affect the genetic code. Name another type of mutation.
Answer
A mutation is a stable, inheritable change in genetic material. A classical example of point mutation is a change of single base pair in the gene for B-globin chain that results in the change of amino acid residue glutamate to valine. It results into a diseased condition called Sickle-cell anaemia.
The insertion or deletion of one or two bases may change the reading frame from the point of insertion or deletion Insertion or deletion of three or its multiple bases insert or delete one or multiple codon, hence one or multiple amino acid may be formed in polypeptide. The other type of mutation is frameshift insertion or deletion mutation.
View full question & answer
Question 1063 Marks
Explain the role of regulatory gene in lac operon. Why is the regulation of the operon called negative regulation?
Answer
  1. The regulatory gene in lac operon (also called i gene or inhibitor gene) codes for a protein, called repressor; the repressor is synthesised all the time constitutively in the cell.
  2. The repressor has high affinity to the operator; it binds to the operator region and prevents the RNA polymerase from transcribing the structural genes of the operon.
  3. It is called negative regulation because the operon is switched off and transcription is prevented.
View full question & answer
Question 1073 Marks
What is ‘semi-conservative’ DNA replication? How was it experimentally proved and by whom?
Answer
After the completion of replication (of the DNA molecule into two)) each DNA molecule will have one parental strand and newly synthesised strand.Meselson and Stahl
  1. E.coli grown in medium containing ${ }^{15} \mathrm{NH}_4 \mathrm{CL}\left({ }^{15} \mathrm{N}\right.$- heavy Nitrogen$)$ for many generations to ensure that all DNA in the bacteria were heavy,
  2. Heavy E.coli transferred to a medium with normal ${ }^{14} \mathrm{NH}_4 \mathrm{CL}$,(after 20 minutes) DNA of generation I extracted to measure their densities, they were of intermediate density,
  3. After 40 minutes DNA of II generation were extracted and tested for their densities, they were of equal amounts of (hybrid) intermediate DNA, and light DNA/${ }^{14} \mathrm{NH}_4 \mathrm{Cl}$.
View full question & answer
Question 1083 Marks
Explain Hershey - Chase experiment. What was proved through this experiments?
Answer

DNA is genetic material.
View full question & answer
Question 1093 Marks
  1.  
  1. What does central dogma state in molecular biology? How does it differ in some viruses?
Answer
  1.  
A - Replication of DNA.

B - Transcription.

C - Translation.
  1.  
  1. The central dogma of molecular biology states that the genetic information flows from DNA to RNA to proteins.
  2. In some viruses, the flow of information is in reverse direction, i.e. from RNA to DNA and is called reverse transcription.
View full question & answer
Question 1103 Marks
There are certain complexities in eukaryotic transcription. Bring out the significance of such complexities.
Answer
Eukaryote gene structure and function differ from prokaryote gene structure and function in several important ways. Eukaryotes have many more genes and these genes are spread across multiple chromosomes. Prokaryotes have less number of genes and these genes are all located on one chromosome. The greater complexity of the eukaryote genome means that a greater variety and complexity of control mechanisms is necessary. There are also more steps in the transcription and translation process at which control of expression can occur in eukaryote.Because eukaryotes pocess a nuclear membrane thair mRNA must be completely formed and must pass across the nuclear envelope before translation. Eukaryotic mRNA s are modified before they are translated . Introns are removed and the remainng exons are spiced together . A 5'cap and a 3' poly-A tail are added . In a eukaryotic cell, transcription occurs in the nucleus, and translation occurs in the cytoplasm.
View full question & answer
Question 1113 Marks
  1. Name the enzyme responsible for the transcription of tRNA and the amino acid the initiator tRNA gets linked with.
  2. Explain the role of initiator tRNA in initiation of protein synthesis.
Answer
  1. RNA polymerase III, Methionine.
  2. This tRNA (charged with amino acid Methionine) reaches the smaller subunit of ribosome, with its anticodon UAC recognises the codon AUG on mRNA and binds by forming complementary base pairs, leaves the amino acid, initiating protein synthesis.
View full question & answer
Question 1123 Marks
Name the specific enzyme responsible for nucleotide polymerisation in DNA replication. Write two characteristic features of this enzyme. Name the region on E.coli DNA, where this enzyme can initiate replication.
Answer
  1. DNA-dependent DNA polymerase is the enzyme.
  2. Two characteristic features of this enzyme include:
  1. It catalyses the polymerisation of nucleotides in 5' → 3' direction only.
  2. It polymerises approximately 2000bp per second, with high degree of accuracy.
  1. Origin of replication is the region on E.coli DNA, where this enzyme can initiate replication.
View full question & answer
Question 1133 Marks
How would you justify circular flow of genetic information?
Answer
This flow of information is dependent on the genetic code, which defines the relation between the sequence of bases in DNA (or its mRNA transcript) and the sequence of amino acids in a protein. The code is nearly the same in all organisms: a sequence of three bases, called a codon, specifies an amino acid. Codons in mRNA are read sequentially by tRNA molecules, which serve as adaptors in protein synthesis. Protein synthesis takes place on ribosomes, which are complex assemblies of rRNAs and more than 50 kinds of proteins.
View full question & answer
Question 1143 Marks
  1. Name the enzyme that catalyses the transcription of hnRNA.
  2. Why does the hnRNA need to undergo changes? List the changes hn RNA undergoes and where in the cell such changes take place.
Answer
  1. RNA polymerase II.
  2. Has (non functional) introns (Methyl guanosine tri-phosphate is added to 5’end) capping, tailing (Poly A tail at 3’ end added), splicing (introns are removed and exons are joined) Nucleus.
View full question & answer
Question 1153 Marks
Differentiate between a template strand and coding strand of DNA.
Answer
S. No Role/ Strand Template strand Coding strand
(i) Function Codes for the protein molecule. Does not code for anything.
(ii) Polarity 3′ → 5′ 3′ → 5′
View full question & answer
Question 1163 Marks
  1. Identify the polarity of strand in the diagram from a to a', and mention how many more amino acids can be expected to be added to this polypeptide chain.
  2. Mention the DNA sequence coding for serine and the anticodon of tRNA for the same amino acid.
  3. Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide?
Where exactly are they present on mRNA?
Answer
  1. a to d' is 5 → 3'.
No more amino acid will be added.
  1. TCA; anticodon is UCA.
  2. The untranslated regions are required for efficient translation process.
They are present before the initiation codon at the 5' end and after the stop/termination codon, at the 3' end.
View full question & answer