9.6 Exercise Problems

Q1. Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).

Answer: Recombinant proteins are proteins produced as a result of recombinant DNA technology. In this technology, there is the transfer of some specific gene from one organism to another by using molecular tools such as biological vectors, restriction enzymes etc. Some of the proteins produced through RDT and are being used for therapeutic uses are as follows.

\(
\begin{array}{|c|l|l|}
\hline \text { S.No } & \begin{array}{l}
\text { Name of the recombinant } \\
\text { protein }
\end{array} & \begin{array}{l}
\text { Therapeutic use of the } \\
\text { recombinant protein }
\end{array} \\
\hline 1 . & \text { DNAse I } & \text { To treat cystic fibrosis } \\
\hline 2 . & \text { Antithrombin III } & \begin{array}{l}
\text { To prevent the formation of } \\
\text { the blood clot }
\end{array} \\
\hline 3 . & \text { Insulin } & \begin{array}{l}
\text { To treat type I diabetes } \\
\text { mellitus }
\end{array} \\
\hline 4 . & \text { Interferon } & \text { Used for chronic hepatitis C } \\
\hline 5 . & \text { Interferon AZA } & \text { Used for herpes and virus } \\
\hline 6 . & \text { Coagulation factor VIII } & \text { To treat haemophilia A } \\
\hline 7 . & \text { Coagulation factor IX } & \text { To treat haemophilia B } \\
\hline 8 . & \text { Interferon B } & \text { To treat multiple sclerosis } \\
\hline 9 . & \begin{array}{l}
\text { Human growth hormone } \\
\text { recombinant }
\end{array} & \text { To promote growth humans } \\
\hline 10 . & \begin{array}{l}
\text { Tissue plasminogen } \\
\text { activator }
\end{array} & \begin{array}{l}
\text { To treat the myocardial } \\
\text { infection }
\end{array} \\
\hline
\end{array}
\)

Q2.  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.

Answer: The following chart shows the action of the restriction enzyme EcoRI, the substrate DNA on which it acts and the site where it cuts :

Q3. From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?

Answer: Both DNA (deoxyribonucleic acid) and enzymes are macromolecules. DNA is a polymer of deoxyribonucleotides and enzymes are proteins hence these are polymers of amino acids. But DNA is bigger in molecular size as compared to proteins because synthesis of proteins is regulated by a small segment of DNA, called genes and also a large number of proteins can be synthesised by a DNA molecule.

Q4. What would be the molar concentration of human DNA in a human cell? Consult Your Teacher.

Answer: The molar concentration of DNA in human cells will be the total no. of chromosomes multiplied by 1023 . Hence, the molar concentration DNA in each diploid cell in humans is \(2.77 \times 1023\) moles.

Q5. Do eukaryotic cells have restriction endonucleases? Justify your answer.

Answer: No, eukaryotic cells do not possess restriction enzymes. All the restriction endonucleases have been developed and isolated from different strains of bacteria. The bacteria possess these restriction endonucleases as a defense mechanism to restrict the growth of viruses. Their own DNA remains safe from these enzymes because it is methylated. The eukaryotic cell has RNA interference and institutes defense mechanisms against foreign DNA. Thus, eukaryotic cells do not have restriction endonuclease.

Q6. Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?

Answer: The advantages of stirred tank bioreactors over shake flasks are as follows :

• 1. Stirred tank bioreactors are utilised for large-scale production of biotechnological products, unlike the shake flask method which is used for small-scale production of products.
• 2. In a stirred tank bioreactor, a small sample can be taken out for testing.
• 3. Stirred tank bioreactors have foam breakers to control the foam.
• 4.Stirred tank bioreactors have temperature and pH control systems.

Q7. Collect 5 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following the base-pair rule.

Answer: Palindromic sequences in the DNA molecule refer to groups of bases forming the same sequence when read either backwardly or forwardly. The recognition sites of restriction endonucleases are palindromic sequences. Five examples of palindromic DNA sequences are given below;

• ACTAGT/TGATCA
• AAGCTT/TTCGAA
• GGATCC/CCTAGG
• AGGCCT/TCCGGA
• ACGCGT/TGCGCA

Q8. Can you recall meiosis and indicate at what stage a recombinant DNA is made?

Answer: In meiosis, during the pachytene stage of Prophase I, crossing-over takes place and recombinant DNA is formed by combining portions of male and female DNA.

Q9. Can you think about how a reporter enzyme can be used to monitor transformation of host cells by foreign DNA in addition to a selectable marker?

Answer: In recombinant DNA technology selection of transformed and non transformed cells can be done using reporter genes that encode for reporter enzymes. During the RDT experiment, the foreign gene is joined with a reporter gene. The reporter gene should be such that it produces visible expression. For example, the Lac \(Z\) gene which codes for enzyme beta-galactosidase is used as a reporter gene. The activity of this gene is not found in transformed cells as the product formed by its catalyzation is not formed in transformed cells and bacterial colonies appear white. In non-transformed cells, this gene shows its activity and the catalysed product is formed, as a result of this, bacterial colonies appear blue. Thus, reporter enzymes can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker.

Q10. Describe briefly the following:
(a) Origin of replication (b) Bioreactor (c) Downstream processing

Answer: (a) Origin of replication: This refers to the DNA sequence, from where replication of DNA starts. By linking a DNA sequence with the origin of replication, it can be allowed to replicate in the host cells. Origin of replication also controls the copy number of linked DNA sequences.
(b) Bioreactors: These are large vessels (100-1000 litres) that are used for large-scale production of biotechnological products such as proteins, enzymes etc. from raw materials. In a bioreactor, optimum conditions such as temperature, pH , vitamins, oxygen, salts etc. re m int ined. Stirred bioreactors are the most commonly used bioreactors. Stirred bioreactors can be simple stirred tank bioreactors or sparged tank bioreactors.
(c) Downstream processing- The process of separation and purification of biotechnological products is called downstream processing. The processes in downstream processing vary depending on the quality of the product. Before the release of the product, it undergoes clinical trials and quality control testing.

Q11. Explain briefly :
(a) PCR (b) Restriction enzymes and DNA (c) Chitinase

Answer: (a) Polymerase Chain Reaction (PCR) – The molecular technique to amplify a gene and obtain its several copies is referred to as PCR. The process of PCR has certain requirements i.e. a thermostable enzyme called Taq polymerase ( obtained from Thermus aquaticus ), primers (short stretches of DNA ), dNTPs, a template strand etc. The process of PCR takes place in these steps.

• Denaturation- The double- stranded DNA helix is opened up by breaking their H-bonds at high temperature.
• Annealing- The primers are allowed to hybridise to complementary regions of DNA. This step takes place at 45-55 C temperature.
• Extension- The primers are extended with the help of Taq polymerase enzyme and the cycle is repeated several times to obtain the desired number of copies.

(b) Restriction enzymes and DNA- Restriction enzymes are those enzymes which cut DNA at particular places. Restriction enzyme first scans the DNA template and looks for its recognition site. Once it finds the recognition site, it binds at that region of DNA and cuts each of the two strands in their sugar-phosphate backbone. The sites at which restriction enzymes cut DNA are called as recognition sites of DNA. These are palindromic sequences i.e. they read similarly from the backward and forward direction.

(c) Chitinase – The enzyme that catalyses the breakdown of chitin polysaccharide which is usually found in the cell wall of fungi. Chitinase is mainly used during DNA isolation from fungi.

Q12. 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

Answer: (a) The differences between plasmid DNA and chromosomal DNA are as follows :

\(
\begin{array}{|l|l|}
\hline \text { Plasmid DNA } & \text { Chromosomal DNA } \\
\hline \begin{array}{l}
\text { Circular, extra-chromosomal DNA which is } \\
\text { capable of self-replication and is found in } \\
\text { bacteria is called plasmid DNA. }
\end{array} & \begin{array}{l}
\text { The entire DNA (excluding } \\
\text { extrachromosomal DNA) present in the cell } \\
\text { constitutes chromosomal DNA }
\end{array} \\
\hline \text { It is found only in bacteria } & \begin{array}{l}
\text { It is found in both bacteria and other } \\
\text { eukaryotic cells. }
\end{array} \\
\hline
\end{array}
\)

(b) The differences between RNA and DNA are as follows :

\(
\begin{array}{|l|l|}
\hline \text { RNA } & \text { DNA } \\
\hline \begin{array}{l}
\text { RNA is a single stranded } \\
\text { molecule. }
\end{array} & \begin{array}{l}
\text { DNA is a double stranded } \\
\text { molecule. }
\end{array} \\
\hline \text { It contains ribose sugar. } & \text { It contains deoxyribose sugar. } \\
\hline \begin{array}{l}
\text { The pyrimidines in RNA are } \\
\text { adenine and uracil. }
\end{array} & \begin{array}{l}
\text { The pyrimidines in DNA are } \\
\text { adenine and thymine. }
\end{array} \\
\hline \text { RNA cannot replicate itself. } & \begin{array}{l}
\text { DNA molecules have the ability to } \\
\text { replicate. }
\end{array} \\
\hline \begin{array}{l}
\text { It is a component of the } \\
\text { ribosomes. }
\end{array} & \begin{array}{l}
\text { It is a component of the } \\
\text { chromosomes. }
\end{array} \\
\hline
\end{array}
\)

(c) The differences between exonuclease and endonuclease are as follows :

\(
\begin{array}{|l|l|}
\hline \text { Exonucleases } & \text { Endonucleases } \\
\hline \text { Remove nucleotides from the outer ends of the DNA} & \begin{array}{l}
\text { Make cuts at specific positions } \\
\text { within the DNA. } 
\end{array} \\
\hline
\end{array}
\)

Exemplar Section

VERY SHORT ANSWER TYPE QUESTIONS

Q1. How is copy number of the plasmid vector related to yield of recombinant protein?

Answer:  Higher the copy number of vector plasmid, higher the copy number of gene and consequently, protein coded by the gene is produced in high amount.

Q2. Would you choose an exonuclease while producing a recombinant DNA molecule?

Answer: No, as exonuclease acts on the free ends of linear DNA molecule. Therefore, instead of producing DNA fragments with sticky ends, it will shorten or completely degrade the DNA fragment containing the gene of interest, and the circular plasmid (vector) will not get cut as it lacks free ends.

Q3. What does H in’ ‘d’ and ‘III’ refer to in the enzyme Hind III?

Answer: In the enzyme Hind III, ‘H in’ refers to Haemophilus influenza, d refers to the strain of H. influenza, and III refers to the sequence in which this enzyme was discovered.

Q4. Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.

Answer: If the restriction enzymes have more than one recognition site in a vector, than the vector itself will get fragmented on treatment with the restriction enzyme.

Q5. What does ‘competent’ refer to in competent cells used in transformation experiments?

Answer: Competent means bacterial cells, on treatment with \(\mathrm{CaCl}_2\), are made capable of taking up foreign DNA.

Q6. What is the significance of adding proteases at the time of isolation of genetic material (DNA).

Answer: Role of proteases is to degrade the proteins present inside a cell (from which DNA is being isolated). If the proteins are not removed from DNA preparation then they could interfere with any downstream treatment of DNA (such action of restriction endonuclease, DNA ligase etc.).

Q7. While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process?

Answer: If denaturation of double-stranded DNA does not take place, then primers will not be able to anneal to the template, no extension will take place, hence no amplification will occur.

Q8. Name a recombinant vaccine that is currently being used in vaccination program.

Answer: Hepatitis B- vaccine is a recombinant vaccine that is currently in use in vaccination program.

Q9. Do biomolecules (DNA, protein) exhibit biological activity in anhydrous conditions?

Answer: No, biomolecules like DNA and protein cannot exhibit biological activity in anhydrous conditions. Hence, life is unsustainable without water.

Q10. What modification is done on the Ti plasmid of Agrobacterium tumefaciens to convert it into a cloning vector?

Answer: The tumor inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified (disarmed) into a cloning vector which is no more pathogenic to the plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

SHORT ANSWER TYPE QUESTIONS

Q1. What is meant by gene cloning?

Answer: Gene cloning refers to a process in which a gene of interest is ligated to a vector. The recombinant DNA thus produced is introduced in a host cell by transformation. Each cell gets one DNA molecule and when the transformed cell grows to a bacterial colony, each cell in the colony has a copy of the gene. This is precisely gene cloning.

Q2. Both a wine maker and a molecular biologist who had developed a recombinant vaccine claim to be biotechnologists. Who in your opinion is correct?

Answer: Both. As biotechnology is a very wide area which deals with techniques of using a ‘natural’ organism (or its parts) as well as genetically modified organism to produce products and processes useful for mankind. A wine maker employs a strain-of yeast to produce wine by fermentation (a natural phenomenon), while the molecular biologist has cloned gene for the antigen (that is used as vaccine) in an organism which allows the production of the antigen in large amount. 

Q3. A recombinant DNA molecule was created by ligating a gene to a plasmid vector. By mistake, an exonuclease was added to the tube containing the recombinant DNA. How does this affect the next step in the experiment i.e. bacterial transformation?

Answer: The experiment will not likely to be affected as recombinant DNA molecule is circular closed, with no free ends. Hence, it will not be a substrate for exonuclease enzyme which removes nucleotides from the free ends of DNA.

Q4. Restriction enzymes that are used in the construction of recombinant DNA are endonucleases which cut the DNA at ‘specific-recognition sequence’. What would be the disadvantage if they do not cut the DNA at specific-recognition sequence?

Answer: If the restriction enzymes would cut DNA at random sites instead of at specific sites, then the DNA fragments obtained will not have ‘sticky ends’. In the absence of sticky ends, construction of recombinant DNA molecule would not be possible.

Q5. A plasmid DNA and a linear DNA (both are of the same size) have one site for a restriction endonuclease. When cut and separated on agarose gel electrophoresis, plasmid shows one DNA band while linear DNA shows two fragments. Explain.

Answer: It is because plasmid is a circular DNA molecule. When cut with enzyme, it becomes linear but does not get fragmented. Whereas, a linear DNA molecule gets cut into two fragments. Hence, a single DNA band is observed for plasmid while two DNA bands are observed for linear DNA in agarose gel.

Q6. How does one visualise DNA on an agarose gel?

Answer: A compound called Ethidium Bromide stains DNA, which on irradiating with Ultraviolet, fluoresce gives orange light. Hence, DNA fragments appear as orange band in the presence of Ethidium Bromide and UV.

Q7. A plasmid without a selectable marker was chosen as vector for cloning a gene. How does this affect the experiment?

Answer: In a gene, cloning experiment, first a recombinant DNA molecule is constructed, where the gene of interest is ligated to the vector (the step would not be affected) and introduced inside the host cell (transformation). Since, not all the cells get transformed with the recombinant/plasmid DNA, in the absence of selectable marker, it will be difficult to distinguish between transformants and non-transformant, because role of selectable marker is in the selection of transformants.

Q8. A mixture of fragmented DNA was electrophoresed in an agarose gel. After staining the gel with ethidium bromide, no DNA bands were observed. What could be the reason?

Answer: The reasons are as follows:
(i) DNA sample that was loaded on the gel may have got contaminated with nuclease (exo-or endo-or both) and completely degraded.
(ii) Electrodes were put in opposite orientation in the gel assembly that is anode towards the wells (where DNA sample is loaded). Since DNA molecules are-negatively charged, they move towards anode and hence move out of the gel instead of moving into the matrix of gel.
(iii) Ethidium bromide was not added at all or was not added in sufficient concentration and so DNA was not visible.

Q9. Describe the role of \(\mathrm{CaCl}_2\) in the preparation of competent cells?

Answer: \(\mathrm{CaCl}_2\) is known to increase the efficiency of DNA uptake to produce transformed bacterial cells. The divalent \(\mathrm{Ca}^{2+}\) ions supposedly create transient pores on the bacterial cell wall by which the entry of foreign DNA is facilitated into the bacterial cells.

Q10. What would happen when one grows a recombinant bacterium in a bioreactor but forget to add antibiotic to the medium in which the recombinant is growing?

Answer:  In the absence of antibiotic, there will be no pressure on recombinants to retain the plasmid (containing the gene of your interest). Since, maintaining a high copy number of plasmids is a metabolic burden to the microbial cells, will thus tend to lose the plasmid.

Q11. Identify and explain steps ‘A’, ‘B’ and ‘C’ in the PCR diagram given below. 

Answer: A – Denaturation
B – Annealing
C – Extension of primers

Q12. Name the regions marked A, B and C. 

Answer: A-Bam HI, B-Pst I , C -Ampicillin resistance gene \(\left(\mathrm{a m p}^R\right)\)

LONG ANSWER TYPE QUESTIONS

Q1. For selection of recombinants, insertional inactivation of antibiotic marker has been superceded by insertional inactivation of a marker gene coding for a chormogenic substrate. Give reasons. 

Answer: Selection of recombinants due to inactivation of antibiotics is a laborious process as it requires:
(i) a vector with two antibiotic resistance marker
(ii) preparation of two kinds of media plate, with one antibiotic each.
Transformed cells are first plated on that antibiotic plate which has not been insertionally inactivated (ampicillin) and incubated overnight for growth of transformants. For selection of recombinants, these transformants are Replica plated on second antibiotic (tetracycline) plate (which got inactivated due to insertion of gene). Non-recombinants grow on both the plates (one carrying ampicillin and the other carrying tetracycline) while recombinants will grow only on ampicillin plate. This entire exercise is laborious and takes more time (two overnight incubation) as well. However, if we choose the second option (insertional inactivation of a marker that produces colour in the presence of a chromogenic compound), we can distinguish between the recombinants and non-recombinants on a single medium plate (containing one antibiotic and the chromogenic compound) after overnight growth. Hence I would choose a marker which produces a coloured compound but gets inactivated due to insertion of foreign DNA.

Q2. Describe the role of Agrobacterium tumefaciens in transforming a plant cell.

Answer: Agrobacterium tumafaciens harbours a mega plasmid called Ti-plasmid. This has a T-DNA region flanked by left border and right border sequence. The T-DNA gets transferred and integrates with the host plant DNA. This property of Ti-plasmid has been exploited for cloning of gene of interest and stably integrating them in the plant genese. Therefore, by using Ti-plasmid or its derivatives, recombinant plant cells with desired genes of interest stably integrated in the plant genome has been successfully produced.

Q3. Illustrate the design of a bioreactor. Highlight the difference between a flask in your laboratory and a bioreactor which allows cells to grow in a continuous culture system.

Answer: Small volume cultures cannot yield appreciable quantities of products. To produce in large quantities, the development of bioreactors, where large volumes (100-1000 litres) of culture can be processed, was required. Thus, bioreactors can be thought of as vessels in which raw materials are biologically converted into specific products, individual enzymes, etc., using microbial plant, animal or human cells. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen).

Flask	Bioreactor
Flask is used to small laboratory scale testing of a culture.	Bioreactor is used for commercial scale production.
The cells harbouring cloned genes of interest may be grown on a small scale in the laboratory.	The cells can also be multiplied in a continuous culture system wherein the used medium is drained out from one side while fresh medium is added from the other to maintain the cells in their physiologically most active log/exponential phase. This type of culturing method produces a larger biomass leading to higher yields of desired protein.
Small volume cultures cannot yield appreciable quantities of products.	To produce in large quantities, the development of bioreactors, where large volumes ( 100-1000 litres) of culture can be processed, was required.