CBSE Class 12 Biology Molecular Basis Of Inheritance Worksheet Set C

Important Questions for NCERT Class 12 Biology Molecular Basis of Inheritance

Question. During replication large amount of energy get exhausted. The source of this energy is
(a) Deoxy ribonucleotide triphosphophate
(b) Deoxyribonucleoside monoplosphate
(c) Deoxyribonucleoside triphosphate
(d) Both 1 and 2

Answer : C

Question. Regarding to direction of DNA replication select out the correct one
(a) 5' → 3' Template - continuous synthesis
(b) 3' → 5' Template - discontinuous synthesis
(c) 3' → 5' Template - continuous synthesis
(d) 5' → 3' Template - synthesis leadingstrand

Answer : B 

Question. In transcription unit promoter and terminator are determined on the basis of
(a) Coding strand
(b) Template strand
(c) Noncoding strand
(d) Antisense strand

Answer : A

Question. In eukaryotes as well as prokaryotes those DNA sequences that appear in mature or processed RNA are known as
(a) Introts
(b) Exons
(c) Recons
(d) Mutons

Answer : B

Question. Regarding to role of RNA in protein synthesis find out the odd one
(a) m-RNA - provides the template
(b) t-RNA - brings aminoacids
(c) r-RNA - read genetic code
(d) sn-RNA splicing

Answer : C

Question. What is the basis of heredity
(a) Variations
(b) Inheritance
(c) Genetics
(d) Recombination

Answer : B

Question. Change in a single base pair of DNA can be termed as
(a) Chromosomal aberrations
(b) Point mutation
(c) Genomatric mutation
(d) Frame shift mutation

Answer : B

Question. Which of the following exclusive property of trancription found in RNA-polymerase
(a) Initiation
(b) Elongation
(c) Termination
(d) Processing

Answer : B

Question. Which of the folloiwng is not a criteria for determination of genetic material
(a) Ability of replication
(b) Chemically and structurally stable
(c) It should be non mutable
(d) Ability to expres itself in from of Mendelian characters

Answer : C

Question. Which of the folloiwng feature of RNA make it labile and easily degradable
(a) Single stranded nature
(b) 2'-OH group on sugar
(c) Phosphodiester bond
(d) Absence of Hydrogen bond

Answer : B

Question. Polysome is formed by
(a) a ribosome with several subunits
(b) ribosomes attached to each other in a linear arrangement
(c) several ribosomes attached to a single mRNA
(d) many ribosomes attached to a strand of endoplasmic reticulum. 

Answer: C

Question. The two sub-units of ribosome remain united at a critical ion level of
(a) magnesium
(b) calcium
(c) copper
(d) manganese. 

Answer: A

Question. Which antibiotic inhibits interaction between tRNA and mRNA during bacterial protein synthesis?
(a) Tetracycline
(b) Erythromycin
(c) Neomycin
(d) Streptomycin 

Answer: C

Question. Amino acid sequence, in protein synthesis is decided by the sequence of
(a) rRNA
(b) tRNA
(c) mRNA
(d) cDNA. 

Answer: C 

Question. Using imprints from a plate with complete medium and carrying bacterial colonies, you can select streptomycin resistant mutants and prove that such mutations do not originate as adaptation. These imprints need to be used
(a) on plates with and without streptomycin
(b) on plates with minimal medium
(c) only on plates with streptomycin
(d) only on plates without streptomycin. 

Answer: C

Question. Protein synthesis in an animal cell occurs
(a) only on the ribosomes present in cytosol
(b) only on ribosome attached to the nuclear envelope and endoplasmic reticulum
(c) on ribosome present in the nucleolus as well as in cytoplasm
(d) on ribosomes present in cytoplasm as well as in mitochondria. 

Answer: D

Question. During translation initiation in prokaryotes, a GTP molecule is needed in
(a) formation of formyl-met-tRNA
(b) binding of 30S subunit of ribosome with mRNA
(c) association of 30S mRNA with formyl-mettRNA
(d) association of 50S subunit of ribosome with initiation complex. 

Answer: C

Question. The RNA that pick up specific amino acid from amino acid pool in the cytoplasm to ribosome during protein synthesis is called
(a) rRNA
(b) RNA
(c) mRNA
(d) tRNA. 

Answer: D

Question. Which of the following step of translation does not consume a high energy phosphate bond?
(a) Peptidyl transferase reaction
(b) Aminoacyl tRNA binding to A-site
(c) Translocation
(d) Amino acid activation 

Answer: A

Question. Protein synthesis in an animal cell, takes place
(a) in the cytoplasm as well as endoplasmic reticulum
(b) only on ribose attached to nucleon
(c) only in the cytoplasm
(d) in the nucleolus as well as in the cytoplasm.

Answer: D

Question. In protein synthesis, the polymerisation of amino acids involves three steps. Which one of the following is not involved in the polymerisation of protein ?
(a) Termination
(b) Initiation
(c) Elongation
(d) Transcription 

Answer: D

Question. Because most of the amino acids are represented by more than one codon, the genetic code is
(a) overlapping
(b) wobbling
(c) degenerate
(d) generate. 

Answer: C

Question. The process of translation is
(a) ribosome synthesis
(b) protein synthesis
(c) DNA synthesis
(d) RNA synthesis. 

Answer: B

Question. Match the following genes of the Lac operon with their respective products.
(A) i gene (i) b-galactosidase
(B) z gene (ii) Permease
(C) a gene (iii) Repressor
(D) y gene (iv) Transacetylase
Select the correct option.
(A) (B) (C) (D)
(a) (iii) (iv) (i) (ii)
(b) (i) (iii) (ii) (iv)
(c) (iii) (i) (ii) (iv)
(d) (iii) (i) (iv) (ii)

Answer: D

Question. Select the correct match.
(a) Alec Jeffreys - Streptococcus pneumoniae
(b) Alfred Hershey and - TMV Martha Chase
(c) Matthew Meselson - Pisum sativum and F. Stahl
(d) Francois Jacob and - Lac operon Jacques Monod 

Answer: D

Question. All of the following are part of an operon except
(a) an operator
(b) structural genes
(c) an enhancer
(d) a promoter.

Answer: C

Question. Which of the following is required as inducer(s) for the expression of Lac operon?
(a) Lactose
(b) Lactose and galactose
(c) Glucose
(d) Galactose

Answer: A

Question. Gene regulation governing lactose operon of E.coli that involves the lac I gene product is
(a) negative and repressible because repressor protein prevents transcription
(b) feedback inhibition because excess of β-galactosidase can switch off transcription
(c) positive and inducible because it can be induced by lactose
(d) negative and inducible because repressor protein prevents transcription.

Answer: D

Question. Which one of the following is wrongly matched?
(a) Transcription - Writing information from DNA to tRNA.
(b) Translation - Using information in mRNA to make protein.
(c) Repressor protein - Binds to operator to stop enzyme synthesis.
(d) Operon - Structural genes, operator and promoter. 

Answer: A

Question. Which enzyme will be produced in a cell if there is a nonsense mutation in the lac Y gene?
(a) Transacetylase
(b) Lactose permease and transacetylase
(c) β-galactosidase
(d) Lactose permease

Answer: C

Question. DNA has genetic properties was revealed for the first time by –
(a) Avery.
(b) Griffith.
(c) Wilkins.
(d) Chargaff.

Answer: A

Question. In bacteria, the formation of peptide bond during translation is affected by –
(a) Lysozyme.
(b) Nucleosome.
(c) Ribozyme.
(d) Microsome.

Answer: C

Question. Choose the particular process used by Matthew Meselson and Franklin Stahl in order to study the semi -conservative replication of DNA.
(a) Centrifugation.
(b) Chromatography.
(c) Buoyant density centrifugation.
(d) Density gradient centrifugation.

Answer: D

Question. Which one is referred to as „soluble RNA?
(a) mRNA.
(b) rRNA.
(c) tRNA.
(d) ssRNA.

Answer: C

Question. Removal of introns and joining of exons in a defined order in a transcription unit is called –
(a) Tailing.
(b) Transformation.
(c) Capping.
(d) Splicing.

Answer: D

Question. Copying genetic information from one strand of DNA into RNA is –
(a) Translation.
(b) Transcription.
(c) Transformation.
(d) Transduction.

Answer: B

Question. Choose the „wrongly‟ matched following pairs of nitrogenous bases in nucleic acids.
(a) Guanine -Adenine -purines
(b) Adenine -Thymine -purines.
(c) Thymine -Uracil -pyrimidines.
(d) Uracil -Cytosine -pyrimidines.

Answer: B

Question. DNA gyrase that participates in the process of DNA replication is a type of –
(a) Reverse transcriptase.
(b) DNA ligase.
(c) DNA topoisomerase.
(d) DNA polymerase.

Answer: C

Question. Which of the following enzymes is used for transcription?
(a) Amino acid synthetase.
(b) DNA polymerase III.
(c) RNA polymerase.
(d) DNA ligase.

Answer: C

Question. Repressor protein is produced by –
(a) Operator gene.
(b) Structural gene.
(c) Regulator gene.
(d) Promotor gene.

Answer: C

Question. Retrovirus has the genetic material –
(a) DNA only.
(b) RNA only.
(c) Both DNA and RNA.
(d) Either DNA or RNA only.

Answer: B

Question. According to Chargaff’s rule, which one of the following is correct?
(a) A+T =G+C
(b) A+C =G+T
(c) A+G =T+C
(d) Both (a) and (c).

Answer: B

Question. Out of 64 codons, only 61 codons code for the twenty different amino acids. This character of the genetic code is called –
(a) Degeneracy.
(b) Non -ambiguous nature.
(c) Redundancy.
(d) Overlapping.

Answer: A

Question. In the DNA molecule –
(a) Proportion of adenine in relation to thymine varies with the organism.
(b) There are two strands which runs antiparallel -one in the 5‟-3‟ direction and the other in 3‟-5‟ direction.
(c) The total amount of purine nucleotides is not always equal.
(d) There are two strands which runs parallel -both in the 5‟-3‟ direction.

Answer: B

Question. The portion of the DNA which contains the information for an entire polypeptide is called –
(a) Cistron.
(b) Muton.
(c) Recon.
(d) Operon.

Answer: A

Question. What will be the correct gene expression pathway?
(a) Gene -mRNA -Transcription -Translation -Protein.
(b) Transcription -Gene -Translation -mRNA -Protein.
(c) Gene -Transcription -mRNA - Translation -Protein.
(d) Gene -Translation -mRNA -Transcription - Protein.

Answer: C

Question. The main aim of the Human Genome Project is –
(a) To introduce new genes into Humans.
(b) To identify and sequence all the genes present in Human DNA.
(c) To develop better techniques for comparing two different human DNA samples.
(d) To remove disease causing genes from Human DNA.

Answer: B

Question. Which one of the following pairs of codons is correctly matched with its function or a signal for a particular amino acid?
(a) AUG; ACG -Start/ Methionine.
(b) UUA; UCA -Leucine.
(c) GUU; GCU -Alanine
(d) UAG; UGA -Stop.

Answer: D

Question. In genetic fingerprinting, the „probe‟ refers to ….
(a) A radioactively labelled single stranded DNA molecule.
(b) A radioactively labelled single stranded RNA molecule.
(c) A radioactively labelled double stranded RNA molecule.
(d) A radioactively labelled double stranded DNA molecule.

Answer: A

Question. If the percentage of cytosine is 18%, then the percentage of adenine will be –
(a) 64%
(b) 32%
(c) 36%
(d) 23%

Answer: B

Long Answer Type Questions:

Question. (a) Draw a neat and labelled diagram of a „Replicating fork‟ showing „polarity‟. Why does DNA replication occur within such replication fork?
(b) Name the enzyme involved in the process of DNA replication along with their properties.
Answer: (a)

For long DNA molecules, since the two strands of DNA cannot be separated in its entire length (due to very high energy requirement), the replication occur within a small opening of the DNA helix, referred to as replication fork. 

b) a) Transforming Principle
♦ Frederick Griffith (1928), conducted a series of experiments with Streptococcus pneumoniae (bacterium responsible for pneumonia).
♦ When Streptococcus pneumoniae (pneumococcus) bacteria are grown on a culture plate, some produce smooth shiny colonies (S) while others produce rough colonies (R).
♦ This is because the S strain bacteria have a mucous (polysaccharide) coat, while R strain does not.
♦ Mice infected with the S strain (virulent) die from pneumonia infection but mice infected with the R strain do not develop pneumonia. 

Griffith was able to kill bacteria by heating them. He observed that heat-killed S strain bacteria injected into mice did not kill the mice.
♦ When he injected a mixture of „heat-killed S‟ and „live‟ R bacteria, the mice died. Moreover, he recovered living S bacteria from the dead mice. 

He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria. Some „transforming principle‟, transferred from the heat-killed S strain, had
enabled the R strain to synthesise a smooth polysaccharide coat and become virulent.
♦ This must be due to the transfer of the genetic material.
b) Biochemical Characterisation of Transforming Principle:
♦ Earlier (before 1933), the genetic material was thought to be a protein.
♦ Oswald Avery, Colin McLeod and Maclyn McCarthy, worked to determine the biochemical nature of „transforming principle‟ in Griffith's experiment.
♦ They purified biochemicals (proteins, DNA, RNA, etc.) from the heat-killed S cells to see which ones could transform live R cells into S cells 

They discovered that DNA alone from heat killed S strain bacteria caused R bacteria to become transformed
♦ They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation.
♦ Therefore, the transforming substance was not a protein or RNA.
♦ Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation.
♦ They concluded that DNA is the hereditary material, but not all biologists were convinced

Question. One chromosome contains one molecule of DNA. In Eukaryotes, the length of the DNA molecule is enormously large. Explain how such a long molecule fits into the tiny chromosomes seen during metaphase.
Answer:
♦ One chromosome contains one molecule of DNA. In Eukaryotes, the length of the DNA molecule is enormously large. Explain how such a long molecule fits into the tiny
chromosomes seen during metaphase.
♦ The distance between two consecutive base pairs = 0.34 nm i.e. (0.34×10^-9m).
♦ If the length of DNA double helix in a typical mammalian cell = 6.6 × 109 bp.
♦ The length of the DNA molecule = 6.6 × 10⁹ bp × 0.34 × 10^-9m/bp =2.2 meters -a length that is far greater than the dimension of a typical nucleus (approximately 10^–6 m).
♦ In eukaryotes, there is a set of positively charged, basic proteins called histones.
♦ A protein acquires charge depending upon the abundance of amino acids residues with charged side chains.
♦ Histones are rich in the basic amino acid residues lysine and arginine -both of carry positive charges in their side chains. which
♦ Histones are organised to form a unit of eight molecules called histone octamer (H₂A, H₂B, H₃ and H₄). H1 is found associated with „linker DNA’ -between the nucleosomes.
♦ The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.
♦ A typical nucleosome contains 200 bp of DNA helix.
♦ Nucleosomes constitute the repeating unit of chromatin, threadlike stained (coloured) bodies seen in nucleus. The nucleosomes in chromatin are seen as „beads-on-string‟ structure when viewed under electron microscope.
♦ The packaging of chromatin at h
♦ igher level requires additional set of proteins that collectively are referred to as Non-histone Chromosomal (NHC) proteins.

Question. (a) Describe the series of experiments conducted by Frederick Griffith. Comment on the significance of the results obtained.
(b) State the contributions of Oswald Avery, Colin McLeod and Maclyn McCarthy.
Answer: a)
♦ tRNA has an anticodon loop that has bases complementary to the triplet codon.
♦ It also has an amino acid acceptor end to which it binds to amino acids.
♦ tRNAs are specific for each amino acid.
Clover leaf model of tRNA 

Inverted „L‟ shaped model of tRNA

For initiation, there is another specific tRNA that is referred to as initiator tRNA.
♦ There are no tRNAs for stop codons.
♦ The secondary structure of tRNA has been depicted that looks like a clover-leaf.
♦ In actual structure, the tRNA is a compact molecule which looks like inverted „L’.
b) Francis Crick postulated the presence of an adapter molecule that would on one hand read the code and on other hand would bind to specific amino acids.
♦ tRNA was the ideal molecule that fits the role of „this‟ adaptor molecule
c) ♦ The primary transcripts contain both the exons and the introns and are non-functional.
♦ Hence, it is subjected to a process called splicing -where the introns are removed and exons are joined in a defined order.
♦ hnRNA undergoes additional processing called as capping and tailing.
♦ In capping an unusual nucleotide (7-methyl guanosine triphosphate) is added to the 5'-end of hnRNA. In tailing, adenylate residues (200-300) are added at 3'-end in a template independent manner.
♦ It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation. 

Question. One chromosome contains one molecule of DNA. In Eukaryotes, the length of the DNA molecule is enormously large. Explain how such a long molecule fits into the tiny chromosomes seen during metaphase.
Answer: The distance between two consecutive base pairs = 0.34 nm i.e. (0.34×10^-9m).
♦ If the length of DNA double helix in a typical mammalian cell = 6.6 × 109 bp.
♦ The length of the DNA molecule = 6.6 × 109 bp × 0.34 × 10-9m/bp =2.2 meters -a length that is far greater than the dimension of a typical nucleus (approximately 10^-6 m).
♦ In eukaryotes, there is a set of positively charged, basic proteins called histones.
♦ A protein acquires charge depending upon the abundance of amino acids residues with charged side chains.
♦ Histones are rich in the basic amino acid residues lysine and arginine -both of carry positive charges in their side chains. which
♦ Histones are organised to form a unit of eight molecules called histone octamer (H2A, H2B, H3 and H4). H1 is found associated with „linker DNA‟ -between the nucleosomes.
♦ The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome.
♦ A typical nucleosome contains 200 bp of DNA helix.
♦ Nucleosomes constitute the repeating unit of chromatin, threadlike stained (coloured) bodies seen in nucleus. The nucleosomes in chromatin are seen as „beads-on-string‟ structure when viewed under electron microscope.
♦ The packaging of chromatin at higher level requires additional set of proteins that collectively are referred to as Non-histone Chromosomal (NHC) proteins

Question. (a) Describe the structure and function of the tRNA molecule.
(b) Why is the tRNA referred to as an „adaptor‟ molecule?
(c) Explain the process of „splicing‟ of hnRNA in a Eukaryotic cell.
Answer:
a) (i) Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
(ii) Similarly, viruses grown on radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur. 
♦ Hershey and Chase worked to discover whether it was protein or DNA from the viruses that entered the bacteria.
♦ They grew some viruses on a medium that contained radioactive phosphorus and some others on medium that contained radioactive 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.
b) (i) 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.
(ii) 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. 

Diagram Based Questions:

1. Study the schematic representation of the genes involved in lac operon given below and answer the questions that follow: 

Question. Name the inducer molecule and the products of the genes‟ „z‟ and „y‟ of the operon. State the functions of these gene products. 
Answer: Lactose is the substrate for the enzyme beta-galactosidase and it regulates switching on and off of the operon. Hence, it is termed as inducer.
♦ The „z‟ gene codes for beta-galactosidase (b-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose.
♦ The „y‟ gene codes for permease, which increases permeability of the cell to b-galactosides.
The „a‟ gene encodes a transacetylase an enzyme that catalyzes the transfer of an acetyl group from Acetyl CoA to another molecule (galactosides, lactosides and glucosides) during
metabolism of lactose.
♦ Hence, all the three gene products in lac operon are required for metabolism of lactose.

Question. Identify and name the regulatory gene in this operon. Explain its role in „switching off‟ the operon. 
Answer: i. The lac operon consists of one regulatory gene (the i gene – here the term i does not refer to inducer, rather it is derived from the word inhibitor).
ii. The repressor of the operon is synthesised (all-the-time – constitutively) from the i gene.
The repressor protein binds to the operator region of the operon and prevents RNA polymerase from transcribing the operon.

Question. Why is the lac operon’s regulation referred to as „negative regulation‟? 
Answer: Repressor proteins have an inhibiting property whose „absence‟ or „presence‟ controls the switching „on‟ or „off‟ of the operon respectively. Thus, regulation of lac operon is referred to as „negative regulation‟.

2. Study the following illustration and answer the questions below:

Question. Explain the function of each of the labelled part.
Answer: All the reference point while defining a transcription unit is made with coding strand.
(i) ‘a’ = Terminator –The terminator is located towards 3'-end (downstream) of the coding strand and it usually defines the end of the process of transcription.
(ii) ‘b’ = Coding strand –the strand which has the polarity (5'®3') and the sequence same as RNA (except thymine at the place of uracil), is displaced during transcription. Strangely,
this strand (which does not code for anything), is referred to as coding strand.
(iii) ‘c’ = Template strand –the strand that has the polarity 3'®5' acts as a template, and is also referred to as template strand.
(iv) ‘d’ = Promotor -The promoter is said to be located towards 5'-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand). It is a
DNA sequence that provides binding site for RNA polymerase, and it is the presence of a promoter in a transcription unit that also defines the template and coding strands.

Question. Name the enzyme involved in the process.
Answer: DNA-dependant RNA polymerase -catalyse the polymerisation in only one direction, that is, 5' to 3'.

Question. Considering that information from strand ‘X’ is to be transcribed, identify and name the parts labelled as „a‟, „b‟, „c‟ and „d‟. 
Answer: „a‟ =Terminator (ii) „b‟ =Coding strand (iii) „c‟ =Template strand (iv) „d‟ =Promotor

Case Based Questions:
 

1. A Representative Diagram of the Human Genome Project: 

Question. Name a free living, non-pathogenic nematode, the DNA of which has been completely sequenced. 
Answer: Caenorhabditis elegans

Question. Summarize the methodology adopted in the Human Genome Project. 
Answer: Expressed Sequence Tags (ESTs): The approach focused on identifying all the genes that are expressed as RNA.
Sequence Annotation: The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning
different regions in the sequence with functions.

Question. In the history of Biology, the Human Genome Project led to the development of:
(a) Biotechnology
(b) Biomonitoring
(c) Bioinformatics
(d) Biosystematics. 
Answer: Biotechnology

Question. Mention at least four salient features of the Human Genome Project.
Answer: (i) The human genome contains 3164.7 million bp.
(ii) The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being dystrophin at 2.4 million bases.
(iii) The total number of genes is estimated at 30,000–much lower than previous estimates of 80,000 to 1,40,000 genes. Almost all (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. Less than 2 per cent of the genome codes for proteins.
(v) Repeated sequences make up very large portion of the human genome.
(vi) Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions,
but they shed light on chromosome structure, dynamics and evolution.
(vii) Chromosome 1 has most genes (2968), and the Y has the fewest (231).
(viii) Scientists have identified about 1.4 million locations where single base DNA differences (SNPs – 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. What are SNPs‟? How are they useful in human genomics?
Answer: Scientists have identified about 1.4 million locations where single base DNA differences (SNPs – 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.
♦ An example of an SNP is the substitution of a C for a G in the nucleotide sequence AACGAT, thereby producing the sequence AACCAT. The DNA of humans may contain many SNPs, since these variations occur at a rate of one in every 100–300 nucleotides in the human genome.
♦ Single nucleotide polymorphism (SNP) technologies can be used to identify disease-causing genes in humans and to understand the inter-individual variation in drug response.
♦ By establishing an association between the genetic make-up of an individual and drug response it may be possible to develop a genome-based diet and medicines that are more
effective and safer for each individual.
♦ SNPs can be used to understand the molecular mechanisms of sequence evolution. It is possible that disease-associated SNPs (or pathology) and evolution can be related to one
another.

2. Two blood samples of suspects „A‟ and „B‟ were sent to the Forensic Department along with the sample „C‟ from the crime scene. The Forensic Department assigned the responsibility of running the samples and matching the samples of the suspects with that of the sample from the scene of the crime and thereby identify the culprit. 

Question. How does polymorphism arise in a population? 
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.
♦ That is, if an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism

Question. State the steps involved in DNA Fingerprinting in a sequential manner.
Answer: Steps involved in DNA Fingerprinting:
(i) Isolation of DNA.
(ii) Digestion of DNA by restriction endonucleases.
(iii) Separation of DNA fragments by electrophoresis.
(iv) Transferring (blotting) of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon.
(v) Hybridisation using labelled VNTR probe, and
(vi) Detection of hybridised DNA fragments by autoradiography.

Question. In genetic fingerprinting the „probe‟ refers to –
(a) A radioactively labelled double stranded RNA molecule.
(b) A radioactively labelled double stranded DNA molecule.
(c) A radioactively labelled single stranded DNA molecule.
(d) A radioactively labelled single stranded RNA molecule.
Answer: A radioactively labelled single stranded DNA molecule.

Question. What does „minisatellite‟ and „microsatellite‟ mean in relation to DNA Fingerprinting. 
Answer: The main difference between microsatellite and minisatellite:
Microsatellite - the repeating unit consists of 2-6 base pairs. Microsatellite array contains 5-200 repeats.
(b) Minisatellite - the repeating unit consists of 10-100 base pairs. Minisatellite array contains 10-1,500 repeats.