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NCERT Solutions · Class 12 Biology Biotechnology : Principles and Processes

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Exercises 9.1–9.12

  1. Exercise 9.1

    Can you list $\displaystyle 10$ recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).

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    A recombinant protein is any protein whose gene has been expressed in a heterologous host — that is the chapter's own definition, and it is the route by which every protein below is made.
    Ten recombinant proteins in medical practice, with their therapeutic use:
    Human insulin — diabetes mellitus.
    Human growth hormone (somatotropin) — pituitary dwarfism, growth failure.
    Erythropoietin (EPO) — anaemia of chronic kidney disease and of cancer therapy.
    Blood clotting Factor VIII — haemophilia A.
    Blood clotting Factor IX — haemophilia B.
    Tissue plasminogen activator (tPA) — dissolving clots in heart attack and stroke.
    Streptokinase — clot buster after myocardial infarction.
    Interferon-alpha — hepatitis B and C, and some cancers.
    Interferon-beta — multiple sclerosis.
    Hepatitis B surface antigen — the hepatitis B vaccine.
    Three more worth knowing: recombinant FSH (infertility treatment), DNase (cystic fibrosis) and asparaginase (leukaemia).
    All of them are produced by the sequence the chapter sets out: clone the gene in a suitable vector, transform a host cell, grow the host in a bioreactor, then recover and purify the protein by downstream processing.

  2. Exercise 9.2

    Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate DNA on which it acts, the site at which it cuts DNA and the product it produces.

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    The chart to make has four columns: Restriction enzyme / Substrate DNA / Site of the cut / Product. Taking the chapter's own example, EcoRI:
    Restriction enzyme: EcoRI, a restriction endonuclease isolated from Escherichia coli RY13. In the name, E is the genus, co the species, R the strain, and the Roman I the order in which it was isolated from that strain.
    Substrate DNA: any double-stranded DNA carrying EcoRI's six-base-pair palindromic recognition sequence — a sequence that reads the same on both strands when both are read in the same direction:
    Top strand: $\displaystyle 5$' - G A A T T C - $\displaystyle 3$'
    Bottom strand: $\displaystyle 3$' - C T T A A G - $\displaystyle 5$'
    Site of the cut: EcoRI cuts the sugar-phosphate backbone of both strands between G and A — a little away from the centre of the palindrome, but between the same two bases on the opposite strands. So the top strand is cut just after its first G, and the bottom strand just before its last G. The cut is therefore staggered, not straight across.
    Product: two DNA fragments, each ending in a single-stranded four-base overhang, $\displaystyle 5$'-AATT, called a sticky end:
    Left fragment: top strand ends $\displaystyle 5$'-...G-$\displaystyle 3$'; bottom strand runs on as $\displaystyle 3$'-...C T T A A-$\displaystyle 5$', leaving AATT unpaired.
    Right fragment: top strand begins $\displaystyle 5$'-A A T T C...-$\displaystyle 3$'; bottom strand begins only at G, again leaving AATT unpaired.
    The ends are called sticky because they form hydrogen bonds with their complementary cut counterparts. Any other DNA cut with the same enzyme carries the same sticky ends, so source DNA and vector DNA can be joined end-to-end by DNA ligase to give recombinant DNA.
    To draw it: rule a four-column table across the page. In column $\displaystyle 2$, draw the DNA as two long horizontal parallel lines with the letters G A A T T C written along the upper line and C T T A A G below, joined by short vertical dashes for the hydrogen bonds; label $\displaystyle 5$' at the left of the upper strand and $\displaystyle 3$' at its right, and the reverse on the lower strand. In column $\displaystyle 3$, redraw the same duplex and mark the cut with a pair of scissors or arrows — one arrow pointing down between the first G and the following A of the upper strand, one pointing up between the last A and G of the lower strand — joined by a stepped (Z-shaped) cut line. In column $\displaystyle 4$, draw the two fragments pulled apart, each showing its projecting single-stranded AATT tail, and label the tails 'sticky ends'. Under the chart add one arrow labelled DNA ligase showing two such sticky ends pairing to form recombinant DNA.

  3. Exercise 9.3

    From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?

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    DNA is very much bigger than an enzyme.
    A gene is only a segment of the DNA, and an enzyme is the product of one such gene — so the whole DNA molecule, which carries hundreds or thousands of genes, must be far larger than any single protein it codes for.
    The chapter says genes lie on long molecules of DNA intertwined with proteins such as histones; a restriction endonuclease, by contrast, recognises a stretch of just six base pairs and 'inspects the length' of the DNA to find it. Only a small molecule can scan along a large one in this way.
    Isolated DNA, precipitated with chilled ethanol, appears as a collection of fine threads that can be spooled out and seen — no enzyme solution ever shows anything to the naked eye.
    DNA must first be cut into fragments by restriction enzymes and then sorted by size on an agarose gel before it can be handled at all; an enzyme needs no such fragmentation.

  4. Exercise 9.4

    What would be the molar concentration of human DNA in a human cell? Consult your teacher.

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    Roughly \(\displaystyle 10^{-10}\) molar — a vanishingly small concentration, because a cell holds only a few dozen DNA molecules.
    Working: a human diploid cell has $\displaystyle 46$ chromosomes, i.e. $\displaystyle 46$ DNA molecules.
    Number of moles \(\displaystyle = 46 \div 6.022 \times 10^{23} \approx 7.6 \times 10^{-23}\) mol.
    Volume of a typical human cell \(\displaystyle \approx 10^{-12}\) litre (about $\displaystyle 1000$ cubic micrometres).
    Molar concentration \(\displaystyle = 7.6 \times 10^{-23} \div 10^{-12} \approx 7.6 \times 10^{-11}\) M, i.e. of the order of \(\displaystyle 10^{-10}\) M.
    The point of the exercise: DNA is present in an extremely low molar concentration yet is enormous in molecular size — which is why it has to be amplified by PCR before it can be worked with.

  5. Exercise 9.5

    Do eukaryotic cells have restriction endonucleases? Justify your answer.

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    No — eukaryotic cells do not have restriction endonucleases.
    These enzymes were discovered in bacteria, as the agents 'responsible for restricting the growth of bacteriophage in Escherichia coli'. They are a bacterial defence against invading phage DNA, and eukaryotes are not attacked by bacteriophages.
    Every one of the more than $\displaystyle 900$ restriction enzymes known has been isolated from over $\displaystyle 230$ strains of bacteria. The naming convention itself takes the genus and species 'of the prokaryotic cell from which they were isolated' — e.g. EcoRI from Escherichia coli RY13.
    A bacterium can afford such an enzyme only because it also carries a partner enzyme that adds methyl groups to its own DNA, protecting it from being cut. The chapter notes that the two enzymes isolated in $\displaystyle 1963$ were exactly this pair. Eukaryotic cells have no such restriction–modification system, so a restriction endonuclease would destroy their own genome.
    Eukaryotes do possess other nucleases — exonucleases and endonucleases used in DNA repair and turnover — but none that recognises a specific palindromic sequence and cuts there.

  6. Exercise 9.6

    Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?

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    Scale. A bioreactor processes $\displaystyle 100$–$\displaystyle 1000$ litres of culture, whereas small-volume flask cultures 'cannot yield appreciable quantities of products'.
    Control of the growth conditions. It provides the optimum temperature, pH, substrate, salts, vitamins and oxygen — a shake flask offers none of these as regulated inputs. Specifically it has a temperature control system and a pH control system.
    Foam control system, which a flask has no way of managing.
    Sampling ports, so small volumes of the culture can be withdrawn periodically to monitor the run without opening or contaminating the vessel.
    Continuous culture. Spent medium can be drained from one side while fresh medium is added from the other, keeping the cells in their physiologically most active log/exponential phase. This produces a larger biomass and hence a higher yield of the desired protein.
    Sterile, contamination-free ambience can be maintained throughout — the requirement the chapter calls bioprocess engineering.

  7. Exercise 9.7

    Collect $\displaystyle 5$ examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.

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    A palindromic DNA sequence is a stretch of base pairs that reads the same on the two strands when both are read in the same orientation ($\displaystyle 5$' to $\displaystyle 3$').
    Rule for building one: take any sequence of even length whose second half is the reverse complement of the first half. Write the complementary strand by the base-pairing rules (A with T, G with C); the two strands will then read alike.
    Five examples, each written $\displaystyle 5$' to $\displaystyle 3$' on the top strand with its complement below:
    GAATTC / CTTAAG — recognised by EcoRI (the chapter's own example).
    GGATCC / CCTAGG — recognised by BamHI.
    AAGCTT / TTCGAA — recognised by HindIII.
    CTGCAG / GACGTC — recognised by PstI.
    GTCGAC / CAGCTG — recognised by SalI.
    Creating your own: pick a triplet, say TGC; its reverse complement is GCA; join them to get TGCGCA. Check it — the complement of TGCGCA is ACGCGT, and read backwards that is TGCGCA again, so both strands read alike. A shorter one made the same way is GATC, and a longer one GTTAAC.

  8. Exercise 9.8

    Can you recall meiosis and indicate at what stage a recombinant DNA is made?

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    Recombinant DNA is made naturally at the pachytene stage of prophase I of meiosis I.
    At pachytene the homologous chromosomes are fully paired as bivalents, and crossing over takes place — an exchange of corresponding segments between non-sister chromatids of a homologous pair.
    The enzyme recombinase catalyses the exchange. The chromatid that results carries DNA from two different chromosomes, i.e. it is recombinant DNA.
    The points of exchange become visible as chiasmata at the next stage, diplotene, which is why the recombination is dated to pachytene rather than to when it can first be seen.
    How this differs from the chapter's recombinant DNA: crossing over joins DNA from two homologous chromosomes of the same species, in vivo, without any restriction enzyme. Recombinant DNA technology joins DNA from different genomes in vitro, using restriction endonucleases to cut and DNA ligase to seal.

  9. Exercise 9.9

    Can you think and answer how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable

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    A reporter enzyme shows, by a change of colour, whether a host cell carries a recombinant DNA — so recombinants can be picked off a single plate by eye.
    The reporter used is beta-galactosidase. The foreign DNA is deliberately inserted within the coding sequence of the beta-galactosidase gene of the vector.
    If an insert is present, the gene is disrupted — insertional inactivation — no functional enzyme is made, the chromogenic substrate in the medium is not cleaved, and the colony stays colourless (white). These are the recombinants.
    If there is no insert, the gene is intact, the enzyme acts on the chromogenic substrate and the colony turns blue. These are the non-recombinants.
    Why this is needed in addition to a selectable marker: a selectable marker such as ampicillin resistance only tells you that the cell has been transformed — it does not tell you whether the plasmid it took up carries the insert.
    Selecting recombinants by inactivation of a second antibiotic resistance gene (e.g. the tetracycline gene of pBR322) does work, but the chapter calls it cumbersome, because it needs simultaneous plating on two plates carrying different antibiotics. The reporter enzyme replaces two plates and two rounds of plating with one plate and a colour difference.

  10. Exercise 9.10

    Describe briefly the following:
    (a)
    Origin of replication
    (b)
    Bioreactors
    (c)
    Downstream processing

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    (a) Origin of replication (ori)
    A specific DNA sequence from which replication starts.
    Any piece of DNA linked to this sequence can be made to replicate within the host cell — so an alien DNA must be joined to an ori before it can be multiplied, i.e. cloned.
    It also controls the copy number of the linked DNA. To recover many copies of a target gene, clone it into a vector whose ori supports a high copy number.
    It is the first of the features every cloning vector must carry.
    (b) Bioreactors
    Vessels in which raw materials are biologically converted into specific products — individual enzymes and other proteins — using microbial, plant, animal or human cells.
    They are needed because small-volume cultures cannot yield appreciable quantities; a bioreactor handles $\displaystyle 100$–$\displaystyle 1000$ litres of culture.
    A bioreactor provides the optimum growth conditions: temperature, pH, substrate, salts, vitamins and oxygen.
    The commonest type is the stirred-tank reactor, cylindrical or with a curved base to help mixing. Its stirrer gives even mixing and oxygen availability throughout; alternatively sterile air is sparged (bubbled) through it.
    Its parts are an agitator system, an oxygen delivery system, a foam control system, a temperature control system, a pH control system, and sampling ports for withdrawing small volumes of culture periodically.
    (c) Downstream processing
    The series of processes a product must go through after the biosynthetic stage and before it is ready for marketing as a finished product.
    It consists chiefly of separation and purification of the product from the culture.
    The product is then formulated with suitable preservatives.
    Such a formulation has to undergo thorough clinical trials, as in the case of drugs.
    Strict quality control testing is required for each product.
    Both the downstream processing and the quality control testing vary from product to product.

  11. Exercise 9.11

    Explain briefly
    (a)
    PCR
    (b)
    Restriction enzymes and DNA
    (c)
    Chitinase

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    (a) PCR
    Polymerase Chain Reaction — a method of synthesising multiple copies of a gene or DNA segment of interest in vitro.
    It needs two sets of primers (small chemically synthesised oligonucleotides complementary to the two regions flanking the DNA of interest), free nucleotides, the genomic DNA as template, and the enzyme DNA polymerase.
    Each cycle has three steps: denaturation (the double-stranded DNA is separated at high temperature), primer annealing, and extension of the primers by DNA polymerase.
    Repeating the cycle many times amplifies the segment about a billion times, i.e. \(\displaystyle 10^{9}\) copies.
    This is possible only because a thermostable DNA polymerase, isolated from the bacterium Thermus aquaticus, is used — it stays active through the high-temperature denaturation step.
    The amplified fragment can then be ligated with a vector for further cloning.
    (b) Restriction enzymes and DNA
    Restriction enzymes are the 'molecular scissors' of genetic engineering; they belong to a larger class of enzymes called nucleases.
    Nucleases are of two kinds: exonucleases, which remove nucleotides from the ends of DNA, and endonucleases, which cut at specific positions within the DNA. Restriction enzymes are restriction endonucleases.
    Each one 'inspects' the length of a DNA, and on finding its own palindromic recognition sequence it binds and cuts the sugar-phosphate backbone of both strands.
    The cut is made a little away from the centre of the palindrome, but between the same two bases on the opposite strands. This staggered cut leaves single-stranded overhangs called sticky ends, which form hydrogen bonds with their complementary counterparts.
    Vector DNA and source DNA cut with the same restriction enzyme therefore carry the same sticky ends, and can be joined end-to-end by DNA ligase to give recombinant DNA. Unless both are cut with the same enzyme, the recombinant vector cannot be created.
    Naming: the first letter is the genus and the next two the species of the bacterium, followed by the strain and a Roman numeral for the order of isolation — EcoRI from Escherichia coli RY13. Hind II was the first to be characterised, recognising a specific sequence of six base pairs; more than $\displaystyle 900$ are now known from over $\displaystyle 230$ bacterial strains.
    (c) Chitinase
    An enzyme used to break open fungal cells so that their DNA is released.
    Fungal cell walls are made of chitin, and chitinase digests it.
    It is used at the very first step of recombinant DNA technology — isolation of the genetic material — alongside lysozyme for bacteria and cellulase for plant cells.
    Once the cell is broken, the DNA comes out mixed with RNA, proteins, polysaccharides and lipids; ribonuclease removes the RNA and protease the proteins, and purified DNA finally precipitates out on adding chilled ethanol, seen as fine threads.

  12. Exercise 9.12

    Discuss with your teacher and find out how to distinguish between
    (a)
    Plasmid DNA and Chromosomal DNA
    (b)
    RNA and DNA
    (c)
    Exonuclease and Endonuclease

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    (a) Plasmid DNA and Chromosomal DNA
    Plasmid DNA: a small, circular, extra-chromosomal DNA that floats freely in the cytoplasm of certain bacterial cells.
    It replicates autonomously, independent of the control of the chromosomal DNA. Some plasmids have only one or two copies per cell, others $\displaystyle 15$–$\displaystyle 100$ or more.
    It carries accessory, dispensable genes — such as antibiotic resistance — and the cell survives without it.
    It can be removed from one cell and reinserted into another, which is why it is used as a cloning vector (pBR322, the Ti plasmid of Agrobacterium tumefaciens).
    Chromosomal DNA: the cell's main genome, very much larger — the 'coding strand of DNA' of the bacterium; in eukaryotes long molecules intertwined with proteins such as histones.
    Its replication is tied to the cell's own division, from its own origin of replication, giving one copy per genome.
    It carries the essential genes; the cell cannot live without it, and it is not used as a vector.
    (b) RNA and DNA
    DNA — deoxyribonucleic acid; sugar is deoxyribose; bases A, T, G, C; usually double stranded; chemically the more stable of the two; it is the genetic material in the majority of organisms.
    RNA — ribonucleic acid; sugar is ribose; uracil replaces thymine; usually single stranded; less stable; in most organisms it works in the expression of genes rather than as the genetic material, though in some viruses it is the genetic material.
    In the laboratory the difference is exploited directly: while isolating DNA, the RNA is destroyed by treatment with ribonuclease and the proteins by protease, so that pure DNA is left to precipitate with chilled ethanol.
    (c) Exonuclease and Endonuclease
    Both belong to the class of enzymes called nucleases, which cut DNA.
    An exonuclease removes nucleotides from the ends of the DNA, chewing inwards from a free end.
    An endonuclease makes cuts at specific positions within the DNA.
    Restriction enzymes are endonucleases — each recognises a specific palindromic sequence, binds there, and cuts both strands a little off the centre of it, leaving sticky ends.
    An exonuclease cannot be used in recombinant DNA technology for this purpose: it recognises no internal sequence and produces no sticky ends.