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Multiple choice question for engineering

Set 1

1. DNA replication is
a) Conservative
b) Non-conservative
c) Semi-conservative
d) None

View Answer

Answer: c [Reason:] Each DNA strand serves as a template for the synthesis of a new strand, producing two new DNA molecules, each with one new strand and one old strand.

2. Semi-conservative DNA replication was first demonstrated in
a) Drosophila melanogaster
b) Escherichia coli
c) Streptococcus pneumonae
d) Drosophila melanogaster

View Answer

Answer: a [Reason:] Semi-conservative DNA replication was first demonstrated in E. coli.

3. Eukaryotes differ from prokaryote in mechanism of DNA replication due to
a) Use of DNA primer rather than RNA primer
b) Different enzyme for synthesis of lagging and leading strand
c) Discontinuous rather than semi-discontinuous replication
d) Unidirectional rather than semi-discontinuous replication

View Answer

Answer: c [Reason:] In eukaryotes one strand of DNA is synthesized continuously but the other one is made of Okazaki fragments.

4. Which of the following is true about DNA polymerase?
a) It can synthesize DNA in the 5’ to 3’ direction
b) It can synthesize DNA in the 3’ to 5’ direction
c) It can synthesize mRNA in the 3’ to 5’ direction
d) It can synthesize mRNA in the 5’ to 3’ direction

View Answer

Answer: a [Reason:] DNA pol can synthesize only a new DNA strand not m-RNA. It can do this in the 5’ to 3’ direction.

5. The reaction in DNA replication catalyzed by DNA ligase is
a) Addition of new nucleotides to the leading strand
b) Addition of new nucleotide to the lagging strand
c) Formation of a phosphodiester bond between the 3’-OH of one Okazaki fragment and the 5’-phosphate of the next on the lagging strand
d) Base pairing of the template and the newly formed DNA strand

View Answer

Answer: c [Reason:] DNA ligase catalyzes the formation of a phosphodiester bond between 3’-OH of one Okazaki fragment and 5’-phosphate of the next.

6. Which of the following reactions is required for proofreading during DNA replication by DNA polymerase III?
a) 5’ to 3’ exonuclease activity
b) 3’ to 5’ exonuclease activity
c) 3’ to 5’ endonuclease activity
d) 5’ to 3’ endonuclease activity

View Answer

Answer: b [Reason:] The 3’ → 5’ exonuclease activity removes the mispaired nucleotide and the polymerase begins again. This activity is known as proofreading.

7. Which of the following enzymes remove supercoiling in replicating DNA ahead of the replication fork?
a) DNA polymerases
b) Helicases
c) Primases
d) Topoisomerases

View Answer

Answer: d [Reason:] Strand separation creates topological stress in the helical DNA structure which is relieved by the action of topoisomerases.

8. DNA unwinding is done by
a) Ligase
b) Helicase
c) Topoisomerase
d) Hexonuclease

View Answer

Answer: b [Reason:] These enzymes move along the DNA and separate the strands using chemical energy from ATP.

9. Which of the following enzymes is the principal replication enzyme in E. coli?
a) DNA polymerase I
b) DNA polymerase II
c) DNA polymerase III
d) None of these

View Answer

Answer: c [Reason:] Only DNA pol III is the principal replication enzyme in E. coli.

10. The enzyme used to join bits of DNA is
a) DNA polymerase
b) DNA ligase
c) Endonuclease
d) Primase

View Answer

Answer: b [Reason:] DNA ligase can be used to join the nicked sites.

Set 2

1. The phenomenon of twisting around itself by a molecule to relieve helical stress is
a) Supercoiling
b) Coiling
c) Elongation
d) Compression

View Answer

Answer: a [Reason:] The phenomenon of twisting around itself by a molecule to relieve helical stress is supercoiling.

2. When there is no net bending of the DNA axis upon itself, the DNA is said to be in a
a) Elongated state
b) Compressed state
c) Relaxed state
d) Denatured state

View Answer

Answer: c [Reason:] When there is no net bending of the DNA axis upon itself, the DNA is said to be in a relaxed state.

3. Over-twisting of a molecule results in
a) Negative supercoiling
b) Positive supercoiling
c) Elongation
d) Compression

View Answer

Answer: b [Reason:] Extra helical twisting leads to positive supercoiling, while subtractive twisting leads to negative supercoiling.

4. Changing twist from relaxed state requires adding energy and increases the
a) Forces in molecule
b) Stress along molecule
c) Strain over molecule
d) None of the above

View Answer

Answer: b [Reason:] Relaxed state is the condition at which there is no net DNA bending, changing twist from this state requires additional energy and increases the stress along molecule.

5. The enzyme responsible for the removal of supercoiling in replicating DNA ahead of the replication fork is
a) Topoisomerase
b) Primase
c) DNA polymerase
d) Helicase

View Answer

Answer: a [Reason:] The enzymes that increase or decrease the extent of DNA under-winding are topoisomerases.

6. The type of topoisomerases which generally relaxes DNA by removing negative supercoiling is
a) Type I
b) Type II
c) Type III
d) Type IV

View Answer

Answer: a [Reason:] Type II enzymes can introduce negative supercoils.

7. The type of topoisomerases that can introduce negative supercoils is
a) Type I
b) Type II
c) Type III
d) Type IV

View Answer

Answer: b [Reason:] Type I generally relaxes DNA by removing negative supercoiling.

8. The type I enzyme(s) is/are
a) Topoisomerases I and III
b) Topoisomerases II
c) Topoisomerase IV
d) Topoisomerases I and IV

View Answer

Answer: a [Reason:] Generally type I enzymes are topoisomerases I, III and type II enzymes are topoisomerase II.

9. Under-winding is measured in terms of
a) Supercoiling
b) Linking difference
c) Positive supercoiling
d) Negative supercoiling

View Answer

Answer: b [Reason:] Under-winding is measured in terms of specific linking difference, σ also called super helical density.

10. Which type of supercoiling takes the form of extended right-handed coils?
a) Plectonemic supercoiling
b) Solenoidal supercoiling
c) Negative supercoiling
d) Positive supercoiling

View Answer

Answer: a [Reason:] Solenoidal (negative) supercoiling takes the form of tight left-handed turns about an imaginary tube-like structure whereas plectonemic supercoiling takes the form of right-handed coils.

Set 3

1. A chlorophyll-like accessory pigment lacking Mg+2 is
a) Pheophytin
b) Cytochrome c
c) Plastocyanin
d) Ferredoxin

View Answer

Answer: a [Reason:] Pheophytin is a chlorophyll-like accessory pigment lacking Mg+2.

2. A soluble Cu-containing electron transfer protein is
a) Pheophytin
b) Cytochrome c
c) Plastocyanin
d) Ferredoxin

View Answer

Answer: c [Reason:] Plastocyanin is a soluble Cu-containing electron transfer protein.

3. Photosystem II is a
a) Pheophytin-quinone type
b) Ferredoxin type
c) Plastocyanin type
d) Plastocyanin-ferredoxin type

View Answer

Answer: a [Reason:] Photosystem II is a pheophytin-quinone type of system.

4. Which of the following describes the complete route by which electrons flow H2O to NADP+?
a) Z scheme
b) A scheme
c) B scheme
d) X scheme

View Answer

Answer: a [Reason:] Z scheme describes the complete route by which electrons flow H2O to NADP+.

5. Plastoquinol formed in PSII is oxidized by
a) Cytochrome b6f
b) Cytochrome c
c) Cytochrome f
d) None

View Answer

Answer: a [Reason:] Plastoquinol formed in PSII is oxidized by cytochrome b6f complex.

6. A mobile, lipid soluble carrier of two electrons present in chloroplasts is
a) PQB
b) Q
c) Cytochrome c
d) Plastocyanin

View Answer

Answer: a [Reason:] Q is present in mitochondria; cytochrome c and plastocyanin are water-soluble protein that carries one electron.

7. A mobile, lipid soluble carrier of two electrons present in mitochondria is
a) PQB
b) Q
c) Cytochrome c
d) Plastocyanin

View Answer

Answer: b [Reason:] PQB is present in chloroplasts; cytochrome c and plastocyanin are water-soluble protein that carries one electron.

8. A water-soluble protein that carries one electron present in mitochondria is
a) PQB
b) Q
c) Cytochrome c
d) Plastocyanin

View Answer

Answer: c [Reason:] PQB and Q are lipid-soluble protein; plastocyanin is present in chloroplast.

9. A water-soluble protein that carries one electron present in chloroplast is
a) PQB
b) Q
c) Cytochrome c
d) Plastocyanin

View Answer

Answer: d [Reason:] PQB and Q are lipid-soluble protein; cytochrome c is present in mitochondria.

10. Cyanobacteria does not use which of the following for both oxidative and photophosphorylation?
a) Cytochrome c6
b) Cytochrome b6f
c) Plastoquinone
d) Plastocyanin

View Answer

Answer: d [Reason:] Cyanobacteria use cytochrome c6, cytochrome b6f and plastoquinone for both oxidative and photophosphorylation.

Set 4

1. Which of the following is true about Michaelis-Menten kinetics?
a) Km, the Michaelis constant, is defined as that concentration of substrate at which enzyme is working at maximum velocity
b) It describes single substrate enzymes
c) Km, the Michaelis constant is defined as the dissociation constant of the enzyme-substrate complex
d) It assumes covalent binding occurs between enzyme and substrate

View Answer

Answer: b [Reason:] Km is defined as that concentration of substrate at which enzyme is working at half of maximum velocity. It is also a measure of the affinity that the enzyme has for its substrate. Michaelis-Menten kinetics assumes non-covalent binding between enzyme and substrate.

2. When the velocity of enzyme activity is plotted against substrate concentration, which of the following is obtained?
a) Hyperbolic curve
b) Parabola
c) Straight line with positive slope
d) Straight line with negative slope

View Answer

Answer: a [Reason:] At low substrate concentration, the rate of a reaction is determined by the rate of formation of an enzyme-substrate complex.

3. Which of the following is the correct Line weaver-Burk equation?

View Answer

Answer: a [Reason:]

4. Which of the following statements is true about competitive inhibitors?
a) It is a common type of irreversible inhibition
b) In the presence of a competitive inhibitor, the Michaelis-Menten equation becomes

c) The apparent Km decreases in the presence of inhibitor by a factor α
d) The maximum velocity for the reaction decreases in the presence of a competitive inhibitor

View Answer

Answer: b [Reason:] Competitive inhibition is a common typ of reversible inhibition. The apparent Km increases in the presence of inhibitor by a factor α. The maximum velocity for the reaction remains same in the presence of a competitive inhibitor.

5. Which of the following statements is true about uncompetitive inhibitors?
a) They bind covalently at a site distinct from the substrate active site
b) In the presence of a uncompetitive inhibitor, the Michaelis-Menten equation becomes

c) They increase the measured Vmax⁡
d) Apparent Km also increases

View Answer

Answer: b [Reason:] They bind non-covalently at a site distinct from the substrate active site. They decrease the measured Vmax⁡ and also apparent Km.

6. The rate determining step of Michaelis-Menten kinetics is
a) The complex dissociation step to produce products
b) The complex formation step
c) The product formation step
d) None of the above

View Answer

Answer: a [Reason:] The breakdown of ES complex is the rate determining step of Michaelis Menten kinetics.

7. The molecule which acts directly on an enzyme to lower its catalytic rate is
a) Repressor
b) Inhibitor
c) Modulator
d) Regulator

View Answer

Answer: b [Reason:] An inhibitor is a substance which interferes with the substrate-active site binding and slows down the catalytic rate.

8. Which of the following is an example for irreversible inhibitor?
a) Disulfiram
b) Oseltamivir
c) Protease inhibitors
d) DIPF

View Answer

Answer: d [Reason:] Disulfiram, Oseltamivir and protease inhibitors are reversible inhibitors.

9. Which of the following is an example of reversible inhibitor?
a) DIPF
b) Penicillin
c) Iodoacetamide
d) Protease inhibitors

View Answer

Answer: d [Reason:] DIPF, Penicillin and Iodoacetamide are irreversible inhibitors.

10. Where does inhibitor binds on enzyme in mixed inhibition?
a) At active site
b) Allosteric site
c) Does not bind on enzyme
d) Binds on substrate

View Answer

Answer: b [Reason:] The inhibitor binds at a place different from active site allosterically.

11. The catalytic efficiency of two distinct enzymes can be compared based on which of the following factor?
a) Km
b) Product formation
c) Size of the enzymes
d) pH of optimum value

View Answer

Answer: a [Reason:] Km is the substrate concentration. Increased substrate concentration increases the rate of reaction.

12. What is the general mechanism of an enzyme?
a) It acts by reducing the activation energy
b) It acts by increasing the activation energy
c) It acts by decreasing the pH
d) It acts by increasing the pH

View Answer

Answer: a [Reason:] For the reaction to occur at a faster rate, activation energy should be less.

Set 5

1. By what factor chymotrypsin enhances the rate of peptide bond hydrolysis?
a) 107
b) 108
c) At least 109
d) 106

View Answer

Answer: c [Reason:] Chymotrypsin enhances the rate of peptide bond hydrolysis by a factor of at least 109.

2. The active site of chymotrypsin consists of a catalytic triad of which of the following amino acid residues?
a) Serine, histidine and aspartate
b) Serine, histidine and glutamate
c) Threonine, histidine and aspartate
d) Methionine, histidine and aspartate

View Answer

Answer: a [Reason:] Aspartate holds the histidine side chain in the correct direction to accept proton from serine.

3. Which of the following statements are true about the reactions at the active center of chymotrypsin?
a) The aspartate residue gives an electron to histidine
b) The aspartate residue gives a proton to histidine
c) The aspartate residue keeps the histidine in the correct direction
d) A proton moves from the aspartate to serine to histidine in the catalytic triad of chymotrypsin

View Answer

Answer: c [Reason:] Aspartate residue accepts a proton from serine.

4. The polypeptide chains in chymotrypsin are linked by
a) Hydrogen bonds
b) Ionic bonds
c) Disulfide bond
d) SH-SH bond

View Answer

Answer: c [Reason:] The protein consists of three polypeptide chains linked by disulfide bonds.

5. Which of the following is false about chymotrypsin?
a) Hydrolytic cleavage of a peptide bond by chymotrypsin has two phases
b) It is activated in the presence of trypsin
c) It is synthesized in the thyroid gland
d) Polypeptide chains in chymotrypsin are linked by S-S bonds

View Answer

Answer: c [Reason:] Chymotrypsin is synthesized in the pancreas.

6. Which of the following is true about the structure of hexokinase?
a) U-shaped
b) T-shaped
c) E-shaped
d) G-shaped

View Answer

Answer: a [Reason:] Hexokinase has a U-shaped structure.

7. Which of the following is true?
a) Xylose is stereo chemically similar to glucose but one carbon shorter
b) Xylose binds to hexokinase in a position where it can be phosphorylated
c) Addition of xylose increases the rate of ATP hydrolysis
d) The binding of xylose is sufficient to induce a change in hexokinase to its active conformation

View Answer

Answer: b [Reason:] Xylose binds to hexokinase at a position where it cannot be phosphorylated.

8. Which of the following catalyzes the reversible degradation of 2-phosphoglycerate to phosphoenolpyruvate?
a) Chymotrypsin
b) Hexokinase
c) Enolase
d) Trypsin

View Answer

Answer: c [Reason:] Enolase catalyzes the reversible degradation of 2-phosphoglycerate to phosphoenolpyruvate.

9. Which of the following catalyzes the reversible reaction of β-D-Glucose to glucose 6-phosphate?
a) Chymotrypsin
b) Hexokinase
c) Enolase
d) Trypsin

View Answer

Answer: b [Reason:] Hexokinase catalyzes the reversible reaction of β-D-Glucose to glucose 6-phosphate.

10. Which of the following is false about lysozyme?
a) It is an antibacterial agent found in tears and egg white
b) The substrate of lysozyme is peptidoglycan
c) Lysozyme cleaves (β1 → 4) glycosidic C-O bonds between two types of sugar residue in the molecule NAM and NAG
d) It is a bisubstrate enzyme

View Answer

Answer: d [Reason:] Hexokinase is the bisubstrate enzyme.

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