Please refer to Molecular Basis of Inheritance HOTs Class 12 Biology provided below with Molecular Basis of Inheritance. All HOTs for Class 12 Biology with answers provided below have been designed as per the latest syllabus and examination petter issued by CBSE, NCERT, KVS. Students of Standard 12 Biology should learn the solved HOTS for Class 12 Biology provided below to gain better marks in examinations.
Molecular Basis of Inheritance Class 12 Biology HOTs
Translation
Very Short Answer Type Questions
Question. Which one of the two sub-units of ribosome encounters an mRNA?
Answer. Smaller sub-unit of ribosome encounters mRNA during initiation of protein synthesis.
Question. Write the two specific codons that a translational unit of mRNA is flanked by one on either sides.
Answer.The two specific codons are initiation codon on one side (AUG or GUG) and termination codon (UAA, UAG or UGA) on the other side of mRNA.
Short Answer Type Questions
Question. State the functions of ribozyme and release factor in protein synthesis respectively.
Answer. Ribozyme (catalytic RNA) is present in ribosome and joins the amino acids together by peptide bond formation to form protein chains. Release factor (RF) is GTP dependent. It binds to the stop codon, terminates translation and release the complete polypeptide from the ribosome.
Question. How do the initiation and termination of translation process occur in bacteria? Where are untranslated regions located in an mRNA? Mention their role.
Answer.
Initiation : When the small subunit of ribosome encounters the mRNA, translation begins. The mRNA binds to the small subunit of ribosome catalysed by initiation factor. There are two sites on the larger subunit of ribosome, the P-site and the A-site. The small subunit attaches to the large subunit in such a way that the initiation codon (AUG) comes on the P-site. The initiator tRNA (methionine-tRNA) binds to the P-site.
Termination : The synthesis of polypeptide chain is terminated when stop/terminator codons UAA, UAG or UGA comes at the A-site of ribosome. These codons do not code for any amino acid.
Guanosine triphosphate (GTP) dependent release factors cleaves the polypeptide chain from the terminal tRNA, releasing the polypeptide from the ribosome.
An mRNA has some additional sequences that are not translated and are referred as untranslated regions (UTR). The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon). They are required for eficient translation process.
Question. Where does peptide bond formation occur in a bacterial ribosome and how?
Answer. Peptide bond formation occurs in the peptidyl transferase centre present in the larger sub-unit of ribosome. Peptide bond formation occurs between the nascent polypeptide chain and the new amino acid resulting in the elongation of polypeptide chain. An aminoacyl tRNA complex reaches the A-site and attaches to mRNA codon next to initiation codon with the help of its anticodon. The step requires GTP and an elongation factor. A peptide bond (—CO —NH—) is established between the carboxyl group (—COOH) of amino acid attached to tRNA at P-site and amino group (—NH 2 ) of amino acid attached to tRNA at A-site. The reaction is catalysed by enzyme peptidyl transferase which is an RNAenzyme. A lot of energy is consumed in protein synthesis. For every single amino acid incorporated the peptide chain one ATP and two GTP molecules are used.
Question. What is aminoacylation? State its significance.
Answer. Aminoacylation or charging of the tRNA is the process during which the amino-acyl-adenylateenzyme complex reacts with tRNA specific for the amino acid to form aminoacyl-tRNA complex. Enzyme and AMP are released. tRNA complexed with amino acid is sometimes called charged tRNA. The amino acid is linked to 3-OH-end of tRNA through its -COOH group,
AA ∼ AMP — E + tRNA → AA -tRNA + AMP + E
Aminoacyl adenylate enzyme
The aminoacyl-tRNA complex specific for the initiation codon reaches the P-site to initiate the process of protein synthesis.
Question.
Study the mRNA segment given above which is complete to be translated into a polypeptide chain.
(a) Write the codons ‘a’ and ‘b’.
(b) What do they code for?
(c) How is peptide bond formed between two amino acids in the ribosome?
Answer. (a) ‘a’ – AUG
‘b’ – UAA/UAG/UGA.
(b) AUG codes for methionine. UAA/UAG/UGA does not code for any amino acid, but brings about termination of polypeptide synthesis.
(c) Peptide bond is formed between carboxyl group of amino acid attached to tRNA at P site and amino group of amino acid attached to tRNA at A site catalysed by the enzyme peptidyl transferase.
Question.
(a) 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.
(b) Mention the DNA sequence coding for serine and the anticodon of tRNA for the same
amino acid.
(c) Why are some untranslated sequence of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA?
Answer.
(a) a = 5′
a′ = 3′
No more amino acids will be added in the given polypeptide chain as a stop codon (UAA) has been encountered.
(b) The DNA sequence coding for serine is TCA and the anticodon of tRNA for serine is UCA.
(c) Untranslated regions (UTRs) are present in the leading region (before the initiation codon, 5′ UTR) and in the trailer region (after the termination codon, 3′ UTR) of the mRNA. 5′ UTR contains a sequence to which ribosome binds and initiates translation. 3′ UTR helps in termination of translation process.
Question. How is the translation of mRNA terminated? Explain.
Answer. Polypeptide chain synthesis is terminated when a nonsense codon of mRNA reaches the A-site of ribosome during protein synthesis. There are three nonsense codons-UAA, UAG and UGA. These codons do not specify any amino acid. So, translation is stopped. GTP dependent release factor, cleaves the polypeptide from the terminal tRNA releasing the product from the ribosome.
Question. Mention the role of ribosomes in peptide-bond formation. How does ATP facilitate it?
Answer.
When a small sub-unit of ribosome encounters an mRNA, the process of translation of mRNA to protein begins. There are two sites in the large subunit for the subsequent amino acids to bind to and thus, be close enough to each other for the formation of peptide bond. The ribosome also acts as a catalyst (23 S rRNA in bacteria is ribozyme) for the formation of peptide bond.Formation of a peptide bond requires energy.
Therefore, in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA – a process called charging of tRNA.
Long Answer Type Questions
Question. Explain the process of translation.
Answer. The process of decoding of the message from mRNA to protein with the help of tRNA, ribosome and enzyme is called translation (protein synthesis).Protein synthesis occurs over ribosomes.
The 4 main steps in protein synthesis (translation) are : activation, initiation, elongation and termination of polypeptide chain. The newly synthesised mRNA joins the smaller subunit of ribosome at 5′ end. mRNAs carry the codon and tRNAs carry the anticodon for the same codon. Activation of amino acid is catalysed by the enzyme aminoacyl tRNA synthetase in the presence of ATP. In presence of ATP an amino acid combines with its specific amino acyl-tRNA synthetase to produce aminoacyl adenylate enzyme complex. This reacts with tRNA to form aminoacyl-tRNA complex. Activated tRNA is taken to ribosome mRNA complex for initiation of protein synthesis. Initiation of protein synthesis is accomplished with the help of initiation factor which are 3 (IF3, IF2, IF1) in prokaryotes and 9 in eukaryotes (eIF2, eIF3, eIF1, eIF4A, eIF4B, eIF4C, eIF4D, eIF5, eIF6). The ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA. A polypeptide chain forms as tRNAs deliver amino acids to the ribosome. Large ribosomal subunit binds the initiation complex forming two (A and P) binding site for tRNA molecules. The first site is P site or peptidyl site which is occupied by tRNAmet. The second site is A or amino acyl site and is positioned over the second codon. The ribosome proceeds to the elongation phase of protein synthesis. During this stage, complexes composed of an amino acid linked to tRNA, sequentially bind to the appropriate codon in mRNA by forming complementary base pairs with the tRNA anticodon. The ribosome moves from codon to codon along the mRNA. Amino acids are added one by one, translated into polypeptide sequences dictated by DNA and represented by mRNA. The enzyme peptidyl synthetase catalyses the formation of peptide bond between the carboxylic group of amino acid at P site and amino group of amino acid at A site. Enzyme translocase brings about the movement of mRNA by one codon.
The termination of protein synthesis occur when a non-sense codon reaches at A site of ribosome. The chain detaches from the ribosome. A release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome. Two subunits of ribosomes dissociate with the help of dissociation factor.
Question. How do m-RNA, t-RNA and ribosomes help in the process of translation?
Answer.
(a) mRNA – Messenger RNA bring coded information from DNA and takes part in its translation by bringing amino acids in a particular sequence during the synthesis of polypeptide. However, the codons of mRNA are not recognised by amino acids but by anticodons of their adapter molecules (tRNAs → aa-tRNAs). Translation occurs over the ribosomes. The same mRNA may be reused time and again. In the form of polysome, it can help synthesise a number of copies simultaneously.
(b) tRNAs – They are transfer or soluble RNAs which pick up particular amino acids (at CCA or 3′ end) in the process called charging. The charged tRNAs take the same to mRNA over particular codons corresponding to their anticodons. A tRNA can pickup only a specific amino acid though an amino acid can be specified by 2-6 tRNAs. Each tRNA has an area for coming in contact with ribosome (T ψ C) and the enzyme amino acyl tRNA synthetase (DHU).
(c) Ribosomes – Protein synthesis occurs over the ribosomes, Ribosomes are, therefore, also called protein factories. Each ribosome has two unequal parts, small and large. The larger subunit of ribosome has a groove for pushing out newly formed polypeptide and protecting the same from cellular enzymes. The smaller subunit its over the larger one like a cap but leaves a tunnel for mRNA. The two subunits come together only at the time of protein formation. Mg2+ is essential for it. Soon after the completion of protein synthesis, the subunits separate.
Regulation of Gene Expression
Very Short Answer Type Questions
Question. Differentiate between the following: Inducer and repressor in lac operon.
Answer. Inducer is a chemical (substrate, hormone or some other metabolite) which after coming in contact with the repressor, changes the latter into non-DNA binding state so as to free the operator gene. The inducer for lac-operon of Escherichia coli is lactose (actually allolactose, or metabolite of lactose.)
Repressor is a regulator protein meant for blocking the operator gene so that the structural genes are unable to form mRNAs.
Short Answer Type Questions
Question. 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.Inducer is a chemical (substrate, hormone or some other metabolite) which after coming in contact with the repressor, changes the latter into non-DNA binding state so as to free the operator gene.
The inducer for lac-operon of Escherichia coli is lactose (actually allolactose, or metabolite of lactose). In the presence of inducer (lactose), the repressor gets inactivated due to its interaction with it. This allows RNA polymerase to access the promoter and tranciption proceeds. Hence, the lac operon is switched on.
When lactose is digested, glucose and galactose are formed. Then, the lac operon will stop due to the accumulation of glucose and galactose in the cell as they cannot be used as an inducer for lac operon.
Question. Study the figure given below and answer the question:
(a) How does the repressor molecule get inactivated?
(b) When does the transcription of lac mRNA stop?
(c) Name the enzyme transcribed by the gene ‘Z’?
Answer.(a) When the inducer (e.g., lactose) comes in contact with repressor, repressor gets inactivated.
(b) When lactose is absent or lacking, the transcription of lac mRNA stops.
(c) b-galactosidase.
Question.
(a) Name the molecule ‘M’ that binds with the operator.
(b) Mention the consequences of such binding.
(c) What will prevent the binding of the molecule ‘M’ with the operator gene? Mention the event that follows.
Answer.(a) M is repressor protein.
(b) Binding of repressor (M) with operator (O) switches off the lac operon.
(c) Presence of inducer i.e., lactose will prevent the binding of the molecule M with the operator gene. Inducer will bind to the repressor, change the latter into non-DNA binding state so as to free the operator gene and switch on the lac operon.
Question.
(a) Name the molecule ‘X’ synthesised by ‘i’ gene. How does this molecule get inactivated?
(b) Which one of the structural genes codes for b-galactosidase?
(c) When will the transcription of this gene stop?
Answer.(a) The molecule ‘X’ is repressor. It gets inactivated when lactose (inducer) binds with it.
(b) Z-gene codes for b-galactosidase.
(c) Transcription of the gene stops when lactose is absent and thus repressor is free to bind with the operator.
Question. How are the structural genes inactivated in lac operon in E.coli ? Explain.
Answer. Lac operon in E.coli consists of structural genes, an operator gene, a promoter gene, a regulator gene, a repressor and an inducer. The structural genes are inactivated in the absence of an inducer (i.e. lactose). It is because in the absence of an inducer, the repressor binds to the operator gene making it nonfunctional. RNA polymerase enzyme cannot move over it to reach the structural genes. Thus, structural genes are inactivated and transcription cannot take place.
Question. How would lac operon operate in E. coli growing in a culture medium where lactose is present as source of sugar?
Answer. When lactose is present in the culture medium, then the lac operon in E. coli is switched on. It is because the inducer (lactose) binds to the repressor protein thereby inactivating it. It prevents binding of repressor to the operator. Consequently, RNA polymerase gets access to the promoter and transcription of structural genes proceeds.
Question. Given below is a schematic representation of lac operon:
(a) Identify i and p.
(b) Name the ‘inducer’ for this operon and explain its role.
Answer.(a) i = Regulator gene
p = Promoter gene
(b) ‘Inducer’ for the given operon is ‘lactose’. Its role is to bind with repressor, change the latter into non- DNA binding state so as to free the operator gene and switch on the lac operon.
Question. Explain the role of regulatory gene in a lac operon. Why is regulation of lac operon called negative regulation?
Answer.
Regulatory gene (i gene) produces a repressor. In the absence of an inducer (i.e. lactose), the repressor binds to the operator gene making it nonfunctional. RNA polymerase enzyme cannot move over it to reach the structural genes. Thus, structural genes are inactivated and transcription cannot take place.
As regulatory gene exerts a negative control over the working of structural genes, therefore regulation of lac operon is called negative regulation
Question.
Given above is a schematic representation of the lac operon in E.coli. What is the significant role of ‘i’ gene in switching on or off the operon?
Answer.‘i’ gene is regulator gene. It produces a repressor,which binds to the operator gene and stops its working. This ‘i’ gene exerts a negative control over the working of structural genes.
Long Answer Type Questions
Question. Sketch a schematic diagram of lac operon in switched on position. How is the operon switched off? Explain.
Answer. Schematic diagram of lac operon in ‘switched on’ position is as follows:
The operon gets switched ‘off’ in the absence of lactose (inducer). The repressor molecule binds with the operator region of the operon and prevents RNA polymerase from transcribing the operon.
Question. Explain the role of lactose as an inducer in a lac operon.
Answer. An operon is a part of genetic material (or DNA) which acts as a single regulated unit having one or more structural genes, an operator gene, a promoter gene, a regulator gene, a repressor and an inducer or corepressor (from outside).
Lactose acts as an inducer in lac operon. The repressor molecule coded by i gene is inactivated by interaction with the lactose. This allows RNA polymerase to access the promoter and transcription proceeds. The operon gets switched ‘off’ in the absence of lactose as the repressor molecule binds with the operator region of the operon and prevents RNA polymerase from transcribing the operon.
Diagrammatic representation of lac operon is as follows:
Question. (a) Who proposed the concept of lac operon?
(b) Draw a labelled schematic representation of a lac operon.
(c) Explain how does this operon get switched ‘on’ and ‘off ’.
Answer.(a) Jacob and Monod (1961) proposed the concept of lac operon.
(b)
(c) Lac operon consists of structural genes, an operator gene, a promoter gene, a regulator gene, a repressor and an inducer. Lac operon gets switched ‘on’ in the presence of lactose. The repressor molecule coded by i gene is inactivated by its interaction with the inducer (lactose). This allows RNA polymerase access to the promoter and transcription proceeds. The operon gets switched ‘off’ in the absence of lactose as the repressor molecule binds with the operator region of the operon and prevents RNA polymerase from transcribing the operon.
Human Genome Project
Very Short Answer Type Questions
Question. Mention the contribution of genetic maps in human genome project.
Answer. Genetic maps have helped in gene sequencing,DNA fingerprinting, tracing human history, etc.
Question. Write the scientific importance of single nucleotide polymorphism identified in human genome.
Answer. Single nucleotide polymorphism (SNPs or snips) help in finding chromosomal locations for disease associated sequences and tracing human history.
Question. Mention any two ways in which Single Nucleotide Polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical sciences.
Answer.
(a) In finding chromosomal locations for disease associated sequences
(b) In tracing human history
Short Answer Type Questions
Question. Expand ‘BAC’ and ‘YAC’. Explain how they are
used in sequencing human genome.
Answer.BAC = Bacterial Artificial Chromosome.
YAC = Yeast Artificial Chromosome.
BAC and YAC are the vectors into which DNA fragments are inserted to form rDNA (recombinant DNA) using recombinant DNA technology and are then multiplied in suitable host.
Question. Which human chromosome has (a) maximum number of genes, and which one has (b) fewest genes?
Answer.(a) Chromosome 1
(b) Chromosome Y
Question. In human genome which one of the chromosomes has the most genes and which has the fewest?
Answer.Chromosome 1 has 2968 genes while Y-chromosome has 231 genes. They are maximum and minimum genes for human chromosomes respectively.
Question. (a) What do ‘Y’ and ‘B’ stand for in ‘YAC’ and ‘BAC’ used in Human Genome Project (HGP).
Mention their role in the project.
(b) Write the percentage of the total human genome that codes for proteins and the percentage of discovered genes whose functions are known as observed during HGP.
(c) Expand ‘SNPs’ identified by scientists in HGP.
Answer.
(a) In human genome project, ‘Y’ stands for yeast in YAC (yeast artificial chromosomes) and ‘B’ stands for bacterial in BAC (bacterial artificial chromosomes). These are specialised vectors used during sequencing in human genome project.
(b) In human genome, less than 2 percent of the genome codes for proteins and functions of only 50% of discovered genes are known.
(c) Human genome has SNPs at 1.4 million locations. Expanded form of SNPs is Single Nucleotide Polymorphism.
DNA Fingerprinting
Very Short Answer Type Questions
Question. How does DNA polymorphism arise in a population?
Answer.DNA polymorphism in a population arise due to mutations.
Question. A burglar in a huff forgot to wipe off his bloodstains 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 is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. DNA fingerprints can be prepared from extremely minute amounts of DNA sample from blood, semen, hair bulb or any other cells of the body.
Question. How is repetitive/satellite DNA separated from bulk genomic DNA for various genetic experiments?
Answer.Repetitive/satellite DNA can be separated from bulk genomic DNA by using density gradient centrifugation.
Short Answer Type Questions
Question. Following the collision of two trains a large number of passengers are killed. A majority of them are beyond recognition. Authorities want to hand over the dead to their relatives. Name a modern scientific method and write the procedure that would help in the identification of kinship.
Answer. By using DNA fingerprinting technique, the kinships can be identified.
DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. It identifies a person on the basis of his/her DNA specidicity.
The major steps in DNA fingerprinting are :
(i) DNA is extracted from the cells.
(ii) DNA is amplified by making many copies of it using polymerase chain reaction.
(iii) Digestion of DNA by restriction endonucleases.
(iv) Separation of DNA fragments by electrophoresis.
(v) Transferring of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon.
(vi) Hybridisation using VNTR probe.
(vii) Detection of hybridised DNA fragments by autoradiography.
Question. (a) Explain DNA polymorphism on the basis of genetic mapping of human genome.
(b) State the role of VNTR in DNA fingerprinting.
Answer.
(a) Polymorphism is variation at genetic level which arises due to mutations. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA finger printing. If an inheritable mutation is observed in a population at high frequency, it is referred to as DNA polymorphism.
(b) Short nucleotide repeats in the DNA are very specific in each individual and vary in number from person to person but are inherited. These are the ‘Variable Number Tandem Repeats’ (VNTRs). These are also called “minisatellites”. Each individual inherits these repeats from his/her parents which are used as genetic markers in a personal identity test. For example, a child might inherit a chromosome with six tandem repeats from the mother and the same tandem repeated four time in the homologous chromosome inherited from the father. One half of VNTR alleles of the child resemble that of the mother and other half with that of the father.
Question. “A very small sample of tissue or even a drop of blood can help determine paternity”. Provide a scientific explanation to substantiate the statement.
Answer.
DNA fingerprinting helps in determining the paternity from a small sample of tissue or a drop of blood. DNA fingerprinting is a technique for identifying individuals, generally using repeated sequences in the human genome that produce a pattern of bands which is unique for every individual. Important for DNA fingerprinting are short nucleotide repeats that vary in number from person to person, but are inherited. These are the Variable Number of Tandem Repeats or VNTRs. The VNTRs of two persons may be of the same length and sequence at certain sites, but vary at others. DNA fingerprints can be prepared from extremely minute amounts of blood, semen, hair bulb or any other cells of the body.
By using DNA fingerprinting technique, the kinships can be identified.
DNA fingerprinting is a technique of determining nucleotide sequences of certain areas of DNA which are unique to each individual. It identifies a person on the basis of his/her DNA specidicity.
The major steps in DNA fingerprinting are :
(i) DNA is extracted from the cells.
(ii) DNA is amplified by making many copies of it using polymerase chain reaction.
(iii) Digestion of DNA by restriction endonucleases.
(iv) Separation of DNA fragments by electrophoresis.
(v) Transferring of separated DNA fragments to synthetic membranes, such as nitrocellulose or nylon.
(vi) Hybridisation using VNTR probe.
(vii) Detection of hybridised DNA fragments by autoradiography.
Question. Write the full form of VNTR. How is VNTR different from ‘Probe’?
Answer.VNTR stands for Variable Number of Tandem Repeats.
VNTRs are short nucleotide repeats in DNA that are specific to each individual and vary in number from person to person. DNA probes, are radioactive, have repeated base sequence complementary to VNTRs .
Question. Explain the significance of satellite DNA in DNA fingerprinting technique.
Answer. Satellite DNA are very specific in each individual, vary in number from person to person and are inherited. These sequences show high degree of polymorphism. Each individual inherits the satellite DNA from, his/her parents which are used as genetic markers in DNA finger printing.
Question. What are satellite DNA in a genome? Explain their role in DNA fingerprinting.
Answer. A small stretch of DNA sequences repeats many a time, shows a high degree of polymorphism are called as satellite DNA. Short nucleotide repeats in DNA are very specific in each individual and vary in number from person to person but are inherited. Each individual inherits these repeats from his/her parents which are used as genetic markers in DNA fingerprinting.
Question. The following is the flow chart highlighting the steps in DNA fingerprinting technique. Identify a, b, c, d, e and f.
Isolation of DNA from blood cells
↓
Cutting of DNA by ‘a’
↓
Separation of DNA fragments by electrophoresis using ‘b’
↓
Transfer (blotting) of fragments to ‘c’ gel
↓
DNA splits into single strand
↓
Introduction of labelled ‘d’ probe
↓
‘e’ of single strands with ‘d’
↓
Detection of banding pattern by ‘f ’
Answer.a → Restriction endonuclease
b → Agarose gel
c → Nitro cellulose membrane
d → VNTR
e → Hybridisation
f → Autoradiography
Long Answer Type Questions
Question. (a) Differentiate between repetitive and satellite DNA.
(b) How can satellite DNA be isolated? Explain.
(c) List two forensic applications of DNA fingerprinting.
Answer.(a) Difference between repetitive and satellite DNA are as follows :
(b) Satellite DNA can be isolated from bulk genomic DNA by density gradient centrifugation.
(c) DNA fingerprinting is very useful in detection of crime and legal pursuit.