Category: Objective Questions

Molecular Biology Multiple Choice Questions on “Repeated Sequences are a Feature of Eukaryotic DNA”.

1. Which of the following is not a part of a gene?
A. Ori
B. Promoter
C. Operon
D. Terminator

Answer: A
Explanation: Ori is the site for the origin of replication. This does not take part in any kind of gene expression and thus is not a part of any gene. Its function is to govern genome duplication.

2. What is the correct order for increasing gene density?
A. Bacteria, Virus, Fruit fly, Human
B. Fruit fly, Bacteria, Virus, Human
C. Human, Fruit fly, Bacteria, Virus
D. Virus, Bacteria, Fruit fly, Human

Answer: C
Explanation: As the complexity of the organism increases the gene density decreases. Thus, gene density is inversely proportional to organism complexity. Therefore, viruses being the simplest organisms have the highest gene density and humans the lowest.

3. Which of these factors contribute to a decrease in gene density?
A. Operon
B. Intergenic sequence
C. Exon sequence
D. Cell size

Answer: B
Explanation: As the genome size increases the DNA between two consecutive sequences increase, known as the Intergenic sequences. Thus, with the increase in genome complexity, regulatory sequences are required for the proper functioning of the different mechanisms. These regulatory sequences are stored in these Intergenic sequences.

4. What do you think is the requirement of Intergenic DNA in higher organisms?
A. Just genetic load
B. To avoid viable mutations
C. Helps in regulation of transcription
D. Helps in genome organization

Answer: B
Explanation: With the increase in genome size the gene density decreases thus the rate of mutation at a viable point also decreases. This is why viable mutations in bacteria and viruses are more frequent than in humans.

5. Which of the following is not a character of unique Intergenic DNA?
A. Mutant genes
B. Pseudogenes
C. Palindromes
D. Nonfunctional gene fragments

Answer: C
Explanation: Palindromes are specialized sequences identified by different enzymes and proteins. They are unique structures but are not found in unique Intergenic regions of DNA. These are generally found in the highly functioning DNA.

6. Which of the following is responsible for the formation of Pseudogenes?
A. Polymerase
B. Reverse transcriptase
C. Klenow fragment
D. Wrong primer

Answer: B
Explanation: Pseudogenes arises from the action of enzyme reverse transcriptase. This enzyme copies RNA into double stranded DNA and are found in certain RNA viruses. Thus, when infected by these specific viruses the reverse transcriptase might copy the host cellular mRNA into a double stranded copy DNA. This copy DNA might get incorporated into the host genome giving rise to Pseudogenes.

7. With respect to microsatellite DNA which of the following is correct?
A. Tandem repeats
B. Dinucleotide repeats
C. 100 bp units
D. Inaccurate duplicating

Answer: C
Explanation: Microsatellites are dinucleotide repeats such as “ATATATATATAT”. Thus, they are of a small stretch of DNA and are very short and less than 13 base pairs.

8. Genome – wide repeat unit are generally transposable elements.
A. True
B. False

Answer: A
Explanation: Genome – wide repeat unit are greater than 100 bp in length. They are found in either as single copies dispersed throughout the genome or in clusters. They are of numerous classes but their most common feature is that all are forms of transposable elements.

9. Repeated DNA can be referred to as junk DNA.
A. True
B. False

Answer: B
Explanation: Although it is tempting to refer to repeated DNA as junk DNA, they are stably maintained over the hundreds and thousands of years. This suggests that Intergenic DNA confers a positive value to the host organisms.

10. The simpler the organism the more likely they are to have transposable elements.
A. True
B. False

Answer: B

Molecular Biology Multiple Choice Questions on “Properties of DNA Polymerase – 1”.

1. Which of the following types of DNA polymerase has 3’→5’ exonuclease activity?
A. DNA polymerase I
B. DNA polymerase II
C. DNA polymerase III
D. DNA polymerase IV

Answer: A
Explanation: DNA polymerases I, II, III and IV all has 5’→3’ exonuclease activity. DNA polymerases I is the only polymerase to have the 3’→5’ exonuclease activity which is the proof reading activity of DNA polymerase.

2. Choose the odd one out with respect to DNA polymerase III.
A. dna E
B. dna N
C. dna Q
D. dna B

Answer: D
Explanation: DNA polymerase III is the product of dna E, dna N, dna Q and 7 other structural genes. Only dna B is the odd one in the provided group as it is not the structural gene for any of the subunits of DNA polymerase III.

3. Which of the following is used in prokaryotic replication?
A. DNA polymerase I
B. DNA polymerase II
C. DNA polymerase III
D. DNA polymerase δ

Answer: C
Explanation: Prokaryotic replication is polymerized by the enzyme DNA polymerase III. DNA polymerase I, DNA polymerase II and DNA polymerase δ all take part in the eukaryotic replication.

4. DNA polymerase II can polymerase upto ______________ nucleotides per minute at 37˚C.
A. 50
B. 500
C. 5000
D. 50000

Answer: A
Explanation: DNA polymerase II has the lowest affinity for dNTPs and can polymerase upto 50 nucleotides per minute at 37˚C. DNA polymerase III has the highest affinity for dNTPs and can polymerase upto 15000 nucleotides per minute at 37˚C.

5. All mutation in polymerases is lethal except the mutation of polymerase III.
A. True
B. False

Answer: B
Explanation: Mutations in both DNA polymerase I and DNA polymerase III are lethal. Only exception to this is polymerase III, mutation to which is not lethal.

6. Polymerase I has the highest affinity for dNTPs.
A. True
B. False

Answer: B
Explanation: DNA polymerase I have a low affinity for dNTPs and can polymerase upto 1000 nucleotides per minute at 37˚C. DNA polymerase III has the highest affinity for dNTPs and can polymerase upto 15000 nucleotides per minute at 37˚C.

7. Polymerase I has two fragments. The residue of larger fragment consists from ____________ residues.
A. 1 – 234
B. 234 – 658
C. 324 – 928
D. 234 – 928

Answer: C
Explanation: Polymerase I has two fragments. The residue of larger fragment consists from 324 – 928 residues is known as the Klenow fragment which has the polymerase activity as well as the 5’→3’ exonuclease activity.

8. Which of the following is false about klenow fragment?
A. Polymerization activity
B. 3’→5’ exonuclease activity
C. 5’→3’ exonuclease activity
D. 324 – 928 residue of polymerase I

Answer: B
Explanation: The residue of larger fragment consists from 324 – 928 residues is known as the klenow fragment which has the polymerase activity as well as the 5’→3’ exonuclease activity. The 3’→5’ exonuclease activity or the proofreading activity is present in the small fragment of DNA polymerase I.

9. DNA polymerase III has ___________ subunits.
A. 4
B. 6
C. 8
D. 10

Answer: D
Explanation: DNA polymerase II has 10 subunits. They are αa, ϵa, θa, τ, γb, δb, δ’b, χb, Ψb and β the structural genes of which are pol C, dna Q, hot E, dna xc, dna χc, hot A, hot B, hot C, hot D and dna N respectively.

10. Which of the following types of DNA polymerase does not take part in DNA repair?
A. DNA polymerase I
B. DNA polymerase II
C. DNA polymerase III
D. DNA polymerase IV

Answer: C

Molecular Biology Multiple Choice Questions on “RNA is Processed in Several Ways”.

1. mRNA of which of the following organism does not undergo processing?
A. Human
B. Yeast
C. Bacteria
D. Fungi

Answer: C
Explanation: Most newly synthesized RNAs must be modified in various ways to convert to their functional forms. Bacterial mRNAs are an exception; they are used immediately as templates for protein synthesis while still being transcribed.

2. rRNA and tRNA of prokaryotes does not undergo any post transcriptional processing.
A. True
B. False

Answer: B
Explanation: rRNA and tRNA of prokaryotes do undergo post transcriptional processing such as splicing and structural modification. Only bacterial mRNAs are an exception; they are used immediately as templates for protein synthesis while still being transcribed.

3. Splicing of mRNA occurs after they arrive the cytoplasm.
A. True
B. False

Answer: B
Explanation: Primary transcript of eukaryotic mRNAs undergo extensive modifications, including the removal of introns by splicing, before they are transported from the nucleus to the cytoplasm to serve as templates for protein synthesis. Regulation of these processing steps provides an additional level of control of gene expression, as does regulation of the rates at which different mRNAs are subsequently degraded within the cell.

4. Which of the following is not a type of RNA processing?
A. Polyadenylation at the 3’ end
B. Capping of 5’ end
C. Removal of exons
D. Splicing

Answer: C
Explanation: The processing events include the following:
i) Capping of the 5’ end of RNA.
ii) Splicing for removing the introns.
iii) Polyadenylation of the 3’ end of the RNA.

5. Elongation of transcription and RNA processing are all interconnected.
A. True
B. False

Answer: B
Explanation: One elongation factor (hSPT5) also recruits and stimulates the 5’ capping enzyme. In another case, elongation factor TAT – SF1 recruits components of the splicing machinery. Thus it seems that elongation, termination of transcription, and RNA processing are interconnected to ensure their proper coordination.

6. The first RNA processing event is __________
A. Capping
B. Tailing
C. Splicing
D. Editing

Answer: A
Explanation: The first RNA processing event is capping. This involves the addition of the modified guanine base to the 5’ end of the RNA.

7. Capping is done by the addition of __________
A. Methylated A
B. Methylated T
C. Methylated G
D. Methylated C

Answer: C
Explanation: The addition of methylated guanine to the RNA transcript by an unusual 5’ – 5’ linkage involving three phosphates. The 5’ cap is created in three enzymatic steps. In the first step, a phosphate group is moved from the 5’ of the transcript. Then, the GTP is added. And in the final step, that nucleotide is modified by the addition of a methyl group.

8. Capping occurs after the mRNA synthesis is complete.
A. True
B. False

Answer: B
Explanation: The RNA is capped when it is still only some 20 – 40 nucleotides long – when the transcription cycle has progressed only to the transition between the initiation and elongation phases. After capping, dephosphorylation of Ser5 within the tail repeats leads to dissociation of the capping machinery and further phosphorylation causes recruitment of the machinery needed for RNA splicing.

9. As the polymerase reaches the end of RNA which of the following event does not occur as a response?
A. Transfer of Polyadenylation enzyme
B. Cleavage of the RNA
C. Addition of poly A at the 3’ end
D. Termination of transcription

Answer: C
Explanation: Once polymerase reaches the end of a gene, it encounters specific sequences that, after being transcribed into RNA, trigger the transfer of the polyadenylation enzymes to that RNA. This leads to three events:
i) Cleavage of the RNA
ii) Addition of many adenine residues at the 5’ end
iii) Subsequent termination of transcription by a polymerase.

10. About how many “A” are added to the nascent RNA in the 5’ end during Polyadenylation?
A. 100
B. 200
C. 300
D. 0

Answer: D
Explanation: Polyadenylation is mediated by an enzyme called poly – A polymerase, which adds about 200 adenines to the RNA’s 3’ end product by the cleavage. This enzyme uses ATP as a precursor and adds the nucleotides using the same process as RNA polymerase, but does not require a template.

11. The long tail of As is unique to the RNA synthesized by the RNA polymerase __________
A. I
B. II
C. III
D. IV

Answer: B

Tricky Molecular Biology Questions on “Differences Between Eukaryotic and Prokaryotic Protein Synthesis, Mistranslation in Protein Synthesis”.

1. The subunit association with the mRNA occurs before tRNA recruitment in the case of both eukaryotes and prokaryotes.
A. True
B. False

Answer: B
Explanation: The subunit association with the mRNA occurs before tRNA recruitment only occurs in case of prokaryotic translation. In case of eukaryotes, the small subunit is already associated with an initiator tRNA when it is recruited to the capped 5’ end of the mRNA.

2. Eukaryotic mRNA generally codes for a single protein because the eukaryotic translation includes ____________
A. A single ORF
B. Monocistronic codons
C. Polycistronic codons
D. Starts at the first triplet codon

Answer: B
Explanation: Eukaryotic mRNA generally codes for a single protein because the eukaryotic translation includes monocistronic codons. As the tRNA ribosome complex scans the mRNA thus they begin translation as soon as they encounter the start codon.

3. The same number of auxiliary proteins is used to drive the initiation process of both prokaryotes and eukaryotes.
A. True
B. False

Answer: B
Explanation: Eukaryotic cells require many more auxiliary proteins to drive the initiation processes than do the prokaryotes. Remarkably, more than 30 different polypeptides are involved in the initiation of translation in eukaryotes.

4. The first amino acid to be incorporated in the eukaryotic polypeptide is ___________
A. Methionine
B. Valine
C. N-formyl methionine
D. N-acyl valine

Answer: A
Explanation: The first amino acid to be incorporated in case of prokaryotic translation is N-formyl methionine. But in case of eukaryotic translation the first amino acid to be incorporated is methionine and is represented as Met-tRNAiMet.

5. Which of the following eukaryotic proteins is an analog of IF2-GTP?
A. eIF2B-GTP
B. eIF2-GTP
C. eIF5B-GTP
D. eIF4G

Answer: C
Explanation: In case of nomenclature the eukaryotic analog of IF2-GTP is eIF5B-GTP. This factor associates with the small subunit in an eIF1A-dependent manner.

6. Recognition of mRNA by 43S pre-initiation complex is mediated by the enzyme ___________
A. eIF2-GTP
B. eIF4F
C. eIF5B-GTP
D. eIF1

Answer: B
Explanation: Recognition of mRNA by 43S pre-initiation complex is mediated by the enzyme begins with the recognition of the 5’ cap found at the end of most eukaryotic mRNAs. This recognition process is mediated by the three subunit protein called eIF4F.

7. Which of the following protein interacts with the eIF4F for recruiting the pre-initiation complex to the mRNA?
A. eIF1
B. eIF2
C. eIF3
D. eIF4

Answer: C
Explanation: The eIF4F/B bound unstructured mRNA recruits the 43S pre-initiation complex to the mRNA. This is facilitated through the interactions between eIF4F/B and eIF3.

8. mRNA scanning is a cost-effective mechanism.
A. True
B. False

Answer: A
Explanation: Once assembled at the 5’ end of the mRNA, the small subunit and its associated factors move along the mRNA in a 5’→3’ direction. This is an ATP dependent process that is driven by the eIF4F-associated RNA helicase.

9. The initiation factors are closely associated with the 3’ end of mRNA through the initiation between the eIF4F and the ___________
A. Poly A tail
B. Poly A binding proteins
C. Small subunit of ribosome
D. Large subunit of ribosome

Answer: B
Explanation: The initiation factors are closely associated with the 3’ end of mRNA through the initiation between the eIF4F and the poly A tail. This interaction is mediated by the interaction between The initiation factors are closely associated with the 3’ end of mRNA through the initiation between the eIF4F and the poly-A binding proteins that coats the poly A tail.

10. IREs for the process of translation is present in viruses. Their function resembles that of eukaryotic ribosome.
A. True
B. False

Answer: B
Explanation: IREs or internal ribosome entry sites are RNA sequences that function like the prokaryotic ribosome binding site. They recruit the small subunit to bind and initiate translation at an internal site in the mRNA. Thus these mRNA are polycistronic and produce a variable amount of proteins unlike the eukaryotic mechanism in which the mRNA is monocistronic and often produces a single polypeptide.

11. How many types of point mutations are able to alter the Genetic code?
A. 1
B. 2
C. 3
D. 4

Answer: C
Explanation: There are four types of point mutations that are able to alter the Genetic code. They are missense mutation, nonsense mutation and frame shift mutation.

12. The alterations that change a codon specific for one amino acid to a codon specific for another amino acid is known as ___________
A. Missense mutation
B. Nonsense mutation
C. Frame shift mutation
D. Reverse Mutation

Answer: A
Explanation: The alterations that change a codon specific for one amino acid to a codon specific for another amino acid is known as a missense mutation. As a consequence, a gene bearing a missense mutation produces a protein product in which a single amino acid has been substituted for another.

13. Which of the following disease is caused by missense mutation?
A. Thalassemia
B. Sickle cell anemia
C. Hemophilia
D. Cystic fibriosis

Answer: B
Explanation: A classic example of a disease caused by missense mutation is sickle cell anemia. In this disease, the amino acid glutamate at position 6 in the β-globin subunit of hemoglobin has been replaced with a valine.

14. Nonsense mutation is also known as ___________
A. Missense mutation
B. Frame shift mutation
C. Stop mutation
D. Reverse mutation

Answer: C

Molecular Biology Multiple Choice Questions on “Plasmid DNA Replicates by Two Alternative Methods – 1”.

1. Transformation involves the uptake of _____________
A. Free DNA
B. Free RNA
C. Free tRNA
D. Free mRNA

Answer: A
Explanation: Transformation involves the uptake of the free DNA molecules. These DNA are released from one bacterium and taken up by the other to complete the process of transformation.

2. What is conjugation?
A. Transformation of DNA from a donor to the recipient cell
B. Interchanging of two types of DNA
C. Transfer of DNA from recipient to donor
D. Uptake of DNA from hosts

Answer: A
Explanation: Conjugation occurs between two bacteria, one donor and one recipient cell. It involves the direct transfer of DNA from a donor cell to a recipient cell containing the F plasmid.

3. What is transduction?
A. Bacterial genes are transferred from donor to recipient cells
B. Bacterial genes are transferred from recipient to donor cells
C. Bacterial genes are carried from a donor cell to a recipient cell via bacteriophage
D. Uptake of naked DNA by hosts

Answer: C
Explanation: In transduction process bacteria is lysed. Thus the bacterial genes are carried from a donor cell to a recipient cell by a bacteriophage that undergoes a lytic lifecycle inside the host.

4. In which year Frederic Griffith discovered transformation?
A. 1920
B. 1928
C. 1936
D. 1944

Answer: B
Explanation: Frederick Griffith discovered transformation in Streptococcus pneumoniae in 1928. Pneumococci exhibited genetic variability in his experiments of causing pneumonia in rats.

5. During Frederick Griffith’s experiments what type of colonies were generated by the capsulated pneumococci?
A. Type R
B. Type L
C. Type S
D. Type B

Answer: C
Explanation: During the experiment, when grown in blood agar media in petridishes. Pneumococci with capsules form large, smooth colonies. These colonies are designated as type S colonies.

6. What is the function of the capsule of the bacterial cells?
A. Virulence
B. Protection
C. Adherence
D. Both Virulence and Protection

Answer: D
Explanation: The polysaccharide capsule of the bacterial cell is required for virulence. It also provides protection to the bacterial cell from destruction by the host immune mechanism.

7. How many types of bacterial capsule are present for pneumococcus depending on the genotype of the cell?
A. 1
B. 2
C. 3
D. Infinite

Answer: D
Explanation: The polysaccharide capsule may be of several different antigenic types. This depends on the specific molecular composition of the polysaccharide and it also depends on the genotype of the cell.

8. The genetic information is stored in DNA rather than protein.
A. True
B. False

Answer: A
Explanation: The genetic information is stored in DNA. This was first discovered by Oswald Avery, Colin Macleod and Maclyn McCarty in 1944.

9. In the following which bacteria only take up their own DNA?
A. S. pneumoniae
B. B. subtilis
C. N. gonorrbacae
D. Mycoplasma

Answer: C
Explanation: S. pneumoniae and B. subtilis take up DNA from any source. Only H. influenzae and N. gonorrbacae takes up DNA from their own or closely related species.

10. How many copies of special short nucleotide pair sequences for identification are present in H. influenzae and N. gonorrbacae?
A. 100 copies
B. 600 copies
C. 500 copies
D. 1000 copies

Answer: B
Explanation: H. influenzae and N. gonorrbacae only takes up DNA from their own or closely related species. This is facilitated by the presence of a special short nucleotide pair sequences for identification which is present roughly about 600 copies in their respective genomes.

11. What do you mean by competent bacteria?
A. Able to take up genetic material from surrounding
B. Able to transfer genetic material to the surrounding
C. Takes up only its own genetic material
D. Can transfer genetic material from one bacterium to another

Answer: A
Explanation: There are some bacterial species that have the ability to take up DNA from their surroundings. These bacteria encode certain proteins that facilitate the process of this type of transformation. Such bacteria are known as competent bacteria.

12. Competence in a bacteria ___________
A. DNA
B. RNA
C. Protein
D. Polysaccharides

Answer: C

Energy Engineering Multiple Choice Questions on “Cool and Ash Handling System – 1”.

1. What is the role of breaker house in coal feeding?
A. To break the coal into smaller pieces
B. To separate different sizes of coal
C. To separate the light dust from the coal
D. To powder the coal
View Answer

Answer: C
clarification: Because of the brittle nature of the coal, it is a common nature of coal to emit light dust/coal dust during transportation, mining and machine handling. This dust needs to be cleared out and it is performed by coal breaker.

2. When coal is being burnt how much % of ash is formed compared to the whole amount?
A. 10-20%
B. 40-50%
C. 25-35%
D. 4-10%

Answer: A
clarification: The coal available in nature already contains some percent of ash and when it is burnt. Due to its brittle nature more amount of ash is produced. And coal is one of the largest types of industrial waste generated. For environmental benefits, this coal ash is reused as a type of by-product in different types of industries.

3. Why is it important to prefer ash handling systems?
A. Coal ash produced destroys the machinery by entering into them
B. Coal ash produced annually accounts for thousands of tones
C. Coal ash can be reutilized for some other purpose
D. Coal ash affects the health of people working at plants
View Answer

Answer: B
clarification: Considering the large coal burning capacity plant of modern times, the amount of ash produced when the coal is burnt is in thousands of tones. It could have an effect on other subjects too if the proper ash handling methods are not followed. And for different environmental, economical and product benefits the coal ash is reused by different types of industries in different ways of its necessity.

4. Large amount of coal is transported by ________
A. railway
B. sea or river ways
C. road transportation
D. by airlifting
View Answer

Answer: A
clarification: The railway is preferred since it is quite economical as well as loading and unloading of coal is easy. And at stretch tons of coal can be transported by goods train from one place to another place. Coal is considered as a bulk commodity value which falls in category of minerals and ores. If roadways are preferred only minimal amount of load can be transported. But by railway, long distances with huge amount of coal can transported to any region/part of the country.

5. The coal is fed to the furnace through _________
A. conveyor belt
B. wagon tipper
C. hopper
D. crane

Answer: C
clarification: Hopper is the conical shaped slow coal dispenser to the furnace. It is placed right above the furnace and a live feeder mechanism is set at end of hopper for a controlled flow. There is no requirement of any external power/force since this works on gravity force.

6. Which system consumes less power out of all ash handling systems?
A. Mechanical ash handling system
B. Pneumatic ash handling system
C. Hydraulic ash handling system
D. Steam jet ash handling system

Answer: A
clarification: The mechanical ash handling system consumes less amount of power. Since the power is required by the conveyor belt to transfer the ash from boiler furnace to over head bunker which is located at end of the conveyor belt. And in case of pneumatic there is high power requirement to draw and blow the air at high velocities and high pressures.

7. What is the function of cyclone separators in pneumatic ash handling system?
A. To separate the lighter dust particles
B. To force up the movement of ash through pipes or tubes
C. To draw out the dust from furnace
D. To separate minute coal particles

Answer: A
clarification: Cyclone separators use air to swirl around the ash that has been dispensed into them. Due to centrifugal action heavier ash settles down, where as lighter dust/ash particles is collected in hopper and dumped out. The air flows in helical pattern which makes easy for the heavier dust particle to settle down easily without interrupting the airflow.

8. Which medium is used to carry ash in the pneumatic ash handling system?
A. Conveyor belt
B. Water trough
C. Air
D. Chain belt
View Answer

Answer: C
clarification: In pneumatic ash handling system, Air is used to carry ash to long distance at a capacity of 5 to 30 tonnes per hour. And the air used for this purpose is easily cleanable and can be exhausted back into atmosphere after the complete filtration processes.

9. Which system is noisy out of all the following ash handling systems?
A. Steam jet ash handling system
B. Mechanical ash handling system
C. Pneumatic ash handling system
D. Hydraulic ash handling system
View Answer

Answer: C
clarification: The air is made to pass at very high pressure in order to carry out the ash for long distance. Since the air is moving at high speed at high velocity in the conveying pipes, it tends to create a lot of noise by hitting the walls of pipe at swift turns and curves.

10. Which medium is used to carry ash in hydraulic system?
A. Air
B. Water
C. Steam
D. Conveyor

Answer: B
clarification: Water is used as the medium to carry ash at high velocity. Depending on water pressure the system is divided as Low pressure system and High pressure system. In low pressure system, sloped sumps are used to move the ash at low velocity and in high pressure system nozzle sprays used to ram up the speed of ash flow.

11. What would be the amount of distance that a low pressure system could carry the ash?
A. 25m
B. 500m
C. 150m
D. 800m
View Answer

Answer: B
clarification: Low pressure system moves the ash mixed in water at a distance of 3 to 5 m/s in a sloped pump made of reinforced constituents and this movement is continuous. So, it has the ability to carry the ash for such long distance. There is no requirement any auxiliary source to move the ash mixed with water.

12. What is the capacity of low pressure hydraulic ash handling system?
A. 80 tonnes/hour
B. 22 tonnes/hour
C. 50 tonnes/hour
D. 10 tonnes/hour

Answer: C
clarification: The capacity of low pressure hydraulic ash handling system is 50tonnes/hour at a speed of 3m/s. Since the ash produced is mixed in water and dumped, the water has the ability to dissolve and intake more amount of ash. And this mixture is spread throughout the sump.