Category: Objective Questions

Basic Molecular Biology Questions on “The tRNA Occupies Three Sites During Elongation of the Polypeptide”.

1. For the correct addition of amino acid to the growing polypeptide chain which of the following does not play any role?
A. Loading of initiator tRNA to the P site
B. Loading of correct aminoacyl tRNA to the A site
C. Formation of peptide bond between the existing amino acid and the incoming amino acid
D. The formed tRNA with the growing polypeptide is translocated from the A site to the P site

Answer: A
Explanation: The correct addition of amino acid involves only three steps:
i) Loading of correct aminoacyl tRNA to the A site
ii) Formation of peptide bond between the existing amino acid and the incoming amino acid
iii) The formed tRNA with the growing polypeptide is translocated from the A site to the P site
Loading of the initiator tRNA is the part of initiation of translation process.

2. The events controlling the correct addition of amino acid is controlled by ____________ proteins.
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: The events controlling the correct addition of amino acid are controlled by 3 proteins known as elongation factors. They are IF1, IF2 and IF3.

3. The auxillary protein required for the elongation of polypeptide is energy independent.
A. True
B. False

Answer: B
Explanation: The auxillary protein required for the elongation of polypeptide is energy dependent processes. All these factors use energy of the GTP binding and hydrolysis to enhance the rate and accuracy of ribosome function.

4. Aminoacyl tRNA needs an escort to carry them to the ribosome bound mRNA.
A. True
B. False

Answer: A
Explanation: Aminoacyl-tRNAs do not bind to the ribosome on their own. Instead, they are escorted to the ribosome by the elongation factor EF-Tu.

5. Once the tRNA is aminoacylated, EF-Tu binds to the tRNA at the ____________
A. 5’ end of the tRNA
B. 3’ end of the tRNA
C. Amino acid
D. Variable loop of tRNA

Answer: B
Explanation: Once the tRNA is aminoacylated, EF-Tu binds to the tRNA at the3’ end of the tRNA, masking the coupled amino acid. This interaction prevents the bound aminoacyl-tRNA from participating in peptide bond formation until it is released from EF-Tu.

6. The activation of GTPase activity of the enzyme EF-Tu occurs after its binding to the aminoacylated tRNA.
A. True
B. False

Answer: B
Explanation: The trigger h the factor binding center after the tRNA is loaded into that activates the EF-Tu GTPase id known as factor binding center. EF-Tu only interacts with the factor binding center after the tRNA is loaded into the A site and a correct codon-anticodon match is made. At this point, EF-Tu hydrolyzed its bound GTP and is released from the ribosome.

7. The only fidelity testing of the translation is by observing the energy difference involved in the correct and incorrect base pairing.
A. True
B. False

Answer: B
Explanation: The energy difference between a correctly formed codon-anticodon pair and that of a near match is not equal but still cannot account for the high level of accuracy required for the translation process. Thus there are three more mechanisms that contribute to this specificity. In each case, these mechanisms select against incorrect codon-anticodon pairings.

8. How many mechanisms are involved in the maintenance of the fidelity of translation?
A. 1
B. 2
C. 3
D. 4

Answer: D
Explanation: A total of 4 mechanisms are involved in the maintenance of the fidelity of translation. In addition to the codon-anticodon interactions, the ribosome exploits minor grove interactions and two phases of proof reading to ensure that a correct aminoacyl-tRNA binds in the A site.

9. Which part of the large subunit helps in the formation of the peptide bond?
A. 5S rRNA
B. Proteins
C. 23S rRNA
D. 18S rRNA

Answer: C
Explanation: Once the correctly charged tRNA has been placed in the A site and has rotated into the peptidyl transferase center, peptide bond formation takes place. This reaction is catalyzed by RNA, specifically the 23S rRNA component of the large subunit.

10. Which of the following steps is not required for the process of translocation?
A. Frame shift in hybrid states
B. Movement of P site tRNA to E site
C. Movement of A site tRNA to P site
D. Movement of ribosome by three nucleotides

Answer: A
Explanation: For a new round of peptide chain elongation to occur, the P site tRNA must move to the E site and the A site tRNA must move to the P site. At the same time, the mRNA must move by 3 nucleotides to expose the next codon. These movements are coordinated within the ribosome and are collectively referred to as translocation.

11. The completion of translocation requires the action of the factor _________________
A. EF-Tu
B. EF-G
C. eIF2
D. eIF4G

Answer: B

Molecular Biology Multiple Choice Questions on “General Properties of Plasmids”.

1. Plasmid replication is dependent on the host cell.
A. True
B. False

Answer: B
Explanation: The plasmid is an autonomous replicating genetic material. It has its own origin of replication and complete replicating machinery thus can replicate freely and is thus independent of the replication of the host genome.

2. How the plasmid clones can be screened?
A. By selectable markers
B. By bacterial resistance gene
C. For restriction site
D. By ARS sequence

Answer: A
Explanation: Plasmid contains a selectable marker that allows cells that contain the vector to be easily identified. Thus selectable markers are used to screen clones.

3. How many restriction sites are contained by a plasmid?
A. 1
B. 2
C. 3
D. More than 1

Answer: D
Explanation: Plasmid has one or more than one site for one or more restriction enzymes. Artificial plasmids contain a single restriction site for one or more restriction enzymes. This allows DNA fragments to be inserted at a definite position.

4. Who were the scientists who discovered the plasmid pBR322?
A. Rodriguez and Bolivar
B. Joller smith
C. Herbert Boyer
D. Stanley Cohen and Joller smith

Answer: A
Explanation: Plasmids are most commonly used as vector DNA. pBR322 is a plasmid vector discovered by Rodriguez and Bolivar in 1977.

5. Under relaxed conditions how many copies of plasmid are present in the cell?
A. 10 – 100 copies
B. 100 – 500 copies
C. 1 – 300 copies
D. 10 – 700 copies

Answer: D
Explanation: Most plasmids used in molecular cloning are generally present under relaxed conditions. They are normally present in 10 to as many as 700 copies per cell.

6. What is the length of the polylinker segment of plasmid?
A. Less than 100 base pair
B. Less than 10 base pair
C. Less than 70 base pair
D. Less than 50 base pair

Answer: A
Explanation: Plasmid vector contains a strategically located short, less than 100 base pair long segments of DNA. This segment of DNA is known as the polylinker segment.

7. The full for of pUC is polylinker university cloning.
A. True
B. False

Answer: B
Explanation: The E. coli derived plasmid.

8. The host bacterium takes up a plasmid in presence of ______________
A. Monovalent cations
B. Monovalent anions
C. Divalent cations
D. Divalent anions

Answer: C
Explanation: The host bacteria can take up a plasmid from its surrounding. This process is greatly enhanced by the presence of divalent cation such as Ca²⁺.

9. What is the temperature at which bacteria can takes up the plasmid?
A. 42˚C
B. -42˚C
C. 40˚C
D. -40˚C

Answer: B
Explanation: Bacteria efficiently take up the plasmid DNA at -42˚C. This increases cell membrane permeability to DNA.

10. Which gene in the pUC18 vector confers antibiotic resistance to the transformed cells?
A. LacZ
B. PvuII
C. AmpR
D. AntII

Answer: C
Explanation: pUC18 is a well-known cloning vector. It contains the AmpR gene that confers resistance to the antibiotic ampicillin which is used as a selectable marker.

11. What is the characteristic of lacZ gene of pUC18 vector among the following?
A. Encodes for antibiotic resistance
B. Encodes for β-galactosidase enzyme
C. Encodes for β-lactamase enzyme
D. Encodes for β-galactoside transferase enzyme.

Answer: B
Explanation: The lacZ gene of the pUC18 vector is a gene of the lac operon encoding for the β-galactosidase enzyme. This enzyme cleaves galactoside present in the medium as a carbon source and liberates a blue coloured product.

12. Which one of the following is the first engineered plasmid vector?
A. pBR322
B. pBR320
C. pUC18
D. pSC101

Answer: D
Explanation: Although pBR322 is the most commonly used vector in scientific research. pSC101 is the first engineered plasmid vector. The pUC family vectors are the derivative of the pBR family of vectors.

13. What is the expanded form of pBR in pBR322?
A. Plasmid Boliver and Rodriguez
B. Plasmid Baltimore and Rodriguez
C. Plasmid bacterial recombination
D. Plasmid bacterial replication

Answer: A
Explanation: pBR322 was the first artificial cloning vector developed in 1977 by Boliver and Rodriguez. Thus pBR represents its creators leading to the expansion- Plasmid Boliver and Rodriguez.

14. What is incorrect about plasmid?
A. Helps in reproduction
B. Contains stress resistant genes
C. Serves as the transformation vehicle
D. They are the genetic material of the bacteria

Answer: D
Explanation: Plasmids are the extra genetic materials that are found in the bacterial cell along with the genetic component. They are autonomously replicating cyclic double strand DNA molecules used as vectors for gene transfer and also for replication.

15. The repressor for the β-galactosidase gene is encoded by ___________
A. XmaI
B. PstI
C. Lac I
D. PvuII

Answer: C

Energy Engineering Multiple Choice Questions on “Necessity of a Steam Plant”.

1. Apart from geographical location, the amount of power generated in a country depends on ___________
A. Number of power producing plants
B. Annual consumption of power
C. Utilization of natural resources
D. Quantity of requirement

Answer: C
clarification: Since most of the power produced is from the natural resources, the estimation of amount of power generated in a country is made by the utilization of its natural resources. The ‘Annual consumption of power ‘would give you the details of power utilized for necessities, it is the amount of power utilized out of wholesome amount of total power produced in the country.

2. Total power generated is usually contributed by power generated through ____________
A. Hydel power plant, Thermal power plant and solar plant
B. Ocean thermal energy, Wind energy and Hydel power plant
C. Hydel power plant, Geo-thermal plant and Nuclear power plant
D. Hydel power plant, Thermal power plant and nuclear power plant

Answer: D
clarification: The energy or power produced from hydel power plant, thermal power plant and nuclear power plant is very abundant compared to any other combinations of power producing plants. All these three plants have ability to produce the power in thousands of megawatts in its own standards.

3. On what factors does hydel plant entirely depend?
A. Vegetation
B. Tropical cycle
C. Amount of Rainfall
D. Hydrological cycle
View Answer

Answer: D
clarification: Hydrological cycle is an explanation of the continuous movement of water above and below earth surface, where as rainfall is not reliable since it varies period to period. The amount of water above the earth surface and below the earth surface, both are responsible for the hydrological cycle. ‘Vegetation’ is info about the assemblage of plant species irrespective of their geographic characteristics.

4. The steam power plant serves as a base plant for ________
A. Nuclear power plant
B. Geothermal power plant
C. Thermal power plant
D. Diesel plant

Answer: A
clarification: Since power is generated by nuclear power plant and the nuclear plant needs power to perform its operations. For this purpose steam power plant is used as base power plant to generate this power. The power produced base plant acts as a fuel to run the nuclear power plant. And all the cost estimation to produce the electricity is made by considering base load expenses too.

5. What is the primary objective of a steam power plant?
A. To convert one form of energy into another form
B. To produce electricity
C. To provide employment
D. To serve as a base load plant to hydel plant or nuclear plant
View Answer

Answer: B
clarification: The primary objective of the steam power plant is to produce electricity and then serving as base load plant to hydel or nuclear power plant comes as second priority. Steam power plants produce 86% of electricity. And the efficiency of steam power plant is typically 33%-48%.

6. A steam power plant works on ___________ cycle.
A. Otto
B. Brayton
C. Hydrological
D. Rankine

Answer: D
clarification: Rankine cycle is a thermodynamic cycle of constant pressure engine that is to convert heat energy into mechanical work and from that following parts like adjoined blades and shafts are made to run to produce electricity. Otto cycle is used in automobile engine and Brayton cycle is used in heat engines & air jet engine.

7. Coal crushers are also known as__________
A. Lather
B. Coal combers
C. Feeder breakers
D. Coal washer
View Answer

Answer: C
clarification: Coal crushers are also known as feeder breakers since it is elaborated by the word itself. ‘Feeder’ depicts the following component coal being fed by the hoppers and ‘Breaker’ stands for breaking off into smaller pieces. And this synonym is rarely used.

8. Road transportation of coal is preferred for what type of usage?
A. Small capacity plant
B. Medium capacity plant
C. Large capacity plant
D. Domestic usage areas

Answer: A
clarification: Road transportation of coal is ideal transporting coal directly to point of consumption. These small capacity plants are usually located in the middle of land. Trucks and tippers are used to supply coal for this purpose. And also when the plant doesn’t have railway or shipway accessibility in such areas roadways are only possible means of transportation.

9. Which is the more economical way of transporting coal?
A. Sea or River ways
B. Railways
C. Road transporting
D. By Airlifting

Answer: B
clarification: We do know shipways are cheaper. But we need another mode of transport to transfer that coal to the plant area. But in case of railways the tracks can be made to directly pass through the plant. Hence the railway is more economical compared to any other means of transport.

Energy Engineering Questions and Answers for Experienced people focuses on “Type of Stokers – 2”.

1. Where is the place at which the volatile matter and fine particles burnt in the spreader stoker?
A. Over the grate
B. In steam gauge
C. In the suspension
D. In the coal bed

Answer: C
clarification: The air supplied by the forced draught fan enters the furnace through the openings provided in the grate. A portion of this air is used to burn the fuel on the grate and remaining air is utilized to burn the volatile matter and fine particles in suspension.

2. From where is the over fire or secondary air is supplied through in the sprinkler stoker?
A. Air duct
B. Nozzle
C. Air damper
D. Tuyere’s

Answer: B
clarification: Over fire or secondary air is supplied through nozzle. The secondary air creates high turbulence and completes the combustion of volatile matter and fine particles of the coal. The unburnt coal and ash are deposited on the grate which should be removed periodically.

3. In which manner is the coal supplied from feeder to spreader in spreader stoker?
A. Feeding in alternate bulks
B. Feeding at regular intervals
C. Feeding in continuous stream
D. Feeding whenever necessary

Answer: C
clarification: The feeder is a slow speed rotating drum on which large numbers of small blades are mounted. It supplies coal to the spreaders in continuous stream. The speed of the feeder can be adjusted as per the load on plant. The feeders are operated with variable speed drive to control the combustion as per the requirement. The feeders may be reciprocating ram, endless belt or spiral worm.

4. Why are the blades of shaft in spreader stoker twisted?
A. For supporting combustion by supplying more amount of air
B. To draw in large amount of coal
C. For uniform distribution of coal
D. To restrict the supply of coal

Answer: C
clarification: Spreader consists of rapidly rotating shaft carrying blades. These blades are twisted to provide uniform distribution of the coal over the grate. The fast rotating blades hit the coal particles coming from the feeder and throw into the furnace. The distribution of coal over the grate depends on the rotating speed of the spreader and on the size of the coal.

5. How much capacity of heat is possible to produce by the travelling grate stoker?
A. 80 * 10⁶ kJ/m².hr
B. 40 * 10⁶ kJ/m².hr
C. 25 * 10⁶ kJ/m².hr
D. 65 * 10⁶ kJ/m².hr

Answer: A
clarification: The spreader stoker has wide applications with respect to the fuels used as well as to the boiler sizes. A wide variety and poor quality coal can be burnt efficiently with this type of stoker. This type of stoker can be used for boiler capacities of 80 tons to 150 tons of steam per hour. The heat release rate of 80 * 10⁶ kJ/m² hr is possible with travelling grate.

6. Which mechanism is used to supply coal in the single retort stoker?
A. Hopper
B. Chain bucket
C. Screw conveyor
D. Air blowers

Answer: C
clarification: The fuel is placed in the large hopper on the front of the furnace. Then it is further fed by reciprocating ram or screw conveyor into the bottom of the horizontal trough. The ash formed after burning of coal is collected at other end.

7. What are provided for the supply of air in single retort stoker?
A. Nozzles
B. Air ducts
C. Tuyere’s
D. Air blowers
View Answer

Answer: C
clarification: The air is supplied through the tuyeres provided along the upper edge of the grate. The ash and clinkers are collected on the ash plate provided with dumping arrangement. The coal feeding capacity of a single retort stoker varies from 100 to 2000 kg per hour.

8. Which mechanism is used for the uniform distribution of the coal in multi retort stoker?
A. Reciprocating ram
B. Nozzle
C. Screw conveyor
D. Steam jet blowers

Answer: A
clarification: the multi retort stoker consists of a series of alternate retorts and tuyere boxes for supply of air. Each retort is fitted with a reciprocating ram for feeding and pusher plates for the uniform distribution of coal. The coal falling from the hopper is pushed forward during the inward stroke ram.

9. The low pressure air entering into the extension grate is supplied to ______________ in the multi retort stoker.
A. At incandescent zone
B. Thinner fuel bed
C. At inlet dampers
D. At extension grate

Answer: B
clarification: The primary air is supplied from the main wind box to the fuel bed situated below the stoker. The partly burnt coal moves on to the extension grate. The low pressure air entering into the extension grate, wind from main wind box is supplied to the thinner fuel bed on the extension grate. The quantity supplied can be regulated by air damper.

10. What is the reason for using forced draft in multi retort stoker?
A. Maintaining combustion for longer periods
B. Helps in blowing away the ash and clinkers
C. Causes rapid combustion
D. Maintains the intensity of fire
View Answer

Answer: C
clarification: Forced draught causes rapid combustion. It is also necessary to introduce ‘over fire’ when high volatile coals are used to prevent the smoke formation. Combustion control is introduced into the stoker drive either by varying the ram stoke or by changing the rate of reciprocation.

Energy Engineering Questions and Answers for Entrance exams focuses on “Hydrographs – 2”.

1. Cusec is___________
A. A unit of flow equal to one cubic feet per sec
B. A unit of flow equal to one centimeter cube per sec
C. A unit of flow equal to one meter per sec
D. A unit of flow equal to one cubic foot per sec
View Answer

Answer: A
clarification: Cusec is a unit of flow especially water which is equal to one cubic feet per sec. And there is also use of Cumec which is one cubic meter per second. One cubic feet per second is equal to 28.317 liters per second.

2. Find the power available if overall efficiency of plant is 80%, flow rate is 4.42cumecs and head 400m?
A. 6.52MW
B. 8.18MW
C. 11.255MW
D. 13.875MW

Answer: D
clarification: Power available:
P = wQHn_o X 10^-3 KW
= 9810 X 4.42 X 400 X 0.8 X 10^-6 MW
P = 13.875MW.

3. What would be the pondage factor for if hydropower plant is used for 10 hours?
A. P.F = 2.4
B. P.F = 1.2
C. P.F = 20
D. P.F = 0.4166
View Answer

Answer: A
clarification: Pondage factor = T1/T2 = (Total number of hours in one day)/(Total number of hours plant is running)
P.F = 24/10 = 2.4.

4. Determine the capacity of hydro power plant to be used 10 hours peaking plant assuming daily flow in a river to be constant at 20m³/s. and overall efficiency is 80%?
A. 1.8835 MW
B. 5.5 MW
C. 3.25 MW
D. 1.0 MW
View Answer

Answer: A
clarification: Capacity of plant:
P = wQHn_o X 10^-3 KW
P = 9810 X 20 X 12 X 0.80 X 10^-6
= 1.8835 MW.

5. Determine the flow rate of water, if the catchment area of hydroelectric power is 2500 km², with an average rainfall of 160cm. the percolation and evaporation losses account for 19%?
A. 9639.8 M³/s
B. 42.8 M³/s
C. 859.63 M³/s
D. 2342 M³/s

Answer: A
clarification: Amount of water available for power generation,
Q_a = A X H X (1-y)
= 2500 X 10⁶ X 160/100 (1-0.19)
= 3.04 X 10¹¹m³
Flow rate of water,
Q = Q_a/(365 X 24 X 60 X 60)
= (3.04 X 10^-6)/(365 X 24 X 60 X 60) = 9639.8 M³/s.

6. Determine the power developed, IF given data is H = 150m, n_g = 0.91, n_t = 0.86 and Q is 9639.8?
A. 74MW
B. 75MW
C. 76MW
D. 78MW

Answer: A
clarification: Power developed = wQHn_o X 10^-3 KW
= 9810 X 9639.8 X 0.86 X 0.91 X 10^-6 MW
= 74 MW.

7. Determine the pondage factor if the plant is working at peak time of 16 hrs?
A. 1.5
B. 0.75
C. 2.5
D. 0.3
View Answer

Answer: A
clarification: Pondage factor = T1/T2 = (Total number of hours in one day)/(Total number of hours plant is running)
P.F = 24/16 = 1.5.

8. Find out the total flow volume in day sec meter for the average daily stream flow for 7 days?

A. 820 day sec meter
B. 95 day sec meter
C. 200 day sec meter
D. 524 day sec meter

Answer: A
clarification: Total flow volume for 7 days:
= 24 X 3600 X (100+300+200+120+50+30+20)
= 70848 X 10³ m³
= 70.848 million m³
= 70848 X 10³/86400
= 820 day sec meter.

9. Determine the pondage factor if the plant is working at time of 8 hrs?
A. 2.5
B. 3.8
C. 1
D. 3

Answer: D
clarification: Pondage factor = T1/T2
= (Total number of hours in one day)/(Total number of hours plant is running)
P.F = 24/8 = 3.

10. What AEP stand for in hydrology?
A. Annual exceedance probability
B. Annual energy production
C. Annual exceedance period
D. Automatic engagement panel

Answer: A
clarification: Annual exceedance probability refers to the probability of a flood event occurring in any year. The probability is expressed as a percentage. The probability that a given rainfall total accumulated over a given duration will be exceeded in any one year.

11. What is the volume of rainfall in day sec-meters if 6.5cm rainfall occurs over an area of 2400 sq.km?
A. 1805.56 day sec meter
B. 1225 day sec meter
C. 895 day sec meter
D. 1555.22 day sec meter

12. A lake behind a dam has a capacity of 30000km²-m approximately. For how many days would this water supply be sufficient to a city of 10⁶ populations if daily requirement per person is 500 liters?
A. 60,000 days
B. 950 days
C. 25000 days
D. 8000 days

Answer: A
clarification: Per day requirement: 500 X 10³ liter = 500 X 10³ m³
Available water in the dam = 30000 X 10⁶ m³
No of days water supplied = 30000 X 10⁶/500 X 10³ = 60000 days.

13. Determine the capacity of hydro power plant to be used 8 hours peaking plant assuming daily flow in a river to be constant at 65m³/s. and overall efficiency is 80% and head 12m?
A. 6.1214 MW
B. 5.5 MW
C. 31.25 MW
D. 22.0 MW

Answer: A
clarification: Capacity of plant:
P = wQHn_o X 10^-3 KW
P = 9810 X 65 X 12 X 0.80 X 10^-6
= 6.1214MW.

14. A hydel plant is supplied from a reservoir of 5 X 10⁶ m³ capacity at a head of 75m. Determine the number of electrical units produced (KWh) during the year if the load factor is 0.6 and overall efficiency of generation is 72%?
A. 441.45MWh
B. 300.22MWh
C. 235MWh
D. 182MWh
View Answer

Answer: A
clarification: The power capacity of plant in KW is given as
P = mgH/1000 X n{overall}
= (5 X 106 X 1000 X 9.81 X 75 X 0.72) / (365 X 24 X 3600 X 1000)
= 83.99Kw
Energy produced in kWh = P X Load factor X (365 X 24)
= 83.99 X 0.6 X 365 X 24
= 4441451.44kwh
= 441.45 Mwh.

15. The graph of the cumulative values of water quantity against time is known as __________
A. Flow curve
B. Power curve
C. Mass curve
D. Load curve

Answer: C
clarification: The graph of the cumulative values of water quantity against time is known as mass curve. A mass curve of the hydrograph which expresses the area under the hydrograph from one time to another.