Category: Objective Questions

Molecular Biology Multiple Choice Questions on “Fundamentals of PCR”.

1. PCR is a DNA amplifying in vivo method.
A. True
B. False

Answer: B
Explanation: A powerful method for amplifying a particular DNA segment is the polymerase chain reaction. This procedure is carried out entirely biochemically, that is invitro.

2. Which of the following is a mismatch?
A. Polymerase – Taq polymerase
B. Template – double stranded DNA
C. Primer – oligonucleotide
D. Synthesis – 5’ to 3’ direction

Answer: B
Explanation: PCR uses the enzyme DNA polymerase that directs the synthesis of DNA from deoxynucleotide substrates on one stranded DNA template. DNA polymerase synthesizes DNA in a 5’→3’ direction and can add nucleotides in the 3’ end of a custom – designed oligonucleotide.

3. Annealing of primer is facilitated by complementary region.
A. True
B. False

Answer: A
Explanation: A synthetic oligonucleotide is annealed to a single stranded DNA template that contains a complementary region. DNA polymerase thus uses this annealed primer to elongate the new strand.

4. Primer used for the process of polymerase chain reaction are ______________
A. Single stranded DNA oligonucleotide
B. Double stranded DNA oligonucleotide
C. Single stranded RNA oligonucleotide
D. Double stranded RNA oligonucleotide

Answer: A
Explanation: Two single stranded, oligonucleotide is synthesized and used as a primer. The primers thus produces binds to the denatured DNA and polymerase starts its synthesis in the 5’ to 3’ direction.

5. Polymerase used for PCR is extracted from _____________
A. Escherichia coli
B. Homo sapiens
C. Thermus aquaticus
D. Saccharomyces cerevisiae

Answer: C
Explanation: Amplification is usually carried out by the DNA polymerase I enzyme from Thermus aquaticus. As this organism lives in hot springs, Taq polymerases are thermostable thus they are resistant to high temperature.

6. How many DNA duplex is obtained from one DNA duplex after 4 cycles of PCR?
A. 4
B. 8
C. 16
D. 32

Answer: C
Explanation: After each cycle the number of duplexes doubles itself thus after the first cycle there are 2 DNA duplexes. A duplex has 2 DNA strands thus after second cycle there will be 4 Duplexes, after third cycle there will be 8 DNA duplex. Lastly, after the 4th cycle there will be 16 duplexes.

7. At what temperature do denaturation of DNA double helix takes place?
A. 60˚
B. 54˚
C. 74˚
D. 94˚

Answer: D
Explanation: The polymerase chain reaction is started by heating the mixture at 94˚C. At this temperature the hydrogen bonding between the nucleotide bases melts thus separating the two DNA strands.

8. At what temperature do annealing of DNA and primer takes place?
A. 42˚
B. 54˚
C. 74˚
D. 96˚

Answer: D
Explanation: After the denaturation of the two strands the temperature is decreased to 50 – 60˚C. At this temperature the primers anneal to their complementary segments. Thus as 54˚ lies within this range, 54˚ is the correct option.

9. The temperature at which DNA synthesis takes place is 74˚C.
A. True
B. False

Answer: A

Molecular Biology Multiple Choice Questions on “Manufacturing the Message – 1”.

1. Without using primers for RNA synthesis the polymerase produces small stretches of RNA of _____________ nucleotides.
A. 8
B. 10
C. 12
D. 14

Answer: B
Explanation: Without using primers for RNA synthesis the polymerase produces small stretches of RNA of 10 nucleotides. These transcripts are released from the polymerase which without dissociating from the template begins the synthesis of RNA again.

2. To facilitate binding of σ⁷⁰ the melting of the template and non – template strands occurs between the positions ____________
A. -11 to +3
B. -9 to +5
C. -10 to +1
D. -10 to +2

Answer: A
Explanation: The transition from the closed to open complex involves structural changes in the enzyme and opening of the DNA duplex to reveal the template and the non-template strands. This melting occurs between positions -11 to +3, in relation to the transcription start site.

3. The isomerisation of σ⁷⁰ of bacterial polymerase is an energy dependent process.
A. True
B. False

Answer: B
Explanation: The isomerisation of σ⁷⁰ of bacterial polymerase is not an energy dependent process. Instead isomerisation is the result of a spontaneous conformational change in the DNA-enzyme complex to a more energetically favorable form.

4. With respect to the formation of a complex between polymerase and DNA template which of the following is correct?
A. Open: reversible; Closed: irreversible
B. Open: irreversible; Closed: reversible
C. Open and closed: reversible
D. Open and closed: irreversible

Answer: B
Explanation: Isomerisation is essentially irreversible thus open complexes are irreversible and once forms transcription will be initiated. But the closed complex is readily reversible and the polymerase can easily dissociate from the promoter and is not bound to transcribe the DNA.

5. How many channels are found in the RNA polymerase clamp?
A. 3
B. 4
C. 5
D. 6

Answer: C
Explanation: The RNA polymerase clamp has 5 channels. They are NTP uptake channel, NT channel, T channel, RNA-exit channel and downstream DNA channel.

6. The channel that allows the exit of the coding strand is known as ____________
A. NTP channel
B. NT channel
C. T channel
D. Downstream DNA channel

Answer: B
Explanation: The channel that allows the exit of the coding strand is known as the NT channel. NT channel is the non-template exit channel. Non-template strand is also known as the coding strand.

7. Region 1.1 of the σ factor mimics the DNA and is negatively charged.
A. True
B. False

Answer: A
Explanation: Yes, the region 1.1 of the σ factor mimics the DNA and is negatively charged. The space in the active center cleft which is either occupied by the region 1.1 or by DNA, is highly positively charged.

8. RNA polymerase adds the nucleotides de novo. The first nucleotide added by the polymerase is ____________
A. ATP
B. TTP
C. GTP
D. CTP

Answer: A
Explanation: The first nucleotide added by the polymerase is always a purine. In this case, adenine is more common than guanine thus ATP is the first nucleotide added.

9. The ternary complex is formed by the polymerase, DNA duplex and the nascent RNA.
A. True
B. False

Answer: B
Explanation: The ternary complex is formed by the polymerase, single stranded DNA template and the nascent RNA. This is formed after the polymerization of atleast 10 nucleotides which stabilizes the structure.

10. After the binding of polymerase to the DNA template the σ factor is ejected.
A. True
B. False

Answer: B

Molecular Biology Questions on “Overview of Protein Synthesis – 2”.

1. Eukaryotic ribosome is recruited to the mRNA by _____________
A. Randomly
B. Shine – Dalgarno sequence
C. 5’ capping
D. 3’ tailing

Answer: C
Explanation: Prokaryotic ribosome is recruited to the mRNA by the Shine – Dalgarno sequence. In the case of eukaryotes ribosome is recruited to the mRNA by the 5’ cap.

2. After bounding to the mRNA eukaryotic ribosome scans the entire mRNA until it encounters a start codon.
A. True
B. False

Answer: A
Explanation: The eukaryotic ribosome is recruited to the mRNA by the 5’ cap. The guanine residue of the 5’ cap is connected to the 5’ end of the mRNA through three phosphate groups that recognizes and recruits the ribosome. Once bound to the mRNA, the ribosome moves in a 5’→3’ direction until it encounters a 5’…….AUG…….3’ start codon by a process called scanning.

3. The end of all tRNAs is ___________
A. 5’ ACC 3’
B. 5’ CCA 3’
C. 3’ CAC 5’
D. 3’ GAG 5’

Answer: B
Explanation: All tRNAs end at the 3’ terminus with the sequence 5’ CCA 3’. This is the site that is attached to the cognate amino acid by the enzyme aminoacyl tRNA synthetase.

4. How many unusual bases are observed in a tRNA molecule?
A. 2
B. 3
C. 4
D. 5

Answer: D
Explanation: Five unusual bases, produced by enzymatic modification of the usual bases, may be observed in a tRNA molecule. These are pseudouridine, dihyrdouridine, hypoxanthine, thymine and methylguanine. Of these five, unusual bases pseudouridine and dihyrdouridine are most commonly found in the tRNA molecules.

5. In pseudouridine the attachment of Uracil base to ribose through ___________ of uracil.
A. N at position 3
B. C at position 6
C. C at position 5
D. N at position 1

Answer: C
Explanation: Pseudouridine is produced by the enzymatic modification of the regular base uridine through isomerization. In the modification process of uridine to pseudouridine the attachment of base to the ribose sugar is switched from nitrogen at ring position 1 to the carbon at ring position 5.

6. Dihydrouridine is the result of oxidation of the uracil base.
A. True
B. False

Answer: B
Explanation: Dihydrouridine is a derivative of uridine formed by enzymatic reduction. The process of reduction forms the double bond between the carbons at positions 5 and 6.

7. How many loops are seen in the tRNA?
A. 1
B. 2
C. 3
D. 4

Answer: D
Explanation: Four different loops are observable in the tRNA 3˚ motif. They are the ΨU loop, D loop, variable loop and the anticodon loop. The ΨU loop is named because of the presence of the pseudouridine base. The D loop contains the dihyrdouridine base. The anticodon loop contains the anticodon and the variable loop is named so because its length is variable.

8. How many double helix regions are observed in a tRNA molecule?
A. 4
B. 3
C. 2
D. 1

Answer: A
Explanation: The tRNA molecules consist of 4 double helix regions. They are the acceptor stem and the stem of the three loops, the ΨU loop, D loop and the anticodon loop.

9. The variable loop is found between the ___________
A. Acceptor stem and ΨU loop
B. Acceptor stem and D loop
C. Anticodon loop and ΨU loop
D. Anticodon loop and D loop

Answer: C
Explanation: The variable loop sits between the anticodon loop and ΨU loop. Also as the name implies the size of the variable loop varies from 3 to 21 bases.

10. The 3 dimension structure of an adaptor molecule of translation is revealed to be ___________
A. Y shaped
B. L shaped
C. K shaped
D. X shaped

Answer: B
Explanation: The adaptor molecule of translation is the tRNA molecule which acts as the bridge between the mRNA code and the respective amino acid. X-ray crystallography reveals an L-shaped tertiary structure in which the terminus of the acceptor stem is at one end of the molecule and the anticodon loop is about 70 Å away at the other end.

11. Which of the following does not take part in stabilizing the cloverleaf model of the tRNA?
A. Base stacking
B. Base and sugar-phosphate backbone interaction
C. Ionic bond
D. Hydrogen bond

Answer: C

Molecular Biology Questions on “Methylation of Eukaryotic DNA Controls Gene Expression – 2”.

1. The CpG islands generally consist of how many GC bases?
A. 100
B. Less than 100
C. More than 1000
D. 1000

Answer: D
Explanation: At certain sites, CpG dinucleotide occurs at a higher frequency and is referred to as CpG islands. CpG islands are approximately 1000 bases long and show an elevated G+C base composition.

2. In humans CpG islands are generally located in the ____________
A. Operator region
B. Promoter region
C. Transcript region
D. Terminator region

Answer: B
Explanation: CpG islands are often associated with the transcription start site, that is, the promoter region. About 60% of human genes have CpG islands at their promoter regions.

3. The Intergenic CpGs are more commonly methylated than in the intragenetic region.
A. True
B. False

Answer: A
Explanation: CpG islands in the genome are located in both intragenic and intergenic regions. In vertebrates over 80% of the methylated cytosine residues are found in the intragenetic regions. In contrast CpGs within the CpG islands are generally either not methylated or have relatively low levels of methylation.

4. Major methylation of CpG islands does not include?
A. Operator
B. Terminator
C. Introns
D. Enhancer

Answer: D
Explanation: The methylation of the genome is persistent throughout and is missing only in regions such as CpG islands within the promoter and enhancer regions. Methylation in these two regions may lead to the silencing of the gene itself.

5. De-novo methylation leads to the phenomenon of genetic imprinting in the offspring.
A. True
B. False

Answer: B
Explanation: Maintenance methylation leads to the phenomenon of genetic imprinting in the offspring. This phenomenon controls the expression of certain genes involved in the development of mammalian embryos.

6. Both the imprinted genes inherited from either parent are expressed in an offspring.
A. True
B. False

Answer: B
Explanation: In an offspring only one of the imprinted genes inherited from either parent is expressed. This phenomenon is known as allelic exclusion.

7. The phenomenon of allelic exclusion leads to the inactivation of one allele of a gene. The active allele is known as ____________
A. Homozygote
B. Homolog
C. Hemizygote
D. Hemilog

Answer: C
Explanation: In the phenomenon of allelic exclusion the active allele is known as hemizygote. And the phenomenon that is exhibited by it is known as functional hemizygosity.

8. Gene of which of the following protein/enzyme/hormone is an example of paternal Hemizygote?
A. Igf-1
B. Igf-2
C. β-globin
D. Insulin

Answer: B

Molecular Biology Multiple Choice Questions on “Site-Directed Mutagenesis & Overview of Recombination”.

1. Which of the following properties is improved by site directed mutagenesis?
A. Physical property
B. Chemical property
C. Kinetic property
D. Integrity

Answer: C
Explanation: Site directed mutagenesis is a process used to achieve protein engineering. Protein engineering improves the kinetic property of the protein by altering the amino acid structure and sequence.

2. What is the formula for annealing temperature for polymerase chain reaction primer?
A. Tm = 81.5 + 0.41 (%AT) – (675/N)
B. Tm = 81.5 + 0.41 (%GC. – (675/N)
C. Tm = 81.2 + 0.41 (%GC. – (672/N)
D. Tm = 81.5 + 0.42 (%AT) – (675/N)

Answer: B
Explanation: For calculating the annealing temperature for a PCR primer is Tm = 81.5 + 0.41 (%GC. – (675/N). Here the % of GC is the total amount of G and C present in the oligonucleotide primer and the target of interested and N is the length of the oligonucleotide primer.

3. What is the function of Dpm I endonuclease in Tm method of site directed mutagenesis?
A. Joining of blunt ends
B. Addition of dNTPs
C. Sensitive to cleavage of methylated DNA
D. Breaking of DNA strand

Answer: C
Explanation: In Tm method the template DNA was deprived from an E. coli cell with an intact restriction modification system. This strand is sensitive to restriction by the Dpn I endonuclease.

4. Which phage is used in oligonucleotide directed mutagenesis?
A. M13
B. Cosmid
C. Phagemid
D. λ – phage

Answer: A
Explanation: M13 phage is single stranded phage. It is used for cloning a specific DNA sequence which will be mutated.

5. Write down the name of scientist who has discovered the method of site directed mutagenesis?
A. Bostein Shortle
B. Craik
C. Grait
D. Joller Smith

Answer: D
Explanation: In 1985 Joller Smith discovered site directed mutagenesis technique. By this technique the nucleotide sequence of a cloned DNA fragment may be changed by site directed mutagenesis using synthetic oligonucleotide.

6. Which two genes are absent in the E. coli strain CJ236?
A. dut-/Ung-
B. Rec-/RecB-
C. duB-/Ung-
D. dut-/Umg-

Answer: A
Explanation: CJ236 which lacks functional dut-phase and uracil N-glycosylase. These are used for generating uracilated single stranded DNA.

7. Which polymerase is used in PCR based mutagenesis?
A. Deep vent R polymerase
B. pfu polymerase
C. Taq polymerase
D. DNA polymerase

Answer: B
Explanation: In PCR based mutagenesis pfu polymerase is used. After 12 – 15 cycles of amplification new strands will be generated.

8. Homologous recombination in germ cells occur in the ____________ phase.
A. Leptotene
B. Zygotene
C. Pachytene
D. Diplotene

Answer: C
Explanation: Homologous recombination in germ cells occurs in the pachytene phase of meiosis I. This exchange is classically termed as crossing over and occurs via the formation of a synaptonemal complex between the two homologous chromosomes.

9. The frequency of crossing over between two genes on the same chromosome is independent of their position.
A. True
B. False

Answer: B
Explanation: The frequency of crossing over between two genes on the same chromosome is dependent of their locus. The more distance between the two genes the more will be the probability of recombination.

10. When two genes on a chromosome are inherited without any record of recombination the genes are said to be ____________
A. Silenced
B. Joined
C. Coordinated
D. Linked

Answer: D
Explanation: When two genes on a chromosome is inherited without any record of recombination the genes are said to be linked. This is because they are placed so close together that no recombination is able to take place. The phenomenon is known as linkage.

11. Homologous recombination does not provide ____________
A. Genetic variation
B. Sequence retrieval
C. Restart of stalled replication
D. Random base incorporation

Answer: D
Explanation: The homologous recombination provides three functions:
i) Provides genetic variation
ii) Allows the retrieval of sequences lost through DNA damage
iii) Provides mechanism for restarting stalled or damaged replication forks.

12. Recombination regulates the function of genes.
A. True
B. False

Answer: A
Explanation: Special types of recombination regulate the expression of some genes. For example, by switching specific segments within chromosomes, cells can put otherwise dormant genes into sites where they are expressed.

13. What is strand invasion?
A. Fork collapse
B. Nick in template strand
C. Formation of lesion
D. Complementary strand pairing

Answer: D
Explanation: The pairing between short regions occurs when a single stranded region of DNA originating from one molecule with its complementary strand in the homologous DNA duplex. This event is known as the strand invasion process.

14. What is the holiday junction?
A. The site of strand break
B. The site of heteroduplex DNA formation
C. Formation of a crossing over complex
D. The site of strand invasion

Answer: C

Energy Engineering Interview Questions and Answers for freshers focuses on “Calorific Value of Fuels – 2”.

1. Which is the common method to relate higher calorific value to lower calorific value?
A. HCV = LCV + H_V (n_{H2O, out} / n_{fuel, in})
B. LCV = HCV + H_V (n_{H2O, out} / n_{fuel, in})
C. HCV = LCV + H_V (n_{fuel, in}/ n_{H2O, out})
D. LCV = HCV + H_V (n_{fuel, in}/ n_{H2O, out})

Answer: A
clarification: H_V – heat vaporization of water.
n_{H2O, out} – moles of water vaporized.
n_{fuel, in} – number of moles of fuel combusted.
High calorific value is equal to low calorific value plus, product of heat of vaporization of water and moles of water vaporized by moles of fuel combusted.

2. Based on what basis are fuels compared?
A. Fire point value
B. High calorific value
C. Flash point value
D. Low calorific value

Answer: D
clarification: On basis of low calorific value the fuels are compared. Low calorific value is the amount of heat evolved when a unit weight of fuel is completely burnt and water vapor leaves with combustion products.

3. Which value is determined by bringing all products of combustion back to original pre-combustion temperature?
A. Higher calorific value
B. Low calorific value
C. Flash point value
D. Fire point value
View Answer

Answer: A
clarification: High calorific value is determined, as all fuels contain hydrogen; they produce water vapor during combustion. When the products of combustion containing water vapor are cooled back to initial temperature, then all water vapors formed condense and evolve latent heat. This adds up to the heat liberated by burning the fuel, producing maximum amount of heat per kg of fuel. This heat is known as the higher calorific value of fuel, and it is denoted by HCV.

4. Which calorific value is same as the thermodynamic heat of combustion?
A. Net calorific value
B. Flash point value
C. Gross calorific value
D. Fire point value

Answer: D
clarification: Gross calorific value is as same as the thermodynamic heat of combustion since the enthalpy change for the reaction assumes a common temperature of the compounds before and after combustion, in which case the water produced by combustion is condensed to a liquid, hence yielding its latent heat of vaporization.

5. Which value is determined by subtracting the heat of vaporization of the water from the higher heating value?
A. Gross calorific value
B. Net calorific value
C. Ignition temperature
D. Fire point temperature
View Answer

Answer: B
clarification: Net calorific value determined. In most of the combustion processes, the products of combustion cannot be cooled to its initial temperature. Thus water vapors don’t condense and hence the latent heat of water vapor is lost to the atmosphere. The resultant heat liberated by the fuel which excludes the latent heat of evaporation of water vapors is known as lower calorific value of fuel.

6. Which formula is used to determine higher calorific value of fuel?
A. Rayleigh’s formula
B. Lamme’s equation
C. Dulongs’s formula
D. Cauchy’s formula

Answer: C
clarification: Higher calorific value of the fuel can be determined by using Dulongs’s formula. Let C, H, O and S represent the percentage by weight of carbon, carbon, oxygen and sulfur respectively.
HCV = 1/100 [33900 + 144000(H – (O/8)) + 9295 S] kJ/Kg
This formula gives gross heating value in terms of the weight fractions of carbon, hydrogen, oxygen and sulfur from the ultimate analysis.

7. Lower calorific value can be determined by equation:
A. LCV = HCV – m*2466
B. LCV = HCV + (m/2466)
C. LCV = HCV – (m/2466)
D. LCV = HCV + (m*2466)
View Answer

Answer: A
clarification: Lower calorific value can be determined equation [LCV = HCV – m*2466]. The latent heat lost to the atmosphere depends on evaporation pressure and the amount of water vapors formed. Due to difficulty in measuring the evaporation pressure, it is assumed that evaporation takes place at a saturation temperature of 15°C. The latent heat corresponding to this saturation is 2466 kl/kg.
m = mass of water vapour formed per kg of fuel burnt.

8. Which fuel has higher calorific value among given fuels?
A. Natural gas
B. Gasoline
C. Diesel
D. Fuel oil

Answer: B
clarification: Gasoline also called as petrol, has the highest calorific value. Gasoline is a transparent petroleum derived liquid that is used primarily as a fuel in internal combustion engines. It consists of mostly of organic compounds obtained by the fractional distillation of petroleum, enhanced with variety of additives.

9. What is amount of minimum air required per kg of liquid fuel for complete combustion using carbon, oxygen, hydrogen and sulfur?
A. 1/23 [8/3 C + 8(H – (O/8)) + S]
B. 1/100 [8/3 C + 8(H – (O/8)) + S]
C. 1/100 [8/3 C + 8(H – (0/8))]
D. 1/23 [8/3 C + 8(H – (0/8))]

Answer: A
clarification: Let C, H, O and S represent percentage by mass of carbon(C), Hydrogen (H₂), oxygen and sulfur respectively.
The mass of oxygen required for complete combustion of fuel is given by,
= 1/100 [8/3 C + 8H – O + S]
= 1/100 [8/3C + 8(H – (0/8)) + S]
As air contains 23% of oxygen by mass, minimum air required for burning one kg of liquid fuel completely is given by,
Min. air required = 1/100 [8/3 C + 8(H – (O/8)) + S] 100/23
= 1/23 [8/3 C + 8 (H – (O/8)) + S].

10. What is minimum amount of air required per m³ of gaseous fuel for complete combustion?
A. 1/21 [(H₂/2) + (CO/2) + 2CH₄ + 3C₂H₄] m³/m³ of fuel
B. 1/100 [(H₂/2) + (CO/2) + 2CH₄ + 3C₂H₄] m³/m³ of fuel
C. 1/21 [(H₂/2) + (CO/2) + 3C₂H₄] m³/m³ of fuel
D. 1/100 [(H₂/2) + (CO/2) + 3C₂H₄] m³/m³ of fuel

Answer: A
clarification: Volumetric analysis of fuels hydrogen, carbon monoxide, methane, ethane, carbon dioxide and nitrogen is done and required minimum amount of oxygen is found for one m³ of gaseous fuel:
O₂ required/m³ of fuel = 1/100[(H₂/2) + (CO/2) + 2CH₄ + 3C₂H₄] m³
As atmospheric air contains 21% of O₂ by volume, minimum air required t burn one m³ of gaseous fuel is given by,
Minimum volume of air required (cm³/m³ of fuel):
= 1/100 [(H₂/2) + (CO/2) + 2CH₄ + 3C₂H₄] 100/21
= 1/21 [(H₂/2) + (CO/2) + 2CH₄ + 3C₂H₄] m³/m³ of fuel.