Molecular Biology (Chapter 1-5, 20-23)
These are the powerpoint questions he asked I'll add some randoms one later when I get through reading all the chapters. Good luck.
- 1.
Some genes located on the same chromosome do not show perfect linkage because?
- A.
They are too close together
- B.
The are interrupted by other genes
- C.
Crossover events occur between homologous genes
- D.
Some genes are deleted
- E.
Too many alleles are present
Correct Answer
C. Crossover events occur between homologous genesExplanation
Correct answer is: (C) crossover events occur between homologous genesRate this question:
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- 2.
During an experiment with drosophilia, it was discovered that all females had red eyes. A conclusing based on this observation would be?
- A.
The gene for eye color is located on the X chromosome
- B.
The eye color phenotype is sex-linked
- C.
The females are homozygotes
- D.
The gene for eye color is located on an autosome
- E.
Both (A) and (B) are correct
Correct Answer
E. Both (A) and (B) are correctExplanation
Correct answer is: (E) both (A) and (B) are correctRate this question:
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- 3.
Which of the following statements is not true regarding genetic mapping?
- A.
Genes are arragned in a linear order on chromosomes
- B.
Recombination occurs between chromosomes
- C.
There is a mathematical relationship between the distance separtating two genes and the recombination frequency
- D.
Mapping is used to establish the location of genes relative to each other
- E.
Mapping is used to determine the composition of a gene.
Correct Answer
E. Mapping is used to determine the composition of a gene.Explanation
Correct answer is: (E) Mapping is used ot determine the compositon of a gene.Rate this question:
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- 4.
Evidence that physical recombination occurs between two chromosomes was provided by?
- A.
McClintock and Creighton
- B.
Watson and Crick
- C.
Avery, McLeod and McCarthy
- D.
Hershey and Chase
- E.
Miescher and Morgan
Correct Answer
A. McClintock and CreightonExplanation
Correct answer: (A) McClintock and CreightonRate this question:
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- 5.
The chromosomal theory of inheritance holds that
- A.
Chromosomes vary in size
- B.
The number of chromosomes vary in individuals
- C.
Chromosomes contain histone protiens
- D.
Genes are arranged in a linear order on chromosomes
- E.
Chromosomes condense during mitosis
Correct Answer
D. Genes are arranged in a linear order on chromosomesExplanation
Correct answer is: (D) genes are arranged in a linear order on chromosomesRate this question:
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- 6.
Which of the following was demonstrated by Frederick and Griffiths?
- A.
Nuclein was the possible genetic material
- B.
A genetic trait can be transferred from one organism to another
- C.
Genes are linked
- D.
Corssing over occurs between chromosomes
- E.
Protein is a major component of nuclein.
Correct Answer
B. A genetic trait can be transferred from one organism to anotherExplanation
Correct answer is: (B) a genetic trait can be transferred from one organsim to anotherRate this question:
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- 7.
Who first conducted the X-ray diffraction studies of DNA?
- A.
Franklin and Wilkins
- B.
Miescher and McClintock
- C.
Wason and Crick
- D.
Meselson and Stahl
- E.
Avery and Beadle
Correct Answer
A. Franklin and WilkinsExplanation
Correct answer is: (A) Franklin and WilkinsRate this question:
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- 8.
Which of the follwing scientists provided definitive evidence for the chromosomal theory?
- A.
Mendel
- B.
Morgan
- C.
McClintock
- D.
Miescher
- E.
Avery
Correct Answer
B. MorganExplanation
Correct answer is: (B) MorganRate this question:
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- 9.
The relationship between genes and enzymes was shown by?
- A.
Watson and Crick
- B.
Beadle and Tatum
- C.
Avery and McLeod
- D.
Morgan and Mendel
- E.
Miescher and Stern
Correct Answer
B. Beadle and TatumExplanation
Correct answer is: (B) Beadle and TatumRate this question:
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- 10.
The notion of one-gene/one-enzyme is true or not? If not, it is because of which of the following?
- A.
An enzyme can be composed of more than one polypeptide
- B.
Many genes contain the information for making polypeptides that are not enzymes.
- C.
The end products of some genes are not polypeptides
- D.
All A-C are correct
- E.
Only A and B are correct
Correct Answer
D. All A-C are correctExplanation
Correct answer is: (D) All A-C are correctRate this question:
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- 11.
Which of the following is not a product of transcription?
- A.
RNA
- B.
DNA
- C.
RRNA
- D.
TRNA
- E.
MRNA
Correct Answer
B. DNAExplanation
Correct answer is: (B) DNARate this question:
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- 12.
Which of the following is true about wild-type organisms?
- A.
They are always foudn in the wild.
- B.
They are free of all mutations.
- C.
They are also referred to as standard-types
- D.
They carry no recessive genes.
- E.
None of the choices
Correct Answer
C. They are also referred to as standard-typesExplanation
Correct answer is: (C) They are also referred to as standard-typesRate this question:
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- 13.
The correct structure of DNA components can be presented as
- A.
Phosphate–base–sugar
- B.
Phosphate-sugar-base
- C.
Base-phosphate-sugar
- D.
Phosphate-sugar-phosphate-base
- E.
Base-sugar-phosphate-base
Correct Answer
B. Phosphate-sugar-baseExplanation
Correct answer is: (B) phosphate-sugar-baseRate this question:
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- 14.
Which of the following is a characteristic of double-stranded DNA?
- A.
2 nanometers in width
- B.
10 base pairs per turn
- C.
0.34 nanometers per basepair
- D.
A and B
- E.
A, B and C
Correct Answer
E. A, B and CExplanation
Correct answer is: (E) A, B and CRate this question:
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- 15.
Which of the following DNA molecules is the most stable?
- A.
CTGCATAC GACGTATG
- B.
GAAATTTC CTTTAAAG
- C.
AGTCGAAT TCAGCTTA
- D.
GCGTGCAC CGCACGTG
- E.
GGATCCTG CCTAGGAC
Correct Answer
D. GCGTGCAC CGCACGTGExplanation
Correct answer is: (D)Rate this question:
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- 16.
Double-stranded DNA consists of two antiparallel strands, meaning that one strand is oriented in the 5' to 3' direction, while the other is oriented in the 3' to 5' direction.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Correct answer is: (A) TrueRate this question:
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- 17.
Asymmetrical spacing of the backbones of the DNA double helix generates major and minor grooves.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Correct Answer is: (A) TrueRate this question:
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- 18.
In experiments conducted by Griffitsh, the inabilityy of the avirulent strain to produce a lethal infection was due to
- A.
The presence of a protective
- B.
Destruction by the host's white blood cells
- C.
The presence of a gene for a polysaccharide coat
- D.
The high body temperature of the host
- E.
None of the choices are correct
Correct Answer
B. Destruction by the host's white blood cellsExplanation
Correct answer is: (B) Destruction by the host's white blood cellsRate this question:
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- 19.
Experiments conducted by Frederick Griffiths laid the foundation for?
- A.
Elucidation of mRNA structure
- B.
DNA as the genetic material
- C.
The virulence of S. pneumoniae
- D.
The discovery of the capsule of S. pneumoniae
- E.
Elucidaiton of tRNA structure.
Correct Answer
B. DNA as the genetic materialExplanation
Correct answer is: (B) DNA as the genetic materialRate this question:
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- 20.
Which of the following would be the substance of choice to destroy the DNA in a solution?
- A.
Ribonuclease
- B.
Trypson
- C.
Chymotrypsin
- D.
Deoxyribonuclease
- E.
None of the choices are correct
Correct Answer
D. DeoxyribonucleaseExplanation
Correct answer is: deoxyribonucleaseRate this question:
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- 21.
The analytic tools used to show that DNA was the transforming substance include all of the following except
- A.
Ultracentrifugation
- B.
Electrophoresis
- C.
UV absorption spectrophotometry
- D.
Heat inactivation
- E.
Chemical analysis
Correct Answer
D. Heat inactivationExplanation
Correct answer is: (D) heat inactivationRate this question:
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- 22.
All of the following are found in RNA except
- A.
Adenine
- B.
Deoxyribose
- C.
Cytosine
- D.
Guanine
- E.
Uracil
Correct Answer
B. DeoxyriboseExplanation
Correct answer is: deoxyriboseRate this question:
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- 23.
Which of the following is not found in RNA and DNA?
- A.
Nitrogen
- B.
Oxygen
- C.
Carbon
- D.
Sulfur
- E.
Hydrogen
Correct Answer
D. SulfurExplanation
Correct answer is: (D) sulfurRate this question:
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- 24.
The bonds that connect adenine to thymine in a DNA double helix are
- A.
Ionic
- B.
Hydrophobic
- C.
Hydrogen
- D.
Covalent
- E.
One of the choices are correct
Correct Answer
C. HydrogenExplanation
The correct answer: (C) hydrogenRate this question:
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- 25.
Phosphodiester bonds are examples of
- A.
Hydrophobic interactions
- B.
Hydrogen bonds
- C.
Convalent bonds
- D.
Noncovalent bonds
- E.
Ionic bonds
Correct Answer
C. Convalent bondsExplanation
Correct answer is: convalent bondsRate this question:
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- 26.
The ability of two polynucleotide strands to hybridize is based on which of the following principles?
- A.
Double helical nature of DNA
- B.
The presence of phosphodiester bonds
- C.
Complementary base pairing
- D.
The presence of 3' and 5' ends
- E.
The presence of covalent bonds
Correct Answer
C. Complementary base pairingExplanation
The correct answer is: (C) complementary base pairingRate this question:
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- 27.
Rapid cooling of DNA following heating at high temperature will cause
- A.
The strands to break
- B.
Rapid renaturation of the strands
- C.
The strands to remain separated
- D.
The double helix to fomr rapidly
- E.
Mismatching between the strands
Correct Answer
C. The strands to remain separatedExplanation
The correct answer: (C) the strands to remain separted. a common procedure to ensure that denatured DNA stays denatured is to brunge the hot dna solution into ice also called quenching.Rate this question:
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- 28.
Which of the following features is common to DNA, RNA, and proteins?
- A.
They contain amino acidds
- B.
They contain nucleotides
- C.
Hey contain uracil
- D.
They are polymers.
- E.
Hey are found in vesicles
Correct Answer
D. They are polymers.Explanation
The correct answer is: (D) They are polymersRate this question:
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- 29.
The messenger hypothesis states that
- A.
Messenger RNAs are very stable and are not destroyed
- B.
Ribosomal RNAs carry the inforamtion for making proteins
- C.
Messenger RNAs carry genetic inforamtion to ribosomes
- D.
New ribosome are made to trasmit new genetic inforamtion
- E.
None of the choices are correct
Correct Answer
C. Messenger RNAs carry genetic inforamtion to ribosomesExplanation
The correct answer is: (C) messenger RNAs carry genetic information to ribosomesRate this question:
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- 30.
In an experiment conducted to toest the messenger hypothesis, whch states that messenger RNA carries the genetic information 35S was used instead of 32P. What is one possible outcome of this experiment?
- A.
The phage RNA would be label along the entire molecule
- B.
No label would be found in the RNA
- C.
The DNA of the host bacteria would be labeled.
- D.
The ribosomal RNA would be labeled
- E.
All fo the choices are possible outcomes
Correct Answer
B. No label would be found in the RNAExplanation
The correct answer is: (B) No label would be found in the RNARate this question:
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- 31.
Predict the consequence of treating the 70s ribosomal with urea:
- A.
The RNA portion would be degraded
- B.
The unit would be undisturbed
- C.
The protein component would be degraded
- D.
The unit would dissociate from RNA and protein components.
- E.
The treatment would lead to tighter association between RNA components.
Correct Answer
D. The unit would dissociate from RNA and protein components.Explanation
The correct asnswer is: (D) the unit would dissociate from RNA and protein componentsRate this question:
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- 32.
A new mutant cell line was accidentall created in the lab. This mutant was found to be deficient in the enzyme aminacyl-tRNA synthetase. Which of the following would most likely to observe in this cell line?
- A.
No tRNA molecules would be present
- B.
TRNA molecules would contain no anticodons
- C.
Most tRNA molecules would not be able to attach to amino acids
- D.
Protein synthesis would increase
- E.
None of the choices are correct
Correct Answer
C. Most tRNA molecules would not be able to attach to amino acidsExplanation
The correct answers is: (C) Most tRNA molecules would not be able to attach to amino acidsRate this question:
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- 33.
Which is the following is not requried during the elongation phase of translation in bacteria?
- A.
EF-Tu
- B.
Peptidyl transferease
- C.
GTP
- D.
70 S ribosome
- E.
Ribonuclease
Correct Answer
E. RibonucleaseExplanation
The correct answer is: (E) ribonucleaseRate this question:
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- 34.
Which of the following is likely to contribute to a shift in the reading frame in an mRNA molecule?
- A.
Removal a triplet code
- B.
Removal of nine bases
- C.
Removal of a single base
- D.
Removal of six bases
- E.
Noneo fthe choices are correct
Correct Answer
C. Removal of a single baseExplanation
The correct answer is: (C) removal of a single baseRate this question:
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- 35.
A gene can accept occasional changes, which allows organisms to evolve
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
Genes can indeed accept occasional changes, which is known as genetic variation. This genetic variation is crucial for the process of evolution as it allows organisms to adapt to their environment over time. These changes can occur through various mechanisms such as mutations, gene flow, and genetic recombination. Therefore, the statement "A gene can accept occasional changes, which allows organisms to evolve" is true.Rate this question:
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- 36.
MRNA encodes a gene
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
mRNA does not encode a gene. mRNA is a molecule that is transcribed from a gene and carries the genetic information from the DNA to the ribosomes, where it serves as a template for protein synthesis. Genes are segments of DNA that contain the instructions for making a specific protein or functional RNA molecule. Therefore, mRNA is a product of gene expression, not the gene itself.Rate this question:
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- 37.
The noncoding strand of a gene is the same as the antisense strand
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
The noncoding strand of a gene is not the same as the antisense strand. The noncoding strand is the template strand that is transcribed into RNA, while the antisense strand is the complementary strand to the coding (sense) strand.Rate this question:
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- 38.
An amino group, a carboxyl group, and a hydrogen are found in all amino acids
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
All amino acids have a common structure consisting of an amino group, a carboxyl group, and a hydrogen atom. The amino group contains a nitrogen atom bonded to two hydrogen atoms, while the carboxyl group consists of a carbon atom double-bonded to an oxygen atom and single-bonded to a hydroxyl group. These functional groups are essential for the formation of peptide bonds and the overall structure and function of proteins. Therefore, it is true that all amino acids contain an amino group, a carboxyl group, and a hydrogen atom.Rate this question:
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- 39.
A segments of DNA is considered upstream when t is located on the 5' end of the start site.
- A.
True
- B.
False
Correct Answer
A. TrueExplanation
The statement is true because in DNA, the 5' end refers to the end of the DNA strand that has a phosphate group attached to the 5th carbon of the sugar molecule. The start site is where transcription begins, and if a segment of DNA is located on the 5' end of the start site, it is considered upstream.Rate this question:
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- 40.
Restriction enzymes were discoverd by
- A.
Boyer and Cohen
- B.
Linn and Arber
- C.
Watson and Crick
- D.
Gurdon and Linn
- E.
Cohen and Gurdon
Correct Answer
B. Linn and ArberExplanation
The correct answer is: (B) Linn and ArberRate this question:
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- 41.
The complementary sequence of oneof the following is a palindrome. Which one?
- A.
GAATTC
- B.
CTGACT
- C.
ATCCTC
- D.
GGCCAA
- E.
CCTTTC
Correct Answer
A. GAATTCExplanation
The correct answer is: (A) GAATTCRate this question:
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- 42.
Select the component that is most critical to the function of as plasmid?
- A.
Tetracycline resistance
- B.
Ampicillin resistance
- C.
EcoRI site
- D.
Origin of replication
- E.
Ligase
Correct Answer
D. Origin of replicationExplanation
The origin of replication is the component that is most critical to the function of a plasmid. It is responsible for initiating the replication process and allows the plasmid to be replicated and passed on to daughter cells. Without the origin of replication, the plasmid would not be able to replicate and would not be maintained in the cell population. The other components listed, such as tetracycline resistance, ampicillin resistance, EcoRI site, and ligase, are important for certain applications or functions of the plasmid but are not essential for its basic replication and maintenance.Rate this question:
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- 43.
An experimenter is able to select recombinant colonies from a bacterial culture plate based on the color of the colonies. Which of the following is a correct statement.
- A.
Colonies that contain recombinant plasmids are red
- B.
Colonies that contain recombinant plasmids are blue due to the disruption of the lacZ gene the insert
- C.
Colonies that contain recombinants are colorless due to the distruption of the lacZ gene by the insert
- D.
Colonies are blue due to the disruption of them ampicillin gene
- E.
All of the choices are correct
Correct Answer
C. Colonies that contain recombinants are colorless due to the distruption of the lacZ gene by the insertExplanation
The correct statement is that colonies that contain recombinants are colorless due to the disruption of the lacZ gene by the insert. This is because the lacZ gene is responsible for producing an enzyme called beta-galactosidase, which is required for the breakdown of a colorless substrate called X-gal into a blue product. When the lacZ gene is disrupted by the insert, the colonies are unable to produce the enzyme and therefore do not turn blue.Rate this question:
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- 44.
One approach to prevent religation of the ends of a restricted plasmid DNA is to
- A.
Store the plasmid in DNase
- B.
Store the plasmid in RNAase
- C.
Treat the plasmid with alkaline phosphatase to remove phosphate for the sticky ends
- D.
Treat the plasmid with a protease to remove all proteins
- E.
Treat the plasmid with BamHI to create gaps
Correct Answer
C. Treat the plasmid with alkaline phosphatase to remove phosphate for the sticky endsExplanation
Treating the plasmid with alkaline phosphatase is an effective approach to prevent religation of the ends of a restricted plasmid DNA. This is because alkaline phosphatase removes the phosphate groups from the sticky ends of the DNA, which are necessary for the ends to rejoin and religate. By removing the phosphate groups, the sticky ends become blunt ends, making it difficult for the DNA to reconnect and religate. Therefore, treating the plasmid with alkaline phosphatase helps to prevent the unwanted rejoining of the DNA ends and maintains the integrity of the restricted plasmid.Rate this question:
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- 45.
A recombinant plasmid was constructed with several restriction enzyme sites and an ampicillin resistance marker. This recombinant was used to transform a host bacteria that also has the ampicillin resistant gene. Which of the following statements is true regarding this experiment?
- A.
All transformed bacteria will grow on a plate containing ampicillin
- B.
All transformed and untransformed bacteria will grow on a plate containing ampicillin
- C.
It will be difficult to screen for a colony containing a plasmid.
- D.
A bacterial host that is ampicillin-sensitive should be sued for the screening
- E.
All of the choices are true
Correct Answer
E. All of the choices are trueExplanation
All of the choices are true. Since the recombinant plasmid contains an ampicillin resistance marker, all transformed bacteria, including both the transformed and untransformed ones, will be able to grow on a plate containing ampicillin. Additionally, the presence of multiple restriction enzyme sites in the plasmid makes it difficult to screen for a colony containing the plasmid. Therefore, it is true that it will be difficult to screen for a colony containing a plasmid. Lastly, it is recommended to use an ampicillin-sensitive bacterial host for screening purposes.Rate this question:
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- 46.
How many enzymes should be used in a plasmid DNA and an insert if there is one desired site to ligate the insert in the vector to a specific orientation?
- A.
1
- B.
2
- C.
3
- D.
4
- E.
5
Correct Answer
B. 2Explanation
If there is one desired site to ligate the insert in the vector to a specific orientation, two enzymes should be used in the plasmid DNA and insert. One enzyme will cut the plasmid DNA at the desired site, creating a linear vector. The other enzyme will cut the insert at a compatible site, creating sticky ends that can base pair with the sticky ends of the linear vector. This will allow the insert to be ligated into the vector in the desired orientation.Rate this question:
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- 47.
A bacterial sample was contaminated with an unknown preparation of vector DNA. In order to identify the vector, the bacteria was streaked on a plate and incubated overnight. Examination of the plate revealed at least 120 clear plaques. Which of the following is a plausible conclusion?
- A.
The vector could be a plasmid
- B.
The vector could be a phage
- C.
The vector is a Ti
- D.
The vector is puc18
- E.
The vector is a cosmid.
Correct Answer
B. The vector could be a phageExplanation
Based on the information provided, the presence of at least 120 clear plaques indicates that the vector could be a phage. Phages are viruses that infect bacteria and can form clear plaques on a bacterial lawn. Plasmids, Ti plasmids, puc18, and cosmids are all types of vectors, but they do not typically form clear plaques. Therefore, the most plausible conclusion is that the vector is a phage.Rate this question:
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- 48.
In constructing a cDNA library, which of the following would you use to generate rare restriction sites on the cDNA strands?
- A.
Restriction enzymes
- B.
PCR with multiple cycles
- C.
BamHI restriction followed by ligation
- D.
Ligate cDNA to specific linkers
Correct Answer
D. Ligate cDNA to specific linkersExplanation
To generate rare restriction sites on the cDNA strands, one would use the technique of ligating cDNA to specific linkers. Linkers are short DNA sequences that contain rare restriction sites. By ligating these linkers to the cDNA strands, rare restriction sites can be introduced. This allows for the identification and isolation of specific DNA fragments within the cDNA library using the corresponding restriction enzymes. Restriction enzymes, PCR with multiple cycles, and BamHI restriction followed by ligation are not methods specifically used to generate rare restriction sites on the cDNA strands.Rate this question:
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- 49.
Which of the following enzymes cannot catalyze the formation of a phosphodiester bond?
- A.
Endonuclease
- B.
RNA polymerase
- C.
DNA polymerase
- D.
Ligase
- E.
None of the choices are correct
Correct Answer
A. EndonucleaseExplanation
Endonucleases are enzymes that cleave phosphodiester bonds within a DNA or RNA molecule. They do not catalyze the formation of phosphodiester bonds. Therefore, the correct answer is endonuclease.Rate this question:
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- 50.
A disadvantage of using a prokaryotic expression systems for eukaryotic proteins is that the proteins are
- A.
Highly phosphorylated after translation
- B.
Improperly folded
- C.
Highly soluble
- D.
Heavily glycosylated
- E.
Over expressed
Correct Answer
B. Improperly foldedExplanation
Using a prokaryotic expression system for eukaryotic proteins can lead to the proteins being improperly folded. Prokaryotic cells lack the machinery and post-translational modifications required for proper folding of eukaryotic proteins. As a result, the proteins may not adopt their correct three-dimensional structure and may be functionally inactive or unstable. This can affect their biological activity and overall usefulness in research or industrial applications.Rate this question:
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