Studied by 4 people
In a DNA strand, the nucleotides are linked together by
Phosphodiester bonds. The nucleotides are linked together by 3’–5’ phosphodiester linkage to form a dinucleotide
Nucleotide components
A nitrogenous base
Pentose Sugar (ribose in RNA deoxyribose in DNA)
Phosphate group derived from phosphoric acid.
a nucleoside differs from a nucleotide. It lacks the base.
Phosphate group
Nucleosides in DNA and RNA
DNA | RNA |
Deoxyadenosine | Adenosine |
Deoxyguanosine | Guanosine |
Deoxycytidine | Cytidine |
Deoxythimidine | Uridine |
Euchromatin v/s heterochromatin
Euchromatin | Heterochromatin |
1. It is loosely packed chromatin | 1. It is more densely packed chromatin |
2. It is lightly stained region of the chromatin. | 2. It is darkly stained region of the chromatin. |
3. It is a transcriptionally active chromatin. | 3. It is a transcriptionally inactive chromatin. |
4. It is found in eukaryotes and prokaryotes. | 4. It is found in eukaryotes only. |
Packaging of DNA Helix in prokaryotes
Prokaryotes do not have a defined nucleus ,DNA is not scattered throughout the cell. DNA being negatively charged is held with some proteins that have positive charges in a region termed as 'nucleoid' and is organised in large loops held by proteins.
Packaging of DNA Helix in eukaryotes/nucleosome
In eukaryotes, the organisation is much more complex. There is a set of positively charged, basic proteins called histones that are positively charged proteins rich in the basic amino acid residues lysines and arginines. Histones are organised to form a unit of eight molecules called as histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome. A typical nucleosome contains 200 bps of DNA helix. Chromatin is a thread-like stained (coloured) bodies seen in nucleus that constitute the nucleosomes and are seen as beads on string. And is packaged to form chromatin fibers that are further coiled and condensed at metaphase stage of cell division to form chromosomes.
Salient features of DNA
It is made of two polynucleotide chains, where the backbone is constituted by sugar- phosphate, and the bases project inside.
The two chains have anti-parallel polarity. → one chain has the polarity 5'3', the other has 3' 5'.
The bases in two strands are paired through hydrogen bonds. (A-T double bond) (G-C triple bond)
purine comes opposite to a pyrimidine. This generates approximately uniform distance between the two strands of the helix.
The two chains are coiled in a right-handed fashion. The pitch is 3.4 nm. number of base pairs in each turn is roughly 10 base pairs. The distance between 2 consecutive base pairs is 0.34 nm.The diameter of the DNA double helix is 2 nm.
plane of one base pair stacks over the other in double helix. This confers stability of the helical structure.
What confers stability to the DNA Helix?
The bases in two strands are paired through hydrogen bonds. (A-T double bond) (G-C triple bond). purine comes opposite to a pyrimidine. This generates approximately uniform distance between the two strands of the helix.
Plane of one base pair stacks over the other in double helix.
Describe the Griffith transformation experiment.
Frederick Griffith studied the principle of transformation Streptococcus pneumoniae. While 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 which is virulent die from pneumonia infection but mice infected with the R strain does not.
Griffith was able to kill bacteria by heating them. He observed that heat- killed S strain bacteria injected into mice did not kill them. When he injected a mixture of heat-killed S and live R bacteria, the mice died. He found living S bacteria from the dead mice.
Griffith concluded that transfer of genetic material by the dead cells of the s-strain transformed the R-type to S-type, enabled the R-strain to synthesise a smooth polysaccharide coat and become virulent.
Griffith’s experiment showed stability as a property of genetic material.
Heat killed the bacteria but did not destroy some properties of genetic material.
Biochemical characterisation of transforming principal. Avery, Macleod, McCarty experiment.
1. Oswald Avery, Colin MacLeod and Maclyn McCarty (1933-44), worked to determine the biochemical nature of 'transforming principle' in Griffith's experiment.
2. 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.
3. They discovered that DNA alone from S bacteria caused R bacteria to become transformed.
4. They also discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, so the transforming substance was not a protein or RNA.
5. Digestion with DNase did inhibit transformation, suggesting that the DNA caused the transformation.
6. They concluded that DNA is the hereditary material.
Who revealed the biochemical nature of The transforming principle?
Oswald Avery
Colin Mcleod
Maclyn McCarty
Who proved that the genetic material is DNA?
Alfred Hershey and Martha Chase
Describe the Hershey and chase experiment
They grew some viruses on a medium containing radioactive phosphorus (32P) and some other on radioactive sulphur(35S).
Viruses grown in radioactive phosphorus contain radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
Viruses grown in radioactive phosphorus contain radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not.
Viruses grown in radioactive sulphur contain radioactive protein.
Radioactive phage were allowed to attach to E. coli bacteria and infected it by transferring genetic material.
The viral coats were removed from the bacteria by educating them in a blender. The virus particles were separated from the bacteria by spinning them in a centrifuge.
Only radioactive 32P was found to be associated with the bacterial cell, whereas radioactive 35S was only found in surrounding medium and not in the bacterial cell. This indicates that only DNA and not protein coat entered the bacterial cell.
This proves that DNA is the genetic material which is passed from virus to bacteria and not protein. If both DNA and proteins contained phosphorus and sulphur, the result might change.
Why is DNA preferred genetic material than RNA?
RNA is single-stranded while DNA exists in double standard form, and having a complimentary strand resist changes by evolving a process of repair.
2’-OH group is a reactive group absent in DNA, but present at every nucleotide in RNA and makes it liable and easily degradable.
DNA is chemically less reactive and structurally more stable, if the strands are separated by heating become together when appropriate conditions were provided, while RNA being a catalyst is chemically more reactive and unstable.
The presence of thymine instead of uracil in DNA confers stability.
RNA Being unstable mutates at a faster rate, while DNA is stable and undergoes a slow change that is essential for evolution.
Properties of genetic material
Generate its replica
Chemically and structurally stable
Provides scope for slow mutations that are required for evolution
Able to express itself in the form of Mendelian characters
Watson and crick scheme for DNA replication
The two strings will separate and it is a template for the synthesis of new complimentary strand. After completion of replication, each DNA will have one parental and one newly synthesised strand.
Semiconservative DNA replication
Messelson and Stahl DNA replication experimental proof
. (i) Meselson and Stahl performed the experiments on E. coli to prove that DNA replication is semi- conservative. Use of 15N helped to prove the semi- conservative replication of DNA.
(ii) They first grew E. coli bacteria in a medium containing 15NH4Cl (in which 15N is the heavy isotope of nitrogen) for many generations.
(iii)Then they transferred the cells into a medium with normal 14NH4Cl (in which 14 N is the lighter isotopes) and took the samples at various definite time intervals as the cells multiplied.
(iv) The extracted DNA were centrifuged and measured to get their densities. The DNA extracted from the culture after one generation of transfer from then ^ 15 N medium to ^ 14 N medium (i.e., after 20 min, E. coli divides every 20 min.) showed an intermediate hybrid density (i.e., both heavy and light nitrogen) which proved the semi-conservative nature of DNA.
how is a nitrogenous base linked to the OH of 1'C of pentose sugar
A nitrogenous base is linked to the OH of 1' C pentose sugar through a N-glycosidic linkage to form a nucleoside, such as adenosine or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.
How is a nitrogenous base linked to the OH of 1'C of pentose sugar
A nitrogenous base is linked to the OH of 1' C pentose sugar through a N-glycosidic linkage to form a nucleoside, such as adenosine or deoxyadenosine, guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.
in which organism is the central dogma reversed
virus ( RNA --> DNA)
what are non-histone chromosomal proteins
The packaging of chromatin at higher level requires additional set of proteins that collectively are referred to as NHC proteins
limitations of Griffith's experiment
the biochemical nature of genetic material was not defined from his experiments.
who performed a similar experiment on Vicia faba
Very similar experiments involving use of radioactive thymidine to detect distribution of newly synthesised DNA in the chromosomes was performed on Vicia faba (faba beans) by Taylor and colleagues in 1958.
explain the machinery and enzymes used in the process of DNA replication
1. The main enzyme is referred to as DNA-dependent DNA polymerase, since it uses a DNA template to catalyse the polymerisation of deoxynucleotides.
2. 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.
3. The DNA-dependent DNA polymerases catalyse polymerisation only in one direction, that is 5’- 3'. This creates some additional complications at the replicating fork.
4. Consequently, on one strand (the template with polarity 3'- 5'), the replication is continuous, while on the other (the template with polarity 5'-3'), it is discontinuous.
5. The discontinuously synthesised fragments are later joined by the enzyme DNA ligase
6. The DNA polymerases on their own cannot initiate the process of replication.
give a brief account of the efficiency of DNA polymerase
1. highly efficient enzymes as they have to catalyse polymerisation of a large number of nucleotides in a very short time. E. coli that has only 4.6 ×10^6 bp (human diploid content is 6.6 × 10^9 bp), completes the process of replication within 18 minutes; that means the average rate of polymerisation has to be approximately 2000 bp per second.
2. Not only do these polymerases have to be fast, but they also have to catalyse the reaction with high degree of accuracy.
3. Any mistake during replication would result into mutations.
DNTP
Deoxyribosenucleoside triphosphate
what is the function of dexoyribose nucleoside triphosphates in the process of replication
1. Deoxyribonucleoside triphosphates serve dual purposes.
2. In addition to acting as substrates, they provide energy for polymerisation reaction (the two terminal phosphates in a deoxynucleoside triphosphates are high-energy phosphates, same as in case of ATP)
what is meant by origin of replication
1. replication does not initiate randomly at any place in DNA. There is a definite region in E. coli DNA where the replication originates.
2. Such regions are termed as origin of replication.
3. It is because of the requirement of the origin of replication that a piece of DNA if needed to be propagated during recombinant DNA procedures, requires a vector. The vectors provide the origin of replication
where does DNA replication take place in eukaryotes
S phase of the cell cycle
a failure in cell division after DNA replication results into
polypoidy - a chromosomal anomaly
define transcription
The process of copying genetic information from one strand of the DNA into RNA is termed as transcription
what governs the principle of transcription
the principle of complementarity governs the process of transcription, except the adenosine complements now forms base pair with uracil instead of thymine.
what is a difference between the process of replication and transcription
unlike in the process of replication, which once set in, the total DNA of an organism gets duplicated, in transcription only a segment of DNA and only one of the strands is copied into RNA. This necessitates defining the boundaries that would demarcate the region and the strand of DNA that would be transcribed.
why aren't both the strand copied during transcription
1. First, if both strands act as a template, they would code for RNA molecule with different sequences (Remember complementarity does not mean identical), and in turn, if they code for proteins, the sequence of amino acids in the proteins would be different. Hence, one segment of the DNA would be coding for two different proteins, and this would complicate the genetic information transfer machinery.
2. Second, the two RNA molecules if produced simultaneously would be complementary to each other, hence would form a double stranded RNA. This would prevent RNA from being translated into protein and the exercise of transcription would become a futile one.
what are the components of a transcription unit
(i) A Promoter
(ii) The Structural gene
(iii) A Terminator
what is the convention in defining the 2 strands of the DNA in the structural gene of a transcription unit
1. Since the two strands have opposite polarity and the DNA-dependent RNA polymerase also catalyse the polymerisation in only one direction, that is, 5'→3', the strand that has the polarity 3'→5' acts as a template, and is also referred to as template strand.
2. The other 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.
3. All the reference point while defining a transcription unit is made with coding strand
explain the role of promoter terminator and structural gene in a transcription unit
1. The promoter and terminator flank the structural gene in a transcription unit.
2. 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).
3. 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.
4. By switching its position with terminator, the definition of coding and template strands could be reversed.
5. The terminator is located towards 3' -end (downstream) of the coding strand and it usually defines the end of the process of transcription.
6. There are additional regulatory sequences that may be present further upstream or downstream to the promoter
define a gene
A gene is defined as the functional unit of inheritance. The DNA sequence coding for tRNA or rRNA molecule also define a gene.
what is meant by split genes
1. by defining a cistron as a segment of DNA coding for a polypeptide, the structural gene in a transcription unit could be said as monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes).
2. In eukaryotes, the monocistronic structural genes have interrupted coding sequences - the genes in eukaryotes are split.
3. The coding sequences or expressed sequences are defined as exons.
4. Exons are said to be those sequence that appear in mature or processed RNA.
5. The exons are interrupted by introns. Introns or intervening sequences do not appear in mature or processed RNA.
what are the types of RNA found in bacteria
1. In bacteria, there are three major types of RNAs: mRNA (messenger RNA), tRNA (transfer RNA), and rRNA (ribosomal RNA).
2. All three RNAs are needed to synthesise a protein in a cell.
3. The mRNA provides the template, tRNA brings aminoacids and reads the genetic code, and rRNAs play structural and catalytic role during translation.
explain the process of transcription
1. There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria.
2. RNA polymerase binds to promoter and initiates transcription (Initiation).
3. It uses nucleoside triphosphates as substrate and polymerises in a template depended fashion following the rule of complementarity.
4. It also facilitates opening of the helix and continues elongation. Only a short stretch of RNA remains bound to the enzyme.
5. Once the polymerases reaches the terminator region, the nascent RNA falls off, so also the RNA polymerase. This results in termination of transcription
how is the RNA polymerase able to catalyse all 3 steps of initiation, elongation and termination
1. The RNA polymerase is only capable of catalysing the process of elongation.
2. It associates transiently with initiation-factor (σ) and termination-factor (ρ) to initiate and terminate the transcription, respectively.
3. Association with these factors alter the specificity of the RNA polymerase to either initiate or terminate
how can the translation begin much before
In bacteria, since the mRNA does not require any processing to become active, and also since transcription and translation take place in the same compartment (there is no separation of cytosol and nucleus in bacteria), many times the translation can begin much before the mRNA is fully transcribed. Consequently, the transcription and translation can be coupled in bacteria.
what are the complexities in the process of transcription in euakryotes
i) There are at least three RNA polymerases in the nucleus (in addition to the RNA polymerase found in the organelles). There is a clear cut division of labour. The RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S), whereas the RNA polymerase III is responsible for transcription of tRNA, 5srRNA, and snRNAs (small nuclear RNAs). The RNA polymerase II transcribes precursor of mRNA, the heterogeneous nuclear RNA (hnRNA).
(ii) The second complexity is that 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 (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 into cytoplasm for translation
who proposed the genetic code
1. George Gamow, a physicist, who argued that since there are only 4 bases and if they have to code for 20 amino acids, the code should constitute a combination of bases.
2. He suggested that in order to code for all the 20 amino acids, the code should be made up of three nucleotides.
3. This was a very bold proposition, because a permutation combination of 43 (4 × 4 × 4) would generate 64 codons; generating many more codons than required.
Har Gobind Khorana's contribution to the genetic code
the chemical method developed by Har Gobind Khorana was instrumental in synthesizing RNA molecules with defined combinations of bases (homopolymers and copolymers)
Marshall Nirenberg's contribution
Marshall Nirenberg's cell-free system for protein synthesis finally helped the code to be deciphered.
how was Severo Ochoa enzyme helpful
Severo Ochoa enzyme (polynucleotide phosphorylase) was also helpful in polymerising RNA with defined sequences in a template independent manner (enzymatic synthesis of RNA).
salient features of the genetic code
(i) The codon is triplet. 61 codons code for amino acids and 3 codons do not code for any amino acids, hence they function as stop codons.
(ii) Some amino acids are coded by more than one codon, hence the code is degenerate.
(iii) The codon is read in mRNA in a contiguous fashion. There are no punctuations.
(iv) The code is nearly universal: for example, from bacteria to human UUU would code for Phenylalanine (phe). Some exceptions to this rule have been found in mitochondrial codons, and in some protozoans.
(v) AUG has dual functions. It codes for Methionine (met) , and it also act as initiator codon.
(vi) UAA, UAG, UGA are stop terminator codons.
example of point mutation
change of single base pair in the gene for beta globin chain that results in the change of amino acid residue glutamate to valine. It results into a diseased condition called as sickle cell anemia
explain frameshif mutations or deletion mutations
Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
what is tRNA give its functions
tRNA has an anticodon loop that has bases complementary to the code, and it also has an amino acid acceptor end to which it binds to amino acids.
tRNAs are specific for each amino acid.
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 looks like a clover-leaf.
In actual structure, the tRNA is a compact molecule which looks like inverted L.
define translation
Translation refers to the process of polymerisation of amino acids to form a polypeptide
the order and sequence of amino acids is defined by
sequence of bases in the mRNA
what is meant by charging of tRNA
in the first phase itself amino acids are activated in the presence of ATP and linked to their cognate tRNA - a process commonly called as charging of tRNA or aminoacylation of tRNA
how does charging of tRNA help
1. If two such charged tRNAs are brought close enough, the formation of peptide bond between them qould be favoured energetically.
2. The presence of a catalyst would enhance the rate of peptide bond formation
give the function of the cellular factory responsible for synthesizing proteins
1. ribosomes
2. The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits; a large subunit and a small subunit.
3. When the small subunit encounters an mRNA, the process of translation of the mRNA to protein begins.
3. There are two sites in the large subunit, for subsequent amino acids to bind to and thus, be close enough to each other for the formation of a peptide bond.
4. The ribosome also acts as a catalyst (23S rRNA in bacteria is the enzyme- ribozyme) for the formation of peptide bond
define UTRs. where are they present?
1. An mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR).
2. The UTRs are present at both 5' -end (before start codon) and at 3' -end (after stop codon).
3. They are required for efficient translation process.
explain the process of translation form start to finish
1. For initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA.
2. 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.
3. The ribosome moves from codon to codon along the mRNA.
4. Amino acids are added one by one, translated into Polypeptide sequences dictated by DNA and represented by mRNA.
5. At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.
in eukaryotes the regulation of gene could be exerted at
(i) transcriptional level (formation of primary transcript),
(ii) processing level (regulation of splicing),
(iii) transport of mRNA from nucleus to the cytoplasm,
(iv) translational level
explain the concept of operators in prokaryotes
1. In prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.
2. In a transcription unit, the activity of RNA polymerase at a given promoter is in turn regulated by interaction with accessory proteins, which affect its ability to recognise start sites.
3. These regulatory proteins can act both positively (activators) and negatively (repressors).
4. The accessibility of promoter regions of prokaryotic DNA is in many cases regulated by the interaction of proteins with sequences termed operators.
5. The operator region is adjacent to the promoter elements in most operons and in most cases the sequences of the operator bind a repressor protein.
6. Each operon has its specific operator and specific repressor.
what is the lac operon? give more examples
1. In lac operon (here lac refers to lactose), a polycistronic structural gene is regulated by a common promoter and regulatory genes.
2. Such arrangement is very common in bacteria and is referred to as operon.
3. To name few such examples, lac operon, trp operon, ara operon, his operon, val operon, etc.
explain the genes consisting the lac operon
1. The lac operon consists of one regulatory gene (the i gene - derived from the word inhibitor) and three structural genes (z, y, and a).
2. The i gene codes for the repressor of the lac operon.
3. The z gene codes for beta-galactosidase (β-gal), which is primarily responsible for the hydrolysis of the disaccharide, lactose into its monomeric units, galactose and glucose.
4. The y gene codes for permease, which increases permeability of the cell to β-galactosides.
5. The a gene encodes a transacetylase.
6. Hence, all the three gene products in lac operon are required for metabolism of lactose.
Applications of DNA fingerprinting
DNA fingerprinting has application in forensic science which serves as a very useful identification tool since DNA from every tissue from an individual shows the same degree of polymorphism
DNA fingerprinting is the basis of parent, paternity testing in case of dispute, such as since the polymorphism is are in heritable from parents to children.
It has a much wider application as in determining population in genetic diversity.
How does Lac Operon operate in the presence and absence of E.coli
In absence of an inducer, the repressor protein binds to the operator region and prevents RNA polymerase from transcribing the structural gene. Thus, the lac operon gene is inactivated. - Lac operon in presence of an inducer binds with repressor protein and inactivates it.
Start and stop codons
start- AUG
Stop- UAA UGA UAG
In humans, which chromosome has the most number of genes and least
Chromosome 1 has 2968 (most) and Y has 231 (least)
How does the repressor protein bring about switching off of Lac Operon
In the presence of lactose, the repressor of the operon is synthesised from the i gene. the repressor protein, binds to the operator region of the operon and prevents RNA polymerases from transcribing the operon and switches off the lac operon.
Codon AUG has dual function. Derive its N2ous bases and anticodons on tRNA
AUG has two main functions. It is a start codon and it codes for methionine.
AUG is transcribed from bases TAC (thymine, adenine and cytosine)in DNA. Its anticodon will be UAC.
Exons and introns
Exons | Introns |
1) They are coding or expressed sequences. | 1) They are non-coding or intervening sequences |
2) They are seen in both prokaryotes and eukaryotes. | 2) They are seen only in eukaryotes. |
3) They appear in mature or processed RNA in eukaryotes. | 3) They do not appear in mature or processed RNA in eukaryotes. |
Cistron
A cistron is defined as a segment of DNA coding for a polypeptide.
The structural gene in a transcription unit is of 2 types:
a) Monocistronic structural gene (mostly in eukaryotes) or
b) Polycistronic structural gene (mostly in bacteria or prokaryotes).
a) Monocistronic structural genes: | ➤b) Polycistronic structural genes: |
➤ It is mostly seen in eukaryotes. | It is mostly seen in bacteria or prokaryotes. |
➤ Most of eukaryotic structural gene codes for single polypeptide. | ➤ ➤ Most prokaryotic structural gene code for more than one polypeptides. |
The monocistronic structural genes have interrupted coding sequences. | |
➤ ➤ The genes in eukaryotes are split i.e. it contains exons which are interrupted by introns which are not expressed. | ➤ The coding region of structural genes is not split. |
➤ The coding sequences or expressed sequences are known as exons. | |
➤ Exons are said to be those sequence that appear in mature or processed RNA. | |
➤ The exons are interrupted by non-coding or intervening sequences or which are known as introns. | |
➤ Introns do not appear in mature or processed RNA. They are spliced out. | ➤ Prokaryotes have only exons. |
Very low expression of lac operon has to be present in cell. Give reason.
In the absence of a preferred carbon source such as glucose, if lactose is provided in the growth medium of bacteria, the lactose is transported into the cell through action of permease. a very low expression of lac operon has to be present in the cell. All the time otherwise the lactose cannot enter the cell.
Which was the last human chromosome to be completely sequenced
Chromosome 1 (may 2006)
There is a paternity dispute for a child'. Which technique can solve the problem? Discuss the principle involved.
i) DNA finger printing technique is used in solving the paternity dispute for a child. It determines the nucleotide sequences uences of certain areas of DNA which are unique to each individual. [1]
(ii) Principle involved :
(a) The basis of DNA finger printing is DNA polymorphism (any difference in the nucleotide sequence between individuals). Although the DNA from different individuals is more alike than different, there are many regions of the human chromosomes that exhibit a great deal of diversity. Such variable sequences are termed polymorphic. [2]
(b) A special type of polymorphism, called VNTR, is composed of repeated copies of a DNA sequence that lie adjacent to one another on the chromosome. Since, polymorphism is the basis of genetic mapping of human.
Based on your understanding of genetic code, explain the formation of any abnormal haemoglobin molecule.
i) Due to point mutation in beta-globin chain of haemoglobin molecule, glutamic acid (Glu) is replaced by valine (Val) at the sixth position.
(ii) Under stress condition, erythrocytes lose their circular shape and become sickle-shaped, As a result, the cells cannot pass through narrow capillaries. Blood capillaries are clogged and thus affect blood supply to different organs.
What is DNA polymorphism? Why is it important to study it?
i) DNA polymorphism states that the variation in DNA arises through mutation at non-coding sequences. VNTR (variable number of tandem repeats) is a special type of polymorphism, which is composed of repeated copies of a DNA sequence that lie adjacent to one another on the chromosome.
(ii) It is important to study because
(a) It is the basis of genetic mapping of human genome and it forms the basis of DNA finger printing too.
(b) The single nucleotide polymorphisms are used in locating diseases and tracing of human history well as in case of paternity testing,
Mention two applications of DNA polymorphism.
Application of DNA polymorphisms is genetic mapping and DNA finger printing.
Give any six features of the human genome.
Human genome is the complete set of genetic information for humans. Salient features of human genome are as follows:
(i) It contains 3,164.7 million nucleotide bases.
(ii) The largest known human gene being dystrophin at 2.4 million bases.
(iii) The total number of protein coding genes is estimated to be 30,000 and 99.9% nucleotide bases are exactly the same in all people.
(iv) The functions are unknown for over 50% of the discovered genes.
(v) Less than 2% of the genome codes for proteins.
(vi) The mutation rate is higher in male meiosis than female meiosis.
(vii) The human genome contains large repeated sequences which are thought to have no direct coding functions but they throw light on structures, dynamics and evolution of chromosomes.
(viii) Chromosome X has most genes (2,968) and the Y has the fewest genes (231).
(ix) Scientists have identified about 1.4 million locations where single-base DNA sequence differences (called SNPs) occur in humans.
Note 
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