Lac operon is the prototype operon in bacteria, which codes for genes responsible for metabolism of lactose. The operon is regulated by the amount of lactose in the medium where the bacteria are grown. Therefore, this regulation can also be viewed as regulation of enzyme synthesis by its substrate. Human genome project was a mega project that aimed to sequence every base in human genome. This project has yielded much new information. Many new areas and avenues have opened up as a consequence of the project. DNA Fingerprinting is a technique to find out variations in individuals of a population at DNA level. It works on the principle of polymorphism in DNA sequences. It has immense applications in the field of forensic science, genetic biodiversity and evolutionary biology.
Transformation and the Transforming Principle
1. Frederick Griffith (1928) conducted a series of experiments with Streptococcus pneumoniae (Pneumococcus), the bacterium that causes pneumonia.
2. He observed two strains of this bacterium; one forming smooth colonies with capsule (S-type) and the other forming rough colonies without capsule (R-type)
3. When live S - type cells were injected into the mice, they produced pneumonia (pathogenic/virulent) and the mice died.
4. When live R - type cells were injected into the mice, the disease was not produced (non-pathogenic/non-virulent)
5. When heat - killed S type cells were injected into the mice, the disease did not appear.
6. When heat - killed S strain cells were mixed with live R - strain cells and injected into the mice, the mice died and he could isolate live S-strain cells from the body of the mice.
7. He concluded that the R-strain bacteria had somehow been transformed by the heat-killed S-strain bacteria, which must be due to the transfer of the genetic material (transforming principle).
8. Later Oswald Avery, Colin Mac Leod and Maclyn Mc Carty (1933-44) discovered that DNA from heat-killed S-strain bacteria caused the transformation of R-strain bacteria.
9. They also discovered that proteases and RNase did not affect transformation, while DNase inhibited the process.
[every organism has three things in genetic material - Protein, DNA, RNA]
DNA is the Genetic Material: -
1. The proof for DNA as the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952), who worked with bacteriophages.
(bacteriophages - viruses that infect bacteria)
2. The bacteriophage on the infection injects only DNA into the bacterial cell and not the protein coat; the bacterial cell treats the viral DNA as its own and subsequently manufactures more virus particles.
3. They made two different preparations of the phage; in one, the DNA was made radioactive with 32P and in the other, the protein coat was made radioactive with 35S .
4. These two phage preparations were allowed to infect the bacterial cells separately.
5. Soon after infection, the cultures were gently agitated in a blender to separate the adhering protein coats of the virus from the bacterial cells.
6. The culture was also centrifuged to separate the viral coat and the bacterial cells.
7. It was found that when the phage containing radioactive DNA was used to infect the bacteria, its radioactivity was found in the bacterial cells (in the sediment) indicating that the DNA has been injected into the bacterial cells.
8. So DNA is the genetic material and not proteins.
Characteristics of Genetic Material: -
1. A molecule that can act as genetic material must have the following properties:
a. it should be able to generate its replica
b. it should be chemically and structurally stable
c. It should provide the scope for slow changes (mutation) that are necessary for evolution
d. It should be able to express itself in the form of Mendelian characters.
2. Nucleic acids i.e. DNA and RNA can replicate, but not protein.
3. The predominant genetic material is DNA, while few viruses like tobacco mosaic virus have RNA as the genetic material
4. The 2'-OH group in the nucleotides of RNA is a reactive group and makes RNA labile and easily degradable; RNA as a catalyst is also more reactive and hence DNA has the property to be genetic material.
RNA - World: -
1. Ribonucleic acid (RNA) was the first genetic material.
2. The 2'-OH group of ribonucleotides is a reactive group, that makes RNA a catalyst.
3. It is evident that essential life processes such as metabolism, translation, splicing etc. have evolved around RNA, even before DNA has evolved as a genetic material.
Structure of a Polynucleotide Chain (RNA): -
1. In RNA also, each nucleotide has three components as in DNA.
2. The nitrogen bases are of two types: -
Purines - (Adenine & Guanine)
Pyrimidines - (Cytosine & Uracil)
3. The sugar is ribose which has an additional OH group on the 2'-position
4. The nucleosides and nucleotides are called ribonucleosides and ribonucleotides respectively.
Structure of t - RNA
types of RNA
There are three types of RNA
Messenger RNA (mRNA)
- It brings the genetic information of DNA transcribed on it for protein synthesis
- it is single stranded
Transfer/ Soluble RNA (tRNA/s RNA)
- It acts as an adapter molecule that reads the code on one hand and binds to the specific amino acid on the other hand.
- tRNA has a clover leaf like secondary structure but actually it is an inverted L - shaped compact molecule.
- It has an 'amino acid acceptor end' (3' - end) and an 'anticodon - loop', where the three bases are complementary to the bases of the codon of the particular amino acid
Ribosomal RNA (rRNA)
- It forms the structure of ribosomes
- It also plays a catalytic role during translation
Replication of DNA: -
- Watson & Crick had proposed a scheme for replication of DNA, when they proposed the double helical structure for DNA
- The scheme suggested that the two strands would separate and each acts as a template for the synthesis of a new complementary strand.
- After complete replication, each DNA molecule would have one parental and one synthesised strand.
- This scheme is termed as semi-conservative DNA replication -
(i) Proof for semi-conservative replication of DNA: -
- Mathew Meselson and Franklin Stahl have performed an experiment using Escherichia coli to prove that DNA replication is semi conservative.
- They grew Escherichia coli in a medium containing in a medium containing 15NH4Cl until 15N was incorporated in the two stands of newly synthesised DNA; this heavy DNA can be separated from the normal 14N DNA by centrifugation in Cesium Chloride (CsCl) density gradient.
- Then they transferred the cells into a medium with normal 14NH4Cl and took out samples at various time intervals and extracted DNAs and centrifuged them to measure their densities.
- The DNA extracted from the cells after one generation to transfer from the 15N medium to 14N medium (i.e. after 20 minutes) had an intermediate/ hybrid density.
- The DNA extracted after two generations (i.e. after 40 minutes) consisted of equal amounts of 'light' DNA and 'hybrid' DNA.
- Similar experiments were conducted by Taylor et al in 1958, by using radioactive thymidine; they proved that DNA on the chromosomes replicates in a semi-conservative manner.
(ii) The process of Replication of DNA: -
- The process involves a number of enzymes/ catalysts of which the main enzyme is DNA-dependent DNA-polymerase, that catalyses the polymerisation of the deoxy-nucleotides at a rate of approximately 2000 base pairs per second.
- The process is also an energy - expensive process; deoxyribo-nucleotide tri-phosphates serve the dual purpose
a. acting as substrate
b. providing energy (from the two terminal phosphates).
- The intertwined strands of DNA separate from a particular point called the origin of replication.
- Since the two strands cannot be separated in its entire length (due to very high energy requirement), replication occurs with small opening of the DNA - helix; the y-shaped structure formed is called replication fork.
- The DNA-dependent DNA - polymerase catalyse polymerisation of the nucleotides only in 5' → 3' direction
- Consequently, on one of the template strands (with 3' → 5' polarity) the synthesis of DNA is continuous, while on the other template stands (with polarity 5' → 3'), the synthesis of DNA is discontinuous i.e. short stretches of DNA are synthesised.
- The discontinuously synthesised strands are later joined together by the enzyme DNA - ligase.
Meselson - Stahl Experiment
replication of DNA
Proteins
Proteins are the building block of alpha amino acids (R-CO-NH2)
R-CO-NH2 - R-CO-NH2 - R-CO-NH2 - R-CO-NH2 -R-CO-NH2 - R-CO-NH2 ..... = Protein
In total we have 20 α-amino acids = 10 are essential + 10 are non - essential
10 essential α-amino acids = Valine, Leucine, Isoleucine, Phenylalanine, Methionine, Tryptophan, Threonine, Lysine, Arginine, Histidine
10 non essential α-amino acids = Glycine, Alanine Serine, Cysteine, Glutamine, Tyrosine, Proline, Aspartic acid, Asparagine, Glutamic acid.
Amino Acids
α-amino acids are the building block of proteins (R-CO-NH2)
R-CO-NH2 - R-CO-NH2 - R-CO-NH2 - R-CO-NH2 -R-CO-NH2 - R-CO-NH2 ..... = Protein
In total we have 20 α-amino acids = 10 are essential + 10 are non - essential
10 essential α-amino acids = Valine, Leucine, Isoleucine, Phenylalanine, Methionine, Tryptophan, Threonine, Lysine, Arginine, Histidine
10 non essential α-amino acids = Glycine, Alanine Serine, Cysteine, Glutamine, Tyrosine, Proline, Aspartic acid, Asparagine, Glutamic acid.
Transcription
- Transcription is also governed by the complementarity of bases as in DNA. (except that uracil in place of thymine is complementary to adenine)
- Only one of the strands of the DNA acts as the template for RNA-synthesis for the following reasons: -
i) If both the strands code for RNA, two different (complementary) RNA molecules and two different proteins would be formed; hence the genetic information- transfer machinery would become complicated
ii) Since the two RNA molecules produced would be complementary to each other, they would wind together to form a double-stranded RNA without carrying out translation; that means the process of transcription would become futile.
- A transcription unit in DNA has three regions: -
i) A promoter ii) Structural gene(s) and iii) A terminator
- The process is catalysed by DNA-dependent RNA-polymerase, which catalyses the polymerisation of the nucleotides only in 5'→3' direction
- The DNA strand with 3'→5' polarity is called 'template strand', while the other strand with 5'→3' polarity is called the 'coding strand'.
- The coding strand is displaced and does not code for RNA, but reference points regarding transcription are made in relation to it.
- The promoter refers to a particular sequence of DNA located towards 5' end (upstream) of the coding strand, where the RNA polymerase becomes bound for transcription.
- The terminator is a sequence of DNA located towards the 3' end (downstream) of the coding strand, where the process of transcription would stop.
(i) Transcription in Prokaryotes (Monera - Bacteria): -
- In prokaryotes, the structural gene are polycistronic and continuous
- In prokaryotes, there is a single DNA-dependent RNA polymerase, that catalyses the transcription of all the three types of RNA (mRNA, tRNA, rRNA)
- RNA polymerase binds to the promoter and initiates the process alongwith certain initiation factor.
- It uses ribonucleoside triphosphates (also called ribonucleotides) for polymerisation on a DNA template following complementarity of bases.
- The enzyme facilitates the opening of the DNA-helix and elongation continues.
- Once the RNA polymerase reaches the terminator, the RNA-polymerase falls off and the nascent RNA separates; it is also called termination of transcription and is facilitated by certain termination factors.
- In prokaryotes the mRNA synthesised does not require any processing to become active and both transcription and translation occur in the same cytosol; translation can start much before the mRNA is fully transcribed, i.e. transcription and translation can be coupled.
Transcription in Eukaryotes: -
- In eukaryotes, the structural genes are monocistronic and 'split'
- They have coding sequences called exons that forms part of mRNA and non - coding sequences, called introns, they do not form part of mRNA and are removed during splicing.
- In eukaryotes, there are at least three different RNA Polymerases in the nucleus, apart from the RNA polymerase in the organelles, which function as follows: -
- RNA polymerase I transcribes rRNAs (26S, 18S, 5.8S)
- RNA polymerase II transcribes the precursor of mRNA (called as heterogenous nuclear RNA (hnRNA)
- RNA - polymerase III catalyses transcription of tRNA.
- The primary script contains both exons and introns and it is subjected to a process called splicing, where the introns are removed and the exons are joined in a definite order to form mRNA.
- The hnRNA undergoes two additional processes called 'capping' and 'tailing'
- In capping, methyl guanosine tri-phosphate is added to the 5'-end of hnRNA
- In tailing, adenylate residues (about 200 - 300) are added at the 3' - end of hnRNA.
- The fully processed hnRNA is called mRNA and is released from the nucleus into the cytoplasm.
Genetic Code
- Genetic code refers to the relationship between the sequence of the nucleotides of mRNA and the sequence of amino acids in the polypeptide.
- It was George Gamow, who suggested that the code must be made up of three bases, in order to code for the twenty different amino acids, with only four bases; this would generate 43 or 4x4x4 = 64 triplet codons.
- Har Gobind Khorana could synthesize RNA molecules with definite combinations of bases (homopolymers and copolymers)
- Marshal Nirenberg made a cell-free system for protein synthesis, that helped in deciphering the code.
- Severo Ochoa discovered enzyme polynucleotide phosphorylate, that could polymerise RNA with definite sequence in template independent manner.
- The checkerboard pattern of genetic code was prepared as given below:
Salient features of genetic code
- The codons are triplets and there are sixty four codons: 61 codons code for 20 amino acids and 3 codons (UAA, UAG, UGA) do not code for any amino acid, but function as stop/ termination codons.
- each codon codes for only one/ particular amino acid and so the genetic code is 'unambiguous' and 'specific'
- since some amino acids are coded by more than one codon, the genetic code is said to be 'degenerate'
- the codons are read in a contiguous manner in a 5'→3' direction and have no punctuations (commaless)
- The genetic code is universal i.e. the codons code for the same amino acid in any organism, be a bacterium or a human being.
- AUG has dual functions of coding for methionine and acting as initiation codon.
Mutations
1. Point Mutation
2. Frameshift Mutation
3. Silent Mutation
1. Point Mutation: - a change in single base pair. A change in single base pair in the gene for beta globin chain of haemoglobin: that results in glutamine in place of valine; it causes a disease called sickle cell anaemia.
Frameshift Mutation
- It is the type of mutation where addition/ insertion or deletion of one or two bases changes the reading from the site of mutation, resulting in a protein with a different set of amino acids.
- This also forms the basis of proof that codon is a triplet and codons are read in a contiguous manner.
- Insertion or deletion of three or its multiples of bases do not alter the reading frame, but one/ more amino acids are coded in the protein translated. Example: - Cystic fibrosis, Crohn's disease.
Silent Mutation: -
- If a base change in a codon does not alter the amino acid coded, the mutation is said to be silent mutation.
- If codon AAA is altered to become AAG, the same amino acid - lysine will be incorporated.
Translation
- In this process, amino acids become joined together by peptide bonds, to form polypeptides
- The formation of peptide bonds requires energy and hence in the first phase, the amino acids are activated: -
Activation of amino acids -
* In this process, a particular amino acid becomes activated and attached to 3' end of a specific tRNA molecule
* The reaction is catalysed by the enzyme amino-acyl-tRNA synthetase.
Amino Acid (AA) + ATP → AA-AMP-ENZ + PPi
AA-AMP-ENZ + tRNA → AA-tRNA + AMP + PPi
(here ATP is Adenosine Triphosphate, AMP is Adenosine Monophosphate, PPi means Protein Protein interaction)
Initiation of polypeptide synthesis: -
- The ribosome, in its inactive state exists as two subunits - a large subunit and a small subunit.
- When the small subunit encounters the mRNA, translation begins
- The mRNA binds to the small subunit of ribosome, following base pair rule between the bases of mRNA and those on rRNA; it is catalysed by 'initiation factors'
- There are two sites on the larger subunit, the P-site and the A-site
- The small subunit (with the tRNA) attaches to the large subunit in such a way that the initiation codon (AUG) comes on the P - site
- The initiator tRNA (methionyl tRNA) binds to the P-site
Elongation of polypeptide chain: -
- A second tRNA charged with an appropriate amino acid binds to the A-site of the ribosome.
- A peptide (CO-NH) bond is formed between the carboxyl group of methionine and the amino group of the second amino acid; this reaction is catalysed by the enzyme peptidyl transferase.
- The ribosome moves from codon to codon along with mRNA in the 5'→3' direction.
- Amino acids are added one by one in the sequence of the codons and become joined together to form polypeptide.
Termination of Polypeptide synthesis: -
- When one of the termination codons comes at the A-site, it does not code for any amino acid and there is no tRNA molecule for it.
- As a result the polypeptide synthesis (or elongation of polypeptide) stops
- The polypeptide synthesized is released from the ribosome, catalysed by a 'release factor'
Regulation of Gene Expression
- In prokaryotes, the gene expression is controlled by the initiation of transcription
- In eukaryotes, the regulation can be exerted at four levels: -
(i) Transcription level (formation of primary transcript)
(ii) Processing level (splicing)
(iii) Transport of mRNA from nucleus to cytoplasm and
(iv) Translational level
- The regulation at transcription level was elucidated by Jacob and Monod
- All the genes controlling one metabolic pathway constitute an operon.
- A few examples are lac-operon, trp-operon, val-operon, bis-operon etc. in Escherichia Coli
- An operon consists of the following components: -
(i) structural gene: - they transcribe the mRNA for the amino acid sequence of protein (enzymes)
(ii) Promoter gene: - The promoter gene is a sequence of DNA, where the RNA polymerase binds and initiates transcription.
(iii) Operator: -
* The operator is a sequence of bases on DNA adjacent to the promoter.
*The accessibility of promoter gene for RNA polymerase is regulated by process, called repressive.
* In most cases a specific, repressor protein binds to the operator.
(iv) Regulator gene: -
* This code for the repressor protein that binds to the operator and 'switches off' the transcription unit
* So the regulatory gene is also represented as 'i' gene (inhibitory gene)
(v) Inducer: -
*The substance/ substrate that prevents the repressor from binding to the operator, is called an inducer; it keeps the switch 'on' and transcription continues.
Lac-operon in Escherichia Coli
- It is an inducible operon where Lactose is the inducer, it is the substance for enzyme β-galactosidase.
- The components of lac-operon and their functions are as follows:
(i) structural genes: -
* there are three structural genes (z,y,a) which transcribe a polycistronic mRNA.
* Gene z codes for β-galactosidase (b-gal), that catalyses the hydrolysis of lactose into galactose and glucose
* Gnee y codes for permease, which increases the permeability of the cell to β-galactosidase (lactose).
* Gene a codes for transacetylase, that catalyse the transacetylation of lactose into its active form
(ii) Promoter: -
* It is a sequence of bases near to the structural gene; near to the structural genes; it is the site where RNA-polymerase binds for transcription
(iii) Operator: -
* It is a sequence of DNA near the promoter, where a repressor always binds
* It functions as a switch for the operon
(iv) Repressor: -
* It is a protein coded by i gene, synthesised all the time constitutively.
* It binds to the operator and prevents the RNA polymerase from transcribing.
(v) Inducer: -
* Lactose is the inducer that inactivates the repressor and prevents it from binding to the operator
* This allows an access for the RNA polymerase to the promoter and transcription
* Thus the substrate, lactose regulates the lac-operon.
Human Genome Project (HGP)
- Human Genome Project was a 13 year project, that was launched in the year 1990 and completed in 2003.
- This project was coordinated by the US Department of Energy and the National Institute of Health
- During the early years of the project, the Wellcome Trust (UK) became a major partner; other countries like Japan, Germany, China and France contributed significantly
- Its aim was to find out the complete DNA sequences of the human genome.
- the two factors that made this possible are:
(i) Genetic Engineering techniques, with which it was possible to isolate and clone any segment of DNA
(ii) Availability of simple and fast techniques, for determining the DNA sequences.
- Human Genome Project was called a mega project for the following facts: -
(i) The Human genome has approximately 3.3 x 109 bp; if the cost of sequencing is US$3 per bp, the approximate cost is about US $9 billions (Rs.900 crores).
(ii) If the sequences obtained were to be stored in typed form in books and if each page contained 1000 letters and each book contained 1000 pages, then 3300 such books would be needed to store the complete information.
(iii) The enormous quantity of data expected to be generated also necessitates the use of high speed computational devices for data storage, retrieval and analysis
- the project was closely associated with a new branch of biology called bioinformatics
Goals of HGP
Some major important goals of HGP are to:
(i) Identify all the genes (approximately 20,000 - 25,000) in human DNA.
(ii) Determine the sequences of the three billion base pairs present in human DNA
(iii) Store this information in databases.
(iv) Improve the tools for data analysis.
(v) Transfer the technologies to other sectors (like industries)
(vi) Address the ethical, legal and social issues (ELSI), that may arise from this project.
Advantages/ Uses of HGP
- Knowledge of the effects of variations of DNA among individuals can revolutionise the ways to diagnose, treat and even prevent a number of diseases/ disorders that affect human beings.
- it provides clues to the understanding of human biology
Methodologies of HGP
- the methods involved two major approaches: -
(i) One approach called as Expressed Sequence Tags (ESTs), focussed on identifying all the genes that expressed as RNA
(ii) Second approach called as Sequence Annotation, was to simply sequence the whole set of genome, that included all the coding and non-coding sequences and later assigning functions to different regions in the sequence.
- total DNA from the cell is isolated and converted into random fragments of relatively smaller sizes.
- these fragments are then cloned in suitable hosts using specialised vectors; the commonly used hosts are bacteria and yeast and the vectors are bacterial artificial chromosomes (BAC) and yeast artificial chromosomes (YAC)
- the fragments are then sequenced using automated DNA sequences, which work on the principle developed by Frederick Sanger.
- The sequences were then arranged on the basis of certain overlapping regions present in them; this required the generation of overlapping fragments for sequencing.
- Specialised computer based programmes were developed for alignment of the sequences.
- These sequences were annotated and assigned to the respective chromosomes.
- The next task was to assign the genetic and physical maps on the genome; this was generated using the information on polymorphism of restriction endonuclease recognition sites and certain repetitive DNA sequences, called microsatellites.
Salient features of Human Genome
- following are some of the salient observations derived from HGP: -
(i) the human genome contains 3164.7 million nucleotides (base pairs)
(ii) the size of the genes varies; an average gene consists of 3000 bases, while the largest gene, dystrophin consists of 2.4 million bases.
(iii) the total number of genes is estimated at 30000 and 99.9% of the nucleotides are the same in all humans.
(iv) The functions of over 50% of the discovered genes are not known.
(v) Only less than 2% of the genome codes for proteins.
(vi) Repetitive sequences make up a large portion of human genome
(vii) Repetitive sequences throw light on chromosome structure and dynamics and evolution, though they are thought to have no direct coding functions
(viii) Chromosome No.1 has 2968 genes (the maximum) and the Y-Chromosome has 231 genes (the least)
(ix) Scientists have identified about 1.4 million locations, where DNA differs in single base in human beings; these are called single nucleotide polymorphisms (SNPs)
Applications/ Future challnges of Human Genome
(i) Having the complete sequence of human genome, will enable a radically new approach to biological research i.e. a systematic approach on a much broader scale.
(ii) All the genes in a genome or all the transcripts in a particular tissue/ organ/ tumor can be studied.
(iii) It will be possible to understand how the enormous number of genes and proteins work together in interconnected networks in the chemistry of life.
DNA-Fingerprinting
Extraction-Amplification-Restriction Digestion-Separation of DNA sequences/restriction fragments-Southern Blotting-Hybridisation-Autoradiography
- The technique of DNA-fingerprinting was developed by Dr. Alec Jeffreys, in an attempt to find out markers for inherited diseases.
- The process is also known as DNA-typing or DNA-profiling
- The most important fact for DNA-fingerprinting are the short nucleotide repeats, called Variable Number Tandem Repeats (VNTRs), that vary in number from person to person and are inherited.
- The VNTRs of two persons may be of same length and sequence at certain sites, but vary at others,
- the procedure of DNA-fingerprinting includes the following major steps: -
(i) Extraction: - DNA is extracted from the cells in a high- speed, refrigerated centrifuge
(ii) Amplification: - Many copies of the extracted DNA are made by polymerase chain reaction (PCR)
(iii) Restriction Digestion: - DNA is cut into fragments with restriction enzymes into precise reproducible sequences.
(iv) Separation of DNA sequences/ restriction fragments: - The cut DNA fragments are introduced and passed through electrophoresis set-up containing agarose polymer gel; the separated fragments can be visualised by staining them with a dye that shows fluorescence under UV radiation.
(v) Southern blotting: - the separated DNA sequences are transferred onto a nitrocellulose or nylon membrane or sheet placed over the gel
(vi) Hybridisation: - the nylon membrane is immersed in a bath and radioactive probes (DNA segments of known sequence) are added; these probes target a specific nucleotide sequence that is complementary to them
(viii) Autoradiography: - The nylon membrane is pressed on to an x-ray film and dark bands develop at the probe sites which resemble the bar-codes.
Uses of DNA fingerprinting technique
(i) To identify criminals in the forensic laboratories
(ii) To determine the real parents i.e. to identify the true biological father / mother in case of disputes
(iii) To verify whether an immigrant is really a close relative (as he/she claims) of the mentioned resident.
(iv) To identify racial groups to rewrite the biological evolution.
Imp Questions
Q1. AUG codes for
a. glycine
b. alanine
c. leucine
d. methionine
Ans. d. methionine
Q2 The transcribed mRNA of a segment of DNA with the codons TAGCATACT is
a. AUCGTAUGA
b. TUCGUAUGA
c. AUCGAUUGA
d. AUCGUAUGA
Ans. d. AUCGUAUGA
Q3 CUAGUAAUA is the transcribed mRNA of one of the following DNA segments
a. GATCATATA
b. CATGATATA
c. TATGATATA
d. GATCATTAT
Ans. d. GATCATTAT
Q4. the complementary strand of DNA synthesised on a template of DNA with the codons TAGCATACT is
a. ATCGTATGA
b. AUCGATTGA
c. ATGGUATGA
d. ATCGTATCA
Ans. a. ATCGTATGA
Q5. the transcribed mRNA of a segment of DNA with the codons GATCATTAT is
a. CUAGUAAUA
b. CUAGUAUAU
c. CAUGUAAUA
d. CUAGAUAUA
Ans. a. CUAGUAAUA
Q6. the non-sense codons are
a. UACUAGUGA
b. UAAUCCUGA
c. UAAUGGUGA
d. UAAUAGUGA
Ans. d.UAAUAGUGA
Q7. the anticodon of tRNA charged with methionine is
a. TAC
b. UAC
c. UTC
d. AUG
Ans. b. UAC
MCQs
1. Which of the following statements is correct?
a. Adenine pairs with thymine through two H bonds
b. Adenine pairs with thymine through one H bonds
c. Adenine pairs with thymine through three H bonds
d. Adenine does not pair with thymine
Ans. a.
2. Purines found both in DNA and RNA are
a. Cytosine and Thymine
b. Adenine and Thymine
c. Adenine and Guanine
d. Guanine and Cytosine
Ans. c.
3. The association of histone H1 with a nucleosome indicates that
a. DNA replication is occurring
b. the DNA is condensed into a chromatin fibre
c. the DNA double helix is exposed
d. transcription is occurring
Ans. b
4. In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are
a. G17%, A33%, T33%
b. G8.5%, A50%, T24.5%
c. G34%, A24.5%, T24.5%
d. G17%, A16.5%, T32.5%
Ans. a
5. The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to C.
DNA --A---mRNA----B---Protein----Proposed by---C
a. A-Transcription, B-Translation, C-Francis Crick
b. A-Translation, B-Extension, C-Rosalind Franklin
c. A-Transcription, B-Replication, C-James Watson
d. A-Translation, B-Transcription, C-Ervin Chargaff
Ans. a
6. DNA ---C----mRNA----B---Protein ---Proposed by ---- A
The figure gives an important concept in the genetic implication of DNA. Fill the blanks A, B and C
a. A-Maurice Wilkins, B-Transcription, C-Translation
b. A-James Watson, B-Replication, C-Extension
c. A-Erwin Chargaff, B-Translation, C-Replication
d. A-Francis Crick, B-Translation, C-Transcription
Ans. d
7. What are the structures called that give an appearance as 'beads-on-string' in the chromosomes when viewed under electron microscope?
a. Genes, b. Nucleotides, c. Nucleosomes, d. Base pairs
Ans. c
8. Which one of the following does not follow the central dogma of molecular biology?
a. Pea, b. Mucor, c. Chlamydomonas, d. HIV
Ans. d
9. The 3'-5' phosphodiester linkages inside a polynucleotide chain serve to join
a. one DNA strand with the other DNA strand, b. one nucleoside with another nucleoside, c. one nucleotide with another nucleotide, d. one nitrogenous base with pentose sugar
Ans. c
10 Which one of the following statements about the particular entity is true?
a. Centromere is found in animal cells, which produces aster during cell division
b. The gene for producing insulin is present in everybody cell
c. Nucleosome is formed of nucleotides
d. DNA consists of core of eight histones.
Ans. b