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Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin
(b) toxin is immature;
(c) toxin is inactive;
(d) bacteria encloses toxin in a special sac.
Answer:
(c) Bacillus thuringiensis forms protein crystals during a particular phase of their growth. These crystals contain a toxic insecticidal protein. Actually, the Bt toxin protein exists as inactive protoxins but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline pH of the insect gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis and eventually cause the death of the insect, but it does not harm Bacillus itself.
Question 2.
What are transgenic bacteria? Illustrate using any one example. (CBSE – 2006)
Answer:
Bacteria carrying foreign genes are called transgenic bacteria. For example, two DNA sequences (A and B chains of human insulin) were introduced into the plasmid of bacteria E.coli. The leans genic bacteria start producing insulin chains.
Question 3.
Compare and contrast the advantages and disadvantages of the production of genetically modified crops.
Answer:
Advantages of GM crops:
Disadvantages of GM crops:
Question 4.
What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit?
Answer:
Cry proteins are protein responsible for killing lepidopteran insect and their larvae (also called Bt toxin). It is secreted by Bacillus thuringienesis. Man exploited gene encoding this toxin, by transferring it into cotton genome with the help of Agrobacterium TDN A as vector.
Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency,
Answer:
It is a collection of methods which allows correction of a gene defect that has been diagnosed in a child or embryo. In gene therapy, normal genes are inserted into a person’s cells or tissues to treat a hereditary defect. Gene therapy is being tried for sickle cell anaemia and Severe Combined Immuno Deficiency (SCID).
In some children, ADA deficiency can be cured by bone marrow transplantation. In others, it can be treated by enzyme replacement therapy, in which functional ADA is given to the patient by injection. However, both of these approaches are not completely curative.
In gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retroviral vector) is then introduced into these lymphocytes, which are subsequently returned to the patient. Because these cells are not immortal, the patient requires a periodic infusion of such genetically engineered lymphocytes. However, if the gene isolated from marrow cells producing ADA is introduced into cells at early embryonic stages, the disease could be cured permanently.
Question 6.
Diagrammatically represent the experimental steps in cloning and expressing a human gene (say the gene for growth hormone) into a bacterium like E. coli?
Answer:
The given diagram represents the experimental steps in cloning and expressing a human gene for growth hormone into a bacterium E. coli.
Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and the chemistry of oil?
Answer:
To remove oil from seeds using recombinant DNA technology would involve:
It will not be very easy because the oils are made up of fatty acids and glycerol. Since fatty acids are important components of the cell membrane system, deleting or switching off of its genes might affect the cell structure itself.
Question 8.
Find out from the internet what is golden rice.
Answer:
Golden rice is transgenic rice having gene coding for vitamin A synthesis enzyme. Golden rice was developed by the Swiss Federal Institute of Technology, rich in vitamin A (beta carotene). The rice grains are golden yellow in colour due to colour it gets from the beta carotene.
Question 9.
Does our blood have proteases and nucleases?
Answer:
No, our blood does not contain enzyme proteases and nucleases. If these two enzymes were there in the blood, it causes the degeneration of blood cells and lining cells of blood cells.
Question 10.
Consult the internet and find out how to make orally active protein pharmaceuticals. What is the major problem to be encountered?
Answer:
Proteinaceous drugs cannot be taken orally because they can be degraded by the proteases of our alimentary canal. To counteract this problem or to make an orally active protein pharmaceutical, it must be coated by a film that is resistant to protein degrading enzymes.
2nd PUC Biology Biotechnology and its Applications One Mark Questions
Question 1.
Name the organism whose genetic material has been altered using genetic engineering techniques.
Answer:
Genetically Modified Organisms (GMO).
Question 2.
Genetically modified cotton is named Bt cotton. What does the prefix ‘Bt’ mean?
Answer:
Cotton containing a toxin gene from Bacillus thuringiensis
Question 3.
Through whom Bt toxin protein originates?
Answer:
By Bacillus thuringiensis.
Question 4.
Biopiracy affects developing countries like India more than industrialized nations because our country is rich in ………………. and …………………. related to bio-resources.
(Bioetics / Biodiversity/Biopatent, Traditional resources/Traditional cultivation/Traditional knowledge)
Answer:
Biodiversity and Traditional knowledge
Question 5.
Full form of RNAi.
Answer:
RNA interference (RNAi).
Question 6.
What GEAC?
Answer:
Genetic engineering Approval committee.
Question 7.
Name the scientific name of bacteria in which be form organism toxin.
Answer:
Bacillus thuringiensis.
Question 8.
A multinational company outside India tried to sell new varieties of turmeric without proper patent rights. What is such an act referred to as?
Answer:
Biopiracy
Question 9.
What is Hirudin?
Answer:
It is a protein which prevent blood dotting.
2nd PUC Biology Biotechnology and its Applications Two Marks Questions
Question 1.
In case of Bt Cotton how does the toxic insecticide protein produced by the bacterium kill the insect pest but not the cells of Bacillus thuringiensis where the toxic protein is generated?
Answer:
The toxin produced by Bacillus thuringiensis is an endotoxin called cry protein. It is crystalline and nontoxic when formed being in the protoxin stage. As it reaches the gut of insects, the cry protein is converted into toxic and soluble state. It attaches the receptors present on the epithelial cells of the gut produces pores and kills the cells resulting in the death of the insects.
Question 2.
What are the 4 main objectives of genetically modified crop plants? (CBSE 2008)
Answer:
Question 3.
Define transgenic organisms (CBSE 2006)
Answer:
They are organisms which have been modified genetically through the introduction of genes of another organism artificially by the technique of genetic engineering instead of conventional hybridisation.
Question 4.
How is early detection of diseases possible using molecular diagnostics?
Answer:
Question 5.
Name any two biological products that are produced in transgenic animals and mention their uses.
Answer:
Question 6.
Bacillus thuringiensis makes our environment pesticide-free. Comment.
Answer:
Bacillus thuringiensis produces a toxin called Bt toxin. Gene for Bt toxin has been cloned from bacteria and been expressed in plants to provide resistance to pests. In this manner, many plants are produced. So in the future, this makes our environment pesticide-free.
Question 7.
What is the cause of adenosine deaminase deficiency in a person?
Answer:
It is due to the deletion of the gene coding for the enzyme adenosine deaminase, this enzyme is crucial for the functioning of the immune system.
Question 8.
Briefly describe the RNA interference process for preventing nematode infestation of plants.
Answer:
RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defense. This method involves the silencing of a specific mRNA due to a complementary double-stranded RNA molecule that binds to it and prevents translation of mRNA. The source of this complementary RNA could be from injection by viruses having RNA genomes or mobile genetic elements.
Question 9.
Why does Bt toxin not kill the bacillus? How does it kill the insect larvae?
Answer:
Question 10.
Write the advantages of recombinant therapeutics?
Answer:
The recombinant therapeutics do not induce any unwanted immunological response like the similar products of non-human origin such therapeutics are highly effective.
Question 11.
How is ELISA used to detect pathogens in the body?
Answer:
2nd PUC Biology Biotechnology and its Applications Three Marks Questions
Question 1.
What is genetically modified (GM) food? Give two examples.
Answer:
Genetically modified food (GM food): The food substances produced from genetically modified crops or transgenic crops is called GM food. This food differs from conventionally developed varieties in the following aspects :
Examples of GM Crops, Food, and Fruits:
1. Flavr Savr Tomato: It is the first food containing genetically engineered DNA. . These tomatoes contain genes for antibiotic resistance for kanamycin.
2. Maize: GM maize has a bacterial gene that increases its resistance to pests and
diseases. It also has a gene for ampicillin resistance which is harmful for us, therefore the introduction of GM maize is opposed by many European countries. ,
3. Rape oilseed: It is a new type of plant that contains genes for resistance to the herbicide Basta. It has more potential, dangers and can become a weed and would be impossible to control with Basta. It could cross-fertilize with relatives such as wild mustard, thus, spreading the resistance to wild plants. Such type of environmental risks could occur with genetically modified rapeseed crop. They might also affect food chains in unpredictable ways.
Question 2.
Mention three reasons for the success of the green revolution in India.
Answer:
Question 3.
Enumerate the fields of application of biotechnology.
Answer:
The applications of biotechnology include
Question 4.
Explain the steps involved in the production of genetically engineered insulin.
Answer:
Insulin used for diabetes was earlier extracted from the pancreas of slaughtered cattle and pigs. Insulin consists of two short polypeptide chains, chain A and chain B that are linked together by disulphide bridge.
In mammals, including humans, insulin is synthesised as a prohormone which contains an extra stretch called the C peptide. This C peptide is not present in the mature insulin and is removed during maturation into insulin. The main challenge for the production of insulin using rDNA techniques was getting insulin assembled into a mature form.
In 1993 Eily Lilly an American company prepared two DNA sequences corresponding to A and B, chain of human insulin and introduced them in the plasmid of E.coli to produce insulin chains, Chains A and B were produced separately, extracted, and combined by creating disulphide bonds to form human insulin.
Question 5.
Name any 6 plants where Bt toxin-producing genes have been included.
Answer:
2nd PUC Biology Biotechnology and its Applications Five Marks Questions
Question 1.
Two of the steps involved in producing nematode-resistant tobacco plants base on the process of RNA I are mentioned below. Write the missing steps in the proper sequence.
Answer:
Question 2.
Describe the application of genetic engineering in the field of Agriculture and Medicine.
Answer:
(A) Application of Genetic engineering or Biotechnology in Agriculture: Genetic engineering is found to be very beneficial in agriculture. Its important use in agriculture are:
1. Increase in photosynthetic efficiency: An increase in photosynthetic efficiency of crop plants can be achieved by introducing suitable Carbon dioxide Fixation Gene (cfx) from any plant into the crop plants.
2. Transfer of nitrogen-fixing ability: Number of symbiotic and non-symbiotic micro-organisms have a capacity of fixing atmospheric nitrogen. Nitrogen fixers are found to possess nitrogen-fixing gene (nif genes) which are located on chromosomes or plasmids. Introduction of nif gene in crop plants results inability in crop plants to fix atmospheric nitrogen and reduction in the use of chemical nitrogen fertilizers.
3. Disease resistance in crop plants: Plant breeders at present are developing high-yield varieties by transferring genes for disease resistance through conventional breeding.
4. Plant tissue in crop improvement: Some of the areas of plant improvement where tissue culture has been applied with success are as follows :
5. VAM (Vesicular-Arbuscular Mycorrhiza) fungi with Rhizobium can boost the yields: Recently there has been a new dimension to this farm practice by the way of increasing Rhizobium inoculation effect by simultaneous inoculating seeds with VAM as well as Rhizobium culture. VAM is a structural modification of hyphae helping in absorption and storage of phosphorus.
(B) Application of Genetic engineering in the Medical field:
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Question 1.
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 used in medical practice as therapeutics are as follows:
Question 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.
Answer:
Question 3.
From what you have learned, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
Human cells vary in size and volume. DNA is fixed in its size. That makes it kind of hard (like impossible) to state a concentration for DNA. Only a range can be given, and that information has little to no practical value and even less meaning.
Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
Molar concentration is the ratio of the number of moles of solute in a solution divided by the volume of the solution expressed in liters.
The average weight of a DNA basepair = 650 daltons (1 dalton equals the mass of a single hydrogen atom or 1.67 x 10-24 grams)
The molecular weight of a double-stranded
DNA molecule = Total no. of basepairs x 650 daltons
The human genome is 3.3 x 109 bp in length.
Hence, the weight of human genome,
= 3.3 x 109 bp x 650 Da
= 3.59 x 10-12 gm.
The molar concentration of DNA can be calculated accordingly.
Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
The stirred tank bioreactor facilitates the mixing and oxygen availability. It controls the temperature and pH inside the bioreactor.
Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors to have over shake flasks?
Answer:
Shake flasks are used for growing and mixing the desired materials on a small scale in the laboratory. Bioreactors are vessels in which raw materials are biologically converted into specific products by microbes, plant and animal cells, and their enzymes. Bioreactors are used for large-scale production of biomass or sell products under aseptic conditions. Here large volumes (100-1000 litres) of culture can be processed. A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions (temperature, pH, substrate, salts, vitamins, oxygen). The most commonly used bioreactors are of stirring type.
A bioreactor is more advantageous than shake flasks. It has an agitator system to mix the contents properly, an oxygen delivery system to make available of oxygen, a foam control system, a temperature control system, a pH control system, and a sampling port to withdraw the small volumes of the culture periodically.
Question 7.
Collect 4 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:
Question 8.
Can you recall meiosis and indicate at what stage recombinant DNA is made?
Answer:
Meiosis I – Pachytene – When recombination nodule appears after synaptonemal complex formation.
Question 9.
Can you think and answer how a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker? Ans: A selectable marker helps to identify transformed host cells and non transformed cells will be eliminated and only transformed cells will grow. Whereas reporter gene is the one whose phenotypic expression can be monitored and thus it reports about activity or change in advance of the effect of the modification, in addition to eliminating non transformed cells by selectable markers. HLDescribe briefly the followings:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Answer:
(a) Origin of replication (on): One of the major components of a plasmid is a sequence of bases where replication starts. It is called the origin of replication (on). This is a specific portion of the plasmid genome that serves as a start signal for self-replication (to make another copy of itself). Any piece of DNA when linked to this sequence can be made to replicate within the host cells. This property is used to make a number of copies of linked DNA (or DNA insert).
(b) Bioreactors: These are vessels in which raw materials are biologically converted into specific products by microbes, plant and animal cells, and their enzymes. They are allowed to synthesise the desired proteins which are finally extracted and purified from cultures. Small volume cultures are usually employed in laboratories for research and production of fewer quantities of products. However, large-scale production of the products is carried out in bioreactors’. The most commonly used bioreactors is a stirring type bioreactor (fermenter) that has a provision for batch culture or continuous culture.
(c) Downstream processing: After the formation of the product in the bioreactors, it undergoes some processes before a finished product is ready for marketing. These processes include separation and purification of products which are collectively called downstream processing. The product is then subjected to quality control testing and kept in suitable preservatives. The downstream process and quality control test are different for different products
Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Answer:
a. The in vitro amplification of DNA by repeated cycles of strand separation and polymerisation is PCR.
b. The nuclease enzyme that cuts the DNA at a unique sequence is called restriction endonuclease. They are also known as molecular knives, molecular scissors, or molecular scalpels.
c. Chitinase is digestive enzymes that break down glycosidic bones in chi tin.
Question 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
Answer:
(a) Plasmid DNA and Chromosomal DNA
| Plasmid DNA | Chromosomal DNA |
| (i) It is extranuclear DNA | It is nuclear DNA |
| (ii) It caries nonvital genes | It possesses vital genes |
| (iii) A bacterial cell may carry one to several plasmid DNAs. | A bacterial cell carries only one chromosome DNA |
(b) RNA and DNA
| RNA | DNA |
| (i) It is ribonucleic acid | Deqxy ribonucleic acid |
| (ii) It is single strandred | Double standard |
| (iii) RNA produced from DNA template | Parental DNA acts as a DNA template |
| (iv) Nitrogen base Uracil is present instead of thymine, which pairs with adenine. | Thymine pairs with adenine |
(c) Exonuclease and Endonuclease
| Exonuclease | Endonuclease |
| (i) It breaks DNA from ends | It cuts DNA from inside |
| (ii) The separated fragments are small nucleotides | These separated fragments are generally large sized |
| (iii) The separated fragments cannot be used in genetic engineering | The desirable separated fragments are used in genetic engineering |
2nd PUC Biology Biotechnology: Principles and Processes One Mark Question
Question 1.
By observing the given pair fill-up the blanks
Answer:
Question. 2.
Which DNA polymerase is active in high temperatures?
Answer:
Taq DNA polymerase is active at high temperatures.
Question 3.
Restriction endonucleases are used to cut DNA at specific sites.. Name the first endonuclease isolated from Escherichia coli.
Answer:
EcoRI
Question. 4.
Write the full form of PCR. Which enzyme is used in?
Answer:
Polymerase Chain Reaction (PCR).
Taq DNA is used in PCR.
Question 5.
Name the technique used for separating DNA fragments in the laboratory
Answer:
Electrophoresis. (Dehli 2005)
Question. 6.
First recombinant DNA was formed in.
Answer:
In Bacteria Salmonella Typhimurium.
Question 7.
Write the function of the restriction enzyme in a bacterial cell?
Answer:
Responsible for restricting the growth of bacteriophage.
Question. 8.
Where does Hind II cut the DNA molecule?
Answer:
It cuts the DNA molecule at a specific 6 base pair seQuestionuence.
Question 9.
Why are plasmids and bacteriophages commonly used as cloning vectors?
Answer:
Plasmids and bacteriophages have the ability to replicate within bacterial cells independently of the chromosomal DNA.
Question. 10.
Which type of charge found in DNA?
Answer:
Negative charge.
Question 11.
What is downstream processing?
Answer:
Downstream processing is the recovery of product from the fully grown genetically modified cells, its purification and preservation
Question. 12.
What is electrophoresis?
Answer:
Electrophoresis is a techniQuestionue used in laboratories in order to separate macromolecules based on size.
Question 13.
What is gene therapy?
Answer:
It is the replacement of a defective gene by normal healthy and functional gene. This method helps to overcome the effect of various disorders like sickle cell anaemia, alkaptonuria, SCID, colour blindness etc.
Question. 14.
Name the technique in which we should be isolated the DNA segment.
Answer:
Electrophoresis.
Question 15.
What is amplification?
Answer:
It is the process of making multiple copies of gene/DNA segments of interest.
Question 16.
What is recombinant protein?
Answer:
It is a biochemical compound or useful protein produced inside the heterologous host cell by recombinant biotechnology method.
Question 17.
What is a bioreactor or fermenter?
Answer:
It is a container in which the biochemical process is carried out by using living cells and their growth medium.
Question 18.
What meant by bioconversion?
Answer:
It is the process by which raw materials are biologically converted into specific products using microbes, plant or animal cells and or their enzymes.
Question 19.
Why are antibiotic resistance genes used as selectable markers for E.Coli?
Answer:
Since E.Coli doesn’t have any of antibiotic resistance genes, antibiotic resistance genes are used from outside as selectable marker.
2nd PUC Biology Biotechnology: Principles and Processes Two Marks Questions
Question 1.
Name the scientists who constructed recombinant DNA. Name the bacterium from which they isolated the gene.
Answer:
Stanley Cohen and Herbert Boyer were the first to construct recombinant DNA. They isolated the gene from the bacterium. Salmonella typhimurium.
Question 2.
Few gaps have been left in the following table showing certain terms and their meanings, fill up the gaps.
Term Meanings
(i) ………….. Non-coding sequence in eukaryotic DNA
(ii) ………….. The technique used in solving paternity disputes
(iii) Restriction endonuclease …………..
(iv) Plasmids …………..
(v) Transgenics …………..
(vi) Nucleotide sequences with single base deficiencies …………….
Question 3.
Name the particular technique in biotechnology whose steps are shown in the figure use the figure to summarize the technique in three steps.
Answer:
Question 3.
Name the particular technique in biotechnology whose steps are shown in the figure use the figure to summarize the technique in three steps.
Answer:
(a) Recombinant technology
(b)
Question 4.
Refer to the diagram and answer the following:
(i) From what T1– plasmid is obtained?
(ii) Name the enzyme which is involved in step I
(iii) What happens in step II
(iv) The plant produced is called hybrid or transgenic
(v) Will the plant produced have other genes along with desired genes? Yes or No explain.
Answer:
(i) Agrobacterium tumefaciens
(ii) Restriction endonuclease
(iii) Incorporation of genes in T1 plasmid in the region of T-DNA.
(iv) Transgenic
(v) Yes. Selectable marker gene which is often an antibiotic resistance gene
Question 5.
Study the linking of DNA fragments shown above
(i) Name “a” DNA and “b” DNA
(ii) Name the restriction enzymes that recognize this palindrome
(iii) Name the enzyme that can link these two DNA fragments (CBSE 2008)
Answer:
(i) (a) – vector DNA
(b) – foreign DNA
(ii) EcoRI
(iii) DNA ligase
Question 6.
Explain the importance of
(a) Ori
(b) amp R and
(c) rop in E. Coli vector shown below (CBSE 2008)
Answer:
Question 7.
An interesting property of restriction enzymes is molecular cutting and pasting Restriction enzymes typically recognize a symmetrical
Notice that the top strand is the same as the bottom strand but reads backward. When the enzyme cut the strand between G and A, it leaves overhanging chains
(A) What is the symmetrical sequence of DNA known as?
(B) What is the significance of these overhanging chains?
(C) Name the restriction enzyme that cuts the strand between G and A.
Answer:
(A) Palindromic sequence
(B) sticky ends
(C) Eco RI
Question. 8.
What is DNA ligase?
Answer:
DNA ligase is a specific type of enzyme, a ligase that facilitates the joining of DNA strands together by catalyzing the formation of a phosphodiester bond.
Question 9.
What are cloning vectors? What functions do these vectors perform?
Answer:
Cloning vectors are those organisms on their DNAs which can multiply independently of the host DNA and increase their copy number along with the alien DNA attached to them. Functions are:
Question. 10.
What is Ti-plasmid?
Answer:
Ti or tumor-inducing plasmid is a plasmid that is a part of the genetic equipment that Agrobacterium tumefacient use to transduce their genetic material to plants.
Question 11.
Palindromic nucleotide sequences have significance in recombinant DNA technology. Explain. Give example for a palindromic DNA sequence.
Answer:
Palindrome in a DNA is a sequence of base pairs that reads same on the two strands when orientation of reading is kept the same. Each restriction endonuclease recognises a specific palindromic nucleotide sequence and cuts the strand of DNA a little away from the centre of the palindrome site, but between the two bases on the opposite strands.
Example of palindrome DNA is
5′ – GAATTC – 3′
3′ – CTTAAG – 5′
Question. 12.
Name the scientist who discovered the artificial DNA synthesizing method.
Answer:
The Nobel prize in physiology in 1968 was awarded jointly to Robert W. Holley, Hargobind Khorana, and M. Nirenberg for their interpretation of the genetic code.
2nd PUC Biology Biotechnology: Principles and Processes Three Marks Question
Question 1.
Given below are the different steps in recombinant DNA technology. Arrange them according to the sequence of occurrence.
a. Transferring the recombinant DNA into the host.
b. Extraction of the product.
c. Fragmentation of DNA by restriction endonucleases
d. Ligation of DNA fragment into a vector.
e. Isolation of DNA.
f. Culturing the host ceils in a medium at large scale.
g. Isolation of the desired DNA fragment.
Answer:
a. Isolation of DNA.
b. Fragmentation of DNA by restriction endonucleases
c. Isolation of the desired DNA fragment.
d. Ligation of DNA fragment into a vector.
e. Transferring the recombinant DNA into the host.
f. Culturing the host cells in a medium at large scale.
g. Extraction of the product.
Question 2.
Represent diagrammatically the E Coli cloning vector pBR 322 showing the restriction site.
Answer:
Question 3.
Draw a labelled diagram of a sparged stirred tank bioreactor.
Answer:
Question 4.
Read the following base sequence of a certain DNA strand and answer the questions that follow.
(i) What is called “Palindromic sequence” in a bNA?
(ii) Write the Palindromic nucleotide sequence shown in the DNA strand given and mention the enzyme that will recognise such a sequence.
(iii) State the significance of enzymes that identify palindromic nucleotide sequences.
(AI – 2008)
Answer:
(i) A palindromic sequence of DNA is a sequence of base pairs that reads the same on the two strands, when orientation of reading is kept the same, i.e. in the 5′ → 3′ direction.
(ii) 5′ GAA TT C – 3′
3′ CTTA AG – 5′
This sequence is restricted by the restriction enzyme Eco RI
(iii) The enzyme that identifies the palindromic nucleotide sequence cuts the strands between the same 2 bases, more often producing sticky ends; hence they are useful in the formation of recombinant DNA.
2nd PUC Biology Biotechnology: Principles and Processes Five Marks Question
Question 1.
Describe in detail the components of a simple stirred tank bioreactor along with a labeled diagram.
Answer:
A stiered tank bioreactor is usually a cylindrical vessel with a curved base to facilitate the mixing of the content. The stirrer facilitates even mixing and oxygen availability throughout the bioreactor.
The bioreactor has an agitator system, an oxygen delivery system along with a foam control system, a temperature control system, pH control system and sampling ports to remove small volumes of culture periodically. It provides optimum growth conditions of pH, temperatures, substrate, oxygen etc. for achieving the desired products.
Question. 2.
What is a clone? Give its preparation, extracted and purified.
Answer:
An organism or cell or group of organisms, produced asexually from an ancestor, to which they are genetical.
1. Gene cloning: Following steps are used by gene cloning:
1. Preparation of gene: DNA extracted from an organism, with the gene of interest is cut into gene size pieces with a restriction enzyme.
2. Insertion into a vector: Bacterial plasmids are cut with the same restriction enzyme. Plasmids are small circles of DNA in bacterial cells that are naturally present in addition to the bacterial other DNA.
3. Transformation of host cells: The recombinant plasmids are then transferred into bacteria using either electrophOration. The plasmid is small enough to pass through the holes into the cells. However, rather than using electricity to create holes in the bacterium, it is done by alternating the temperature between hot and cold. The bacteria are grown on a culture dish and allowed to grow into colonies. All the colonies on all the plates are called a gene library.
2. Plant cloning: Plant tissue culture is a method of propagation that has been sprouting in popularity as an alternative to cloning.
The plant can be cloned artificially using tissue culture. Vegetative propagation works because the end of the cutting forms a mass of nonspecialized cells called a callus, the callus will grow divide and form various specialized cells eventually forming a new plant.
Question 3.
(a) If the restriction enzyme has to cut a DNA, the DNA must be in pure form, i.e. free from the associated RNA and proteins. How is it achieved?
(b) Represent only diagrammatically the steps in the recombinant DNA (r DNA) technology.
Answer:
(a) The RNAs are removed by using enzymes called ribonucleases (RNases) The proteins are removed by using enzyme proteases.
(b)
Question 4.
(a) Show only diagrammatically the three steps in the polymerase chain reaction,
(b) How is repeated amplification achieved using this method?
Answer:
(b) Repeated amplification is achieved by the use of a thermostable DNA polymerase, it is isolated from the bacterium Thermus aquaticus. It remains active during the high temperature used for denaturation of the double-stranded DNA.
Question 5.
Make a diagrammatic representation of showing a restriction enzyme, the substrate DNA on u which it acts the site at which it cuts DNA and the product it produces.
Answer:
You can Download Chapter 13 Kinetic Theory Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.
Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 A.
Answer:
Consider 1 mole of oxygen gas at STP
1 mol = 6.023 × 1023 molecules
= 22400 cm3 of gas
Diameter of oxygen molecule = 3 A°
∴ Radius of oxygen molecule
= \(\frac{3}{2}\) =1.5 A° (1 A° = 10-8cm)
Volume of N oxygen molecules
\(=\frac{4}{3} \pi r^{3} \times N\)
\(=\left[\frac{4}{3} \times 3.142 \times\left(1.5 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}\right] \mathrm{cm}^{3}\)
= 8.514 cm3
∴ fraction of molecular volume to actual volume of oxygen
\(=\frac{8.514}{22,400}\)
= 3.8 × 10-4.
Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 liters.
Answer:
P = 1 atm = 1. 013 × 105 Nm-2
n= 1 mol
R = 8. 314 J mol-1 k-1
T = 0° C = 273. 15 k
The ideal gas state equation is given by
PV = n RT .
⇒ ∴ \(V=\frac{n R T}{P}=\frac{1 \times 8.314 \times 273.15}{1.013 \times 10^{5}}\)
V = 0.0224 m3 = 0.0224 × 106 cm3
= 22400 cm3
V = 22. 4 litres
∴ The volume occupied by 1 mol of gas at 1 atm pressure and 0°C is 22.4 liters.
Question 3.
Figure 13.8 shows plot of PV/T versus P for 1.00 × 10-3 kg of oxygen gas at two different temperatures.
1. What does the dotted plot signify?
2. Which is true: T1 > T2 or T1 < T2?
3. What is the value of PV/T where the curves meet on the y-axis?
4. If we obtained similar plots for 1.00 × 10-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for the low-pressure high-temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of 02 = 32.0 u, = 8.31 J mol-1 K-1.)
Answer:
1. The dotted plot signifies the behavior of equal amount of an ideal gas. Since the ideal gas satisfies the equation PV = nRT,
nR = \(\frac{P V}{T}\) is a constant forgiven amount (n) of an ideal gas.
Thus \(\frac{P V}{T}\) is independent of pressure.
2. The \(\frac{P V}{T}\) curve at temperature T1 is closer to the dotted line (ideal gas) than the \(\frac{P V}{T}\) curve at T2. Since a real gas at higher temperature behaves more like an ideal gas than a real gas at lower temperature, T1 >T2.
3. Since all three curves meet on the y-axis the value of \(\frac{P V}{T}\) = nR, where n is the no of moles of ideal gas.
∴ n = no. of moles of oxygen gas
\(=\frac{\text { Given mass of oxygen }}{\text { gram molecular mass of oxygen }}\)
\(=\frac{1}{32}=0.31\)
∴ \(\frac{P V}{T}\) = nR =0.031 × 8.314
= 0.26 J K-1
4. Since the molecular mass of hydrogen is less than that of oxygen, the number of moles is 1 g of hydrogen is more than that of oxygen. Therefore, the value of \(\frac{P V}{T}\) at the point where the curves meet the Y-axis will not be same for oxygen and hydrogen gases.
n = 0.031 mol of H2 will yield same \(\frac{P V}{T}\)
as that of 1 g of O2 gas.
∴ Mass of H2 gas
= 0. 311 × 2.02 = 0. 0626g
Question 4.
An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm, and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol-1 K-1, the molecular mass of o2 = 32 u).
Answer:
Initial Volume, V1 = 30 litres
= 30 × 10-3 m3
Initial Pressure P1 = 15 atm
= 15 × 101.3 KPa
Initial Tamperature, T1 = 27°C ≃ 300K
Let ‘n’ be the initial number of moles of oxygen.
From ideal gas law,
P1 V1 = n1 RT1
⇒ ∴ n1 = \(\frac{P_{1} V_{1}}{R T_{1}}\)
\(=\frac{\left(15.195 \times 10^{5}\right) \times 30 \times 10^{-3}}{8.314 \times 300}\)
= 18.276
Final volume, V2 = 30 litres = 30 × 10-3
Final pressure, P2 = 11 atm = 11.143 × 105
Final temperature, T2 = 17°C = 290 K
∴ Final no. of moles, n2 is given by.
n2 = \(\frac{P_{2} V_{2}}{R T_{2}}\)
\(=\frac{\left(11.143 \times 10^{5}\right) \times 30 \times 10^{-3}}{8.314 \times 290}\)
= 13.865
∴ Moles of oxygen taken out,
∆n = n1 – n2
= 18.276-13.865
= 4.411
∴ Mass of oxygen taken out
= ∆n × M
where M is the molecular mass of oxygen.
∴ mass of oxygen taken out = 4.411 × 32
= 141.152 g
∴ 141.152 g of oxygen was taken out of cylinder.
Question 5.
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 ° C?
Answer:
Initial volume of air bubble,
V1 = 1.0 cm3
= 10-6m3
Initial temperature, T1 = 12°C = 285K
Initial Pressure, P1= Ps + ρgh
where P1 = depth of air bubble = 40m
Ps = Pressure at lake surface
= 1atm
ρ = density of water = 102kg/cm3
g = acceleration due to gravity
= 9.8 ms-2
∴ P1 = 1.013 × 105+103 × 9.8 × 40
= 493300 Pa
= 4.933 × 105 Pa.
Let final volume of air bubble = V2
Final Pressure = Ps = 1.013 × 105 Pa
Final Temperature, T2 = 35°C = 308 K
From ideal gas law,
PV = n RT
∴ \(\frac{P V}{T}\) = n R
∴ for given n, \(\frac{P V}{T}\) = constant
⇒ \(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
∴ V2 = \(\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}\)
\(=\frac{4.933 \times 10^{5} \times 10^{-6} \times 308}{285 \times 1.013 \times 10^{5}}\)
= 5.263 × 10-6 m3
= 5.263 cm3.
∴ The volume of air bubble grows to 5. 263 cm3 when it reaches the lake surface.
Question 6.
Estimate the total number of air molecules (Inclusive of oxygen, nitrogen, water vapour, and other constituents) in a room of capacity 25.0 m3at a temperature of 27 °C and 1 atm pressure.
Answer:
From ideal gas law
PV = nRT
Here P = atm = 1. 013 × 105 Pa.
V = Volume = 25 m3
T = Temperature = 27°C = 300k
n = no of moles of gas
R = Gas constat = 8. 314 J mol-1 k-1
∴ n = \(\frac{P V}{R T}=\frac{1.013 \times 10^{5} \times 25}{8.314 \times 300}\)
= 0.01015 × 105 =1015 mol
Since 1 mol ≅ 6. 023 × 1023 molecules
N = 1015 mol ≅ 6.023 × 1023 × 1015
N = 6.113 × 1026 molecules.
∴ The total no of molecules in the given room is 6.113 × 1026.
Question 7.
Estimate the average thermal energy of a helium atom at
1. room temperature (27 °C),
2. the temperature on the surface of the Sun (6000 K),
3. the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Answer:
Average thermal energy of an atom
E= \(\frac{3}{2}\) KT
where, K = Boltzmann constant
= 1.38 × 10-23m2kgs-2K-1
T = absolute temperature
1. room temperature (27°C)
T = 27°C = 300k
∴ E= \(\frac{3}{2}\) × 1.38 × 10-23 × 300
= 1.242 × 10-21 J
2. Surface of the sun (6000 K)
T = 6000K
∴ E = \(\frac{3}{2}\) × (1.38 × 10-23) × 6000
E = 1.242 × 10-19 J
3. core of a star (10 million K)
T = 10 × 106 = 107K
∴ E = \(\frac{3}{2}\) × (1.38 × 10-23) × 107
E =2.07 × 10-16J.
Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same In the three cases? If not, in which case is Vrms the largest?
Answer:
Since all vessels are of same capacity and are at same pressure and temperature,
P = P1 = P2 = P3 ……..(1)
V = V1 = V2 = V3 ……..(2)
T = T1 = T2 = T3 ……..(3)
where Pi Vi and Ti are pressure, volume and temperature of ith vessel respectively.
From ideal gas law
PV = nRT
∴ Pi Vi = ni RTi i = 1,2, 3
From equation (1), (2) and (3).
PV = ni RT
or ni = \(\frac{P V}{R T}\) = constant for all the three vessels.
∴ All three vessels have same no of moles of respective gases. From Avagadro’s law, equal mols of gases contain equal no of molecules. The root mean square (RMS) speed of a molecule is given by
\(V_{r m s}=\sqrt{\frac{3 k_{B}}{m}}\)
where kB = Boltzmann constant
T = temperature
m = molecular mass of the gas.
As Vrms α \(\frac{1}{\sqrt{\mathrm{m}}}\) the rms speed of gas molecules in the three vessels are different. Since neon has the least molecular mass of the three gases, it has the highest rms speed.
Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 ° C? (atomic mass of Ar= 39.9 u, of He = 4.0 u).
Answer:
Temperature of helium gas,
T1= – 20°C = 253 k
gram molecular mass of Helium, m1 =4.0
gram molecular mass of Argon, m2 = 39.9 gm2
Let temperature of Argon = T2
The rms velocity of a gas atom is given by
\(\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}\)
where T = Temperature of the gas
R = Universal Gas constant
M = Molar mass of the gas.
Given Vrms(Helium) = Vrms(Argon)
∴ \(\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}\)(Helium) = \(\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}\)(Argon)
\(\frac{T_{1}}{M_{1}}=\frac{T_{2}}{M_{2}}\)
∴ T2 = \(\frac{T_{1} M_{2}}{M_{1}}\)
\(=\frac{253 \times 39.9}{4}\)
= 2523.675 K
∴ At 2523.675 K, the rms speed of an atom of argon will be equal to that of helium at – 20°C.
Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a atm and temperature of 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Answer:
The mean free path is given by
\(\ell=\frac{\mathrm{k} \mathrm{T}}{\sqrt{2} \pi \mathrm{d}^{2} \cdot \mathrm{P}}\)
where k = Bottzmann constant
= 1.38 × 10-23kg2m2s-2k-1
d = diameter of the atom / molecule
= 2 × 1 A° = 2 × 10-10 m
P = Pressure of the gas = 2 atm
= 2.026 × 105 Pa
\(\therefore \ell=\frac{1.38 \times 10^{-23} \times 290}{\sqrt{2} \times \pi \times\left(2 \times 10^{-10}\right)^{2} \times\left(2.026 \times 10^{5}\right)}\)
= 11.12 × 10-8 m
= 1.112 × 10-7 m
∴ mean free path = 1.112 × 10-7 m
RMS speed of nitrogen molecules is given by
Vrms = 508.26 ms-1
∴ collision frequency, f = \(\frac{V_{rms}}{\ell}\)
\(=\frac{508.26}{1.112 \times 10^{-7}}\) = 4.57 × 109s-1
∴ collision time Tc = \(\frac{\mathrm{d}}{\mathrm{V}_{\mathrm{rms}}}\)
\(=\frac{2 \times 10^{-10}}{508.26}\)
= 3.935 × 10-13 s.
Time gap between successive collisions,
Tg = \(\frac{1}{f}\)
\(=\frac{1}{4.57 \times 10^{9}} \mathrm{s}\)
= 2.188 × 10-10s
\(\therefore \frac{T_{9}}{T_{c}}=\frac{2.188 \times 10^{-10}}{3.935 \times 10^{-13}}=555 \simeq 500\)
∴ the time taken between successive collisions is about 500 times the time spent on collision.
Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube Is held vertically with the open end at the bottom?
Answer:
Total length of bore = 1 m = 100 cm.
length of air column = 15cm
length of mercury thread = 76cm.
∴ length of open space = 100 – (15 + 76)
= 9 cm
Initial pressure P1 = 76cm of mercury
Initial volume V1= 15. A cm3
where, A = sectional area of the bore in cm2
Let h cm3 of mercury flow out of tube when held vertically
∴ New length of air column = 15 + 9 + h = 24 + h cm
New length of mercury = 76 – h cm
∴ Final pressure, P2 = 76 – (76-h)
= h cm of mercury
Find volume, V2 = (24 + h) × A cm3
Since temperature is constant and no air escapes the bore,
P1 V1 = P2 V2
∴ 76 × 15A = h (24+h) A
∴ 76 × 15 = h (24+h)
⇒ h = \(\frac{-24 \pm \sqrt{576+4560}}{2}\)
∴ h = \(\frac{-24 \pm 71.67}{2}\)
h = 23. 83, -47.83
Since height can only be positive, the amount of mercury that flows out of bore is 23. 83cm. There fore, mercury left in the bore = 52.2 cm and length of air column = 47.8 cm.
Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s-1. The diffusion of another gas under the same conditions ie measured to have an average rate of 7.2 cm3 s-1. Identify the gas. [Hint: Use Graham’s law of diffusion: R1/R2 = ( M2/M1)1/2 , where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
Answer:
According to graham’s law of diffusion,
\(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right)^{1 / 2}\)
R1 = rate of diffusion of hydrogen
= 28.7cm3 S-1
R2 = Rate diffusion of unknown gas
= 7.2cm3 s-1
M1 = Molecular mass of hydrogen = 2.02
M2= Molecular mass of unknown gas.
∴ M2 = \(M_{1}\left(\frac{R_{1}}{R_{2}}\right)^{2}\)
= \(2.02\left(\frac{28.7}{7.2}\right)\)
= 32.09
The molecular mass of oxygen gas is 32. Hence, The given unknown gas is oxygen.
Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so called law of atmospheres
n2 = n1 exp [ -mg (h2 – h1) /KBT] where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n2 = n1 exp [-mg NA \(\left(\rho-p^{\prime}\right)\) (h2-h1/ \(\boldsymbol{\rho}\) RT)]
where ρ is the density of the suspended particle, and \(\rho^{\prime}\) that of surrounding medium. [NA is Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Answer:
From the law atmospheres , we have n2 = n1 exp (-mg (h2 -h1)/ KBT ] …..(1)
where n2 = number density at height h2
n1 = number density at height h1
m = mass of the particle suspended in the column
Let
ρ = density of suspended particle
\(\rho^{\prime}\) = density of the medium
m2 = mass of medium displaced
From Archimedes’ principle,
Effective weight of the suspended particle = Actual weight of the particle – weight of the medium displaced.
i.e., We = mg – \(m^{\prime}\)g
Let V be the volume displaced.
∴ \(\mathrm{m}^{1}=\mathrm{v}_{\mathrm{f}}^{\prime}\) and \(\mathrm{V}=\left(\frac{\mathrm{m}}{\mathrm{f}}\right)\)
∴ \(\mathrm{m}^{\mathrm{l}}=\left(\frac{\mathrm{m}}{\rho}\right) \rho^{\mathrm{l}}\)
∴ effective weight, We = mg – m \(\left(\frac{\rho^{\prime}}{\rho}\right) \mathrm{g}\)
We = mg [1- \(\rho^{\prime}\)/ρ] ……(2)
We know that
R = NA.kB (by definition)
where R = Universal Gas constant
NA = Avogadro number
kB = Boltzmann constant
∴ kB = R/NA …………(3)
Substituting the effective weight in (2) inplace of mg and using (3) in equation (1),
\(n_{2}=n_{1} \exp \left[m g\left(1-\frac{\rho^{1}}{\rho}\right)\left(h_{2}-h_{1}\right) /\left(R / N_{A}\right) T\right]\)
\(=n_{1} \exp \quad\left[\mathrm{mg} \mathrm{N}_{\mathrm{A}}\left(\rho-\rho^{\prime}\right)^{\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right)} / \rho \mathrm{RT}\right]\)
Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :
| Substance | Atomic mass(u) | Density (103 kg nr3) |
| Carbon (diamond) | 12.01 | 2.22 |
| Gold | 197.00 | 19.32 |
| Nitrogen (liquid) | 14.01 | 1.00 |
| Lithium | 6.94 | 0.53 |
| Fluorine (liquid) | 19.00 | 1.!4 |
[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few A].
Answer:
Assuming all molecules to be spherical, and solids and liquids are tightly packed,
Volume, V= \(\frac{4}{3} \pi r^{3}\)
where r = radius of a molecules
Also, V = \(\frac{\text { mass }}{\text { density }}\)
For 1 mole of a substance,
\(\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3} \cdot \mathrm{N}_{\mathrm{A}}=\frac{\mathrm{M}}{\mathrm{J}}\)
where NA = Avogadro number = 6. 023 × 1023
M = gram atomic mass
ρ = density of the substute
∴ \(r=\left(\frac{3 M}{4 \pi \int N_{A}}\right)^{1 / 3}\)
1. Carbon (diamond):
M = 12.01g = 12.01 × 10-3Kg
f = 2.2 × 103 kgm-3
∴ \(r_{c}=\left(\frac{3 \times 12.01 \times 10^{-3}}{4 \pi \times 2.2 \times 10^{3} \times 6.023 \times 10^{23}}\right)\)
≃ 1.29 × 10-10m
rc ≃ 1.29 A°
2. Gold :
M = 197 g = 197 × 10-3 kg
f= 19.32 × 103 kg m-3
\(\therefore r_{9}=\left(\frac{3 \times 197 \times 10^{-3}}{4 \pi \times 19.32 \times 10^{3} \times 6.027 \times 10^{23}}\right)^{1 / 3}\)
= 1.59 × 10-10m
rg = 1.59 A°
3. Nitrogen (liquid):
M = 14.01g = 14.01 × 10-3kg
f = 103kgm-3
\(r_{n}=\left(\frac{3 \times 6.94 \times 10^{-3}}{4 \pi \times 0.53 \times 10^{3} \times 6.023 \times 10^{3}}\right)^{1 / 3}\)
= 1.73 A°
4. Fluorine (liquid):
M = 19 × 10-3kg
f = 1.14 × 103kgm-3
\(r_{n}=\left(\frac{3 \times 19 \times 10^{-3}}{4 \pi \times 1.14 \times 10^{3} \times 6.023 \times 10^{23}}\right)^{1 / 3}\)
= 1.876 A°
| Substance | Radius |
| Carbon (diamond) | (A°) 1.29 |
| Gold | 1.59 |
| Nitrogen (liquid) | 1.77 |
| Lithium | 1.73 |
| Fluorine (liquid) | 1.88 |
Question 1.
Define mean free path.
Answer:
The average distance covered by a moving molecule between successive collisions is called the mean free path.
Question 2.
State Avagadro’s hypothesis.
Answer:
Avagadro’s hypothesis states that under fixed conditions of temperature and pressure, equal volumes of all gases contain equal number of molecules.
Question 3.
State Boyle’s law.
Answer:
Boyle’s law states that under constant temperature, the pressure of given mass of a gas is inversely proportional to its volume.
Question 4.
State Charle’s low.
Answer:
Charle’s law states that under constant pressure, the volume of a given mass of any gas is directly proportional to its temperature.
Question 5.
State law of equipartition of energy.
Answer:
The law of equipartition of energy states that in thermal equilibrium, the total energy of a particle is equally distributed in all possible modes of energy, with each mode having an average energy equal to
\(\frac{1}{2}\) kB T.
Question 6.
Define the Avogadro number.
Answer:
Avogadro number, NA, is defined as the number of molecules contained in 22.4 litres of any gas at STP, Its value is equal to 6.02 × 1023.
Question 7.
Calculate the number of molecules contained in 448 ml of an ideal gas at STP.
Answer:
22.4 L contains 6.02 × 1023 molecules therefore 448 ml contains
N = \(\frac{448}{22400} \times 6.02 \times 10^{23}\)
∴ N = 12.04 × 1021
= 1.2 × 1022
∴ 448 ml of ideal gas at STP contains 1.2 × 1022molecules.
Question 8.
If the temperature (absolute scale) of an ideal gas is doubled keeping pressure and volume constant, find the increase in its rms speed.
Answer:
For an ideal gas,
\(\mathrm{V}_{\mathrm{ms}} \alpha \sqrt{\mathrm{T}}\)
There fore, if the temperature is doubled, the rms speed increases by a factor of \(\sqrt{2}\).
Question 1.
What is an ideal gas? Mention the ideal gas equation.
Answer:
An ideal gas is a gas that obeys all the gas laws at all temperatures and is given by
PV = µRT
where P = pressure of the gas
V = Volume of the gas
µ = number of mols of the gas
R = Universal gas constant
T = absolute temperature .
Question 2.
A sample of gas is at 300K. Find the temperature at which the r.m.s speed of its molecules is doubled without changing the pressure.
Answer:
Initial temperature, Ti = 300k
let final temperature, Tf = T
let initial rms speed, Vi = v1
∴ final rms speed, Vf = 2V1
For a gas of mass m and temperature T, the rms speed is given by
\(V_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{k}_{\mathrm{B}} \mathrm{T}}{\mathrm{m}}}\)
where kB = Boltzman constant.
∴ \(\frac{2 V_{1}}{V_{1}}=\sqrt{\frac{T}{300}}\)
∴ T = 300× (2)2
∴ T = 1200 k.
Question 3.
The temperature of an ideal gas is increasing from 100K to 500K if the rms speed of molecules at 100k Is V, what Is the rms speed of the molecules at 500K?
Answer:
Initial temperature, Ti = 100K
final temperature of the gas Tf = 500K
Initial rms speed of molecules, Vrms,i = V
let final r.m.s speed of molecules Vrms,f = V’
for ideal gas,
\(V_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{k}_{\mathrm{B}} \mathrm{T}}{\mathrm{m}}}\)
where kB = Boltzmann constant
T = absolute temperature
m = mass of the gas
There fore, the r.m.s speed of molecules of the given ideal gas at 500k is \(\sqrt{5}\) V.
Question 4.
At a given temperature T, the rms speed of Hydrogen molecules is V. Find rms speed of oxygen molecules at the same temperature.
Answer:
Molecular mass of Hydrogen, MH2= 2
Molecular mass of Oxygen, MO2 = 32
rms speed of hydrogen gas VH2 = V
let rms speed of Oxygen gas, Vo2 = v’
The rms speed of gas is given by
∴ rms speed of oxygen at temperature T is \(\frac{\mathrm{V}}{16}\).
Question 5.
A box contains N molecules of an ideal gas at temperature T and pressure P. The number of molecules in the box is increased to two times the original value. Find the final temperature and pressure if the kinetic energy of the gas is kept constant.
Answer:
Initial no. of molecules ni = N
Initial temperature, Ti = T
Initial Pressure, Pi = P
Final no. of molecules, nf= 2N
Let final temperature, Tf=T’
let final pressure, Pf = p’
For ideal gases,
PV = \(\frac{2}{3}\)E ……..(1)
where P = pressure of the gas
V = volume of the gas
E = kinetic energy of molecules of the gas
Since E is constant, P is also a constant for the given case
∴ P’ = p
For ‘n’ mols of an ideal gas of molecular mass M,
\(\frac{1}{2} \mathrm{nM} \mathrm{V}_{\mathrm{rms}}^{2}=\frac{3}{2} \mathrm{nRT}\)
∴ E = \(\frac{3}{2}\) NKBT
Question 6.
Estimate the fraction of molecular volume to the actual volume occupied by Nitrogen gas at STP. Assume the radius of N2 = 300 pm.
Answer:
Consider one mole of N2 gas at STP
∴ Actual volume, vact = 22.4L
vact = 22.4 × 10-3m3
molecular volume, Vmol= NA × \(\frac{4}{3}\) × π ×R3
∴ Vmol = 6.81 × 10-5m3
∴ Ratio of molecular volume to actual volume,
\(\frac{V_{\text {mol }}}{V_{\text {act }}}=\frac{6.81 \times 10^{-5}}{22.4 \times 10^{-3}}\)
= 3.04 × 1 O-3
Question 7.
The molecules of a given mass of an ideal gas have rms speed 100 ms-1at 273K and 100kPa calculate the rms speed of the molecules if the temperature is doubled and pressure heated.
Answer:
Initial temperature of gas, Ti = 273K
Final temperature of gas, Tf = 2 × Ti
= 546K
Initial rms speed of gas
molecules, Vi = 100 m/s
let final rms speed of molecules, Vf = V
For an ideal gas,
∴ V = 100 \(\sqrt{2}\)
∴ V = 141.4 m/s.
Question 8.
State and explain Boyle’s law.
Answer:
Boyle’s law states that at constant temperature, the volume of a given mass of an ideal gas is inversely proportional to its pressure.
Let the volume of the gas be ‘V’ and pressure ‘P’ and temperature, T. Then according to Boyle’s law,
\(\mathrm{V} \propto \frac{1}{\mathrm{P}}\)
or V =\(\frac{k}{P}\) ⇒ V.P = k.
where k = proportionality constant i.e, according to Boyle’s law, if the temperature is kept constant, for a given change in volume the pressure changes such that the product of pressure and volume is a constant.
Question 9.
State and explain Charles law.
Answer:
Charles’ law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature.
consider a gas at pressure ‘P’, temperature ‘T’ having volume ‘V’, Then Charles’ law states that
\(\mathrm{V} \propto \mathrm{T}\)
or V = kT where k is the proportionality constant.
Question 10.
At what temperature the molecules of nitrogen will have the same rms velocity as the molecules of oxygen at 400K?
Answer:
Molecular mass of Nitrogen, M1 =28
Molecular mass of Oxygen, M2 = 32
Temperature of the Oxygen, T2 = 400
let temperature of nitrogen, T1= T
The rms speed of gas molecules given by,
∴ T = 350 K
Question 1.
Derive the ratio of specific heats for a mono atomic and diatomic gases
Answer:
1. Monatomic gas :
No of degrees of freedom = 3 (Translational) of each molecule
∴ Average kinetic energy of a molecule at temperature T, E = 3 \(\frac{1}{2}\) × kBT
where kB = Boltzmonn constant
U = \(\frac{3}{2}\) kBT. NA [Total internal energy of the gas]
∴ U = \(\frac{3}{2}\) R T
The molar specific heat at constant volume,
Cv = \(\frac{\mathrm{d} \mathrm{U}}{\mathrm{d} \mathrm{T}}\)
= \(=\frac{d}{d T}\left(\frac{3}{2} R T\right)=\frac{3}{2} R\) ……..(1)
For ant ideal gas
Cp – Cv = R …………(2)
Using equation (1) in equation (2),
Cp = R + \(\frac{3}{2}\) R = \(\frac{5}{2}\) R
∴ ratio of specific heats,
\(\gamma=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{W}}=\frac{\left(\frac{5}{2}\right) \mathrm{R}}{\left(\frac{3}{2}\right) \mathrm{R}}=\frac{5}{3}\)
2. Diatomic gas:
Case 1:
molecules are rigid
Number of degrees of freedom of each molecules = 5 (3 translational + 2 rotational)
∴ Average energy per molecule
= 5 × \(\frac{1}{2}\) kBT = \(\frac{5}{2}\) kBT
∴ Total internal energy
U = \(\frac{5}{2}\) RT
Cv \(\frac{d}{d T}\left(\frac{5}{2} R T\right)\)
= \(\frac{5}{2}\) R …….(3)
∴ Using equation (3) in equation (2),
Cp – \(\frac{5}{2}\) R = R
∴ Ratio of specific heats for diatomic gas,
case 2:
diatomic molecules are not rigid,
Average energy per molecule
= 5 × \(\frac{1}{2}\) kBT + kBT =\(\frac{7}{2}\) kBT
∴ Internal energy of the gas, U = \(\frac{7}{2}\) RT
\(C_{v}=\frac{d}{d T}\left(\frac{7}{2} R T\right)=\frac{7}{2} R\) ……(4)
Using equation (4) in equation (2),
Cp = \(\frac{9}{2}\) R
∴ Ratio of specific heats,
γ = \(\frac{9}{7}\).
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