250+ TOP MCQs on Batch Reactor Design Equations and Answers

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Batch Reactor Design Equations”.

1. The design equation for Batch reactor in differential form is _______
A. NA0(frac{dXA}{dt}) = -rA V
B. (frac{dXA}{dt}) = -rA
C. (frac{dXA}{dt}) = -rAV
D. NA0(frac{dXA}{dt}) = -rA
View Answer

Answer: A
Explanation: To obtain the time required to achieve a certain conversion. Mole balance is written in terms of conversion and then differentiated to get the above equation.

2. The design equation for Batch reactor in integral form is _____
A. t = NA0(int_0^{XA} frac{dXA}{-r_A} )
B. t = (int_0^{XA} frac{dXA}{-r_A} )
C. t = NA0(int_0^{XA} frac{dXA}{-r_{AV}} )
D. t = (int_0^{XA} frac{dXA}{-r_{AV}} )
View Answer

Answer: C
Explanation: The differential form of the design equation is widely used in interpreting the lab data. Mole balance written in terms of conversion is integrated to get the above equation.

3. The design equation for constant volume batch reactor in terms of partial pressure is (Assuming the gases to be ideal)
A. –ri = (frac{1}{RT}frac{dpi}{dt} )
B. –ri = (frac{V}{RT} frac{dpi}{dt} )
C. –ri = (frac{PV}{RT} frac{dCi}{dt} )
D. –ri = V (frac{dCi}{dt} )
View Answer

Answer: A
Explanation: The design equation for batch system is –ri = (frac{1}{V}frac{dNi}{dt}.) For constant volume system
–ri = (frac{d(frac{Ni}{V})}{dt} = frac{dCi}{dt}) where C = p/RT. Hence, –ri = (frac{1}{RT}frac{dpi}{dt}. )

4. Most suitable reactor for pharmaceutical industry is ______
A. MFR
B. PFR
C. Batch reactor
D. PBR
View Answer

Answer: C
Explanation: When small scale production is desired, high or stringent quality standards to be met batch reactors used extensively.

5. Design equation for varying volume system is ______
A. t = (int_0^{XA} frac{dXA}{-rA εA} )
B. t = (int_0^{XA} frac{dXA}{CA0} )
C. t = (int_0^{XA} frac{dXA}{NA0} )
D. t = CA0 (int_0^{XA} frac{dXA}{-rA(1+XA εA.} )
View Answer

Answer: D
Explanation: The varying volume in a batch reactor is given as V = V0 (1+XA εA) where εA is called the fractional volume change of the system. The above equation is substituted in the general batch reactor design equation.

6. Calculate the value of εA of gas for the following isothermal gas phase reaction.
(Assuming pure A.
A → 4R
A. 1
B. 2
C. 3
D. 4
View Answer

Answer: C
Explanation: εA is fractional volume change between no conversion and complete conversion. It is calculated using the formula given below
εA = (frac{V1-V0}{V0} = frac{4-1}{1}) = 3
where V1 is volume at complete conversion and V0 is volume at no conversion.

7. For the following reaction, calculate εA, containing 50% A and 50% inerts.
A. 0.5
B. 1.5
C. 2.5
D. 0.75
View Answer

Answer: B
Explanation: When X=0, A=0.5 and Inerts=0.5
When X=1, A=2 and Inerts=0.5
εA = (frac{2.5-1}{1}) = 1.5.

8. Constant volume batch reactors are the widely used industrial batch reactors.
A. True
B. False
View Answer

Answer: A
Explanation: Density change for most liquid phase reactions are negligible. Hence, constant volume batch reactors are widely used.

9. Determine the rate law for first order reaction 2A → R at constant pressure with 60% A in the initial reaction mixture and reduces by 15% in 4 minutes.
A. -rA = 0.17 CA
B. -rA = CA
C. -rA = 2CA
D. -rA = 3.6CA
View Answer

Answer: A
Explanation:
2A → R
εA = (frac{.7-1}{1}) = -0.3
(begin{array}{c|c c}
& X_A=0 & X_A = 1 \
hline
A & 0.6 & 0.3 \
I & 0.4 & 0.4 \
hline
& 1 & 0.7 \
end{array}
)
ΔV = V-Vo
= Vo – 0.15 VO – VO = -0.15 VO
-ln(1-(frac{ΔV}{Vo εA})) = kt
=> k = 0.173 min-1.

10. Determine the order of the reaction from the graph.
A. 0
B. 1
C. 2
D. 3
View Answer

Answer: C

250+ TOP MCQs on Solid Catalysed Reactions – Pore Diffusion Resistance and Answers

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Solid Catalysed Reactions – Pore Diffusion Resistance”.

1. Which of the following represents Dispersion number? (Where D is the fluid dispersion coefficient,
L is the length of spread of tracer and u is the fluid velocity)
A. (frac{L}{uD} )
B. (frac{L}{D} )
C. (frac{D}{uL} )
D. (frac{Lu}{D} )
Answer: C
Explanation: Dispersion number is a dimensionless group characterizing spread in the entire reactor vessel. The value of D determines the spread.

2. The value of the Dispersion coefficient for plug flow is ____
A. 1
B. 0
C. ∞
D. 2
Answer: B
Explanation: There is no axial mixing in a PFR. Hence, the dispersion of the fluid in the longitudinal direction is assumed to be zero.

3. Which of the following represents Peclet number?
A. (frac{Lu}{D} )
B. (frac{u}{D} )
C. (frac{u}{DL} )
D. (frac{D}{uL} )
Answer: A
Explanation: Peclet number also defines the Dispersion model. It is the reciprocal of Dispersion number.

4. The value of the Peclet number for CSTR is ____
A. 1
B. 0
C. ∞
D. 2
Answer: B
Explanation: CSTR is characterised by complete mixing and recycling between the reactants and products. Hence, there is high diffusion of molecules within the reactor. As peclet number is inversely proportional to dispersion coeffient, PeCSTR = (frac{Lu}{∞}) = 0.

5. The range of dispersion number for PFR is ____
A. (frac{D}{uL}) < 0.1
B. (frac{D}{uL}) < 0.01
C. (frac{D}{uL}) > 1
D. (frac{D}{uL}) > 10
Answer: B
Explanation: For PFR, (frac{D}{uL}) < 0.01. Due to negligible dispersion in PFR, (frac{D}{uL}) nearly approaches 0.

6. The dispersion model accounts for ____
A. Deviation from ideal PFR
B. Modelling ideal CSTR
C. Combining batch and CSTR
D. CSTRs connected in parallel
Answer: A
Explanation: Dispersion model involves a modification of the ideal PFR. It imposes axial dispersion on plug flow.

7. The the species continuity equation for the axial dispersion model is ____
A. u(frac{∂C_A}{∂z} = frac{∂D_a}{∂z}frac{∂C_A}{∂z}) + (rA)C
B. u(frac{∂C_A}{∂z} = frac{∂D_a}{∂z}frac{∂C_A}{∂z}) + C
C. (frac{∂C_A}{∂z} = frac{∂D_a}{∂z}frac{∂C_A}{∂z}) + (rA)C
D. u(frac{∂C_A}{∂z} = frac{∂D_a}{∂z}frac{∂C_A}{∂z}) + (rA)
Answer: A
Explanation: For statistically stationary flow, the species continuity equation for the axial dispersion model is u (frac{∂C_A}{∂z} = frac{∂D_a}{∂z}frac{∂C_A}{∂z}) + (rA)C
u is taken to be the mean (plug flow) velocity through the vessel, and Da is an axial dispersion coefficient to be obtained by means of experiments.

8. If Da is diffusivity, CT is tracer concentration and UT is overall heat transfer coefficient, then the pulse tracer balance with dispersion is obtained as ____
A. (frac{∂^2 C_T}{∂z^2} – frac{∂U_T}{∂z} = frac{∂C_T}{∂z} )
B. Da(frac{∂^2 C_T}{∂z^2} – frac{∂U_T}{∂z} =frac{∂C_T}{∂z} )
C. Da(frac{∂^2 C_T}{∂z^2} + frac{∂U_T}{∂z} =frac{∂C_T}{∂z} )
D. Da(frac{∂C_T}{∂z^2} – frac{∂U_T}{∂z} =frac{∂C_T}{∂z} )
Answer: A
Explanation: The equation (frac{∂^2 C_T}{∂z^2} – frac{∂U_T}{∂z} = frac{∂C_T}{∂z} ) is obtained by a combination of mole balance on inert tracer and the molar flow rate of tracer by both convection and dispersion.

9. The range of reactor peclet number for open tubes is ____
A. 106
B. 1010
C. 102
D. 103
Answer: A
Explanation: Peclet number for open tubes is greater than that in packed beds. In open tubes, there is no restriction to flow velocity.

10. The dispersion model is a ____
A. Two parameter model
B. One parameter model
C. No parameter model
D. Three parameter model
Answer: B
Explanation: Dispersion model is a one parameter model. The parameter modelling the non – ideal condition is the dispersion coefficient.

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250+ TOP MCQs on Chemical Structures of Nucleic Acids and Answers

Molecular Biology Multiple Choice Questions on “Chemical Structures of Nucleic Acids”.

1. Which macromolecule is not abundantly found though being of critical importance for biological mechanisms?
A. Proteins
B. Lipids
C. Nucleic acids
D. Polysaccharides

Answer: B
Explanation: Though lipids is of essential use in the cell for the formation of cell wall it is a micro molecule and is generally found in association with either phosphate or polysaccharide. In case of the other three they are all macromolecules and are of essential use for the cell. For example, proteins such as enzymes regulates biological functions, nucleic acids carry genetic information and polysaccharides function as either storage of energy or acts as structural polymers.

2. Which of the following is wrongly paired?
A. Proteins – peptide bond
B. Nucleic acid – hydrogen bond
C. Polysaccharide – glycosidic bond
D. Phospholipids –phosphate linkage

Answer: B
Explanation: Nucleic acids, that is, DNA and RNA show phosphodiesterase linkage which is the major type of linkage. Without the phosphodiester bonds between the phosphate and adjacent 3’OH sugar molecule the backbone will not be formed. Therefore, the nucleotides would not be able to attach and bond to form a nucleic acid.

3. With respect to nucleosides which of the following is paired correctly?
A. Purine – Adenosine, Thymidine
B. Purine – Guanosine, Thymidine
C. Pyrimidine – Uridine, Cytidine
D. Pyrimidine – Uridine, Adenosine

Answer: C
Explanation: As we know purines have a fused ring structure with 9 element backbones ring structure. They are of two types Adenosine and Guanine. Again, the other three residues cytidine, Thymidine and Uridine are pyrimidines having a ring structure of 6 elements in the core backbone. Thus the correct pair among the above options is pyrimidine – uridine, cytidine.

4. Which of the following is not a component of the nucleic acid backbone?
A. Pentose sugar
B. Phosphate group
C. Nucleotide
D. Phosphodiesterase bond

Answer: C
Explanation: A nucleic acid backbone is mainly composed of a pentose-phosphate unit which acts as a monomer. The repeat of this monomer is attached together by the phosphodiesterase linkage, thus, giving rise to the backbone. The nucleotides attached to the sugar moieties forms the side chain which gives rise to the hydrogen linkage with its complementary strand.

5. According to Chargaff’s rule the two strands of DNA has ___________
A. Same molecular weight
B. Same amount of A and G
C. Different amount of A and G
D. Different molecular weight

Answer: D
Explanation: According to Chargaff’s rule the two strands have equal number of A and T residues and equal number of G and C residues. Thus when in one strand A is more automatically in the other T is more. Thus, A being a pyrimidine has a higher molecular weight than T, which is a purine, and so the two strands have different molecular weight. The same happens in case of G and C also.

6. In one strand of a double stranded DNA the rate of occurrence of A is 3 times C in consecutive 10 bases. So how many G will be there in 100 base pairs of a DNA duplex?[Consider G=T in one strand].
A. 30
B. 20
C. 40
D. 60

Answer: C
Explanation: Let’s consider C = 1
Therefore, A = 3C = (3*1) = 3
Now, A+C = (3+1) = 4
Again G = T [given]
And A+C+G+T = 10 [given]
Now replacing T with G and putting the value of A+C
We get,
4 + 2G = 10
2G = 6
G = 3, T = 3, A = 3, C = 1
Thus in 100 bases in one strand there are (3*10 = 30) G residues
Now for the complementary strand G = C residues [Chargaff’s rule]
Therefore, C = G = 1 residue in every 10 bases
That is, G = 1*10 = 10 in 100 bases
Therefore total number of G residues = 30 + 10 = 40 in 100 base pairs of a DNA duplex.

7. In a diploid organism with 30,000 bases haploid genome contains 23% A residues. What is the number of G residues in the genome of this organism?
A. 16000
B. 16200
C. 16500
D. 14200

Answer: B
Explanation: By Chargaff’s rule, A = T = 23%
Therefore, G + C = [100 – (A+T)]
G + C = [100 – 46]
G + C = 54
G = C = 27 [By Chargaff’s rule]
Therefore, G = 27%
Now, each cell is diploid thus it contains (2*30,000 = 60,000) bases
So, G = 27% of 60,000
G = 16,200 bases.

8. Which of the following is not a characteristic of nucleotide bases?
A. Planar
B. Heterocyclic
C. Aliphatic
D. Ubiquitous

Answer: C
Explanation: The nucleotide bases have a 6 member ring structure as one component which is common for all five bases. Thus, they are considered as aromatic molecules due to the presence of a benzene ring structure which provides the molecules with an aromatic property.

9. Which of the following factors do not provide to the separation of DNA fragments during electrophoresis?
A. Chargaff’s rule
B. Matrix density
C. Ethidium bromide
D. Size

Answer: C
Explanation: Ethidium bromide only helps in tagging the DNA molecules to make the visible under the UV radiation due to its fluorescence property, but does not take any part in the separation of DNA molecules during electrophoresis.

10. Which one of the following is not a function of a nucleotide?
A. Nucleic acid monomer
B. Ribozyme
C. Energy carrier molecules
D. Receptors

Answer: D

250+ TOP MCQs on Chromatin Centromere and Telomere and Answers

Molecular Biology Multiple Choice Questions on “Chromatin Centromere and Telomere”.

1. Nucleosome was first described in 1974 by ___________
A. William Asbury
B. Rosalind Franklin
C. Roger Kornberg
D. John Crick

Answer: C
Explanation: The basic structural unit of chromatin, the nucleosome, was described by Roger Kornberg in 1974. Two types of experiments led to Kornberg’s proposal of the nucleosome model. First, partial digestion of chromatin with micrococcal nuclease was found to yield DNA fragments approximately 200 bp long. In contrast, similar digestion of naked DNA yielded a continuous smear of randomly sized fragments.

2. The extent of chromosome coiling in non – dividing cells is _________
A. Supercoiled
B. Euchromatin
C. Condensed
D. Heterochromatin

Answer: B
Explanation: In non – dividing or interphase cells most of the chromatin is decondensed and distributed throughout the nucleus. This form of chromatin is known as Euchromatin. During this period of the cell cycle, genes are transcribed and the DNA is replicated in preparation for cell division.

3. The whole length of DNA is transcriptionally active.
A. True
B. False

Answer: B
Explanation: About 10% of interphase chromatin is in a very highly condensed state that resembles the chromatin of cells undergo mitosis. This type of condensed chromatin is known as heterochromatin. Heterochromatin is transcriptionally inactive and contains highly repeated DNA sequences, such as those present at centromeres and telomeres.

4. Why are chromosomes condensed?
A. To facilitate accommodation
B. Always condensed
C. To facilitate cell division
D. To facilitate distribution in daughter cells

Answer: D
Explanation: As the cells enter mitosis, there chromosomes become highly condensed so that they can be distributed to daughter cells. The chromatin in interphase nuclei are organized in loops, which are thought to fold upon themselves to form the compact metaphase chromosomes of mitotic cells in which the DNA has been condensed nearly ten – thousand folds.

5. At which phase do transcription ceases?
A. Interphase
B. Prophase
C. S phase
D. G phase

Answer: B
Explanation: The chromatin in interphase nuclei are organized in loops that fold upon them to form the compact metaphase chromosomes of mitotic cells which are condensed nearly upto ten – thousand folds. Such condensed chromatin can no longer be used as a template for RNA synthesis and thus transcription ceases during mitosis. As prophase marks the beginning of mitosis thus it is the right option.

6. The part that plays a critical role in even distribution of parental DNA during division is ___________
A. Telomere
B. Centromere
C. Spindle fibre
D. Centrioles

Answer: B
Explanation: The centromere plays a critical role in even distribution of parental DNA during cell division. It is a specialized region on the chromosome known as the constricted chromosomal region that holds the sister chromatids together and attaches the chromosomes to the spindle fibres during metaphase of division.

7. With respect to centromere which of the following is wrong?
A. Constricted chromosomal region
B. Holds the sister chromatids together
C. Attaches to spindle fibres
D. Facilitates even distribution

Answer: D
Explanation: As the cell enters mitosis, chromatin condensation leads to the formation of metaphase chromosome consisting of two identical sister chromatids. These sister chromatids are held together at the centromere, which is known as the constricted chromosomal region. As mitosis precedes the microtubules of mitotic spindles attaches to the centromere, and the two sister chromatids separate and move to the opposite poles of the spindle.

8. The protein that binds to the spindle fibres are known as centromeric proteins.
A. True
B. False

Answer: B
Explanation: The proteins associated with centromeres form a specialized structure called the kinetochore. The spindle fibres thus formed attaches to these proteins which finally lead to the segregation of the chromosomes to the daughter cells.

9. Centromeric DNA was initially defined in ___________
A. Bacteria
B. Fungi
C. Yeast
D. Human

Answer: C
Explanation: Centromeric DNA sequences were initially identified in yeasts, where their functions were assayed by following the segregation of plasmids at mitosis. Plasmids that contain functional centromeres segregate like chromosomes and are equally distributed to daughter cells following mitosis.

10. How many satellite sequence elements are present in yeast centromere?
A. 2
B. 3
C. 4
D. 5

Answer: B
Explanation: The centromere sequences of the well studied yeast are contained in approximately 125 base pairs. It has been found to consist of three sequence elements: two short sequences of 8 to 25 bp separated by 78 to 86 bp of very A/T rich DNA.

11. The centromere is A/T rich region.
A. True
B. False

Answer: A
Explanation: Yes the centromere is A/T rich region. For example, the centromere of Drosophila is about 420 kb consisting of two highly repetitive satellite DNAs, AATAT and AAGAG, in addition to a non-repetitive region of A/T rich DNA. Again in Arabidopsis, centromeres consist of 3 million bp of an A/T rich 178 bp satellite DNA. In humans also, this feature is observable which is a 171 bp A/T sequence arranged in tandem repeats spanning around 1–5 million base pairs.

12. Which alternate form of histone is seen in centromeric histones of humans?
A. H2A.z
B. SMC protein
C. H1
D. CENP-A

Answer: D
Explanation: It has been observed that the human chromatin near the centromere has a unique structure. In particular, histone H3 is replaced by its variant CENP-A. CENP-A is uniformly present in centromere of all organisms that have been studied and thus are thought to be of prime requisite for the assembly of the kinetochore proteins needed for centromere function.

13. Telomere is not related to __________
A. Maintenance
B. Chromosome degradation
C. Division
D. Replication

Answer: C
Explanation: The sequences at the ends of eukaryotic chromosomes are called the telomeres. They play a critical role in maintenance and replication of the DNA and also play a part in chromosomal degradation. Degradation of the telomeres leads to ageing.

14. Which of the following nucleotides is rich I telomere of an organism?
A. A, T
B. T, G
C. G, C
D. C, A

Answer: C
Explanation: The telomere DNA sequences of a variety of eukaryotes are similar, consisting of repeats of a simple sequence DNA containing a cluster of G residues on one strand. For example, the repeat sequence of telomere in humans and other mammals is TTAGGG, and in Tetrahymena is TTGGGG. This sequence is repeated over hundreds and thousands of times and is terminated with a 3’ overhanging DNA.

15. How does telomerase activity depend on age?
A. Directly
B. Inversely
C. Remains the same
D. Does not occur

Answer: D
Explanation: Telomerase activity does not occur in a normal somatic cell as it is present in inactive form. Thus the ends of the DNA does not get replicated every time a cell divides which leads to aging.

16. In cancer telomerase activity __________
A. Increases
B. Decreases
C. Remains constant
D. Plays no role

Answer: A

250+ TOP MCQs on Genes Are Expressed by Making RNA – 1 and Answers

Molecular Biology Multiple Choice Questions on “Genes Are Expressed by Making RNA – 1”.

1. How is the genetic material expressed?
A. By replication and transcription
B. By transcription and translation
C. By translation and modification
D. By mutation and transposition

Answer: B
Explanation: Expression of the genetic material is the series of processes of how the sequence of bases in the DNA directs the production of the RNAs and proteins that perform cellular functions and define cellular identity. The basic processes responsible for gene expression are transcription and RNA processing followed by a translation.

2. RNA polymerase requires a DNA primer for RNA synthesis.
A. True
B. False

Answer: B
Explanation: RNA polymerase does not require any primer for RNA synthesis. It can start transcription of the DNA template de novo.

3. The transcription process carried out by the RNA polymerase is very accurate but less accurate than replication.
A. True
B. False

Answer: A
Explanation: Transcription, though accurate, is less accurate than replication. One mistake occurs in every 10,000 nucleotides added in transcription, compared to 10,000,000 nucleotides for replication. This difference reflects the lack of extensive proofreading mechanism for transcription, although two forms of proofreading for RNA synthesis do exist.

4. Which of the following RNA polymerases are responsible for the production of 5S rRNA?
A. RNA polymerase I
B. RNA polymerase II
C. RNA polymerase III
D. RNA polymerase IV

Answer: C
Explanation: Polymerase I is responsible for the transcription of the different types of rRNA except for the 5S rRNA. 5S rRNA is transcribed by polymerase III along with some small nuclear RNA genes and the tRNAs.

5. Which RNA polymerase deals with the production of mRNA?
A. RNA polymerase I
B. RNA polymerase II
C. RNA polymerase III
D. RNA polymerase IV

Answer: A
Explanation: Polymerase I is responsible for the transcription of the different types of rRNA except for the 5S rRNA. 5S rRNA is transcribed by polymerase III along with some small nuclear RNA genes and the tRNAs. Polymerase II deals with the transcription of the mRNAs.

6. The eukaryotic subunit of RNA polymerase homologous to the ω subunit of bacterial RNA polymerase is ___________
A. RPA1
B. RPB2
C. RPC5
D. RPB6

Answer: D
Explanation: The eukaryotic subunit of RNA polymerase homologous to the ω subunit of bacterial RNA polymerase is RPB6. RPA1, RPB2 and RPC5 subunits of eukaryotic RNA polymerase are homologous to the bacterial RNA polymerase subunits β’, β and αI respectively.

7. The E. coli RNA polymerase adds __________ nucleotides per second.
A. 20
B. 30
C. 40
D. 50

Answer: C
Explanation: The E. coli RNA polymerase can synthesize RNA at a rate of 40 nucleotides per second. For proper functioning of the RNA polymerase enzyme Mg2+ ion and an optimum temperature of 37˚C is required.

8. The RNA polymerase holoenzyme has the structural formula of ___________
A. α2ββ’ωσ
B. αβ2β’ωσ
C. α2ββ’ω
D. α2ββ’σ

Answer: A
Explanation: In a complete RNA polymerase, called the holoenzyme there are 5 subunits. Of which two are α and one of the each of the other 4 subunits namely β, β’, ω and σ.

9. Which of the following statements are true about the roles of RNA polymerase subunits?

i. α subunit is encoded by “rpo A” gene, required for core protein assembly.
ii. β subunit is encoded by “rpo C” gene is the catalytic center of RNA polymerase.
iii. ω subunit is encoded by “rpo Z” gene and helps in proper association of all the subunits.
iv. σ factor is encoded by “rpo D” gene and contributes in promoter recognition.

A. i, ii, iii
B. ii, iii, iv
C. i, iii, iv
D. i, ii, iii, iv

Answer: C
Explanation: The β subunit of the RNA polymerase is coded by “rpo B” gene. The β’ subunit is known to be encoded by “rpo C gene.

10. The α subunits of polymerase has a function of ____________
A. Promoter binding
B. Initiation
C. Elongation
D. Termination

Answer: A

250+ TOP MCQs on Alternative Splicing Produces Multiple Forms of RNA and Answers

Molecular Biology Multiple Choice Questions on “Alternative Splicing Produces Multiple Forms of RNA”.

1. How many types of specific domains are present in the SR protein?
A. 1
B. 2
C. 3
D. 0

Answer: B
Explanation: The SR protein has 2 special RNA binding domains. They are RNA-recognition motif (RRM) and the other is RS domain.

2. Which of the following components of the splicosome machinery is recruited by the SR protein?
A. U4
B. U2
C. U2AF
D. U6

Answer: C
Explanation: The SR proteins help efficient recruitment of the components to the different splicosome components to the nearby splice site. Specifically, the SR proteins recruit the U2AF proteins to the 3’ splice site and U1 to the 5’ splice site.

3. SR proteins are not essential for RNA splicing.
A. True
B. False

Answer: B
Explanation: SR proteins are essential for RNA splicing. They not only ensure the accuracy and efficiency of constitutive splicing but also regulate alternative splicing. They come in many varieties, some controlled by physiological signals, others constitutively active.

4. The presence and activity of SR determines the ___________
A. Product formed after splicing
B. Mechanism of splicing
C. Usage of particular splice site
D. Repression of splicing mechanism

Answer: C
Explanation: The SR family protein is large and diverse that has specific roles in regulated alternative splicing by directing the splicing machinery to different splice sites under different conditions. Thus, the presence of activity of a given SR protein can determine whether a particular splice site is used in a particular cell type, or at a particular stage of development.

5. Alternative splicing is regulated by activators and repressors.
A. True
B. False

Answer: A
Explanation: Proteins that regulate splicing bind to the specific sites called exonic or intronic splicing enhancers, or exonic or intronic splicing silencers. The former enhance and the latter repress splicing in nearby splice sites.

6. The Drosophila half-pint protein is _____________
A. Positive regulator
B. Negative regulator
C. Non regulator
D. Not related to fly ovarian protein

Answer: A
Explanation: An activator that promotes a particular alternative splicing event in a specific tissue type is the Drosophila half-pint protein. This activator regulates the alternative splicing of a set of pre-mRNA in the fly ovary.

7. Most silencer proteins are able to bind to pre-mRNA but cannot recruit the splicing machinery.
A. True
B. False

Answer: A
Explanation: Most silencers are recognized by members of the heterogeneous nuclear ribonucleoprotein (hnRNP) family. These bind RNA but lack the RS domains and so cannot recruit the splicing machinery.

8. hrRNPA1 acts as the repressor for the final mRNA form of ____________
A. Adenovirus
B. Insulin
C. SV40
D. HIV

Answer: D
Explanation: The hrRNPA1 binds to an exonic silencer element within an exon of the HIV at pre-mRNA and represses the inclusion of that exon in the final mRNA. Bi binding to its site, the repressor blocks binding of the activator SC35 to a nearby enhancer element.

9. The pre-mRNA binding minor splicosome is also known as ___________
A. AG – CT splicosome
B. AT – GC splicosome
C. AC – AG splicosome
D. AT – AC splicosome

Answer: D
Explanation: The minor splicosome recognizes rarely occurring introns having consensus sequences distinct from the sequences of most pre-mRNA introns. This recently discovered form is known as the AT – AC splicosome, because the termini of the originally identified rare introns contain AU at the 5’ splice site and AC at the 3’ site in RNA or AT and TC in DNA.

10. hnRNPI binds to which segment of the mRNA?
A. 5’ splice site
B. Branch point site
C. 3’ splice site
D. AT – AC splicosome

Answer: C
Explanation: The hnRNPI binds to the pyrimidine tract of the intron. Thus it is also known as the polypyrimidine tract-binding protein. This is a repressor protein which when binds to the mRNA blocks the binding of the splicosome machinery.

11. Alternative splicing of the Troponin T gene produces ____________ alternative forms of mRNA.
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: A region of the Troponin T gene encodes 5 exons. After splicing they recombine in such a way that only two forms of mRNA are generated. One contains exon 1, 2, 4 and 5 known as the β-tropoinin T mRNA whereas the other containing exon 1, 2, 3 and 5 known as the α-tropoinin T mRNA.

12. Which of the following mRNA is an example of having an extended exon?
A. Human slo gene
B. Mammalian muscle Troponin T mRNA
C. T-antigen mRNA of monkey virus SV40
D. Drosophila Dscam mRNA

Answer: C
Explanation: The pre-mRNA of the T-antigen of monkey virus SV40 contains 3 exons interrupted by 2 introns. These are spliced in 5 different alternative pathways – normal, exon skipped, exon extended, intron retained and alternative exons.

13. Alternative splicing is always used to produce multiple isoforms of proteins.
A. True
B. False

Answer: B
Explanation: Some genes that encode only a single functional protein show alternative splicing. In those cases, alternative splicing is used simply as a way of switching expression of the gene on and off.

14. Alternative splicing can be used to switch off a gene’s expression. How many ways are there to achieve it?
A. 1
B. 2
C. 3
D. 4

Answer: B
Explanation: There are 2 alternative splicing can be used to switch off a gene’s expression. Most straightforwardly, alternative splicing determines the presence of a stop codon to be included with the exons as stop codons are extensively found in the intronic regions. The second way is by regulating the use of an intron, which, when retained in the mRNA, ensures that species is not transported out of the nucleus and so is never translated.

15. Alternative splicing was discovered in the year ______________
A. 1977
B. 1958
C. 1964
D. 1982

Answer: A
Explanation: Alternative splicing was discovered in the studies of gene expression in the mammalian adenovirus, where mRNAs are alternatively spliced. It was discovered in the year 1977.

16. Which of the following is not a part of the minor splicosome machinery?
A. U2
B. U4
C. U5
D. U6

Answer: A