Category: Objective Questions

Bioprocess Engineering Multiple Choice Questions on “Disposal of Effluent”.

1. What do you mean by the term “Ammoniacal Nitrogen”?
A. A safe disposal
B. An unsafe disposal
C. A metal
D. A non-toxic substance
Answer: B
Explanation: Ammoniacal nitrogen (NH₃-N), is a measure for the amount of ammonia, a toxic pollutant often found in landfill leachate and in waste products, such as sewage, liquid manure and other liquid organic waste products. Ammonia can directly poison humans and upset the equilibrium of water systems.

2. Disposal in sea and rivers are safe.
A. True
B. False
Answer: B
Explanation: Beyond the standards of disposal in water bodies is toxic. In addition, levels of ammoniacal nitrogen may be stipulated. There are as well, often stringent upper limits for toxic metals and chemicals which might kill the fauna and flora.

3. Lagoon is a part of sea water.
A. True
B. False
Answer: A
Explanation: Lagoon is an area of sea water separated from the sea by a reef (= a line of rocks and sand).

4. What is a Facultative pond?
A. Top layer is anaerobic
B. Bottom layer is aerobic
C. Aerobic throughout depth
D. Top layer is aerobic
Answer: D
Explanation: The most often used ponds in domestic wastewater treatment are the stabilization pond and facultative lagoon. The stabilization pond is designed to be aerobic throughout its depth and the facultative lagoon will be anaerobic at the bottom and aerobic at the top.

5. Deep water ponds are mechanically agitated.
A. True
B. False
Answer: A
Explanation: Achieving good agitation is an important part of this. Failure to properly agitate the manure will result in a continuous buildup of settled solids within the storage, resulting in less and less available storage as time goes by. Good agitation of the manure will re-suspend those settled solids and facilitate their removal from the storage ensuring we maintain that capacity we need. Additionally, agitation of the manure helps homogenize it and provide a more consistent nutrient content as it is applied.

6. Which of the following step is required for the spray irrigation?
A. Waste should be chlorinated
B. Waste should not be chlorinated
C. Land high above 38mm rainfall
D. Land with below 38mm rainfall
Answer: A
Explanation: The waste are initially chlorinated to lower the BOD and reduce unpleasant odors and then sprayed on to land until the equivalent of 38mm of rainfall is reached. The process is repeated at monthly intervals.

7. What is micro spray irrigation?
A. Surface
B. Sprinkler
C. Drip
D. Subsurface
Answer: C
Explanation: Micro irrigation is defined as the frequent application of small quantities of water directly above and below the soil surface; usually as discrete drops, continuous drops or tiny streams through emitters placed along a water delivery line. Drip Irrigation. Simcha Blass, an Israeli hydraulic engineer, is credited with the discovery and introduction of modern drip irrigation in the early 1930’s. Drip irrigation (also known as micro-irrigation) became more common with the introduction of plastics in the 1950’s.

8. Injection wells are used for waste disposal apart from oil and gas production.
A. True
B. False
Answer: A
Explanation: An injection well is a device that places fluid deep underground into porous rock formations, such as sandstone or limestone, or into or below the shallow soil layer. The fluid may be water, wastewater, brine (salt water), or water mixed with chemicals.

9. What do you mean by “Leachate”?
A. A worm
B. A solid
C. A fluid
D. A gas
Answer: C
Explanation: A leachate is any liquid that, in the course of passing through matter, extracts soluble or suspended solids, or any other component of the material through which it has passed.

10. Landfill is produced by microbes.
A. True
B. False
Answer: A
Explanation: Landfill gas is a complex mix of different gases created by the action of microorganisms within a landfill. Landfill gas is approximately forty to sixty percent methane, with the remainder being mostly carbon dioxide.

11. Landfill gas (LFG) contains Nonmethane organic compounds (NMOCs).
A. True
B. False
Answer: A
Explanation: By volume, landfill gas typically contains 45% to 60% methane and 40% to 60% carbon dioxide. Landfill gas also includes small amounts of nitrogen, oxygen, ammonia, sulfides, hydrogen, carbon monoxide, and nonmethane organic compounds (NMOCs) such as trichloroethylene, benzene, and vinyl chloride.

12. Landfills do not decompose.
A. True
B. False
Answer: B
Explanation: The solid waste layer becomes laced with these strips of dirt. Landfills are not designed to break down waste, only to store it. But garbage in a landfill does decompose, albeit slowly and in a sealed, oxygen-free environment.

13. Incineration is a “Cold treatment” process.
A. True
B. False
Answer: B
Explanation: Incineration is a waste treatment process that involves the combustion of organic substances contained in waste materials. Incineration and other high-temperature waste treatment systems are described as “thermal treatment”. Incineration of waste materials converts the waste into ash, flue gas and heat. The ash is mostly formed by the inorganic constituents of the waste, and may take the form of solid lumps or particulates carried by the flue gas. The flue gases must be cleaned of gaseous and particulate pollutants before they are dispersed into the atmosphere. In some cases, the heat generated by incineration can be used to generate electric power.

14. What do you mean by the term “Flue gas”?
A. Dry CO₂
B. Chimney gas
C. Non-combustible gas
D. Cold gas
Answer: B
Explanation: A flue is a duct, pipe, or opening in a chimney for conveying exhaust gases from a fireplace, furnace, water heater, boiler, or generator to the outdoors. Historically the term flue meant the chimney itself. Flue gas is the gas exiting to the atmosphere via a flue, which is a pipe or channel for conveying exhaust gases from a fireplace, oven, furnace, boiler or steam generator. Quite often, the flue gas refers to the combustion exhaust gas produced at power plants.

15. Flue gas does not need pre-treatment.
A. True
B. False
Answer: B
Explanation: Flue Gas Treatment. Waste incineration and many other industrial processes generate flue gases. These often contain pollutants such as sulfur oxides (SO₂ + SO₃), hydrochloric acid (HCl), hydrofluoric acid (HF) as well as heavy metals, dioxins and furans. So it needs to undergo pre-treatment process.

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Chemical Reaction Engineering Questions on “Reaction Rates – 2”.

1. For a homogeneous reaction of order n, the unit of rate constant, k is ____
A. conc^n-1
B. (frac{conc^{-1}}{time} )
C. (frac{conc^{-1}}{time^{1-n}} )
D. (frac{conc^{1-n}}{time^{-1}} )

Answer: D
Explanation: For any order n, (-r_A) = kC_Aⁿ
(frac{mol}{m^3×s} = k.(frac{mol}{m^3})^n )
Hence the unit of k is (frac{conc^{1-n}}{time^{-1}}. )

2. The unit of rate constant, k for a first order reaction is ____
A. (time)^-1
B. mol. time
C. (frac{mol}{m^3} )
D. (frac{mol}{m^3×s} )

3. Which among the following reactions is elementary?
A. A + B → R; (-r_A) = k C_AC_B
B. 2A + B → R; (-r_A) = k C_A²
C. 2A + B → R; (-r_A) = k C_AC_B²
D. A + B → R; (-r_A) = k C_B

Answer: A
Explanation: Elementary reactions are the ones in which rate equation corresponds to stoichiometry. For the reaction A + B → R, 1 mole of A and 1 mole of B take part in the reaction. Hence, the rate equation has to be
(-r_A) = k C_A¹C_B¹.

4. An example of pseudo first order reaction is ____
A. Hydrolysis of an ester to a carboxylic acid using water
B. NO₂ + CO → NO + CO₂
C. N₂O₅ → NO₂ + 1/2 O₂
D. 2NO + H₂ → N₂O + H₂O

Answer: A
Explanation: Pseudo first order reactions are the ones in which the concentration of one of the reactants is used in excess to convert higher order reactions to 1st order. As water is used as solvent in esterification reaction, its concentration is in huge excess compared to the ester. Thus, the reaction is pseudo first order for the ester.

5. The study of rates at which chemical reactions occur and the effect of various parameters on the rate are termed as ____
A. Chemical Technology
B. Chemical Sciences
C. Chemical Kinetics
D. Fluid Mechanics

Answer: C
Explanation: Chemical Kinetics is the study of rate of chemical reaction and the effect of parameters like temperature, pressure, concentration, etc. on reaction rates. Kinetic studies help in obtaining information about reaction mechanism and transition states.

6. The rate of reaction is rapid for which of the following reactions?
A. Cellular reactions
B. Reactions in coal furnaces
C. Reactions in rocket engines
D. Gas reactions in porous catalysts

Answer: C
Explanation: The rate of reaction in rocket engines is in the range 10⁵-10⁸. The order of decrease in rates is, Reactions in rocket engines > Reactions in coal furnaces > Gas reactions in porous catalysts > Cellular reactions.

7. For the reaction 3A+2B → R, which of the following expressions is valid?
A. (frac{-r_A}{3})=(frac{-r_B}{2})=(frac{r_C}{1})
B. (frac{-r_A}{3})=(frac{-r_B}{2})=(frac{r_C}{1})
C. (frac{-r_A}{1})=(frac{-r_B}{1})=(frac{r_R}{1})
D. (frac{-r_A}{3})=(frac{-r_B}{4})=(frac{r_C}{1})

Answer: C
Explanation: For the reaction aA + bB → rR, (frac{-r_A}{a})=(frac{-r_B}{b})=(frac{r_R}{c}). In the reaction 3A+2B → R, a=3, b=2 and r=1.

8. Which among the following reactions is elementary?
A. Ester formation from ethanol and acetic acid
B. Formation of Hydrogen Bromide
C. Decomposition of acetaldehyde to give methane and carbon monoxide
D. Formation of Hydrogen Iodide

Answer: A
Explanation: Ester formation is an elementary reaction. The rate equation for esterification reaction corresponds to the reaction stoichiometry.

9. The molecularity of the reaction A+2B → R, (-r_A) = k C_AC_B² is ____
A. 2
B. 1
C. 3
D. 0

Answer: C
Explanation: In the reaction A+2B → R, 1 molecule of A and 2 molecules of B react. Hence, molecularity = 3.

10. Order of the reaction (-r_A) = kC_A^a.C_B^b is ____
A. a
B. b
C. a-b
D. a+b

Answer: D
Explanation: Sum of the powers of the concentrations of the reactants, A and B is the order of reaction. For the reaction (-r_A) = kC_A^a.C_B^b, order = (a+B..

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Rate Laws and Stoichiometry Definitions”.

1. Hydrogenation of oil is an example of ___________ reactor
A. Homogeneous
B. Heterogeneous
C. Autocatalytic
D. Insufficient data
Answer: B
Explanation: The reactants are of different phases i.e. Hydrogen (Gas phase) and oil (Liquid phase). Ni catalyst (SoliD. is also used.

2. Arrhenius first suggested the temperature dependence of the rate constant.
A. True
B. False
Answer: A
Explanation:
k = (koe^{frac{-E}{RT}})
This is known as Arrhenius equation and it gives the relation between temperature and rate constant.

3. The slope of the line in the graph gives the ___________
A. Activation energy
B. Rate constant
C. Frequency factor
D. Insufficient data
Answer: A
Explanation: The Arrhenius equation k=(koe^{frac{-E}{RT}}) can be modified as
ln⁡(k)=ln⁡(ko) – ((frac{E}{R})frac{1}{T} )
The slope of the line gives E/R from which Activation energy can be determined.

4. Arrhenius equation to determine the activation energy.
A. ln⁡((frac{k1}{k2}) = frac{E}{R} (frac{1}{T2}-frac{1}{T1}) )
B. ln⁡((frac{k1}{k2}) = frac{E}{R} (frac{1}{T2}+frac{1}{T1}) )
C. ln⁡((frac{k1}{k2}) = -frac{E}{R} (frac{1}{T2}-frac{1}{T1}) )
D. ln⁡((frac{k1}{k2}) = frac{E}{R} (frac{T1}{T2}) )
Answer: A
Explanation: When we have two values of k and T
k1 = ko(e^{-frac{E}{RT1}}) and k2 = ko(e^{-frac{E}{RT2}} )
Modifying it gives
ln⁡(k1) = ln⁡(ko) – ((frac{E}{R})frac{1}{T1}) and ln⁡(k2) = ln⁡(ko) – ((frac{E}{R})frac{1}{T2} )
On further simplification we get ln⁡((frac{k1}{k2}) = frac{E}{R} (frac{1}{T2}-frac{1}{T1}). )

5. Temperature dependence of rate constant according to transition state theory is ______
A. k=ko(e^{-frac{E}{RT}} )
B. k=(e^{-frac{E}{RT}} )
C. k=koT(e^{-frac{E}{RT}} )
D. k=ko(e^{-frac{1}{RT}} )
Answer: C
Explanation: According to transition state theory the relationship between temperature and rate constant is given by = koT(e^{-frac{E}{RT}}. )

6. Value of m to obtain the collision theory relation for temperature dependency of rate constant is ________
A. 1
B. 0
C. -1
D. 0.5
Answer: D
Explanation: The general equation for temperature dependency is k=koT^m (e^{-frac{E}{RT}} )
If m = 0 → Arrhenius equation
m = 1 → transition state theory
m = 0.5 → Collision theory.

7. Calculate the activation energy for the following data using transition state theory.

T in deg C	k
0	0.002
80	0.08

A. 30.07 KJ/mol
B. 50.43 KJ/mol
C. 34.37 KJ/mol
D. 82.31 KJ/mol
Answer: C
Explanation: When we have two values of k and T
k1=koT1(e^{-frac{E}{RT1}} ) and k2=koT2(e^{-frac{E}{RT2}} )
Modifying it gives
ln⁡(k1)=ln⁡(ko)+ln⁡(T1) – ((frac{E}{R})frac{1}{T1}) and ln⁡(k2)=ln⁡(ko)+ln⁡(T2) – ((frac{E}{R})frac{1}{T2} )
On further simplification we get ln⁡((frac{k1}{k2}))=ln⁡((frac{T1}{T2}) – frac{E}{R}(frac{1}{T1}-frac{1}{T2}) )
ln⁡((frac{0.002}{0.08}))=ln⁡⁡((frac{273}{353})-frac{E}{8.314}(frac{1}{273} – frac{1}{353}) )
E = 34.377 KJ/mol.

8. The reaction rate of a bimolecular reaction at 300K is 10 times the reaction rate at 150K. Calculate the activation energy using collision theory.
A. 4928 J/mol
B. 5164 J/mol
C. 3281 J/mol
D. 1296 J/mol
Answer: A
Explanation: T₁ = 150K, T₂ = 300K and k₂ = 10k₁
When we have two values of k and T
k1=ko(sqrt{T1}e^{-frac{E}{RT1}}) and k2=ko(sqrt{T2}e^{-frac{E}{RT2}})
Modifying it gives
ln⁡(k1)=ln⁡(ko)+0.5ln⁡(T1) – ((frac{E}{R})frac{1}{T1}) and ln⁡(k2)=ln⁡(ko) – ((frac{E}{R})frac{1}{T2} )
On further simplification we get ln⁡⁡((frac{k1}{k2}) = 0.5ln⁡(frac{T1}{T2}) frac{E}{R} (frac{1}{T1}-frac{1}{T2}) )
ln⁡((frac{k1}{10k1}))=0.5ln⁡((frac{150}{300}) – frac{E}{8.314}(frac{1}{300} – frac{1}{150}) )
E = 4928 J/mol.

9. For the reaction -r_A = (frac{1670[A][Po]}{6+CA} frac{kmol}{ml.hr}) write the units for constants 6 and 1670.
A. kmol/ml and 1/hr
B. 1/hr and 1/hr
C. kmol and kmol
D. data insufficient
Answer: A
Explanation: To get the final units
-r_A = (frac{1670[frac{kmol}{ml}][frac{kmol}{ml}]}{6+frac{kmol}{ml}} )
The units of constants have to be 1/hr for 1670 and kmol/ml for 6.

10. Which theory is the basis for Arrhenius equation?
A. Collision theory
B. Kinetic theory of gases
C. Ideal gas law
D. Charles and Boyles law
Answer: B
Explanation: Arrhenius equation is based on Kinetic theory of gases.

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Chemical Reaction Engineering Problems focuses on “Fluid-Fluid Reactions – Absorption Systems with Chemical Reaction”.

1. A microfluid is the one in which ____
A. Fluid motion is restricted
B. Fluid molecules are free to move
C. Molecules do not take part in reaction
D. Fluid is compressed in due course of reaction
Answer: B
Explanation: A microfluid is characterized by free movement of fluid molecules. Molecules collide and intermix.

2. For early mixing, which of the following reactors/ combination of reactors is used?
A. CSTR followed by PFR
B. PFR
C. Two CSTRs in series
D. Two CSTRs in parallel
Answer: A
Explanation: CSTR is followed by a PFR for early mixing. This combination favours reactions of negative order.

3. For a macrofluid, the average concentration of reactant in the exit stream is _______
A. (frac{overline{C_A}}{C_{A0}}) = (int_0^∞ frac{C_A}{C_{A0}})dt
B. (frac{overline{C_A}}{C_{A0}}) = E.dt
C. (frac{overline{C_A}}{C_{A0}}) = (int_0^∞)E.dt
D. (frac{overline{C_A}}{C_{A0}}) = (int_0^∞ frac{C_A}{C_{A0}})E.dt
Answer: D
Explanation: The average concentration of reactant in the exit stream is the summation of the products of reactant concentration remaining between t and t+dt and the fraction that exits between t and t+dt. The fraction exiting between t and t+dt is given as Edt.

4. State true or false.
Late mixing is favoured for reactions with positive order.
A. True
B. False
Answer: A
Explanation: The reactant is at high concentration and reacts rapidly as n > 1. Hence, PFR is followed by CSTR for late mixing.

5. State true or false.
Macrofluids form large aggregates.
A. True
B. False
Answer: A
Explanation: Macrofluid exhibits segregation. Globules, each consisting of a large number of molecules do not mix well with the other globules.

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Molecular Biology Multiple Choice Questions on “Genome Organization – 1”.

1. In the beads on a string model, the bead is made up of __________
A. 6 histone proteins
B. 8 histone proteins
C. 6 histone proteins and DNA
D. 8 histone proteins and DNA

Answer: B
Explanation: The “beads on a string” model is for the nucleosome. It consists of the 8 histone protein core or the bead and the DNA wound around imitating a string.

2. The unpacked stretches of DNA are the extra chromosomal load found in the eukaryotic genome.
A. True
B. False

Answer: B
Explanation: Linker DNA is the stretches of DNA that are not packed into a nucleosome. Typically these are the regions engaged in gene expression, replication and recombination and are generally associated with non – histone proteins.

3. How many types of histone molecules are found in nature?
A. 3
B. 4
C. 5
D. 6

Answer: C
Explanation: Eukaryotic cells commonly contain five abundant histone molecules. They are named as H1, H2A, H2B, H3 and H4.

4. Nucleosome is made up of __________
A. DNA, histone core protein
B. DNA, histone core protein, linker H1
C. RNA, histone core protein
D. RNA, histone core protein, linker H1

Answer: B
Explanation: The core histone proteins are H2A, H2B, H3 and H4, over which the DNA is wrapped. Histone H1 is not a part of nucleosome core particle, instead it binds to the linker DNA and thus is referred to as linker histone. Thus, the histone core, linker histone and DNA are the components of the nucleosome.

5. Histones have a high content of negatively charged amino acids.
A. True
B. False

Answer: B
Explanation: As histones maintain a constant association with negatively charged DNA thus histone molecules are made up of high content of positively charged amino acid. Greater than 20% of the residues in each histone molecules are either lysine or arginine.

6. With respect to assembly of every core histone which of the following is wrong?
A. A conserved region
B. Histone fold domain
C. Disc shaped structure
D. 2 α helices and an unstructured loop

Answer: D
Explanation: The histone fold is composed of 3 α helices and two unstructured loops. In each of these cases the histone fold mediates the formation of head to tail heterodimers of specific pairs of histone.

7. Which of the following histone pairs forms tetramers in solution?
A. H1, H2A
B. H2A, H2B
C. H2B, H3
D. H3, H4

Answer: D
Explanation: H3 and H4 histone first forms heterodimers then they come together to form a tetramer with two molecules of each. In contrast, H2A and H2B form heterodimers only in the solution and histone H1 only acts as the linker histone.

8. With respect to the “tails” of the histone molecules which of the following is not true?
A. N – terminal extension
B. Lacks defined structure
C. Required for the association of nucleosome
D. Sites for extensive modifications

Answer: C
Explanation: The “tail” of the histone is not required for the association for the DNA with the histone octamer into a nucleosome. This is proved when the nucleosome is treated with the protease, trypsin. Trypsin is known to cleave proteins after positively charged amino acid thus, when the N – terminal tail is removed no structural variation is observable in the nucleosome.

9. How many contacts are observed between the DNA and the histone core protein?
A. 14
B. 21
C. 54
D. 17

Answer: A
Explanation: 14 distinct sites of contact are observed, one for each time the minor grove of the DNA faces the histone octamer. 147 base pairs of DNA is wound around the histone octamer and each minor grove occurs after 10 base pairs thus, 14 contacts are observable.

10. Association of DNA and histone is mediated by _________
A. Covalent bonding
B. Hydrogen bonding
C. Hydrophobic bonding
D. Vander Waals interactions

Answer: B

Molecular Biology Multiple Choice Questions on “General Principles for Chain Termination Sequencing”.

1. Which of the following is not required for DNA sequencing?
A. Restriction digestion
B. Electrophoresis
C. Cloning
D. Polymerase chain reaction

Answer: C
Explanation: DNA sequencing is the process of determining the precise order of nucleotides within a DNA molecule. It includes the methods and technologies of restriction endonuclease, electrophoretic techniques and Polymerase Chain Reaction.

2. End labeled DNA sequencing is known as dideoxy method of sequencing.
A. True
B. False

Answer: B
Explanation: The end labeled DNA sequencing is known as Maxem and Gilbert method. This procedure involves either 3’ or 5’ end labeling thus is also known as end labeled DNA sequencing.

3. The 32P is added at the 3’ end by polynucleotidyl kinase.
A. True
B. False

Answer: B
Explanation: The 32P dNTP is added at the 5’ end by polynucleotidyl kinase. The 32P dNTP is added at the 3’ end by deoxynucleotydil transferase. The end labeling is done in either one of the two ends.

4. The end labeled fragment is cleaved in how many pieces?
A. 2
B. 3
C. 4
D. 5

Answer: A
Explanation: The end labeled fragment is digested with a restriction endonuclease which cleaves it into two unequal lengths of fragments. As a result, only one end of each of the two fragments thus produced will be labeled.

5. Restriction digestion is the only process to achieve sequencing by the Maxem and Gilbert method.
A. True
B. False

Answer: B
Explanation: Restriction digestion is one of the two processes to achieve sequencing by the Maxem and Gilbert method. The alternative method includes denaturation of its two complementary strands separated by gel electrophoresis.

6. The denatured strands cannot be separated.
A. True
B. False

Answer: B
Explanation: The two complementary strands of DNA generally show different mobility during electrophoresis. This is because of their different molecular weight as one of the strands has a higher number of purines and the other pyrimidines.

7. What are the basic base-specific cleavage sites used in Maxem and Gilbert method?
A. A, T, G, C
B. C, T, A+G, T+C
C. A, G, A+T, G+C
D. G, C, A+G, C+T

Answer: D
Explanation: The single end labeled double or single stranded DNA samples produced is subjected to base-specific cleavage. The bases – specific cutters are used to cut at one of the four sites G, C, A+G and C+T.

8. Which step is not involved in base-specific cleavage of DNA fragment?
A. Modification of concerned base
B. Removal of modified base from DNA strand
C. Induction of random strand break
D. End labeled DNA fragments of variable lengths produced

Answer: C
Explanation: The strand breaks is induced in a specific position. This type of break is induced in the position from which a modified base is removed.

9. The process of electrophoresis is the key to sequencing.
A. True
B. False

Answer: A
Explanation: The digests from the four reaction mixtures, that is, G, C, A+G, C+T, are separately subjected to electrophoresis. This separates the fragments according to their sizes. The base sequence is thus determined by the sequential reading of the bands developed in the four lanes of the gel through electrophoresis.

10. The dideoxy method is also known as _____________
A. Maxem and Gilbert method
B. Autosequencing
C. Sanger’s enzymatic sequencing
D. Pyrosequencing

Answer: C
Explanation: The dideoxy method is also known as Sanger’s enzymatic sequencing. It was developed by Fred Sanger and his co-workers in 1070s. It is also known as the chain termination sequencing.

11. Which enzyme is used for the replication in case of Sanger’s method of sequencing?
A. Polymerase I
B. Smaller subunit polymerase I
C. Polymerase III
D. Larger subunit polymerase I

Answer: D
Explanation: One of the two complementary strands is used as a template for DNA replication in the dideoxy method of sequencing. DNA replication is catalyzed by the larger subunit known as the klenow fragment.

12. With respect to Sanger’s enzymatic method of sequencing pick the odd one out.
A. Radioactive dideoxyribonucleotides
B. Primers
C. Klenow fragment
D. Restriction digestion

Answer: D
Explanation: In the reaction system for DNA replication, at least one of the four deoxyribonucleotides is radioactive in order to permit the autoradiographic development of bands after gel electrophoresis. A small primer sequence with a free 3’ – OH group must be provided with the template strand of DNA replication to proceed, since a free 3’ – OH is absolutely essential for the larger subunit of DNA polymerase I to catalyze DNA replication.

13. The four reaction mixtures that were prepared, contained ____________
A. dCTP, dTTP, dGTP, dATP
B. dCTP, ddTTP, dGTP, ddATP
C. ddCTP, dTTP, ddGTP, dATP
D. dCTP, ddTTP, dGTP, dATP

Answer: D
Explanation: Four different reaction mixtures are prepared for the replication of each DNA strand to be sequences. One of the systems contains 2’, 3’ – dideoxycytidine triphosphate in a concentration of about 1/100th of the normal amount of 2’ – dideoxycytidine triphosphate present in the mixture. In each one of the other 3 reaction mixtures using the same DNA an template, 2’, 3’ – dideoxythymidine triphosphate (),2’, 3’ – dideoxyadinosine triphosphate () and 2’, 3’ – dideoxyguanosine triphosphate () is used.

14. ddNTP acts as the chain radioactive markers.
A. True
B. False

Answer: B
Explanation: ddNTP acts as a terminator of the polynucleotide chain being newly synthesized on the template strand. Chain termination by ddNTP is achieved due to the fact that it does not have a 3’ – OH group, as a result of which further nucleotides cannot be added to the new chain.

15. Strand separation for DNA sequencing is done to ____________
A. sort the fragments according to their size
B. sort the fragments according to their molecular weight
C. obtain consecutive sequence as per termination
D. obtain specific lanes of sequencing

Answer: C
Explanation: The four singe stranded sample reaction mixtures are subjected to electrophoresis. This separates the strands according to their size. The smallest is the fastest moving strand that migrates furthest from the negative terminal and vice versa. Therefore, by comparing the bands of the four gels thus obtained, nucleotide sequence of the DNA fragment can be determined.

16. The position of band during electrophoresis indicates the dNTP used as a chain terminator in that mixture.
A. True
B. False

Answer: B