300+ REAL TIME Objective Questions & Answers 2023

300+ TOP MCQs on Gel electrophoresis, Oligonucleotide & Microarrays – 1 and Answers

Genetic Engineering Multiple Choice Questions & Answers (MCQs) on “Gel Electrophoresis, Oligonucleotide & Microarrays – 1”

1. Gel electrophoresis separates nucleic acid molecules based on ____________
a) charge on molecules
b) size of the molecules
c) nature of the molecules i.e. whether DNA or RNA
d) chemical properties of the nucleic acids
Answer: b
Explanation: Gel electrophoresis separates nucleic acid molecules on the basis of their size. The nucleic acid molecules move through the gel because of the force provided by the applied electric field.

2. The charge present on the DNA backbone is negative. The force required to accelerate the molecules towards anode is directly proportional to the number of ____________
a) sugar molecules
b) nitrogenous bases
c) phosphate groups
d) both phosphate group and sugar molecules
Answer: c
Explanation: The DNA backbone is negatively charged because of the phosphate groups. Hence, the force required to move the molecules towards anode is directly proportional to the number of phosphate groups present.

3. Force is defined as mass per unit acceleration. As the number of phosphate molecules increases, the charge also increases which increases the force required. The acceleration is dependent on the size of the molecules.
a) True
b) False
Answer: b
Explanation: As the charge increases, which is because of the number of phosphate molecules, the size also increases. As, the size increases, it also results in increase of mass. Thus, now the force and also the acceleration are independent of size until a retarding force is present. But, in the case of gel electrophoresis, the retarding force is provided by gel and thus separation takes place on the basis of size.

4. Which one of the following will travel fastest through the gel if the amount of DNA present is same in all?
a) Circular
b) Supercoiled
c) Nicked
d) Supercoiled and circular will move at same speed and faster than nicked
Answer: b
Explanation: The supercoiled form of DNA will travel the fastest. It is so because; the movement through the gel is based on the size. The smaller the molecule is the less retarding force it experiences when it moves. Hence, supercoiled which is having the smallest size will move the fastest.

5. How is the size of molecules under analysis measured?
a) By measuring the distance moved through a ruler
b) By measuring the amount of visualising dye used
c) By running a standard molecule, whose size is known in parallel
d) There is no exact criterion for doing so
Answer: c
Explanation: The size of the DNA molecules is measured by a standardized molecule. It is known as DNA ladder. The molecules under analysis are compared with the DNA ladder.

6. Gel matrices are generally of two types Agarose and Polyacrylamide. Which of the statements is correct with respect to agarose gels?
a) Agarose is a polysaccharide which is obtained from red algae
b) Agarose is a polysaccharide obtained from fungus
c) It is composed of glucose residues
d) It is obtained via dissolving in its water by boiling in water and then cooling it
Answer: a
Explanation: Agarose is a polysaccharide which is obtained from red algae. It is composed of galactose residues. Agarose gel is obtained via dissolving in it a buffer by boiling it and then cooling it. It leads to the formation of agar in conjunction with other polysaccharides, agropectin.

7. Polyacrylamide gels are the other types of gels which are commonly used. Which of the following statement is not correct with respect to these types of gels?
a) They are obtained via polymerization between acrylamide and bis-acrylamide
b) The components added for initiating polymerization are ammonium persulphate and TEMED
c) It is casted in horizontal and flat trays
d) TEMED catalyses the formation of free radicals from persulphate ions which leads to initiation of cross-linking
Answer: c
Explanation: Polyacrylamide gels are made up by the polymerization of acryl amide and bis acryl amide units. The polymerization takes place because of TEMED and ammonium persulphate, as TEMED catalyses the formation of free radicals from persulphate. These gels are usually casted in vertical trays whereas the agarose gels are casted in horizontal and flat trays.

8. If the amount of agarose added is more, the molecular under analysis should have the following characteristic:
a) small size
b) large size
c) size has no relation with the amount of agarose
d) the amount of molecules under analysis matters
Answer: a
Explanation: In the case of agrose gels, the separation takes place because of size. If the amount of agarose is more, smaller size molecules will be able to move easily as in comparison to larger size molecules. It is so because; the movement is through the pores. More the amount of agarose, smaller the size of the pore.

9. In the case of electrophoresis of single stranded DNA or RNA, which type of gels are used?
a) Renaturing
b) Denaturing
c) The routine agarose gel
d) The routine polyacrylamide gel
Answer: b
Explanation: In the case of single stranded molecules, denaturing gels are used. It is so because, if denaturing gels are not there, there are chances of formation of secondary structures which hinder the movement of nucleic acids. Thus to avoid this, denaturing agents such as urea, etc. are added.

10. At times, a specific fragment of the molecules of DNA which are analyzed needs to be separated. One of the methods is digesting the gels simply. Which of the following statement is not correct in with respect to it?
a) The gelling temperature has an important role to play
b) For higher gelling temperatures, digestion of DNA is done via the addition of agarase or some chaotropic agents
c) For lower gelling temperatures, simply slicing out of target DNA is done which is followed by melting
d) The DNA can be extracted via the addition of sodium or any positively charged group
Answer: d
Explanation: For the method of solubilising DNA, the gelling temperature is very important. For lower gelling temperatures simply the target DNA can be sliced out and further melting is done. But for higher gelling temperatures, either agarase or chaotropic agents are added. After this, the required DNA is obtained via purification which is done by the addition of silica or solvent extraction can also be done.

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300+ TOP MCQs on Cultured Cells And Bacillovirus and Answers

Genetic Engineering Problems focuses on “Cultured Cells And Bacillovirus”.

1. Autographa californica nuclear polyhedrois virus (AcNPV) is commonly used bacilovirus for infecting cultured cells. The virus is ________ and _______
a) single stranded, linear
b) double stranded, linear
c) double stranded, circular
d) single stranded, circular
Answer: c
Explanation: AcNPV is a commonly used bacilovirus for infecting cultured cells. It is double stranded and circular. It is isolated from caterpillar and infects around 30 species.

2. DNA replication of the AcNPV takes 6 hours after infection. After how many hours of infection new virus particles are produced by budding?
a) 8 hrs
b) 10 hrs
c) 12 hrs
d) 16 hrs
Answer: b
Explanation: After 10 hrs of infection, new virus particles are produced. And these virus particles are termed as extracellular virus particles.

3. The virus particles are held together by polyhedron and p10 protein.
a) True
b) False
Answer: a
Explanation: The virus particles are held together by polyhedron and p10 proteins. Polyhedron protein is of 29kDa. The virus particles are held together as occlusion bodies.

4. The polyhedrin and p10 protein are produced in ____________ amount and other genes are expressed in _________ amount.
a) low, high
b) high, high
c) high, low
d) low, low
Answer: c
Explanation: The polyhedron and p10 protein are expressed in large amounts and other genes are expressed in low amount. The former proteins are up to 20% of protein synthesis.

5. Expression of proteins using nuclear polyhedrosis viruses is advantageous because it gives _______ protein yields and post-translational modifications are _________
a) high, not possible
b) high, possible
c) low, possible
d) low, not possible
Answer: b
Explanation: Expression of proteins using nuclear polyhedrosis viruses is advantageous because it gives high protein yields and post-translational modifications are also possible. These modifications are not possible in prokaryotic systems and yeast or are carried out very less.

6. The size of the viral genome is large and thus poses a problem. This problem can be resolved by using ____________
a) transfer vector
b) co-integration plasmid
c) hybrid vector
d) fusion vector
Answer: a
Explanation: The size of the viral genome is a problem because it is very large. It is solved by using a transfer vector. Transfer vector is composed of sequences that allow propagation of E.coli, the viral promoter, the polyhedrin mRNA polyadenylation signal and the polyhedron gene is flanked by the viral sequences.

7. The gene of interest is inserted into transfer vector and then further into viral genome leading to the formation of modified virus. How can plaques from modified virus differentiated from plaques from wild-type virus?
a) Chemically
b) Visually
c) On the basis of magnetic properties
d) On the basis of longeitivity of plaques
Answer: b
Explanation: Plaques from the recombinant virus and from the wild type virus can be differentiated simply visually.

8. If the DNA is linearized, what is the effect on the recombination frequency?
a) The recombination frequency decreases
b) The recombination frequency increases
c) The recombination may increase or decrease depending on the amount of DNA
d) There is no effect on the recombination frequency
Answer: b
Explanation: If the DNA is linearixed, the recombination frequency increases. It is so because eit makes the DNA more recombinogenic.

9. The recombinant virus can be propagated in E.coli and this is referred to as bacmid.
a) True
b) False
Answer: a
Explanation: The recombinant virus can be propagated in E.coli and this is called as bacmid. This arrangement can be used as shuttle vector.

10. For the generation of recombinant baculovirus, recombinants can be selected by ____________
a) blue-white screening
b) antibiotic resistance
c) either antibiotic resistance or blue white screening
d) radioactivity
Answer: a
Explanation: For the generation of recombinant baculovirus, recombinants can be selected by blue-white screening. After this, the recombinant baculovirus is introduced into insect cells for expression.

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250+ TOP MCQs on Sequence Homology Versus Sequence Similarity and Identity and Answers

Bioinformatics test focuses on “Sequence Homology Versus Sequence Similarity and Identity”.

1. Which of the following is incorrect regarding pair wise sequence alignment?
A. The most fundamental process in this type of comparison is sequence alignment
B. It is an important first step toward structural and functional analysis of newly determined sequences
C. This is the process by which sequences are compared by searching for common character patterns and establishing residue-residue correspondence among related sequences
D. It is the process of aligning multiple sequences

Answer: D
Explanation: Pair wise sequence alignment is the process of aligning two sequences and is the basis of database similarity searching and multiple sequence alignment. As new biological sequences are being generated at exponential rates, sequence comparison is becoming increasingly important to draw functional and evolutionary inference of a new protein with proteins already existing in the database.

2. Which of the following is incorrect about evolution?
A. The macromolecules can be considered molecular fossils that encode the history of millions of years of evolution
B. The building blocks of these biological macromolecules, nucleotide bases, and amino acids form linear sequences that determine the primary structure of the molecules
C. DNA and proteins are products of evolution
D. The molecular sequences barely undergo changes

Answer: D
Explanation: During this time period, the molecular sequences undergo random changes, some of which are selected during the process of evolution. As the selected sequences gradually accumulate mutations and diverge over time, traces of evolution may still remain in certain portions of the sequences to allow identification of the common ancestry.

3. The presence of evolutionary traces is because some of the residues that perform key functional and structural roles tend to be preserved by natural selection; other residues that may be less crucial for structure and function tend to mutate more frequently.
A. True
B. False

Answer: A
Explanation: The residues that perform key functional and structural roles tend to be preserved by natural selection. For example, active site residues of an enzyme family tend to be conserved because they are responsible for catalytic functions. Therefore, by comparing sequences through alignment, patterns of conservation and variation can be identified.

4. The degree of sequence variation in the alignment reveals evolutionary relatedness of different sequences, whereas the conservation between sequences reflects the changes that have occurred during evolution in the form of substitutions, insertions, and deletions.
A. True
B. False

Answer: B
Explanation: The degree of sequence conservation in the alignment reveals evolutionary relatedness of different sequences, whereas the variation between sequences reflects the changes that have occurred during evolution in the form of substitutions, insertions, and deletions. Identifying the evolutionary relationships between sequences helps to characterize the function of unknown sequences. When a sequence alignment reveals significant similarity among a group of sequences, they can be considered as belonging to the same family.

5. If the two sequences share significant similarity, it is extremely ______ that the extensive similarity between the two sequences has been acquired randomly, meaning that the two sequences must have derived from a common evolutionary origin.
A. unlikely
B. possible
C. likely
D. relevant

Answer: A
Explanation: Sequence alignment provides inference for the relatedness of two sequences under study. Regions that are aligned but not identical represent residue substitutions; regions which residues from one sequence correspond to nothing in the other represent insertions or deletions that have taken place on one of the sequences during evolution.

6. Sometimes, it is also possible that two sequences have derived from a common ancestor, but may have diverged to such an extent that the common ancestral relationships are not recognizable at the sequence level.
A. True
B. False

Answer: A
Explanation: There are examples of such paralogous genes that have distinct functions but similar origin. In that case, the distant evolutionary relationships have to be detected using other methods.

7. Which of the following is incorrect regarding sequence homology?
A. Two sequences can homologous relationship even if have do not have common origin
B. It is an important concept in sequence analysis
C. When two sequences are descended from a common evolutionary origin, they are said to have a homologous relationship
D. When two sequences are descended from a common evolutionary origin, they are said to share homology

Answer: A
Explanation: Homologous relationships are more certain when the sequences have common evolutionary origin. A related but different term is sequence similarity, which is the percentage of aligned residues that are similar in physiochemical properties such as size, charge, and hydrophobicity.

8. Sequence similarity can be quantified using ________ homology is a ______ statement.
A. percentages, quantitative
B. percentages, qualitative
C. ratios, qualitative
D. ratios, quantitative

Answer: B
Explanation: Similarity is a direct result of observation from the sequence Alignment. For example, one may say that two sequences share 40% similarity. It is incorrect to say that the two sequences share 40% homology. They are either homologous or nonhomologous.

9. Shorter sequences require higher cutoffs for inferring homologous relationships than longer sequences.
A. True
B. False

Answer: A
Explanation: For determining a homology relationship of two protein sequences, for example, if both sequences are aligned at full length, which is 100 residues long, an identity of 30% or higher can be safely regarded as having close homology. If their identity level falls between 20% and 30%, determination of homologous relationships in this range becomes less certain.

10. Sequence similarity and sequence identity are synonymous for nucleotide sequences and protein sequences as well.
A. True
B. False

Answer: B
Explanation: Sequence similarity and sequence identity are synonymous for nucleotide sequences. For protein sequences, however, the two concepts are very different. In a protein sequence alignment, sequence identity refers to the percentage of matches of the same amino acid residues between two aligned sequences. Similarity refers to the percentage of aligned residues that have similar physicochemical characteristics and can be more readily substituted for each other.

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300+ TOP MCQs on Drosophila and Answers

Genetic Engineering Multiple Choice Questions on “Drosophila”.

1. Transgenic Drosophila can be created by microinjection of DNA into the embryos. These embryos are at which stage of development?
a) One-cell stage
b) Pre-blastoderm stage
c) Blastoderm
d) Morula
Answer: b
Explanation: Transgenic Drosophila can be created by microinjection of DNA into the embryos. The embryo should be at pre-blastodermal stage. Other developmental stages such as blastoderm and morula follow the pre-blastodermal stage.

2. Syncytium is a layer of ______ that have not been separated into individual cells.
a) nuclei
b) mitochondria
c) cytoplasm
d) nuclei and cytoplasm
Answer: a
Explanation: Syncytium is a layer of nuclei which is not divided into separate individual cells. This is found at pre-blastodermal stage of development.

3. In Drosophila, only nuclei cells give rise to germline cells.
a) True
b) False
Answer: a
Explanation: In Drosophila, only nuclei cells give rise to germline cells. These germline cells are located at one end of the embryo and acquire DNA stably.

4. ______ integration systems are used for the transfer of DNA in Drosophila and it is composed of ______
a) Artificial, P elements
b) Artificial, S elements
c) Natural, P elements
d) Natural, S elements
Answer: c
Explanation: In Drosophila, the natural integration systems are used for the transfer of DNA and it is composed of transposable elements. These transposable elements are known as P elements.

5. Only the _____ part of Drosophila is transgenic and the rest is not. This is known as ______
a) germline cell, mosaic
b) nurse cells, mosaic
c) germline cells, hybrid
d) nurse cells, hybrid
Answer: a
Explanation: Only the germline cells portion (but not necessarily all germline cells) is modified and the rest is non-transgenic. This is collectively known as mosaic.

6. rosy gene is used as a selectable marker for transformation in Drosophila, it produces an enzyme required for the synthesis of ____________
a) wing pigment
b) eye pigment
c) both eye and wing pigment
d) thorax pigment
Answer: b
Explanation: Rosy gene is used as a selectable marker for transformation in Drosophila, it produces enzyme xanthine dehydrogenase and it is required for pigments of the eye.

7. Wild type flies have crimson red eyes whereas rosy mutant have brick-red eyes.
a) True
b) False
Answer: b
Explanation: Wild type genes have brick-red eyes whereas rosy mutant have crimson red eyes. Thus, rosy genes can be used as selectable marker.

8. Wild-type Drosophila flies are _____ to ethanol supplied in food.
a) resistant
b) non-resistant
c) resistant at low concentration and non-resistant at higher concentration
d) resistant at higher concentrations and non-resistant at lower concentration
Answer: a
Explanation: The wild type flies are resistant to ethanol supplied in the food. This property can be exploited as a selectable marker and which is in the form of alcohol dehydrogenase gene.

9. Cis-acting sites should be present in the vector for ________
a) replication
b) selecting recombinants by acting as a marker
c) transposition
d) providing high copy number
Answer: c
Explanation: Apart from selectable markers, cis acting sites should also be present. These are required for transposition.

10. The host should ____ P elements, these elements lead to _______
a) not have, instability
b) have, stability
c) not have, increase the time taken for integration
d) have, reduces the time taken for integration
Answer: a
Explanation: The host should not have P elements, these elements lead to instability. It is so because they are responsible for transposition.

11. Where do P elements integrate into the genome?
a) At specifically defined sites
b) Randomly
c) Only at the ends of the genome
d) Integration depends on reaction conditions
Answer: b
Explanation: P elements integrate into the genome randomly. This makes it non-feasible to be used directly for gene disruption.

12. Transient gene silencing can be carried out by microinjecting ______
a) single stranded RNA
b) double stranded DNA
c) double stranded RNA
d) either double stranded DNA or RNA
Answer: c
Explanation: Transient gene silencing is carried out by microinjecting double stranded RNA. Long term gene silencing can be carried out by synthesizing an inverted copy of the target gene.

300+ TOP MCQs on Gel Electrophoresis, Oligonucleotide & Microarrays – 2 and Answers

Genetic Engineering Questions and Answers for Freshers focuses on “Gel electrophoresis, Oligonucleotide & Microarrays – 2”.

1. A piece of DNA that is to be separated is crushed and soaked into the buffer. The majority of DNA diffuses, how can it be separated?
a) Filtration and centrifugation
b) Filtration or centrifugation
c) Allowing to sediment
d) By passing through a silica column
Answer: b
Explanation: As the target DNA is crushed and soaked into the buffer, the majority of it diffuses into the gel. It can be separated via filtration or separation. It is simply termed as recovering the DNA from the gel by diffusion.

2. Another method of DNA recovery is termed as freeze-squeeze. The correct statement for it is?
a) The slice containing DNA is cut and frozen into liquid oxygen
b) Freezing doesn’t have any effect on the DNA structure
c) After freezing centrifugation is carried out through glass wool plug
d) The substance used for centrifugation allows the gel to pass through but the liquid is retained
Answer: c
Explanation: It is also a method for obtaining DNA from gels. The slice containing DNA is cut and frozen into liquid nitrogen. Freezing in liquid nitrogen leads to disruption of structure. As the centrifugation by glass wool plug is carried out after freezing out, the gel is retained by it. The liquid which is containing dissolved DNA will be allowed to pass through.

3. If further electrophoresis is used for recovery of DNA from gels, the method is termed as ____________
a) secondary electrophoresis
b) recovery electrophoresis
c) denaturing electrophoresis
d) electro-elution
Answer: d
Explanation: If electrophoresis is used further for recovery, the method is termed as electro-elution. There are different electro-elution methods which are used such as dialysis tube or membrane is used.

4. The DNA is sliced and is placed into the dialysis tube containing buffer. Further electrophoresis is performed. After it, the DNA moves out of the gel but is retained by buffer.
a) True
b) False
Answer: a
Explanation: As further electrophoresis is carried out, the DNA moves out of the gel but is retained by the buffer. From the buffer, it can be obtained via pipetting.

5. UV shadowing is used at times for visualising the DNA. Which of the statements is correct for it?
a) In this method, the DNA bands are visualized because they fluoresce under UV
b) The DNA bands won’t fluoresce but will absorb the UV light
c) It is not suitable for visualizing single stranded molecules
d) It is less preferred over the use of ethidium bromide
Answer: b
Explanation: In this method, the DNA bands won’t fluoresce but will absorb the UV light. This leads to the formation of a non-fluorescent shadow on the screen. It is suitable for visualising single stranded species because the mode of action is not intercalating as in the case of use of ethidium bromide. It is preferred over it because ethidium bromide is often difficult to remove.

6. The greatest separation is obtained in which portion of the gel?
a) Lower portion where the anode is
b) Lower portion where the cathode is
c) The separation is uniform all over
d) It varies according to the quantity of the size of the molecules to be separated
Answer: a
Explanation: The greatest separation is obtained in the lower portion of the gel where the anode is. Thus the smaller molecules are having the greatest separation because they are the one which travels farther in the gel.

7. Which of the statements is correct for non uniform separation obtained other than the conventional one?
a) It can be obtained via applying strong field towards the cathode
b) It can be obtained via applying the same field on both the sides
c) The separation is greater on the side of stronger field
d) Its is greater towards the weaker field
Answer: c
Explanation: Conventionally, separation is greater towards the lower portion i.e. where the anode is. In order to obtain greater separation towards the cathode, the stronger field is applied here. As the field is weaker on the anode side, the molecules retard as they move towards anode and are less separated.

8. How can gradient in the field strength be obtained?
a) It can be obtained via varying the amount of current, keeping the resistance constant
b) It can be obtained via varying the amount of resistance, keeping the current constant
c) Both can be varied, but resistance is having more effect
d) Buffer-gradient gels can be used with decreasing concentration of buffer at the bottom
Answer: b
Explanation: The gradient field strength can be obtained via varying the resistance because the current flowing throughout the gel needs to remain constant. To vary the resistance, often buffer-gradient gels and wedge gels can be used. Wedge gels are thick and thus are having lower resistance at the bottom. Whereas buffer-gradient gels with increasing concentration at the bottom are used. As the concentration increases at the bottom, resistance decreases.

9. If the molecules to be separated are larger than the size for conventional electrophoresis, which type of gels can be used?
a) Wedge gels
b) Buffer-gradient gels
c) Pulse field gels
d) Varying the amount of agarose will carry out the separation
Answer: c
Explanation: In the case, if the molecules to be separated are very large, say the chromosomes are to be separated. In this case, pulse-field gels are used. It is termed as pulse field gel electrophoresis (PFGE).

10. Which of the following statement is correct for the method of separation of large molecules?
a) The large DNA molecules can pass through the matrix in any direction, not necessarily in the direction parallel to the movement
b) The separation can be carried out if the fields are at right angles to each other
c) The time of pulse should be less than that of the reorientation time
d) Separation is not of the megabase level
Answer: b
Explanation: For the separation of large DNA molecules, the electric fields are often applied at right angles to each other. Usually, the large molecules pass through the matrix only in the direction of the movement. In rest directions, the movement is blocked by the gel matrix. The time of the pulse is greater than that of the reorientation time. Thus, smaller molecules have greater time for movement. DNA of megabase size can also be separated.

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250+ TOP MCQs on Dynamic Programming Algorithm for Sequence Alignment and Answers

Bioinformatics Multiple Choice Questions on “Dynamic Programming Algorithm for Sequence Alignment”.

1. Use of the dynamic programming method requires a scoring system for the comparison of symbol pairs, and a scheme for GAP penalties.
A. True
B. False

Answer: A
Explanation: Once those parameters have been set, the resulting alignment for two sequences should always be the same. Hence, the use of the dynamic programming method requires a scoring system for the comparison of symbol pairs (nucleotides for DNA sequences and amino acids for protein sequences), and a scheme for insertion/deletion (GAP) penalties.

2. After the derivation, the outputs of the dynamic programming are the ratios are called even scores.
A. True
B. False

Answer: B
Explanation: After the derivation, the outputs of the dynamic programming are the ratios are called odd scores. The ratios are transformed to logarithms of odds scores, called log odds scores, so that scores of sequential pairs may be added to reflect the overall odds of a real to chance alignment of an alignment. This happens in Dayhoff PAM250 and BLOSUM62.

3. The matrices PAM250 and BLOSUM62 contain _______
A. positive and negative values
B. positive values only
C. negative values only
D. neither positive nor negative values, just the percentage

Answer: A
Explanation: These matrices contain positive and negative values, reflecting the likelihood of each amino acid substitution in related proteins. Using these tables, an alignment of a sequential set of amino acid pairs with no gaps receives an overall score that is the sum of the positive and negative log odds scores for each individual amino acid pair in the alignment.

4. The higher is the score in the alignment _________
A. the more significant is the alignment
B. or the less it resembles alignments in related proteins
C. the less significant is the alignment
D. the less it aligns with the related protein sequence

Answer: A
Explanation: In the scoring system, the higher this score, the more significant is the alignment, or the more it resembles alignments in related proteins. Also, the score given for gaps in aligned sequences is negative, because such misaligned regions should be uncommon in sequences of related proteins. Such a score will reduce the score obtained from an adjacent, matching region upstream in the sequences.

5. Gaps are added to the alignment because it ______
A. increases the matching of identical amino acids at subsequent portions in the alignment
B. increases the matching of or dissimilar amino acids at subsequent portions in the alignment
C. reduces the overall score
D. enhances the area of the sequences

Answer: A
Explanation: In the alignment process, gaps are added to the alignment in a manner that increases the matching of identical or similar amino acids at subsequent portions in the alignment. Ideally, when two similar protein sequences are aligned, the alignment should have long regions of identical or related amino acid pairs and very few gaps. As the sequences become more distant, more mismatched amino acid pairs and gaps should appear.

6. Which of the following is not a description of dynamic programming algorithm?
A. A method of sequence alignment
B. A method that can take gaps into account
C. A method that requires a manageable number of comparisons
D. This method doesn’t provide an optimal (highest scoring) alignment

Answer: D
Explanation: The method of sequence alignment by dynamic programming provides an optimal (highest scoring) alignment as an output. The quality of the alignment between two sequences is calculated using a scoring system that favors the matching of related or identical amino acids and penalizes for poorly matched amino acids and gaps.

7. Which of the following is not a site on internet for alignment of sequence pairs?
A. BLASTX
B. BLASTN
C. SIM
D. BCM Search Launcher

Answer: A
Explanation: BLASTP is used under BLAST 2 sequence alignment. Also, The BLAST algorithm normally used for database similarity searches can also be used to align two sequences. SIM is known as Local similarity program for finding alternative alignments.

8. Dayhoff PAM matrices, are based on an evolutionary model of protein change, whereas, BLOSUM matrices, are designed to identify members of the same family.
A. True
B. False

Answer: A
Explanation: There is a very large number amino acid scoring matrices in use, some much more popular than others, and these scoring matrices are designed for different purposes. Some, such as the Dayhoff PAM matrices, are based on an evolutionary model of protein change, whereas others, such as the BLOSUM matrices, are designed to identify members of the same family. Alignments between DNA sequences require similar kinds of considerations.

9. A feature of the dynamic programming algorithm is that the alignments obtained depend on the choice of a scoring system for comparing character pairs and penalty scores for gaps.
A. True
B. False

Answer: A
Explanation: For an algorithm, the output depends on the choice of a scoring system. For protein sequences, the simplest system of comparison is one based on identity. A match in an alignment is only scored if the two aligned amino acids are identical. However, one can also examine related protein sequences that can be aligned easily and find which amino acids are commonly substituted for each other.

10. Which of the following is untrue regarding dynamic programming algorithm?
A. The method compares every pair of characters in the two sequences and generates an alignment
B. The output alignment will include matched and mismatched characters and gaps in the two sequences that are positioned so that the number of matches between identical or related characters is the maximum possible
C. The dynamic programming algorithm provides a reliable computational method for aligning DNA and protein sequences
D. This doesn’t allow making evolutionary predictions on the basis of sequence alignments

Answer: D
Explanation: Optimal alignments provide useful information to biologists concerning sequence relationships by giving the best possible information as to which characters in a sequence should be in the same column in an alignment, and which are insertions in one of the sequences (or deletions on the other). This information is important for making functional, structural, and evolutionary predictions on the basis of sequence alignments.

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